Waterloo Euclid preparation center

Prepare for the Euclid Contest with a complete training system

Q: What is the Euclid Mathematics Contest?

The Euclid Contest is the University of Waterloo CEMC’s senior secondary-school mathematics contest. It uses ten written questions, lasts 2.5 hours, and awards marks for both correct results and clearly communicated reasoning.

Q: Who can write the Euclid Contest?

It is aimed at students in their final year of secondary school and CÉGEP students, but motivated younger students may also participate. Students must write through a participating school rather than registering individually.

Q: When is the 2027 Euclid Contest?

The official 2027 date is April 6 in North and South America and April 7 outside those regions. The school ordering deadline is March 11, 2027. Students should confirm arrangements with their school because administration details can change.

Q: How should I use past Euclid papers?

Begin with an untimed diagnostic, classify every lost mark, then practise weak topics. In the final month, write complete official papers under the 150-minute limit and compare your reasoning with the official solutions. Use CalculatorVillage’s original sets between official papers so you do not memorize an archive.

Q: What score should I target?

There is no passing score. In 2026 the official mean was 52.2 and the Certificate of Distinction cutoff was 66, but both vary by year. A useful target is to improve your own timed baseline while protecting partial credit on every attempted question.

Q: Are calculators allowed?

The CEMC states that some calculators are permitted. Check the current contest instructions before test day. A calculator may support numerical work, but full-solution marks still require the mathematical steps and justification to be written.

Q: Does Waterloo require the Euclid Contest for admission?

Participation is not required for admission to Waterloo’s Faculty of Mathematics. The university directs students to its current scholarship information for details about how contest results may be considered.

Q: Are these official Euclid questions?

No. CalculatorVillage’s questions, drills, and practice sets are our own preparation material designed around the published format and topic areas, not copied from CEMC papers. The bank holds 224 practice questions: 32 authored multi-step problems with marking checkpoints and 192 generated fluency variants that practise one skill each. The drills and 150-minute practice sets reuse questions from this one bank. Official contests and solutions remain on the CEMC website and are linked from this guide.

Learn the official format, master all eight CEMC preparation areas, practise complete written solutions, and measure progress with topic drills and 150-minute practice sets.

Start practice Follow the 12-week plan

224

practice questions

topic drills

150-minute practice sets

official topic areas

32 authored multi-step problems and 192 generated fluency variants

Last Updated: November 11, 2026

Independent preparation resource · Contest facts sourced from Waterloo CEMC

Official 2027 snapshot

Know the paper before training for it

Euclid is CEMC’s senior written mathematics contest. The paper has ten questions, lasts 2.5 hours and is marked out of 100. Questions mix final-answer work with full solutions, so a correct numerical result may not earn full marks when the reasoning is missing or poorly organized.

CEMC reports that 23,985 students wrote the 2026 contest. Most content is based on mathematics through the final year of secondary school, although a problem may require an unfamiliar combination of ideas. The challenge is choosing and communicating a method, not racing through a list of advanced formulas.

Dates, deadline, format and fee above: checked November 2026 against the official CEMC Euclid page. Confirm them there before you rely on them.

2027 contest date: April 6 in North and South America; April 7 elsewhere
Ordering deadline: March 11, 2027; students participate through a school
Paper format: 10 written questions mixing final-answer and full-solution work
Time and score: 2.5 hours; 100 marks total
Intended audience: Final-year secondary and CÉGEP students; motivated younger students are welcome
Current fee: C$18 per participant, plus applicable school order and shipping fees

What the recent score data means

For the 2026 contest, CEMC reported a mean of 52.2, a Certificate of Distinction cutoff of 66, a school medal cutoff of 52 and 23,985 contestants. These are historical reference points, not promises for 2027; cutoffs change with each paper and cohort. The useful lesson is that meaningful partial credit matters. A student does not need to finish every final part to produce a strong result.

There is no pass or fail. Every participant in the top 25% receives a Certificate of Distinction, while other awards and honour-roll rules depend on official-participant status and CEMC’s published criteria. Check the official page for the current age, residency and paperwork rules rather than relying on an old summary.

2026 mean, cutoffs and contestant count: checked November 2026 against the official CEMC Euclid page.

Start with evidence

Run a diagnostic before choosing what to study

1. Write a baseline

Choose one recent official paper you have not studied. Use 150 uninterrupted minutes, paper and only the materials allowed by the current instructions.

2. Classify every loss

Label each lost mark: missing concept, wrong setup, algebra, arithmetic, incomplete reasoning, misread condition or time.

3. Pick two priorities

Select one content weakness and one process weakness. A narrow target produces better practice than “do more papers.”

4. Retest transfer

After two weeks of focused drills, attempt new problems and one timed section. Improvement must appear on unfamiliar work.

A score is not a diagnosis

Two students can earn 55 for different reasons. One may understand nearly every topic but write too little for full credit. Another may present well but lack trigonometry and counting tools. Their next month should look completely different. Keep an error log with the question source, topic, first incorrect line, lost-mark category and a specific correction task.

Review the log weekly. If “algebra” appears repeatedly, split it into precise causes such as sign errors, invalid cancellation or an untested extraneous root. Training can address a cause; it cannot address a vague label.

Use a three-pass review

Pass 1 — repair: Find the earliest wrong or missing step and finish the question without copying.
Pass 2 — compare: Read the official or worked solution and identify a more efficient representation or theorem.
Pass 3 — retrieve: Close the solution and rewrite the full argument one to three days later.

A question is not learned when its solution feels familiar. It is learned when the method can be reconstructed later on a blank page and transferred to a new problem.

Complete syllabus map

The eight official preparation areas, turned into a study system

CEMC organizes its senior contest preparation material into the eight areas below. They are not a guarantee that every paper uses the same distribution, and a single Euclid question may connect several modules. Use the map to find missing tools, then practise mixed work so you learn to select the tool without being told the topic.

Exponents and logarithms

This module is less about memorizing a list of laws than recognizing which form makes the structure visible. Euclid problems can move between exponential, logarithmic, polynomial and geometric descriptions. A student should be comfortable changing bases, matching exponents, using negative and rational powers, and checking the domain before taking a logarithm. The strongest solutions name the law being used and preserve exact values until the final line.

What to master

Exponent laws, including zero, negative and rational powers
Logarithm product, quotient and power laws
Solving equal-base equations and simple exponential equations
Domain restrictions and the one-to-one property of exponential functions

Training method

Begin by rewriting every term with a common base when possible. If that fails, isolate the exponential expression before taking logarithms. For a full solution, explicitly state why the base is valid and why equal powers imply equal exponents. Train with short transformations first, then add word problems involving repeated growth, decay or geometric scaling.

Readiness checkpoint: Can you solve an exponential equation two ways, identify an extraneous logarithmic result, and explain which form is most efficient?

Trigonometry

Euclid trigonometry rewards diagram control. Many lost marks come from assigning the wrong side to an angle, rounding too early, or quoting a formula without connecting it to the figure. Preparation should cover right-triangle ratios, exact special-angle values, sine and cosine laws, area formulas, identities and periodic behavior. Trigonometry often appears inside a geometry or coordinate problem rather than as an isolated calculation.

What to master

Exact values for 30°, 45° and 60°
Sine law, cosine law and triangle area using sine
Basic identities and transformations of trigonometric expressions
Angles of elevation, bearings and multi-triangle diagrams

Training method

Redraw the figure and label sides relative to the angle you are actually using. Keep radicals and fractions exact. Before applying the sine law, identify an opposite side-angle pair; before applying the cosine law, identify the included angle or the three known sides. On a calculator-permitted question, the written solution must still show the equation that produced the numerical value.

Readiness checkpoint: Can you decide between right-triangle ratios, the sine law and the cosine law without starting a calculation?

Functions, equations and polynomials

This is the algebraic engine of the contest. Questions may ask for a value without requiring every root, hide a useful substitution inside a symmetric expression, or turn a geometric condition into a polynomial. Students need flexible manipulation rather than expansion by default. Factoring, the remainder theorem, Vieta’s formulas, function composition and the relationship between graphs and equations are especially useful.

What to master

Factoring, completing the square and discriminants
Remainder and factor theorems
Vieta’s formulas and symmetric expressions in roots
Function notation, composition, inverses and transformations

Training method

Write down the requested quantity before solving. If the question asks for r²+s², use (r+s)²−2rs instead of finding both roots. If a polynomial is divided by x−a, test P(a) before long division. Check denominators and square roots for domain restrictions, and substitute every candidate into the original equation when operations could introduce an extraneous result.

Readiness checkpoint: Can you use coefficients to evaluate a symmetric expression in the roots without applying the quadratic formula?

Analytic geometry

Coordinates turn geometry into equations, but an efficient Euclid solution moves in both directions. A slope can establish parallelism or perpendicularity; a distance equality can identify a perpendicular bisector; a circle equation can reveal a centre and radius. Students should be able to choose a coordinate system that simplifies the figure rather than accepting awkward coordinates unnecessarily.

What to master

Slope, midpoint and distance formulas
Equations of lines in multiple forms
Circle equations, tangency and intersections
Coordinate proofs of collinearity, parallelism and perpendicularity

Training method

Choose the origin and axes deliberately when the problem permits it. Put a symmetric object around the origin or align a major side with an axis. When proving a geometric fact, translate the calculation back into a sentence: equal slopes show parallel lines; slopes with product −1 show perpendicular lines when both are defined; equal distances place a point on a perpendicular bisector.

Readiness checkpoint: Can you prove a quadrilateral property using coordinates and explain what each algebraic result means geometrically?

Sequences and series

Sequence problems test whether a student can move among a recursive rule, an explicit formula and a sum. Euclid may wrap the sequence in a counting process, a geometric diagram or repeated operation. Arithmetic and geometric sequences are the foundation, but telescoping sums, recurrence relations and pattern justification are what separate a complete solution from an observed list of terms.

What to master

Arithmetic and geometric nth-term and sum formulas
Recursive definitions and iteration
Telescoping sums and cancellation
Recognizing and proving a pattern rather than extrapolating it

Training method

Calculate the first few terms neatly, then ask what changes and what stays fixed. If fractions have consecutive factors, try partial fractions to expose cancellation. When proposing a formula, verify the starting case and explain why the rule carries one term to the next. A table is useful evidence, but it is not by itself a proof that a pattern continues forever.

Readiness checkpoint: Can you derive a proposed nth-term rule from a recurrence and verify it beyond the first few examples?

Euclidean geometry

Geometry questions reward a disciplined diagram and a short chain of justified facts. The useful move may be drawing an altitude, joining a centre to a point of tangency, extending a side, or identifying similar triangles. Students should know standard theorems well enough to name them, yet remain willing to use coordinates or trigonometry when a synthetic route becomes long.

What to master

Triangle congruence and similarity
Circle theorems, chords, tangents and cyclic quadrilaterals
Area ratios, Pythagorean theorem and right-triangle structure
Angle chasing with parallel lines, isosceles triangles and polygons

Training method

Mark equal lengths and angles directly on a fresh diagram. Look for two triangles that share an angle or contain parallel-line angle pairs. State the similarity criterion before writing a proportion, and keep corresponding vertices in the same order. For circle problems, identify the intercepted arc or the radius-to-tangent right angle instead of relying on how the picture looks.

Readiness checkpoint: Can another reader trace every claimed angle or ratio back to a named theorem or a previously established fact?

Counting and probability

Counting problems are difficult when the same outcome can be produced more than once or when cases overlap. Probability adds the need for a correct sample space. Euclid solutions should make clear whether order matters, whether repetition is allowed, and why the cases are exhaustive and disjoint. Small examples often reveal the right structure, but the final count needs a general argument.

What to master

Permutations, combinations and the multiplication principle
Casework, complementary counting and inclusion-exclusion
Conditional probability and sampling without replacement
Paths, arrangements and simple recurrence-based counting

Training method

Before calculating, write three questions: What is one outcome? Does order matter? Can the same outcome appear in two cases? Use a complement when the forbidden condition is simpler than the allowed condition. For probability, define equally likely outcomes before dividing. If sequential probabilities are used, update both the numerator and denominator after each draw.

Readiness checkpoint: Can you explain why your cases cover every possibility exactly once and why your probability denominator matches the sample space?

Properties of numbers

Number theory questions often look experimental at first. The aim is to replace a list of trials with divisibility, parity, prime factorization, modular arithmetic or bounds. Euclid does not require advanced university number theory, but it does reward students who can express “same remainder” as a congruence and who recognize that integer conditions sharply restrict possible values.

What to master

Prime factorization, divisor counts, gcd and lcm
Parity, divisibility tests and integer equations
Remainders, modular cycles and last digits
Perfect squares, factor pairs and bounding integer solutions

Training method

Factor early and record the domain. For large powers, calculate a short remainder cycle rather than the number itself. For an integer equation, rearrange it into a product when possible and list factor pairs systematically, including negative pairs when allowed. A proof by parity or congruence should state the possible residue classes and show that every case has been covered.

Readiness checkpoint: Can you turn a brute-force divisibility search into a short argument using factorization or congruences?

12-week preparation plan

Move from diagnosis to full-paper confidence

A plan should change when the evidence changes. The schedule below assumes a student can study three or four times per week. Shorten each phase if the contest is closer, but preserve the sequence: diagnose, build tools, isolate weaknesses, mix topics, simulate, review and taper.

Time	Focus	Work to complete
Week 1	Diagnostic and scoring map	Write one recent official paper without help, but do not worry about the final score. Record time spent and confidence for every part. Mark with the official solution, then classify lost marks as concept, setup, algebra, communication, arithmetic or time. This produces the baseline that controls the rest of the plan.
Weeks 2–3	Algebraic foundation	Rebuild exponents, logarithms, functions, equations and polynomial methods. Complete two CalculatorVillage drills from each algebraic module and rewrite every full solution after checking it. The goal is clean symbolic work and reliable domain checking, not speed alone.
Weeks 4–5	Geometry in two languages	Alternate Euclidean geometry, analytic geometry and trigonometry. Solve each geometry problem first with the most natural method, then ask whether a coordinate or trigonometric approach offers a useful check. Keep a one-page theorem sheet written from memory.
Weeks 6–7	Discrete methods	Study sequences, number properties, counting and probability. Practise defining cases before counting and writing congruence arguments in full sentences. Complete at least one mixed drill under time and one untimed drill where presentation is graded strictly.
Week 8	First full simulation	Write a 150-minute practice set under test conditions. Spend the next session on review, not another test. For every lost mark, write the earliest line where the solution went wrong and the smallest practice action that would prevent a repeat.
Weeks 9–10	Weakness repair and official papers	Use two focused drills for the weakest module, then write one official archived paper. Compare the structure of your solution with the official solution rather than comparing only the final answer. Build a personal checklist of missing justifications.
Week 11	Second and third simulations	Write two full papers several days apart: one CalculatorVillage practice set and one official paper. Test a deliberate pacing plan. Attempt accessible early parts across the paper before investing heavily in a final challenge. Use the same marking rubric both times.
Week 12	Taper and test-day routine	Review the error log, theorem sheet and two previously missed questions from each weak area. Complete short confidence-building sets rather than exhausting daily papers. Confirm school logistics and permitted materials, then protect sleep and arrive ready to write clearly.

Three-session weekly rhythm

Session A learns or reviews a tool. Session B applies it in a focused drill. Session C uses a mixed timed set and updates the error log. Add a fourth session for solution rewriting or a full paper.

Use time in layers

Start new material untimed, then use a generous limit, then use official pace. Timing an unfamiliar method too early can train hurried mistakes instead of fluency.

Define improvement broadly

Track correct answers, recovered partial marks, fewer repeated errors, cleaner diagrams, shorter solutions and better skip decisions. A total score alone hides useful progress.

Written-solution training

Turn correct thinking into marks another person can award

A Euclid solution is mathematical communication. The marker should not need to infer why two triangles are similar, where a probability denominator came from, or whether an omitted case was considered. Clear writing also exposes errors while there is still time to correct them.

Use rough paper to explore, then write the submitted solution as a short argument. Include enough intermediate work to preserve partial credit. Do not bury the key idea under every failed attempt, but do not reduce the solution to a calculator value or an unsupported equation.

Part	What a complete response shows	Common mark loss
Setup	Defines variables, records conditions and identifies the requested quantity.	Do not begin with unexplained equations or assumptions hidden in a diagram.
Method	Names or clearly applies the theorem, identity, formula or counting principle.	A calculator output or unexplained pattern is not a mathematical justification.
Logical chain	Connects each statement to the previous one and handles all cases.	Watch for division by zero, square-root signs, overlapping cases and unproved converses.
Conclusion	States the answer with units or restrictions and checks it against the question.	For “determine all,” verify that every candidate works and that no other case remains.

A repeatable full-solution template

Given and goal: define variables and state what must be found or proved.
Key fact: name the theorem, identity or counting principle that makes progress possible.
Execution: show algebra, cases or deductions in a readable order.
Validity: address restrictions, equality conditions, all cases or candidate checks.
Conclusion: answer the exact question with units and a final sentence.

Self-mark without being generous

Cover the worked solution and reread your response as if it belonged to someone else. Underline every claim whose reason is not on the page. Circle every place where you divided, squared, took a root or discarded a case. Then compare methods. If your answer is correct but the central justification is absent, record a communication error rather than awarding full credit.

When an official marking scheme is unavailable, use a 10-mark practice split: 2 for setup, 4 for the central method, 3 for correct execution and 1 for a valid conclusion. This is a training rubric, not a claim about CEMC’s exact marking allocation.

Private browser workspace

Turn practice into a repeatable training loop

Assess your topics, build a custom set, save difficult questions, record errors and score a mock. Everything stays in this browser. CalculatorVillage does not create an account or send this study record to a server.

0 of 224 questions completed

questions saved for review

error-log entries

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average topic confidence

Topic confidence diagnostic

Choose the description that matches what you can do on an unfamiliar problem—not what felt familiar while reading a solution.

Exponents and logarithmsTrigonometryFunctions and polynomialsAnalytic geometrySequences and seriesEuclidean geometryCounting and probabilityProperties of numbers

Recommended next focus

Start with Exponents and logarithms, then Trigonometry. Complete one untimed authored problem, one 35-minute drill and one mixed set before raising the confidence level.

Custom practice-set builder

Build a custom, printable set from the complete question bank.

TopicDifficultyQuestions

1
Reciprocals of polynomial roots 9
The roots of x²−11x+28=0 are r and s. Determine 1/r+1/s.
Functions and polynomials · · Full solution
Check answer and solution
Answer: 11/28
Vieta's formulas give r+s=11 and rs=28. Therefore 1/r+1/s=(r+s)/(rs)=11/28=11/28.
2
Reciprocals of polynomial roots 8
The roots of x²−9x+18=0 are r and s. Determine 1/r+1/s.
Functions and polynomials · · Full solution
Check answer and solution
Answer: 1/2
Vieta's formulas give r+s=9 and rs=18. Therefore 1/r+1/s=(r+s)/(rs)=9/18=1/2.
3
Reciprocals of polynomial roots 7
The roots of x²−7x+10=0 are r and s. Determine 1/r+1/s.
Functions and polynomials · · Full solution
Check answer and solution
Answer: 7/10
Vieta's formulas give r+s=7 and rs=10. Therefore 1/r+1/s=(r+s)/(rs)=7/10=7/10.
4
Reciprocals of polynomial roots 10
The roots of x²−13x+40=0 are r and s. Determine 1/r+1/s.
Functions and polynomials · · Full solution
Check answer and solution
Answer: 13/40
Vieta's formulas give r+s=13 and rs=40. Therefore 1/r+1/s=(r+s)/(rs)=13/40=13/40.
5
Reciprocals of polynomial roots 6
The roots of x²−12x+32=0 are r and s. Determine 1/r+1/s.
Functions and polynomials · · Full solution
Check answer and solution
Answer: 3/8
Vieta's formulas give r+s=12 and rs=32. Therefore 1/r+1/s=(r+s)/(rs)=12/32=3/8.
6
Reciprocals of polynomial roots 11
The roots of x²−15x+54=0 are r and s. Determine 1/r+1/s.
Functions and polynomials · · Full solution
Check answer and solution
Answer: 5/18
Vieta's formulas give r+s=15 and rs=54. Therefore 1/r+1/s=(r+s)/(rs)=15/54=5/18.
7
Reciprocals of polynomial roots 5
The roots of x²−18x+77=0 are r and s. Determine 1/r+1/s.
Functions and polynomials · · Full solution
Check answer and solution
Answer: 18/77
Vieta's formulas give r+s=18 and rs=77. Therefore 1/r+1/s=(r+s)/(rs)=18/77=18/77.
8
Reciprocals of polynomial roots 12
The roots of x²−17x+70=0 are r and s. Determine 1/r+1/s.
Functions and polynomials · · Full solution
Check answer and solution
Answer: 17/70
Vieta's formulas give r+s=17 and rs=70. Therefore 1/r+1/s=(r+s)/(rs)=17/70=17/70.
9
Reciprocals of polynomial roots 4
The roots of x²−16x+60=0 are r and s. Determine 1/r+1/s.
Functions and polynomials · · Short answer
Check answer and solution
Answer: 4/15
Vieta's formulas give r+s=16 and rs=60. Therefore 1/r+1/s=(r+s)/(rs)=16/60=4/15.
10
Polynomial remainder 11
Find the remainder when P(x)=x³+2x²−3x+14 is divided by x−4.
Functions and polynomials · · Full solution
Check answer and solution
Answer: 98
By the remainder theorem, the remainder is P(4). Substitution gives 4³+2(4²)−3(4)+14=98.

Error log and retry queue

Record the earliest incorrect or missing step. “Careless” is not a useful category; identify the action that must change.

Problem referenceError category

Specific correction

Your error log is empty. Add an entry after a drill or mock.

100-mark mock score analyzer

Enter a 0–10 self-score for each question. This is a private training summary, not an official CEMC score prediction or percentile.

Q1Q2Q3Q4Q5Q6Q7Q8Q9Q10

Practice total

0/100

Rebuild the most common concept and setup errors before another full mock.

Review rule: spend at least as long reviewing a mock as writing it. Rewrite every incomplete full solution without looking at the model answer, then compare again.

Interactive preparation tools

Build a 150-minute Euclid pacing plan

Set checkpoints before a mock and use the full-paper timer when you are ready. The tools run only in your browser and do not create an account or send student activity to a server.

Contest pacing planner

QuestionsMinutesAttempt goal

Allow about 13.5 minutes per attempted question.

Quarter-time checkpoint: about 3 attempted.
Half-time checkpoint: about 5 attempted.
Three-quarter checkpoint: about 8 attempted.
Protect the final 15 minutes for returns, units, signs, and answer transfer.

Practice timer

30:00

When time ends, circle unfinished questions before opening any solutions. That separates time-management errors from concept errors.

Practice library

Search every Euclid question, drill and practice set

The library contains 224 practice questions: 32 authored multi-step problems and 192 generated fluency variants. It also has 32 topic drills and 12 practice sets of ten questions and 150 minutes, built from the same question bank. Only the authored problems carry practice marks (10 each); generated fluency questions carry none. Drills and sets reuse questions, so one question can appear in several. Search the problem statement, answer, method, topic or skill. These are CalculatorVillage practice questions, not copied CEMC papers.

Filter the complete Euclid library

Search questions and solutionsTopicDifficultyAnswer format

268 matching resources

32 matching topic drills

Use a 35-minute drill to isolate one weakness. Score the first attempt, correct it in another colour, then repeat a different drill from the same module within seven days.

Topic drill

Exponents and logarithms drill 1

35 minutes

Six exponents and logarithms practice questions. Work the set without notes, then compare every line of a full-solution response with the explanation.

6 practice questions · 2 authored problems worth 20 marks

1A sum of related powers
2Quadratic structure in an exponential equation
3Equal-base exponent equation 3
4Equal-base exponent equation 6
5Equal-base exponent equation 9
6Equal-base exponent equation 12

Open all questions and solutions

Exponents and logarithmsCoreFull solution10 marks

A sum of related powers

Solve for the real number x: 2^x+2^(x+1)+2^(x+2)=448.

Skill: Factoring a common exponential term

Show answer and worked solution

Answer: x=6

Factor out the smallest power: 2^x(1+2+4)=448. Thus 7·2^x=448, so 2^x=64=2^6. Therefore x=6.

Practice marking checkpoints

• 2 marks for identifying a common factor of 2^x
• 4 marks for reducing the equation to 2^x=64
• 3 marks for solving x=6
• 1 mark for a clear conclusion

Common mistake: Adding the exponents across a sum. Exponent laws apply to products, not directly to addition.

Alternative route: Set y=2^x. The equation becomes y+2y+4y=448, so 7y=448 and y=64.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Exponents and logarithmsChallengeFull solution10 marks

Quadratic structure in an exponential equation

Determine all real numbers x satisfying 9^x−10·3^x+9=0.

Skill: Recognizing a quadratic in an exponential expression

Show answer and worked solution

Answer: x=0 or x=2

Let y=3^x, which is positive. Since 9^x=(3^x)²=y², the equation becomes y²−10y+9=0=(y−1)(y−9). Thus y=1 or y=9. Therefore 3^x=1 gives x=0, and 3^x=9 gives x=2.

Practice marking checkpoints

• 2 marks for a valid substitution
• 3 marks for the quadratic equation
• 2 marks for factoring
• 2 marks for both x-values
• 1 mark for noting positivity or checking

Common mistake: Treating 9^x as 9x or failing to recognize that 9^x=(3^x)².

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Exponents and logarithmsFluency practiceShort answer

Equal-base exponent equation 3

Solve for x: 5^(x+1)=5^6.

Skill: Equating exponents with a common base

Show answer and worked solution

Answer: x=5

Because the common base 5 is positive and not 1, equal powers have equal exponents. Thus x+1=6, so x=5.

Exponents and logarithmsFluency practiceFull solution

Equal-base exponent equation 6

Solve for x: 4^(x+1)=4^9.

Skill: Equating exponents with a common base

Show answer and worked solution

Answer: x=8

Because the common base 4 is positive and not 1, equal powers have equal exponents. Thus x+1=9, so x=8.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Exponents and logarithmsFluency practiceFull solution

Equal-base exponent equation 9

Solve for x: 3^(x+1)=3^12.

Skill: Equating exponents with a common base

Show answer and worked solution

Answer: x=11

Because the common base 3 is positive and not 1, equal powers have equal exponents. Thus x+1=12, so x=11.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Exponents and logarithmsFluency practiceFull solution

Equal-base exponent equation 12

Solve for x: 2^(x+1)=2^15.

Skill: Equating exponents with a common base

Show answer and worked solution

Answer: x=14

Because the common base 2 is positive and not 1, equal powers have equal exponents. Thus x+1=15, so x=14.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Topic drill

Exponents and logarithms drill 2

35 minutes

Six exponents and logarithms practice questions. Work the set without notes, then compare every line of a full-solution response with the explanation.

6 practice questions

1Equal-base exponent equation 2
2Equal-base exponent equation 5
3Equal-base exponent equation 8
4Equal-base exponent equation 11
5Logarithm laws 2
6Logarithm laws 5

Open all questions and solutions

Exponents and logarithmsFluency practiceShort answer

Equal-base exponent equation 2

Solve for x: 4^(x+3)=4^7.

Skill: Equating exponents with a common base

Show answer and worked solution

Answer: x=4

Because the common base 4 is positive and not 1, equal powers have equal exponents. Thus x+3=7, so x=4.

Exponents and logarithmsFluency practiceFull solution

Equal-base exponent equation 5

Solve for x: 3^(x+3)=3^10.

Skill: Equating exponents with a common base

Show answer and worked solution

Answer: x=7

Because the common base 3 is positive and not 1, equal powers have equal exponents. Thus x+3=10, so x=7.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Exponents and logarithmsFluency practiceFull solution

Equal-base exponent equation 8

Solve for x: 2^(x+3)=2^13.

Skill: Equating exponents with a common base

Show answer and worked solution

Answer: x=10

Because the common base 2 is positive and not 1, equal powers have equal exponents. Thus x+3=13, so x=10.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Exponents and logarithmsFluency practiceFull solution

Equal-base exponent equation 11

Solve for x: 5^(x+3)=5^16.

Skill: Equating exponents with a common base

Show answer and worked solution

Answer: x=13

Because the common base 5 is positive and not 1, equal powers have equal exponents. Thus x+3=16, so x=13.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Exponents and logarithmsFluency practiceShort answer

Logarithm laws 2

Evaluate log_4(4^3)+2log_4(4^5).

Skill: Applying logarithm power and product laws

Show answer and worked solution

Answer: 13

Use log_b(b^k)=k. The expression is 3+2(5)=13. Equivalently, combine it as log_4(4^13).

Exponents and logarithmsFluency practiceFull solution

Logarithm laws 5

Evaluate log_4(4^2)+2log_4(4^4).

Skill: Applying logarithm power and product laws

Show answer and worked solution

Answer: 10

Use log_b(b^k)=k. The expression is 2+2(4)=10. Equivalently, combine it as log_4(4^10).

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Topic drill

Exponents and logarithms drill 3

35 minutes

Six exponents and logarithms practice questions. Work the set without notes, then compare every line of a full-solution response with the explanation.

6 practice questions

1Equal-base exponent equation 7
2Equal-base exponent equation 10
3Logarithm laws 1
4Logarithm laws 4
5Logarithm laws 7
6Logarithm laws 10

Open all questions and solutions

Exponents and logarithmsFluency practiceFull solution

Equal-base exponent equation 7

Solve for x: 5^(x+2)=5^11.

Skill: Equating exponents with a common base

Show answer and worked solution

Answer: x=9

Because the common base 5 is positive and not 1, equal powers have equal exponents. Thus x+2=11, so x=9.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Exponents and logarithmsFluency practiceFull solution

Equal-base exponent equation 10

Solve for x: 4^(x+2)=4^14.

Skill: Equating exponents with a common base

Show answer and worked solution

Answer: x=12

Because the common base 4 is positive and not 1, equal powers have equal exponents. Thus x+2=14, so x=12.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Exponents and logarithmsFluency practiceShort answer

Logarithm laws 1

Evaluate log_3(3^2)+2log_3(3^4).

Skill: Applying logarithm power and product laws

Show answer and worked solution

Answer: 10

Use log_b(b^k)=k. The expression is 2+2(4)=10. Equivalently, combine it as log_3(3^10).

Exponents and logarithmsFluency practiceShort answer

Logarithm laws 4

Evaluate log_3(3^1)+2log_3(3^3).

Skill: Applying logarithm power and product laws

Show answer and worked solution

Answer: 7

Use log_b(b^k)=k. The expression is 1+2(3)=7. Equivalently, combine it as log_3(3^7).

Exponents and logarithmsFluency practiceFull solution

Logarithm laws 7

Evaluate log_3(3^4)+2log_3(3^2).

Skill: Applying logarithm power and product laws

Show answer and worked solution

Answer: 8

Use log_b(b^k)=k. The expression is 4+2(2)=8. Equivalently, combine it as log_3(3^8).

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Exponents and logarithmsFluency practiceFull solution

Logarithm laws 10

Evaluate log_3(3^3)+2log_3(3^5).

Skill: Applying logarithm power and product laws

Show answer and worked solution

Answer: 13

Use log_b(b^k)=k. The expression is 3+2(5)=13. Equivalently, combine it as log_3(3^13).

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Topic drill

Exponents and logarithms drill 4

35 minutes

Six exponents and logarithms practice questions. Work the set without notes, then compare every line of a full-solution response with the explanation.

6 practice questions · 1 authored problem worth 10 marks

1Equal-base exponent equation 12
2Logarithm laws 3
3Logarithm laws 6
4Logarithm laws 9
5Logarithm laws 12
6An exponent inside an exponent

Open all questions and solutions

Exponents and logarithmsFluency practiceFull solution

Equal-base exponent equation 12

Solve for x: 2^(x+1)=2^15.

Skill: Equating exponents with a common base

Show answer and worked solution

Answer: x=14

Because the common base 2 is positive and not 1, equal powers have equal exponents. Thus x+1=15, so x=14.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Exponents and logarithmsFluency practiceShort answer

Logarithm laws 3

Evaluate log_2(2^4)+2log_2(2^2).

Skill: Applying logarithm power and product laws

Show answer and worked solution

Answer: 8

Use log_b(b^k)=k. The expression is 4+2(2)=8. Equivalently, combine it as log_2(2^8).

Exponents and logarithmsFluency practiceFull solution

Logarithm laws 6

Evaluate log_2(2^3)+2log_2(2^5).

Skill: Applying logarithm power and product laws

Show answer and worked solution

Answer: 13

Use log_b(b^k)=k. The expression is 3+2(5)=13. Equivalently, combine it as log_2(2^13).

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Exponents and logarithmsFluency practiceFull solution

Logarithm laws 9

Evaluate log_2(2^2)+2log_2(2^4).

Skill: Applying logarithm power and product laws

Show answer and worked solution

Answer: 10

Use log_b(b^k)=k. The expression is 2+2(4)=10. Equivalently, combine it as log_2(2^10).

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Exponents and logarithmsFluency practiceFull solution

Logarithm laws 12

Evaluate log_2(2^1)+2log_2(2^3).

Skill: Applying logarithm power and product laws

Show answer and worked solution

Answer: 7

Use log_b(b^k)=k. The expression is 1+2(3)=7. Equivalently, combine it as log_2(2^7).

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Exponents and logarithmsChallengeFull solution10 marks

An exponent inside an exponent

Determine all positive real numbers x satisfying x^(log₂x)=16.

Skill: Substitution using a logarithmic variable

Show answer and worked solution

Answer: x=4 or x=1/4

Let y=log₂x, so x=2^y. Then x^(log₂x)=(2^y)^y=2^(y²). Since 16=2⁴, y²=4 and y=±2. Therefore x=2²=4 or x=2^(-2)=1/4. Both are positive and satisfy the equation.

Practice marking checkpoints

• 2 marks for setting y=log₂x
• 3 marks for obtaining 2^(y²)=2⁴
• 3 marks for both values y=±2
• 2 marks for converting back and checking both x-values

Common mistake: Keeping only y=2 after solving y²=4 and losing the valid solution x=1/4.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Topic drill

Trigonometry drill 1

35 minutes

Six trigonometry practice questions. Work the set without notes, then compare every line of a full-solution response with the explanation.

6 practice questions · 2 authored problems worth 20 marks

1Exact cosine of an angle sum
2A median after the cosine law
3Exact right-triangle ratio 3
4Exact right-triangle ratio 6
5Exact right-triangle ratio 9
6Exact right-triangle ratio 12

Open all questions and solutions

TrigonometryCoreFull solution10 marks

Exact cosine of an angle sum

Acute angles A and B satisfy sin A=3/5 and cos B=12/13. Determine cos(A+B) exactly.

Skill: Using exact ratios and the cosine addition identity

Show answer and worked solution

Answer: 33/65

Since A is acute, cos A=4/5. Since B is acute, sin B=5/13. Therefore cos(A+B)=cosAcosB−sinAsinB=(4/5)(12/13)−(3/5)(5/13)=(48−15)/65=33/65.

Practice marking checkpoints

• 2 marks for cosA=4/5
• 2 marks for sinB=5/13
• 3 marks for the correct identity
• 2 marks for 33/65
• 1 mark for exact presentation

Common mistake: Using cos(A+B)=cosAcosB+sinAsinB; the plus sign belongs to cos(A−B).

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

TrigonometryChallengeFull solution10 marks

A median after the cosine law

In triangle ABC, AB=5, AC=7, and ∠A=60°. Determine the exact length of the median from A to side BC.

Skill: Combining the cosine law with a median formula

Show answer and worked solution

Answer: √109/2

First, the cosine law gives BC²=5²+7²−2(5)(7)cos60°=25+49−35=39. If m is the median from A, Apollonius’ theorem gives m²=[2(5²)+2(7²)−BC²]/4=(50+98−39)/4=109/4. Thus m=√109/2.

Practice marking checkpoints

• 3 marks for BC²=39
• 2 marks for a correct median relation
• 3 marks for m²=109/4
• 2 marks for the positive exact length

Common mistake: Assuming the median is also an altitude; that is not true in a general non-isosceles triangle.

Alternative route: Place B=(0,0), C=(√39,0), use the two distance equations to find A, and compute the distance from A to the midpoint of BC.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

TrigonometryFluency practiceShort answer

Exact right-triangle ratio 3

A right triangle has legs 7 and 24. If angle θ is opposite the shorter listed leg, determine sin θ exactly.

Skill: Using exact trigonometric ratios

Show answer and worked solution

Answer: 7/25

The side lengths are a scale-1 copy of the 7-24-25 right triangle. Therefore sin θ=opposite/hypotenuse=7/25=7/25.

TrigonometryFluency practiceFull solution

Exact right-triangle ratio 6

A right triangle has legs 16 and 30. If angle θ is opposite the shorter listed leg, determine sin θ exactly.

Skill: Using exact trigonometric ratios

Show answer and worked solution

Answer: 8/17

The side lengths are a scale-2 copy of the 8-15-17 right triangle. Therefore sin θ=opposite/hypotenuse=16/34=8/17.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

TrigonometryFluency practiceFull solution

Exact right-triangle ratio 9

A right triangle has legs 15 and 36. If angle θ is opposite the shorter listed leg, determine sin θ exactly.

Skill: Using exact trigonometric ratios

Show answer and worked solution

Answer: 5/13

The side lengths are a scale-3 copy of the 5-12-13 right triangle. Therefore sin θ=opposite/hypotenuse=15/39=5/13.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

TrigonometryFluency practiceFull solution

Exact right-triangle ratio 12

A right triangle has legs 9 and 12. If angle θ is opposite the shorter listed leg, determine sin θ exactly.

Skill: Using exact trigonometric ratios

Show answer and worked solution

Answer: 3/5

The side lengths are a scale-3 copy of the 3-4-5 right triangle. Therefore sin θ=opposite/hypotenuse=9/15=3/5.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Topic drill

Trigonometry drill 2

35 minutes

Six trigonometry practice questions. Work the set without notes, then compare every line of a full-solution response with the explanation.

6 practice questions

1Exact right-triangle ratio 2
2Exact right-triangle ratio 5
3Exact right-triangle ratio 8
4Exact right-triangle ratio 11
5Cosine-law side 2
6Cosine-law side 5

Open all questions and solutions

TrigonometryFluency practiceShort answer

Exact right-triangle ratio 2

A right triangle has legs 8 and 15. If angle θ is opposite the shorter listed leg, determine sin θ exactly.

Skill: Using exact trigonometric ratios

Show answer and worked solution

Answer: 8/17

The side lengths are a scale-1 copy of the 8-15-17 right triangle. Therefore sin θ=opposite/hypotenuse=8/17=8/17.

TrigonometryFluency practiceFull solution

Exact right-triangle ratio 5

A right triangle has legs 10 and 24. If angle θ is opposite the shorter listed leg, determine sin θ exactly.

Skill: Using exact trigonometric ratios

Show answer and worked solution

Answer: 5/13

The side lengths are a scale-2 copy of the 5-12-13 right triangle. Therefore sin θ=opposite/hypotenuse=10/26=5/13.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

TrigonometryFluency practiceFull solution

Exact right-triangle ratio 8

A right triangle has legs 6 and 8. If angle θ is opposite the shorter listed leg, determine sin θ exactly.

Skill: Using exact trigonometric ratios

Show answer and worked solution

Answer: 3/5

The side lengths are a scale-2 copy of the 3-4-5 right triangle. Therefore sin θ=opposite/hypotenuse=6/10=3/5.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

TrigonometryFluency practiceFull solution

Exact right-triangle ratio 11

A right triangle has legs 21 and 72. If angle θ is opposite the shorter listed leg, determine sin θ exactly.

Skill: Using exact trigonometric ratios

Show answer and worked solution

Answer: 7/25

The side lengths are a scale-3 copy of the 7-24-25 right triangle. Therefore sin θ=opposite/hypotenuse=21/75=7/25.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

TrigonometryFluency practiceShort answer

Cosine-law side 2

Two sides of a triangle have lengths 5 and 9, and their included angle is 60°. If the opposite side has length c, find c².

Skill: Applying the cosine law with exact values

Show answer and worked solution

Answer: 61

The cosine law gives c²=5²+9²−2(5)(9)cos60°. Since cos60°=1/2, c²=25+81−45=61.

TrigonometryFluency practiceFull solution

Cosine-law side 5

Two sides of a triangle have lengths 8 and 8, and their included angle is 60°. If the opposite side has length c, find c².

Skill: Applying the cosine law with exact values

Show answer and worked solution

Answer: 64

The cosine law gives c²=8²+8²−2(8)(8)cos60°. Since cos60°=1/2, c²=64+64−64=64.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Topic drill

Trigonometry drill 3

35 minutes

Six trigonometry practice questions. Work the set without notes, then compare every line of a full-solution response with the explanation.

6 practice questions

1Exact right-triangle ratio 7
2Exact right-triangle ratio 10
3Cosine-law side 1
4Cosine-law side 4
5Cosine-law side 7
6Cosine-law side 10

Open all questions and solutions

TrigonometryFluency practiceFull solution

Exact right-triangle ratio 7

A right triangle has legs 14 and 48. If angle θ is opposite the shorter listed leg, determine sin θ exactly.

Skill: Using exact trigonometric ratios

Show answer and worked solution

Answer: 7/25

The side lengths are a scale-2 copy of the 7-24-25 right triangle. Therefore sin θ=opposite/hypotenuse=14/50=7/25.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

TrigonometryFluency practiceFull solution

Exact right-triangle ratio 10

A right triangle has legs 24 and 45. If angle θ is opposite the shorter listed leg, determine sin θ exactly.

Skill: Using exact trigonometric ratios

Show answer and worked solution

Answer: 8/17

The side lengths are a scale-3 copy of the 8-15-17 right triangle. Therefore sin θ=opposite/hypotenuse=24/51=8/17.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

TrigonometryFluency practiceShort answer

Cosine-law side 1

Two sides of a triangle have lengths 4 and 7, and their included angle is 60°. If the opposite side has length c, find c².

Skill: Applying the cosine law with exact values

Show answer and worked solution

Answer: 37

The cosine law gives c²=4²+7²−2(4)(7)cos60°. Since cos60°=1/2, c²=16+49−28=37.

TrigonometryFluency practiceShort answer

Cosine-law side 4

Two sides of a triangle have lengths 7 and 6, and their included angle is 60°. If the opposite side has length c, find c².

Skill: Applying the cosine law with exact values

Show answer and worked solution

Answer: 43

The cosine law gives c²=7²+6²−2(7)(6)cos60°. Since cos60°=1/2, c²=49+36−42=43.

TrigonometryFluency practiceFull solution

Cosine-law side 7

Two sides of a triangle have lengths 4 and 5, and their included angle is 60°. If the opposite side has length c, find c².

Skill: Applying the cosine law with exact values

Show answer and worked solution

Answer: 21

The cosine law gives c²=4²+5²−2(4)(5)cos60°. Since cos60°=1/2, c²=16+25−20=21.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

TrigonometryFluency practiceFull solution

Cosine-law side 10

Two sides of a triangle have lengths 7 and 11, and their included angle is 60°. If the opposite side has length c, find c².

Skill: Applying the cosine law with exact values

Show answer and worked solution

Answer: 93

The cosine law gives c²=7²+11²−2(7)(11)cos60°. Since cos60°=1/2, c²=49+121−77=93.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Topic drill

Trigonometry drill 4

35 minutes

Six trigonometry practice questions. Work the set without notes, then compare every line of a full-solution response with the explanation.

6 practice questions · 1 authored problem worth 10 marks

1Exact right-triangle ratio 12
2Cosine-law side 3
3Cosine-law side 6
4Cosine-law side 9
5Cosine-law side 12
6A trigonometric equation on a full cycle

Open all questions and solutions

TrigonometryFluency practiceFull solution

Exact right-triangle ratio 12

A right triangle has legs 9 and 12. If angle θ is opposite the shorter listed leg, determine sin θ exactly.

Skill: Using exact trigonometric ratios

Show answer and worked solution

Answer: 3/5

The side lengths are a scale-3 copy of the 3-4-5 right triangle. Therefore sin θ=opposite/hypotenuse=9/15=3/5.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

TrigonometryFluency practiceShort answer

Cosine-law side 3

Two sides of a triangle have lengths 6 and 11, and their included angle is 60°. If the opposite side has length c, find c².

Skill: Applying the cosine law with exact values

Show answer and worked solution

Answer: 91

The cosine law gives c²=6²+11²−2(6)(11)cos60°. Since cos60°=1/2, c²=36+121−66=91.

TrigonometryFluency practiceFull solution

Cosine-law side 6

Two sides of a triangle have lengths 3 and 10, and their included angle is 60°. If the opposite side has length c, find c².

Skill: Applying the cosine law with exact values

Show answer and worked solution

Answer: 79

The cosine law gives c²=3²+10²−2(3)(10)cos60°. Since cos60°=1/2, c²=9+100−30=79.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

TrigonometryFluency practiceFull solution

Cosine-law side 9

Two sides of a triangle have lengths 6 and 9, and their included angle is 60°. If the opposite side has length c, find c².

Skill: Applying the cosine law with exact values

Show answer and worked solution

Answer: 63

The cosine law gives c²=6²+9²−2(6)(9)cos60°. Since cos60°=1/2, c²=36+81−54=63.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

TrigonometryFluency practiceFull solution

Cosine-law side 12

Two sides of a triangle have lengths 3 and 8, and their included angle is 60°. If the opposite side has length c, find c².

Skill: Applying the cosine law with exact values

Show answer and worked solution

Answer: 49

The cosine law gives c²=3²+8²−2(3)(8)cos60°. Since cos60°=1/2, c²=9+64−24=49.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

TrigonometryChallengeFull solution10 marks

A trigonometric equation on a full cycle

Determine all angles x, in degrees, satisfying 0°≤x<360° and sin(2x)=sin x.

Skill: Factoring a trigonometric equation and covering all cases

Show answer and worked solution

Answer: x=0°, 60°, 180°, or 300°

Use sin(2x)=2sinx cosx. Then 2sinx cosx=sinx, so sinx(2cosx−1)=0. Hence sinx=0, giving x=0° or 180°, or cosx=1/2, giving x=60° or 300°.

Practice marking checkpoints

• 2 marks for the double-angle identity
• 2 marks for correct factoring
• 2 marks for the sinx=0 case
• 3 marks for the cosx=1/2 case
• 1 mark for the complete interval check

Common mistake: Dividing by sinx and losing the solutions x=0° and x=180°.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

12 matching 150-minute practice sets

Each set has ten 10-mark questions and a mix of short-answer and full-solution work. These are practice sets built from the question bank, not copies of CEMC papers, and they reuse questions that also appear in drills. The scope mirrors the eight published CEMC preparation areas while every question is our own rather than copied from a CEMC paper.

Practice set from the question bank

150-minute practice set 1

150 minutes

A practice set built from the question bank: 10 questions, 150 minutes, short-answer and full-solution work across all eight CEMC preparation areas. Authored problems in a set carry 10 practice marks each; generated fluency questions carry no mark value. Sets reuse questions from the same bank, so a question can also appear in drills and other sets.

10 practice questions · 6 authored problems worth 60 marks

1Equal-base exponent equation 1
2Exact right-triangle ratio 2
3Polynomial remainder 3
4Slope between two points 4
5A first-order recurrence
6Ptolemy’s theorem in a cyclic quadrilateral
7Selecting non-consecutive integers
8Three simultaneous congruences
9A sum of related powers
10Recovering an infinite geometric sequence

Open all questions and solutions

Exponents and logarithmsFluency practiceShort answer

Equal-base exponent equation 1

Solve for x: 3^(x+2)=3^5.

Skill: Equating exponents with a common base

Show answer and worked solution

Answer: x=3

Because the common base 3 is positive and not 1, equal powers have equal exponents. Thus x+2=5, so x=3.

TrigonometryFluency practiceShort answer

Exact right-triangle ratio 2

A right triangle has legs 8 and 15. If angle θ is opposite the shorter listed leg, determine sin θ exactly.

Skill: Using exact trigonometric ratios

Show answer and worked solution

Answer: 8/17

The side lengths are a scale-1 copy of the 8-15-17 right triangle. Therefore sin θ=opposite/hypotenuse=8/17=8/17.

Functions and polynomialsFluency practiceShort answer

Polynomial remainder 3

Find the remainder when P(x)=x³+4x²−8x+6 is divided by x−4.

Skill: Using the remainder theorem

Show answer and worked solution

Answer: 102

By the remainder theorem, the remainder is P(4). Substitution gives 4³+4(4²)−8(4)+6=102.

Analytic geometryFluency practiceShort answer

Slope between two points 4

Find the slope of the line through A(-3,-1) and B(3,3).

Skill: Computing and simplifying slope

Show answer and worked solution

Answer: 2/3

Slope is change in y divided by change in x: [3−(-1)]/[3−(-3)]=4/6=2/3.

Sequences and seriesChallengeFull solution10 marks

A first-order recurrence

A sequence satisfies a₁=2 and a_(n+1)=3a_n+1. Determine a₅ and derive an explicit formula for a_n.

Skill: Shifting a recurrence to make it geometric

Show answer and worked solution

Answer: a₅=202 and a_n=(5·3^(n−1)−1)/2

Set b_n=a_n+1/2. Then b_(n+1)=3a_n+3/2=3b_n, and b₁=5/2. Hence b_n=(5/2)3^(n−1), so a_n=(5·3^(n−1)−1)/2. Substituting n=5 gives (5·81−1)/2=202.

Practice marking checkpoints

• 3 marks for a useful shift or equivalent recurrence work
• 3 marks for the explicit formula
• 2 marks for verifying the formula
• 2 marks for a₅=202

Common mistake: Assuming the sequence is geometric even though the added 1 changes every ratio.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Euclidean geometryChallengeFull solution10 marks

Ptolemy’s theorem in a cyclic quadrilateral

A cyclic quadrilateral has consecutive side lengths 2, 3, 4, and 5. Determine the product of the lengths of its diagonals.

Skill: Applying Ptolemy’s theorem

Show answer and worked solution

Answer: 23

Ptolemy’s theorem states that for a cyclic quadrilateral, the product of the diagonals equals the sum of the products of opposite sides. Therefore the product is 2·4+3·5=8+15=23.

Practice marking checkpoints

• 4 marks for stating or correctly applying Ptolemy’s theorem
• 3 marks for identifying opposite side pairs
• 2 marks for 23
• 1 mark for a clear conclusion

Common mistake: Pairing adjacent sides in both products instead of the two pairs of opposite sides.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Counting and probabilityChallengeFull solution10 marks

Selecting non-consecutive integers

How many three-element subsets of {1,2,3,4,5,6,7,8} contain no two consecutive integers?

Skill: Gap removal for selections with forbidden adjacency

Show answer and worked solution

Answer: 20

Write the chosen numbers as a<b<c and define a′=a, b′=b−1, c′=c−2. The non-consecutive condition becomes 1≤a′<b′<c′≤6. Thus choose any three values from {1,…,6}, giving C(6,3)=20.

Practice marking checkpoints

• 3 marks for encoding the gaps
• 3 marks for a bijection with three choices from six
• 3 marks for C(6,3)=20
• 1 mark for explaining reversibility

Common mistake: Starting with C(8,3) and subtracting adjacent pairs without correcting overlaps among forbidden cases.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Properties of numbersChallengeFull solution10 marks

Three simultaneous congruences

Determine the smallest positive integer n satisfying n≡2 (mod 5), n≡3 (mod 7), and n≡4 (mod 9).

Skill: Solving compatible congruences step by step

Show answer and worked solution

Answer: 157

The first two congruences have solution n≡17 (mod 35), so write n=17+35k. Modulo 9 this is 8+8k≡4, so 8k≡5 (mod 9). Since 8≡−1, k≡4 (mod 9). The smallest non-negative k is 4, giving n=17+35(4)=157.

Practice marking checkpoints

• 3 marks for combining the mod 5 and mod 7 conditions
• 3 marks for the modulo 9 equation
• 2 marks for k=4
• 2 marks for 157 and a check

Common mistake: Finding one number that satisfies two congruences and failing to preserve the full arithmetic progression while imposing the third.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Exponents and logarithmsCoreFull solution10 marks

A sum of related powers

Solve for the real number x: 2^x+2^(x+1)+2^(x+2)=448.

Skill: Factoring a common exponential term

Show answer and worked solution

Answer: x=6

Factor out the smallest power: 2^x(1+2+4)=448. Thus 7·2^x=448, so 2^x=64=2^6. Therefore x=6.

Practice marking checkpoints

• 2 marks for identifying a common factor of 2^x
• 4 marks for reducing the equation to 2^x=64
• 3 marks for solving x=6
• 1 mark for a clear conclusion

Common mistake: Adding the exponents across a sum. Exponent laws apply to products, not directly to addition.

Alternative route: Set y=2^x. The equation becomes y+2y+4y=448, so 7y=448 and y=64.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Sequences and seriesCoreFull solution10 marks

Recovering an infinite geometric sequence

An infinite geometric series has first term 12 and sum 20. Determine its common ratio and its fifth term.

Skill: Using an infinite sum to recover a ratio

Show answer and worked solution

Answer: r=2/5 and the fifth term is 192/625

For |r|<1, 20=12/(1−r). Thus 1−r=3/5 and r=2/5. The fifth term is 12(2/5)^4=12·16/625=192/625.

Practice marking checkpoints

• 3 marks for the infinite-sum equation
• 3 marks for r=2/5
• 3 marks for the fifth term
• 1 mark for confirming |r|<1

Common mistake: Using 12r⁵ for the fifth term; the first term corresponds to exponent 0, so the fifth uses r⁴.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Practice set from the question bank

150-minute practice set 2

150 minutes

10 practice questions · 4 authored problems worth 40 marks

1Equal-base exponent equation 2
2Exact right-triangle ratio 3
3Polynomial remainder 4
4Circle from centre and radius 1
5Describing the previous term
6Length of an internal angle bisector
7Committee with a required member 5
8Positive-divisor count 6
9A median after the cosine law
10Selecting non-consecutive integers

Open all questions and solutions

Exponents and logarithmsFluency practiceShort answer

Equal-base exponent equation 2

Solve for x: 4^(x+3)=4^7.

Skill: Equating exponents with a common base

Show answer and worked solution

Answer: x=4

Because the common base 4 is positive and not 1, equal powers have equal exponents. Thus x+3=7, so x=4.

TrigonometryFluency practiceShort answer

Exact right-triangle ratio 3

A right triangle has legs 7 and 24. If angle θ is opposite the shorter listed leg, determine sin θ exactly.

Skill: Using exact trigonometric ratios

Show answer and worked solution

Answer: 7/25

The side lengths are a scale-1 copy of the 7-24-25 right triangle. Therefore sin θ=opposite/hypotenuse=7/25=7/25.

Functions and polynomialsFluency practiceShort answer

Polynomial remainder 4

Find the remainder when P(x)=x³+5x²−3x+7 is divided by x−1.

Skill: Using the remainder theorem

Show answer and worked solution

Answer: 10

By the remainder theorem, the remainder is P(1). Substitution gives 1³+5(1²)−3(1)+7=10.

Analytic geometryFluency practiceShort answer

Circle from centre and radius 1

Write the equation of the circle with centre (-2,-2) and radius 3.

Skill: Writing a circle in centre-radius form

Show answer and worked solution

Answer: (x+2)²+(y+2)²=9

A circle with centre (h,k) and radius r has equation (x−h)²+(y−k)²=r². Substituting h=-2, k=-2, and r=3 gives (x+2)²+(y+2)²=9.

Sequences and seriesCoreFull solution10 marks

Describing the previous term

The sequence begins 1, 11, 21, 1211, 111221. Each term describes the consecutive digits in the previous term. Determine the next term and explain the grouping.

Skill: Interpreting a recursively described digit pattern

Show answer and worked solution

Answer: 312211

The term 111221 consists of three 1s, two 2s, and one 1. Describing those consecutive groups gives 31, 22, 11, which concatenate to 312211.

Practice marking checkpoints

• 3 marks for preserving consecutive groups
• 4 marks for the three correct descriptions
• 2 marks for 312211
• 1 mark for explanation

Common mistake: Counting all occurrences of each digit in the entire term instead of reading consecutive runs from left to right.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Euclidean geometryChallengeFull solution10 marks

Length of an internal angle bisector

In triangle ABC, AB=6, AC=9, and BC=10. The internal bisector of ∠A meets BC at D. Determine AD exactly.

Skill: Combining the angle-bisector theorem with Stewart’s theorem

Show answer and worked solution

Answer: √30

The angle-bisector theorem gives BD:DC=AB:AC=2:3. Since BC=10, BD=4 and DC=6. The angle-bisector form of Stewart’s theorem gives AD²=AB·AC−BD·DC=6·9−4·6=30. Hence AD=√30.

Practice marking checkpoints

• 3 marks for BD=4 and DC=6
• 3 marks for a valid Stewart/angle-bisector relation
• 3 marks for AD²=30
• 1 mark for the positive exact length

Common mistake: Assuming the angle bisector also bisects BC; that occurs only in an isosceles triangle at A.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Counting and probabilityFluency practiceFull solution

Committee with a required member 5

A 5-student committee is chosen from 13 students. If Ava must be on the committee, how many committees are possible?

Skill: Counting selections under a restriction

Show answer and worked solution

Answer: 495

Ava's place is fixed. Choose the remaining 4 members from the other 12 students: C(12,4)=495.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Properties of numbersFluency practiceFull solution

Positive-divisor count 6

The integer 12 has prime factorization 2^2·3^1. How many positive divisors does it have?

Skill: Counting divisors from a prime factorization

Show answer and worked solution

Answer: 6

A divisor independently chooses the exponent of 2 from 0 through 2 and the exponent of 3 from 0 through 1. Therefore the count is (2+1)(1+1)=6.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

TrigonometryChallengeFull solution10 marks

A median after the cosine law

In triangle ABC, AB=5, AC=7, and ∠A=60°. Determine the exact length of the median from A to side BC.

Skill: Combining the cosine law with a median formula

Show answer and worked solution

Answer: √109/2

First, the cosine law gives BC²=5²+7²−2(5)(7)cos60°=25+49−35=39. If m is the median from A, Apollonius’ theorem gives m²=[2(5²)+2(7²)−BC²]/4=(50+98−39)/4=109/4. Thus m=√109/2.

Practice marking checkpoints

• 3 marks for BC²=39
• 2 marks for a correct median relation
• 3 marks for m²=109/4
• 2 marks for the positive exact length

Common mistake: Assuming the median is also an altitude; that is not true in a general non-isosceles triangle.

Alternative route: Place B=(0,0), C=(√39,0), use the two distance equations to find A, and compute the distance from A to the midpoint of BC.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Counting and probabilityChallengeFull solution10 marks

Selecting non-consecutive integers

How many three-element subsets of {1,2,3,4,5,6,7,8} contain no two consecutive integers?

Skill: Gap removal for selections with forbidden adjacency

Show answer and worked solution

Answer: 20

Practice marking checkpoints

• 3 marks for encoding the gaps
• 3 marks for a bijection with three choices from six
• 3 marks for C(6,3)=20
• 1 mark for explaining reversibility

Common mistake: Starting with C(8,3) and subtracting adjacent pairs without correcting overlaps among forbidden cases.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Practice set from the question bank

150-minute practice set 3

150 minutes

10 practice questions · 1 authored problem worth 10 marks

1Equal-base exponent equation 3
2Exact right-triangle ratio 4
3Reciprocals of polynomial roots 1
4Circle from centre and radius 2
5Arithmetic-series sum 5
6Similar-triangle side 6
7Committee with a required member 7
8Positive-divisor count 8
9Line-circle intersections
10Equal-base exponent equation 6

Open all questions and solutions

Exponents and logarithmsFluency practiceShort answer

Equal-base exponent equation 3

Solve for x: 5^(x+1)=5^6.

Skill: Equating exponents with a common base

Show answer and worked solution

Answer: x=5

Because the common base 5 is positive and not 1, equal powers have equal exponents. Thus x+1=6, so x=5.

TrigonometryFluency practiceShort answer

Exact right-triangle ratio 4

A right triangle has legs 3 and 4. If angle θ is opposite the shorter listed leg, determine sin θ exactly.

Skill: Using exact trigonometric ratios

Show answer and worked solution

Answer: 3/5

The side lengths are a scale-1 copy of the 3-4-5 right triangle. Therefore sin θ=opposite/hypotenuse=3/5=3/5.

Functions and polynomialsFluency practiceShort answer

Reciprocals of polynomial roots 1

The roots of x²−10x+21=0 are r and s. Determine 1/r+1/s.

Skill: Combining Vieta’s formulas with symmetric expressions

Show answer and worked solution

Answer: 10/21

Vieta's formulas give r+s=10 and rs=21. Therefore 1/r+1/s=(r+s)/(rs)=10/21=10/21.

Analytic geometryFluency practiceShort answer

Circle from centre and radius 2

Write the equation of the circle with centre (-1,0) and radius 4.

Skill: Writing a circle in centre-radius form

Show answer and worked solution

Answer: (x+1)²+y²=16

A circle with centre (h,k) and radius r has equation (x−h)²+(y−k)²=r². Substituting h=-1, k=0, and r=4 gives (x+1)²+y²=16.

Sequences and seriesFluency practiceFull solution

Arithmetic-series sum 5

An arithmetic sequence has first term 7, common difference 7, and 13 terms. Find the sum of all its terms.

Skill: Finding a finite arithmetic-series sum

Show answer and worked solution

Answer: 637

The last term is 7+(13−1)7=91. Thus S=13(7+91)/2=637.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Euclidean geometryFluency practiceFull solution

Similar-triangle side 6

Two triangles are similar. A side of length 9 in the smaller triangle corresponds to a side of length 36 in the larger triangle. What length corresponds to a different smaller side of length 6?

Skill: Using similarity and scale factors

Show answer and worked solution

Answer: 24

The scale factor from the smaller triangle to the larger is 36/9=4. Multiply the corresponding side by 4: 6×4=24.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Counting and probabilityFluency practiceFull solution

Committee with a required member 7

A 4-student committee is chosen from 15 students. If Ava must be on the committee, how many committees are possible?

Skill: Counting selections under a restriction

Show answer and worked solution

Answer: 364

Ava's place is fixed. Choose the remaining 3 members from the other 14 students: C(14,3)=364.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Properties of numbersFluency practiceFull solution

Positive-divisor count 8

The integer 432 has prime factorization 2^4·3^3. How many positive divisors does it have?

Skill: Counting divisors from a prime factorization

Show answer and worked solution

Answer: 20

A divisor independently chooses the exponent of 2 from 0 through 4 and the exponent of 3 from 0 through 3. Therefore the count is (4+1)(3+1)=20.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Analytic geometryCoreFull solution10 marks

Line-circle intersections

Determine all points where the line y=x+1 intersects the circle x²+y²=25.

Skill: Solving a line-circle system by substitution

Show answer and worked solution

Answer: (3,4) and (−4,−3)

Substitute y=x+1 into the circle: x²+(x+1)²=25, so 2x²+2x−24=0 and x²+x−12=0. Factoring gives (x−3)(x+4)=0, so x=3 or x=−4. Then y=x+1 gives the two points (3,4) and (−4,−3).

Practice marking checkpoints

• 2 marks for substitution
• 3 marks for the quadratic
• 2 marks for both x-values
• 2 marks for both points
• 1 mark for checking

Common mistake: Reporting the x-values as the final answer when the question asks for points.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Exponents and logarithmsFluency practiceFull solution

Equal-base exponent equation 6

Solve for x: 4^(x+1)=4^9.

Skill: Equating exponents with a common base

Show answer and worked solution

Answer: x=8

Because the common base 4 is positive and not 1, equal powers have equal exponents. Thus x+1=9, so x=8.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Practice set from the question bank

150-minute practice set 4

150 minutes

10 practice questions · 1 authored problem worth 10 marks

1Equal-base exponent equation 4
2Cosine-law side 1
3Reciprocals of polynomial roots 2
4Circle from centre and radius 3
5Arithmetic-series sum 7
6Similar-triangle side 8
7Committee with a required member 9
8Positive-divisor count 10
9Ptolemy’s theorem in a cyclic quadrilateral
10Logarithm laws 5

Open all questions and solutions

Exponents and logarithmsFluency practiceShort answer

Equal-base exponent equation 4

Solve for x: 2^(x+2)=2^8.

Skill: Equating exponents with a common base

Show answer and worked solution

Answer: x=6

Because the common base 2 is positive and not 1, equal powers have equal exponents. Thus x+2=8, so x=6.

TrigonometryFluency practiceShort answer

Cosine-law side 1

Two sides of a triangle have lengths 4 and 7, and their included angle is 60°. If the opposite side has length c, find c².

Skill: Applying the cosine law with exact values

Show answer and worked solution

Answer: 37

The cosine law gives c²=4²+7²−2(4)(7)cos60°. Since cos60°=1/2, c²=16+49−28=37.

Functions and polynomialsFluency practiceShort answer

Reciprocals of polynomial roots 2

The roots of x²−12x+32=0 are r and s. Determine 1/r+1/s.

Skill: Combining Vieta’s formulas with symmetric expressions

Show answer and worked solution

Answer: 3/8

Vieta's formulas give r+s=12 and rs=32. Therefore 1/r+1/s=(r+s)/(rs)=12/32=3/8.

Analytic geometryFluency practiceShort answer

Circle from centre and radius 3

Write the equation of the circle with centre (0,2) and radius 5.

Skill: Writing a circle in centre-radius form

Show answer and worked solution

Answer: x²+(y−2)²=25

A circle with centre (h,k) and radius r has equation (x−h)²+(y−k)²=r². Substituting h=0, k=2, and r=5 gives x²+(y−2)²=25.

Sequences and seriesFluency practiceFull solution

Arithmetic-series sum 7

An arithmetic sequence has first term 9, common difference 3, and 15 terms. Find the sum of all its terms.

Skill: Finding a finite arithmetic-series sum

Show answer and worked solution

Answer: 450

The last term is 9+(15−1)3=51. Thus S=15(9+51)/2=450.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Euclidean geometryFluency practiceFull solution

Similar-triangle side 8

Two triangles are similar. A side of length 4 in the smaller triangle corresponds to a side of length 8 in the larger triangle. What length corresponds to a different smaller side of length 8?

Skill: Using similarity and scale factors

Show answer and worked solution

Answer: 16

The scale factor from the smaller triangle to the larger is 8/4=2. Multiply the corresponding side by 2: 8×2=16.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Counting and probabilityFluency practiceFull solution

Committee with a required member 9

A 3-student committee is chosen from 8 students. If Ava must be on the committee, how many committees are possible?

Skill: Counting selections under a restriction

Show answer and worked solution

Answer: 21

Ava's place is fixed. Choose the remaining 2 members from the other 7 students: C(7,2)=21.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Properties of numbersFluency practiceFull solution

Positive-divisor count 10

The integer 6 has prime factorization 2^1·3^1. How many positive divisors does it have?

Skill: Counting divisors from a prime factorization

Show answer and worked solution

Answer: 4

A divisor independently chooses the exponent of 2 from 0 through 1 and the exponent of 3 from 0 through 1. Therefore the count is (1+1)(1+1)=4.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Euclidean geometryChallengeFull solution10 marks

Ptolemy’s theorem in a cyclic quadrilateral

A cyclic quadrilateral has consecutive side lengths 2, 3, 4, and 5. Determine the product of the lengths of its diagonals.

Skill: Applying Ptolemy’s theorem

Show answer and worked solution

Answer: 23

Ptolemy’s theorem states that for a cyclic quadrilateral, the product of the diagonals equals the sum of the products of opposite sides. Therefore the product is 2·4+3·5=8+15=23.

Practice marking checkpoints

• 4 marks for stating or correctly applying Ptolemy’s theorem
• 3 marks for identifying opposite side pairs
• 2 marks for 23
• 1 mark for a clear conclusion

Common mistake: Pairing adjacent sides in both products instead of the two pairs of opposite sides.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Exponents and logarithmsFluency practiceFull solution

Logarithm laws 5

Evaluate log_4(4^2)+2log_4(4^4).

Skill: Applying logarithm power and product laws

Show answer and worked solution

Answer: 10

Use log_b(b^k)=k. The expression is 2+2(4)=10. Equivalently, combine it as log_4(4^10).

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

224 matching practice questions

Attempt each problem on paper before opening the explanation. For a full-solution question, award yourself credit only when another student could follow the argument without guessing a missing step.

Exponents and logarithmsCoreFull solution10 marks

A sum of related powers

Solve for the real number x: 2^x+2^(x+1)+2^(x+2)=448.

Skill: Factoring a common exponential term

Show answer and worked solution

Answer: x=6

Factor out the smallest power: 2^x(1+2+4)=448. Thus 7·2^x=448, so 2^x=64=2^6. Therefore x=6.

Practice marking checkpoints

• 2 marks for identifying a common factor of 2^x
• 4 marks for reducing the equation to 2^x=64
• 3 marks for solving x=6
• 1 mark for a clear conclusion

Common mistake: Adding the exponents across a sum. Exponent laws apply to products, not directly to addition.

Alternative route: Set y=2^x. The equation becomes y+2y+4y=448, so 7y=448 and y=64.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Exponents and logarithmsCoreFull solution10 marks

Changing logarithm bases

Positive numbers a and b satisfy log₂(a)=3 and log_a(b)=2. Determine log₄(b).

Skill: Translating between logarithmic and exponential form

Show answer and worked solution

Answer: 3

From log₂(a)=3, a=2³=8. From log_a(b)=2, b=a²=64. Since 64=4³, log₄(b)=3.

Practice marking checkpoints

• 3 marks for finding a=8
• 3 marks for finding b=64
• 3 marks for evaluating log₄64
• 1 mark for presentation

Common mistake: Multiplying the two given logarithms without first checking that their bases align.

Alternative route: Use change of base: log₄b=(log₂b)/(log₂4). Also log₂b=log₂a·log_ab=3·2=6, so the result is 6/2=3.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Exponents and logarithmsChallengeFull solution10 marks

An exponent inside an exponent

Determine all positive real numbers x satisfying x^(log₂x)=16.

Skill: Substitution using a logarithmic variable

Show answer and worked solution

Answer: x=4 or x=1/4

Let y=log₂x, so x=2^y. Then x^(log₂x)=(2^y)^y=2^(y²). Since 16=2⁴, y²=4 and y=±2. Therefore x=2²=4 or x=2^(-2)=1/4. Both are positive and satisfy the equation.

Practice marking checkpoints

• 2 marks for setting y=log₂x
• 3 marks for obtaining 2^(y²)=2⁴
• 3 marks for both values y=±2
• 2 marks for converting back and checking both x-values

Common mistake: Keeping only y=2 after solving y²=4 and losing the valid solution x=1/4.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Exponents and logarithmsChallengeFull solution10 marks

Quadratic structure in an exponential equation

Determine all real numbers x satisfying 9^x−10·3^x+9=0.

Skill: Recognizing a quadratic in an exponential expression

Show answer and worked solution

Answer: x=0 or x=2

Let y=3^x, which is positive. Since 9^x=(3^x)²=y², the equation becomes y²−10y+9=0=(y−1)(y−9). Thus y=1 or y=9. Therefore 3^x=1 gives x=0, and 3^x=9 gives x=2.

Practice marking checkpoints

• 2 marks for a valid substitution
• 3 marks for the quadratic equation
• 2 marks for factoring
• 2 marks for both x-values
• 1 mark for noting positivity or checking

Common mistake: Treating 9^x as 9x or failing to recognize that 9^x=(3^x)².

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

TrigonometryCoreFull solution10 marks

Exact cosine of an angle sum

Acute angles A and B satisfy sin A=3/5 and cos B=12/13. Determine cos(A+B) exactly.

Skill: Using exact ratios and the cosine addition identity

Show answer and worked solution

Answer: 33/65

Since A is acute, cos A=4/5. Since B is acute, sin B=5/13. Therefore cos(A+B)=cosAcosB−sinAsinB=(4/5)(12/13)−(3/5)(5/13)=(48−15)/65=33/65.

Practice marking checkpoints

• 2 marks for cosA=4/5
• 2 marks for sinB=5/13
• 3 marks for the correct identity
• 2 marks for 33/65
• 1 mark for exact presentation

Common mistake: Using cos(A+B)=cosAcosB+sinAsinB; the plus sign belongs to cos(A−B).

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

TrigonometryCoreFull solution10 marks

Triangle area from an included angle

Two sides of a triangle have lengths 7 and 8, and their included angle is 60°. Determine the exact area of the triangle.

Skill: Applying the sine area formula

Show answer and worked solution

Answer: 14√3

Use area=(1/2)ab sin C. The area is (1/2)(7)(8)sin60°=28(√3/2)=14√3.

Practice marking checkpoints

• 3 marks for choosing the sine area formula
• 3 marks for correct substitution
• 3 marks for simplifying to 14√3
• 1 mark for square units

Common mistake: Using the cosine law to find the third side and then stopping without computing the area.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

TrigonometryChallengeFull solution10 marks

A trigonometric equation on a full cycle

Determine all angles x, in degrees, satisfying 0°≤x<360° and sin(2x)=sin x.

Skill: Factoring a trigonometric equation and covering all cases

Show answer and worked solution

Answer: x=0°, 60°, 180°, or 300°

Use sin(2x)=2sinx cosx. Then 2sinx cosx=sinx, so sinx(2cosx−1)=0. Hence sinx=0, giving x=0° or 180°, or cosx=1/2, giving x=60° or 300°.

Practice marking checkpoints

• 2 marks for the double-angle identity
• 2 marks for correct factoring
• 2 marks for the sinx=0 case
• 3 marks for the cosx=1/2 case
• 1 mark for the complete interval check

Common mistake: Dividing by sinx and losing the solutions x=0° and x=180°.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

TrigonometryChallengeFull solution10 marks

A median after the cosine law

In triangle ABC, AB=5, AC=7, and ∠A=60°. Determine the exact length of the median from A to side BC.

Skill: Combining the cosine law with a median formula

Show answer and worked solution

Answer: √109/2

First, the cosine law gives BC²=5²+7²−2(5)(7)cos60°=25+49−35=39. If m is the median from A, Apollonius’ theorem gives m²=[2(5²)+2(7²)−BC²]/4=(50+98−39)/4=109/4. Thus m=√109/2.

Practice marking checkpoints

• 3 marks for BC²=39
• 2 marks for a correct median relation
• 3 marks for m²=109/4
• 2 marks for the positive exact length

Common mistake: Assuming the median is also an altitude; that is not true in a general non-isosceles triangle.

Alternative route: Place B=(0,0), C=(√39,0), use the two distance equations to find A, and compute the distance from A to the midpoint of BC.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Functions and polynomialsCoreFull solution10 marks

Sum of squares of three roots

The roots of x³−6x²+11x−6=0 are r, s, and t. Determine r²+s²+t² without solving for the individual roots.

Skill: Using Vieta’s formulas in a symmetric expression

Show answer and worked solution

Answer: 14

Vieta’s formulas give r+s+t=6 and rs+rt+st=11. Therefore r²+s²+t²=(r+s+t)²−2(rs+rt+st)=6²−2(11)=14.

Practice marking checkpoints

• 3 marks for both Vieta relationships
• 3 marks for the symmetric identity
• 3 marks for 14
• 1 mark for avoiding unnecessary root solving

Common mistake: Using r²+s²+t²=(r+s+t)² and forgetting the pairwise-product terms.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Functions and polynomialsChallengeFull solution10 marks

A telescoping polynomial difference

A polynomial P satisfies P(0)=0 and P(x)−P(x−1)=3x²−3x+1 for every real x. Determine P(10).

Skill: Recognizing and summing a telescoping difference

Show answer and worked solution

Answer: 1000

Observe that 3x²−3x+1=x³−(x−1)³. For integer x from 1 to 10, P(x)−P(x−1)=x³−(x−1)³. Summing makes both sides telescope: P(10)−P(0)=10³−0³. Since P(0)=0, P(10)=1000.

Practice marking checkpoints

• 3 marks for recognizing the difference of cubes
• 3 marks for setting up the telescoping sum
• 3 marks for P(10)=1000
• 1 mark for using P(0)=0

Common mistake: Trying to guess every coefficient of P before noticing that only the telescoping sum is needed.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Functions and polynomialsChallengeFull solution10 marks

Composing a rational function

Let f(x)=(x−1)/(x+1). Simplify f(f(x)) and state every real x for which the composition is defined.

Skill: Function composition with domain restrictions

Show answer and worked solution

Answer: f(f(x))=−1/x, with x≠−1 and x≠0

Substitute f(x): f(f(x))=[(x−1)/(x+1)−1]/[(x−1)/(x+1)+1]=[-2/(x+1)]/[2x/(x+1)]=−1/x. The inner function requires x≠−1. The outer denominator also requires f(x)≠−1, which gives x≠0.

Practice marking checkpoints

• 2 marks for correct substitution
• 3 marks for simplifying the numerator and denominator
• 2 marks for −1/x
• 3 marks for both excluded values

Common mistake: Reporting only x≠−1 and overlooking the value that makes the outer function undefined.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Functions and polynomialsChallengeFull solution10 marks

A multiplicative functional rule on integers

A function f is defined on the non-negative integers and satisfies f(m+n)=f(m)f(n) for all non-negative integers m and n. If f(0)=1 and f(1)=2, determine f(10) and justify the rule used.

Skill: Deriving a function rule by induction

Show answer and worked solution

Answer: f(10)=1024

For n≥0, f(n+1)=f(n)f(1)=2f(n). Starting with f(0)=1, induction gives f(n)=2^n. Therefore f(10)=2^10=1024.

Practice marking checkpoints

• 3 marks for deriving f(n+1)=2f(n)
• 4 marks for a valid inductive or iterative argument
• 2 marks for 1024
• 1 mark for a clear conclusion

Common mistake: Stating f(n)=2^n from the first few values without explaining why the pattern must continue.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Analytic geometryCoreFull solution10 marks

Equation of a perpendicular bisector

Determine an equation of the perpendicular bisector of the segment joining A(2,3) and B(8,−1).

Skill: Combining midpoint and perpendicular-slope facts

Show answer and worked solution

Answer: y−1=(3/2)(x−5)

The midpoint is ((2+8)/2,(3−1)/2)=(5,1). The slope of AB is (−1−3)/(8−2)=−4/6=−2/3, so a perpendicular line has slope 3/2. Through (5,1), its equation is y−1=(3/2)(x−5).

Practice marking checkpoints

• 3 marks for the midpoint
• 3 marks for both slopes
• 3 marks for a correct line equation
• 1 mark for an equivalent simplified form

Common mistake: Using the negative reciprocal incorrectly; the perpendicular slope to −2/3 is 3/2.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Analytic geometryCoreFull solution10 marks

Tangent to a circle at a point

The circle x²+(y−2)²=25 contains the point P(3,6). Determine an equation of the tangent at P.

Skill: Using the radius as a normal to a tangent

Show answer and worked solution

Answer: 3x+4y=33

The centre is C(0,2). The radius CP has direction vector (3,4), which is normal to the tangent. Thus the tangent has equation 3(x−3)+4(y−6)=0, or 3x+4y=33.

Practice marking checkpoints

• 2 marks for the centre
• 3 marks for the radius direction or slope
• 3 marks for the tangent equation
• 2 marks for simplification or verification at P

Common mistake: Giving the equation of the radius instead of a line perpendicular to it.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Analytic geometryCoreFull solution10 marks

Line-circle intersections

Determine all points where the line y=x+1 intersects the circle x²+y²=25.

Skill: Solving a line-circle system by substitution

Show answer and worked solution

Answer: (3,4) and (−4,−3)

Substitute y=x+1 into the circle: x²+(x+1)²=25, so 2x²+2x−24=0 and x²+x−12=0. Factoring gives (x−3)(x+4)=0, so x=3 or x=−4. Then y=x+1 gives the two points (3,4) and (−4,−3).

Practice marking checkpoints

• 2 marks for substitution
• 3 marks for the quadratic
• 2 marks for both x-values
• 2 marks for both points
• 1 mark for checking

Common mistake: Reporting the x-values as the final answer when the question asks for points.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Analytic geometryChallengeFull solution10 marks

Classifying and measuring a quadrilateral

Points A(1,2), B(7,4), C(10,9), and D(4,7) are joined in order. Prove that ABCD is a parallelogram and determine its area.

Skill: Using diagonal midpoints and a determinant area

Show answer and worked solution

Answer: ABCD is a parallelogram with area 24

The midpoint of AC is (11/2,11/2), and the midpoint of BD is also (11/2,11/2). Since the diagonals bisect each other, ABCD is a parallelogram. Vectors AB=(6,2) and AD=(3,5), so its area is |6·5−2·3|=24.

Practice marking checkpoints

• 3 marks for both diagonal midpoints
• 2 marks for the parallelogram conclusion
• 3 marks for a valid area method
• 2 marks for area 24

Common mistake: Showing only that one pair of opposite sides has equal length, which is not enough by itself.

Alternative route: Show AB=DC and AD=BC as vectors, then use a base-height or shoelace calculation for the area.

For full-solution practice, rewrite this reasoning in complete sentences and state the theorem, formula, or counting principle before substituting values.

Plan every official Euclid paper without copying it

The CEMC archive provides authorized contest papers, solutions and results. This planner links to the official filtered archive and stores only your status and score in this browser.

papers reviewed

papers still unseen

Recommended archive strategy

Reserve: keep two recent papers completely unseen.
Diagnose: use an older paper without a strict time limit.
Repair: complete original drills before returning to similar official work.
Simulate: use a reserved paper for a final 150-minute test.

Official questions and solutions remain on Waterloo’s website. CalculatorVillage records no copies and sends no planner data to a server.

YearStatusScoreOfficial

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Official-paper workflow

Use the CEMC archive without memorizing it

Official past papers are the best source for authentic wording, layout and difficulty progression. They become less useful when a student repeatedly rereads familiar solutions or uses every paper as an unreviewed race. Preserve recent papers for measurement and use older papers plus original questions for skill building.

1. Reserve

Save two recent unseen papers for final simulations. Do not browse their questions or solutions early.

2. Diagnose

Use one older paper untimed. Tag every part by topic, method, confidence and lost-mark cause.

3. Train

Use focused original drills to repair the causes. Return to selected official questions only after the method is understood.

4. Simulate

Write a reserved paper in 150 minutes, mark it carefully and rewrite incomplete solutions from memory.

University of Waterloo CEMC

Official Euclid Contest page

Confirm the current date, deadline, fee, eligibility, calculator policy, awards, shipping and administration rules. Time-sensitive facts in this guide were checked here in November 2026.

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University of Waterloo CEMC

Euclid and CSMC preparation material

Use the official general advice and eight topic modules. The organization of this preparation center follows those published topic areas so students can move between both resources.

Open official resource

University of Waterloo CEMC

Past contests, solutions and results

Download authorized Euclid papers, official solutions and results. Use recent papers for complete simulations and older questions for targeted review.

Open official resource

University of Waterloo CEMC

How CEMC contests work

Understand school ordering and supervision. Students cannot register individually; they must participate through their school or find another local participating school.

Open official resource

For parents, teachers and coaches

Support the process without taking over the mathematics

Protect regular practice

Two or three focused sessions every week are more useful than a last-minute stack of papers. Help schedule the time and create a quiet environment, while leaving the student responsible for selecting and explaining methods.

Ask process questions

Useful prompts include “What do you know?”, “What is the question asking?”, “Why is that step valid?” and “How could you check it?” Avoid revealing the key observation too quickly; productive struggle ends when a hint replaces the discovery.

Celebrate better evidence

Notice a cleaner diagram, a recovered partial solution, a good skip decision or an error that did not recur. Contest scores vary with paper difficulty. Sustainable progress is visible in the quality and independence of the work.

Frequently asked questions

Euclid Contest questions students ask first

What is the Euclid Mathematics Contest?

The Euclid Contest is the University of Waterloo CEMC’s senior secondary-school mathematics contest. It uses ten written questions, lasts 2.5 hours, and awards marks for both correct results and clearly communicated reasoning.

Who can write the Euclid Contest?

It is aimed at students in their final year of secondary school and CÉGEP students, but motivated younger students may also participate. Students must write through a participating school rather than registering individually.

When is the 2027 Euclid Contest?

The official 2027 date is April 6 in North and South America and April 7 outside those regions. The school ordering deadline is March 11, 2027. Students should confirm arrangements with their school because administration details can change.

How should I use past Euclid papers?

Begin with an untimed diagnostic, classify every lost mark, then practise weak topics. In the final month, write complete official papers under the 150-minute limit and compare your reasoning with the official solutions. Use CalculatorVillage’s original sets between official papers so you do not memorize an archive.

What score should I target?

There is no passing score. In 2026 the official mean was 52.2 and the Certificate of Distinction cutoff was 66, but both vary by year. A useful target is to improve your own timed baseline while protecting partial credit on every attempted question.

Are calculators allowed?

The CEMC states that some calculators are permitted. Check the current contest instructions before test day. A calculator may support numerical work, but full-solution marks still require the mathematical steps and justification to be written.

Does Waterloo require the Euclid Contest for admission?

Participation is not required for admission to Waterloo’s Faculty of Mathematics. The university directs students to its current scholarship information for details about how contest results may be considered.

Are these official Euclid questions?

No. CalculatorVillage’s questions, drills, and practice sets are our own preparation material designed around the published format and topic areas, not copied from CEMC papers. The bank holds 224 practice questions: 32 authored multi-step problems with marking checkpoints and 192 generated fluency variants that practise one skill each. The drills and 150-minute practice sets reuse questions from this one bank. Official contests and solutions remain on the CEMC website and are linked from this guide.

Your next useful step is one measured attempt

Choose a diagnostic if you do not know your weak areas. Choose one topic drill if you do. Choose a full 150-minute set only when you are ready to practise pacing and written communication together. Record the evidence, repair one cause and return to a new problem.

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