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2024 Euclid Contest: Top 20 Problems & Solutions

Are you a high school student who loves math but often finds yourself stumped by challenging problems? Do you crave stimulating brain teasers to push your problem-solving skills to the limit? If so, you’re in the right place! Join us on a journey through the 2024 Euclid Contest, where we’ll unravel the top 20 problems and provide clear, step-by-step solutions. Get ready to conquer the world of mathematics with confidence as we delve into this prestigious competition!

2024 Euclid Contest: Top 20 Problems & Solutions - WuKong Education Blog
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Understanding the Euclid Contest

Curious about the 2024 Euclid Contest and how it works? Let’s break it down! The Euclid Contest is organized by the Center for Education in Mathematics and Computing (CEMC) and is renowned as one of Canada’s most prestigious math competitions, alongside the Canadian Senior Mathematics Competition (CSMC) and the Canadian Open Mathematics Challenge (COMC). While the contest primarily targets high school students, it’s recommended for grades 7 to 12, as well as for those considering University of Waterloo applications.

Format and Structure

  • The contest spans 150 minutes and features 10 problems of increasing difficulty.
  • Each problem is worth 10 points and includes multiple sub-problems, indicated by either a “light bulb” or “paper/pencil” icon.
  • Questions marked with a “light bulb” require short answer responses and are typically worth 2-3 points, with partial marks awarded for showing work.
  • Those marked with “paper/pencil” demand a written solution to earn the remaining points. These solutions must be well-presented and include justifications.

Curriculum Coverage

  • The Euclid Contest encompasses a wide array of topics akin to those found in the AMC 10.
  • These include geometry, trigonometry, algebra, number theory, counting, probability, and arithmetic.
  • Additionally, it may touch on other concepts like logarithms.

Top 20  2024 Euclid Contest Problems

Problem 1:

Question Description:

A farmer wants to enclose a rectangular pasture using 400 meters of fencing. What dimensions should the farmer use to maximize the area of the pasture?

Answer:

  • Let the length of the rectangle be \( l \) meters and the width be \( w \) meters.
  • The perimeter of the rectangle, which is the sum of all sides, is given by \( 2l + 2w = 400 \).
  • Solving this equation for one of the variables, we get \( l = 200 – w \).
  • The area \( A \) of the rectangle is given by \( A = l \times w \).
  • Substitute \( l = 200 – w \) into the formula for area to get \( A = (200 – w) \times w \).
  • To maximize the area, we differentiate \( A \) with respect to \( w \), set the derivative equal to zero, and solve for \( w \).
  • After finding \( w \), substitute it back into \( l = 200 – w \) to find the corresponding length.
  • Thus, the dimensions that maximize the area are \( l = 100 \) meters and \( w = 100 \) meters, creating a square pasture.

Problem 2:

Question Description:

Solve the equation \( x^2 + 4x + 4 = 0 \).

Answer:

  • This is a quadratic equation in the form \( ax^2 + bx + c = 0 \).
  • We can solve it by factoring or using the quadratic formula.
  • Factoring the equation \( x^2 + 4x + 4 = (x + 2)^2 = 0 \).
  • Setting \( (x + 2)^2 = 0 \) implies \( x + 2 = 0 \).
  • Therefore, the solutions are \( x = -2 \) with a multiplicity of 2.

Problem 3:

Question Description:

Determine the value of \( \log_2(16) + \log_3(81) \).

Answer:

  • Use the properties of logarithms to rewrite the expression.
  • \( \log_2(16) = 4 \) because \( 2^4 = 16 \).
  • \( \log_3(81) = 4 \) because \( 3^4 = 81 \).
  • Therefore, \( \log_2(16) + \log_3(81) = 4 + 4 = 8 \).

Problem 4:

Question Description:

If \( x + y = 10 \) and \( xy = 24 \), find the value of \( x^2 + y^2 \).

Answer:

  • Use the formula \( (x + y)^2 = x^2 + 2xy + y^2 \).
  • Substitute \( x + y = 10 \) and \( xy = 24 \) into the formula.
  • \( (x + y)^2 = 10^2 = 100 \).
  • Use \( xy = 24 \) to find \( 2xy = 48 \).
  • Thus, \( x^2 + 2xy + y^2 = x^2 + y^2 + 48 \).
  • Since \( (x + y)^2 = x^2 + 2xy + y^2 \), we have \( x^2 + y^2 + 48 = 100 \).
  • Therefore, \( x^2 + y^2 = 100 – 48 = 52 \).

Problem 5:

Question Description:

A car travels 150 km at 60 km/hr and returns at 40 km/hr. What is the average speed for the whole journey?

Answer:

  • Use the formula for average speed: \( \text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} \).
  • The total distance traveled is \( 150 \text{ km} + 150 \text{ km} = 300 \text{ km} \).
  • The time taken for the outbound journey is \( \frac{150 \text{ km}}{60 \text{ km/hr}} = 2.5 \text{ hours} \).
  • The time taken for the return journey is \( \frac{150 \text{ km}}{40 \text{ km/hr}} = 3.75 \text{ hours} \).
  • Therefore, the total time is \( 2.5 \text{ hours} + 3.75 \text{ hours} = 6.25 \text{ hours} \).
  • Thus, the average speed is \( \frac{300 \text{ km}}{6.25 \text{ hours}} = 48 \text{ km/hr} \).

Problem 6:

Question Description:

In a school, 40% of the students study French, 30% study German, and 20% study both languages. If 300 students study at the school, how many study neither language?

Answer:

  • Let \( x \) represent the number of students who study neither French nor German.
  • From the given information, 40% study French, which is \( 0.4 \times 300 = 120 \) students.
  • Similarly, 30% study German, which is \( 0.3 \times 300 = 90 \) students.
  • Additionally, 20% study both languages, which is \( 0.2 \times 300 = 60 \) students.
  • Therefore, the total number of students studying either French or German is \( 120 + 90 – 60 = 150 \).
  • Hence, the number of students who study neither language is \( 300 – 150 = 150 \).

Problem 7:

Question Description:

In triangle ABC, angle A measures \( 60^\circ \) and angle B measures \( 45^\circ \). If AB = 12 cm, find the length of BC.

Answer:

  • Since angles A and B are given, we can find angle C using the fact that the sum of angles in a triangle is \( 180^\circ \).
  • – \( \angle C = 180^\circ – 60^\circ – 45^\circ = 75^\circ \).
  • We can use the law of sines to find the length of side BC.
  • – \( \frac{AB}{\sin A} = \frac{BC}{\sin C} \).
  • Substitute the given values to get \( \frac{12}{\sin 60^\circ} = \frac{BC}{\sin 75^\circ} \).
  • Solve for \( BC \) to find \( BC = \frac{12 \times \sin 75^\circ}{\sin 60^\circ} \).
  • Using trigonometric identities, \( \sin 75^\circ = \sin(45^\circ + 30^\circ) = \sin 45^\circ \cdot \cos 30^\circ + \cos 45^\circ \cdot \sin 30^\circ \).
  • – \( \sin 75^\circ = \frac{\sqrt{6} + \sqrt{2}}{4} \) and \( \sin 60^\circ = \frac{\sqrt{3}}{2} \).
  • Thus, \( BC = \frac{12 \times \frac{\sqrt{6} + \sqrt{2}}{4}}{\frac{\sqrt{3}}{2}} = \frac{6(\sqrt{6} + \sqrt{2})}{\sqrt{3}} \).

Problem 8:

Question Description:

In a geometric sequence, the first term is 3 and the second term is 6. Find the fourth term.

Answer:

  • A geometric sequence has a common ratio between consecutive terms.
  • Let \( r \) represent the common ratio.
  • The second term is the first term multiplied by the common ratio, so \( 6 = 3r \).
  • Solve for \( r \) to find \( r = 2 \).
  • The fourth term is the second term multiplied by the common ratio twice, so \( \text{Fourth term} = 6 \times 2 \times 2 = 24 \).

Problem 9:

Question Description:

A box contains 5 red balls, 6 blue balls, and 4 green balls. If a ball is randomly selected from the box, what is the probability that it is not blue?

Answer:

  • The total number of balls in the box is \( 5 + 6 + 4 = 15 \).
  • The probability of selecting a blue ball is \( \frac{6}{15} = \frac{2}{5} \).
  • Therefore, the probability of not selecting a blue ball is \( 1 – \frac{2}{5} = \frac{3}{5} \).

Problem 10:

Question Description:

A company manufactures two types of calculators: scientific and graphing. The scientific calculator sells for $20 each, and the graphing calculator sells for $40 each. If the company sells a total of 100 calculators and earns $3000 in revenue, how many of each type of calculator did it sell?

Answer:

  • Let \( x \) represent the number of scientific calculators sold and \( y \) represent the number of graphing calculators sold.
  • We can create a system of equations based on the given information.
  • The first equation represents the total number of calculators sold: \( x + y = 100 \).
  • The second equation represents the total revenue earned: \( 20x + 40y = 3000 \).
  • Solve the system of equations to find the values of \( x \) and \( y \).

Problem 11:

Question Description:

A group of friends consisting of Alice, Bob, Cindy, David, and Emily went to a restaurant together. They all sat around a circular table. Alice sat to the left of Bob, Cindy sat to the right of David, and Emily sat between Alice and David. Who sat directly across from Cindy?

Answer:

  • Since Emily sat between Alice and David, Alice and David must be adjacent.
  • Cindy sat to the right of David, so Cindy cannot be directly across from David.
  • Therefore, Cindy must be directly across from Emily.

Problem 12:

Question Description:

A bookshelf contains 5 different novels, 4 different poetry collections, and 3 different plays. In how many ways can you select one novel, one poetry collection, and one play?

Answer:

  • To select one item from each category, we multiply the number of choices in each category.
  • There are 5 ways to select a novel, 4 ways to select a poetry collection, and 3 ways to select a play.
  • Therefore, the total number of ways to select one novel, one poetry collection, and one play is \( 5 \times 4 \times 3 = 60 \).

Problem 13:

Question Description:

Five friends – Alice, Bob, Cindy, David, and Emily – are standing in a line. Bob is not at either end of the line. Alice is to the immediate left of Cindy, and David is to the immediate right of Emily. Who is standing in the middle of the line?

Answer:

  • Bob cannot be at either end of the line, so either Alice or David must be at one end.
  • Alice is to the immediate left of Cindy, and David is to the immediate right of Emily.
  • Therefore, Alice must be at one end of the line, and David must be at the other end.
  • Since Bob cannot be at either end, Cindy must be in the middle of the line.

Problem 14:

Question Description:

A car travels 120 miles at an average speed of 60 mph and then travels 240 miles at an average speed of 80 mph. What is the car’s average speed for the entire journey?

Answer:

  • Use the formula for average speed: \( \text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} \).
  • The total distance traveled is \( 120 \text{ miles} + 240 \text{ miles} = 360 \text{ miles} \).
  • The time taken for the first part of the journey is \( \frac{120 \text{ miles}}{60 \text{ mph}} = 2 \text{ hours} \).
  • The time taken for the second part of the journey is \( \frac{240 \text{ miles}}{80 \text{ mph}} = 3 \text{ hours} \).
  • Therefore, the total time is \( 2 \text{ hours} + 3 \text{ hours} = 5 \text{ hours} \).
  • Thus, the average speed is \( \frac{360 \text{ miles}}{5 \text{ hours}} = 72 \text{ mph} \).

Problem 15:

Question Description:

A rectangular garden measures 20 meters by 15 meters. If the garden is enlarged by 25%, what are the new dimensions of the garden?

Answer:

  • To enlarge the garden by 25%, we multiply both dimensions by \( 1.25 \).
  • The new length is \( 20 \text{ meters} \times 1.25 = 25 \text{ meters} \).
  • The new width is \( 15 \text{ meters} \times 1.25 = 18.75 \text{ meters} \).
  • Therefore, the new dimensions of the garden are 25 meters by 18.75 meters.

Problem 16:

Question Description:

A group of 5 friends plans to sit in a row of 5 seats in a movie theater. If they choose seats randomly, what is the probability that two specific friends sit next to each other?

Answer:

  • Consider the two specific friends as a single entity.
  • There are \(5 – 1 = 4\) ways they can sit next to each other in a row of 5 seats.
  • The total number of ways the 5 friends can sit is \(5!\) (5 factorial).
  • Therefore, the probability is \( \frac{4}{5!} = \frac{4}{120} = \frac{1}{30} \).

Problem 17:

Question Description:

A box contains 4 red balls, 3 blue balls, and 5 green balls. If 3 balls are randomly drawn without replacement, what is the probability of selecting exactly 2 red balls?

Answer:

  • There are \(\binom{12}{3}\) ways to choose 3 balls from the box.
  • To select exactly 2 red balls, we need to choose 2 red balls from 4 and 1 ball from the remaining 8 non-red balls.
  • There are \(\binom{4}{2}\) ways to choose 2 red balls and \(\binom{8}{1}\) ways to choose 1 non-red ball.
  • Therefore, the probability is \( \frac{\binom{4}{2} \times \binom{8}{1}}{\binom{12}{3}} \).

Problem 18:

Question Description:

In a town, there are 6 houses along a street, numbered consecutively from 1 to 6. If 4 houses are randomly selected for a survey, what is the probability that at least one of the selected houses is numbered 5 or 6?

Answer:

  • There are \(\binom{6}{4}\) ways to choose 4 houses from the 6.
  • To find the probability that none of the selected houses is numbered 5 or 6, we need to choose 4 houses from the remaining 4 non-numbered-5-or-6 houses.
  • There are \(\binom{4}{4}\) ways to choose 4 houses from these 4.
  • Therefore, the probability of selecting no houses numbered 5 or 6 is \( \frac{\binom{4}{4}}{\binom{6}{4}} \).
  • Subtract this probability from 1 to find the probability of selecting at least one house numbered 5 or 6.

Problem 19:

Question Description:

A committee of 5 people is to be selected from a group of 8 men and 5 women. If the committee must have at least 2 women, how many different committees can be formed?

Answer:

  • There are \(\binom{13}{5}\) ways to choose 5 people from the total group.
  • To find the number of committees with at least 2 women, we can subtract the number of committees with no women or only one woman from the total.
  • Calculate the number of committees with no women and only one woman, and subtract these from the total.

Problem 20:

Question Description:

A bag contains 6 red balls and 4 blue balls. If 3 balls are drawn without replacement, what is the probability that they are all red?

Answer:

  • There are \(\binom{10}{3}\) ways to choose 3 balls from the bag.
  • To find the probability that all 3 balls drawn are red, we need to choose 3 red balls from the 6 red balls.
  • There are \(\binom{6}{3}\) ways to choose 3 red balls.
  • Therefore, the probability is \( \frac{\binom{6}{3}}{\binom{10}{3}} \).

FAQs For 2024 Euclid Contest Problems & Answers

Q1. How difficult are the 2024 Euclid Contest problems?

The 2024 Euclid Contest problems are known for their high level of difficulty, challenging even the most skilled math students. They often require deep analytical thinking and creative problem-solving strategies.

Q2. Can I use calculators or other aids during the 2024 Euclid Contest?

No, the use of calculators or any other aids is typically not allowed during the 2024 Euclid Contest. Participants are expected to rely solely on their mathematical knowledge and problem-solving skills to tackle the questions.

Q3. Are there any resources available to help prepare for the 2024 Euclid Contest?

Yes, there are various resources available to help students prepare for the 2024 Euclid Contest, including practice problem sets, study guides, and tutoring services. Additionally, reviewing past contest papers and solutions can provide valuable insight into the types of questions that may be asked.

Conclusion:

In this article, we delved into the fascinating realm of the 2024 Euclid Contest problems and answers, offering straightforward explanations and solutions to complex mathematical puzzles. From logical reasoning to probability and combinatorics challenges, we navigated through various problem-solving strategies, making math more accessible and enjoyable. Whether you’re a high school student or a math enthusiast, this article serves as a handy guide to sharpen your problem-solving skills and boost your confidence. Dive into the world of the 2024 Euclid Contest problems and answers, and uncover the secrets to mastering mathematics with ease!