Competitive Math
I taught Competitive Math in 2017-2022 to middle and high school students at various levels. I have a large collection of materials that I assembled during that time, and I’d be happy to share them (feel free to email me!). The collection includes structured problem sets covering almost all topics typical of IMO-style competitions, organized by grade and level.
Some brainteasers based on classic ideas:
Problem 1.
You are a tourist, and you have just arrived to Waldzell, capital of famous Castalia. You are standing in the middle of the street, trying to understand where your hotel is. In front of you, there are three Castalians, looking at you with curiosity.
Your companion tells you that, according to the rules of the Province, each of them must be either a Knight or a Knave. You are allowed to ask them one or two yes-or-no questions, and your goal is to figure out whether you need to go left or right. Each of them will write their answers for the your question(s) in random order on a paper and handle them to you. Your Castalian is good enough so that you can tell a "yes" from "no", but you cannot tell, which word means what.
Suppose that there are at least 1 Knight and 1 Knave among the three. Suggest one question to figure out whether the hotel is to the left or right from you.
Suggest two questions to ask to figure out the correct direction without relying on the additional assumption from part 1.
Clarification. A Knight gives the true answer to any question. A Knave gives the opposite of the true answer to any question. For example, both a Knight and a Knave answer "Yes" to the question "Are you a Knight?".
As usual, you can ask questions containing logical links ("and", "or", implications, etc), and part of the question can refer to the answerer's type.
The only way to extract information is to check whether all the Castalians gave the same answer. Thus, we need to ask a question such that answers of a Knight and a Knave differ if and only if, for example, left is the correct direction. One truth table to corresponding to such a question is the following:
Direction of hotel Left Right
Knight's answer 1 0
Knave's answer 0 0
As usual, "1" means that the agent answers "yes", and "0" corresponds to "no". An example of a question that delivers this truth table:
"Is at least one of the following true: hotel is to the left, or you are a Knave?"
Then the hotel is to the right if and only if all answers coincide.
One way to extract information in this situation is to check whether each agent gives the same answer to the two questions. To solve the problem, we can ask two questions such that any agent gives the same answer to both questions if and only if the correct direction is, for example, right. Let's keep the first question as in part 1. We can pick the second question according to, for example, the following truth table:
Direction of hotel Left Right
Knight's answer 0 0
Knave's answer 1 0
The second question is, for example, the following:
"Are both of the following true: hotel is to the right and you are a Knave?"
All agents give the same answer to the two questions if and only if the hotel is to the right.
P.S. There are, of course, many other question pairs that deliver solution. Another pair of shorter questions: "Is the hotel to the left?" and "Is 1=0 true?".
Problem 2.
How many equilateral triangles can be inscribed in a parabola?
Infinitely many: for any point A on the parabola there exist points B, C on the parabola such that the triangle ABC is equilateral.
WLOG consider the parabola y=x^2.
Continuity-based solution.
Pick a point A = (a, a^2) on the parabola, WLOG assume a > 0. Pick a point B = (b, b^2) from the parabola such that b > a. Perpendicular bisector to AB has two intersections with the parabola, denote the one with negative x-coordinate by C. By construction we have AC = AB for any value of b. Notice that:
when b is close to a, AC - AB > 0
when b >> a, AC - AB < 0
AC - AB is a continuous function of b.
Then by intermediate value theorem there exist a value of b such that AC = AB. This means that for any A on the parabola we can find B, C on the parabola such that the triangle ABC is equilateral. Picture.
Problem 3.
Several boulders together weigh 10 tons. Each boulder weighs at most 1 ton. You have access to 3-ton trucks, and you want to haul all the boulders in a single trip. What is the smallest number of 3-ton trucks needed to transport the entire load at once?
Consider a special case when all the boulders have equal weight. Try different numbers of boulders, starting from 10. Is it always enough to have 4 trucks?
Consider a case of 13 boulders, each weighting 10/13 tons. Observe that 4 trucks is not enough.
5 trucks is always enough: since every stone weights at most 1 ton, you can always load at least 2 tons of stones in each truck.
Problem 4.
What is the shortest path connecting A and B, if rivers (in blue) must be crossed horizontally? See the picture below for the map and an example path.
As usual, the straightest path between 2 points is a straight line. If only you could get the rivers out of the way...
Firstly, observe that the shortest path must look like five consecutive vectors, like on the picture. Secondly, let us notice that two of these vectors (the ones that correspond to rivers) are always horizontal and always have the same lengths of 3 and 2. Let's start in A and firstly move 5 miles to the right, call the point, at which we arrive, C. This counts towards crossing both rivers. But then the shortest path between B and C (which is the line segment BC) must be the same length as the length of the non-river part of the shortest path between A and B. Since BC=5, the shortest path between A and C equals 3+2+5=10.
Top part of the picture shows how to construct the shortest path between A and B: start in B along the vector BC until you reach the river, cross it, continue parallel to BC, cross the other river, and then head straight towards A.
Problem 5.
Consider a toy planet system located on the plane. Sun is the point (0,0). Let's represent Earth by a circle of radius r=3. Suppose that Earth orbits Sun on a circular orbit and distance between centers of Earth and Sun equals R=20. Then the equation below spans, as time t passes, the trajectory of Earth.
Now suppose that Moon is a circle of radius 1 that orbits Earth on circular orbit, and distance between their centers equals 6. Suppose that, just like real Moon, our toy Moon orbits Earth faster than Earth orbits Sun. Suggest an equation that would span trajectory (see the picture below) of the Moon, as time t passes. Just like the equation above, it should depend on x, y, and t.
For example, this equation works:
For every t the equation describes a circumference of unit radius with center A(t) = (20cos t + 6 sin (10t); 20sin t + 6 cos (10t)), which is the center of the toy Moon.
For every t the point A(t) lies on a circumference of radius 6 with a center B(t) = (20 cos t; 20 sin t), which is the center of the toy Earth.
A(t) indeed orbits B(t). To see this, consider coordinate change: x' = x - 20 cos t, y' = y - 20 sin t, in which the equation takes the simple form (x' - 20 cos t)^2 + (y' - 20 sin t)^2 = 1.
B(t) orbits (0,0), as desired.
Finally, the multiplier "N=10" in "sin (10t)" and "cos (10t)" determines how many times per Earth year Moon orbits Earth (see this gif for the trajectory). Compare it to the cases when N=1, N=2, N=3.
Problem 6.
Your neighbor's daughter, who just learned proof by induction, approaches you and claims she can prove that in any herd of horses all horses have the same color.
Base, n=1. All horses in a nerd of size 1 (which is just one horse) have the same color.
Step. Suppose that the statement is true for any herd of size k. Take a herd of size k+1 and enumerate the horses from 1 to k+1. Then all horses from 1 to k have the same color, and so do all horses from 2 to k+1. But then all k+1 horses must have the same color, Q.E.D!
You certainly remember seeing two differently colored horses in your life, so you suspect that her reasoning might have a flaw. Where is it?
The first step, from n=1 to n=2, fails: sets of horses {1} and {2} do have the same color within the sets, but intersection of the sets is empty. If base for n=2 could be established, the whole reasoning would be correct.