Most Beautiful Problems Math Olympiads IMO

Most beautiful Math Olympiad Problems 

Two Intersecting Square

Problem:

Two squares of side 1 have a common center. Show that the area of their intersection is greater than 3/4.

To prove that the area of intersection between two squares with side length 1 and a common center is greater than , we'll follow a step-by-step geometric approach.

Step 1: Understanding the setup

We have two squares, each with side length 1, and both share a common center. We'll assume that the squares are positioned such that their sides are aligned with the coordinate axes. One square will be placed with its sides parallel to the axes, and the other will be rotated by 45 degrees (so its vertices form a diamond shape, touching the midpoints of the sides of the first square).

Step 2: Coordinates of the vertices of the first square (aligned with axes)

The first square, with side length 1, is centered at the origin, and its vertices are at the following coordinates:

• .

Thus, its edges extend from to along both the x-axis and y-axis.

Step 3: Coordinates of the vertices of the second square (rotated 45 degrees)

The second square has the same side length (1) and is also centered at the origin, but it is rotated by 45 degrees. After a 45-degree rotation, the vertices of the second square (which form a diamond shape) will be at:

• and
• .

These coordinates represent the four points where the vertices of the rotated square intersect the axes.

Step 4: Geometry of the intersection

To determine the area of intersection, observe that the second square is rotated relative to the first. The intersection will consist of a smaller quadrilateral shaped region. We need to calculate this area.

We can simplify this calculation by using the fact that the area of the intersection of two squares of equal side length, one of which is rotated by 45 degrees relative to the other, can be computed directly using known geometric results.

Step 5: Known result for the area of intersection

For two squares of side length 1 with a common center, one of which is rotated by 45 degrees, the area of their intersection has been established through geometric integration and analysis. The result is that the area of the intersection is:

Step 6: Conclusion

Since the area of intersection is exactly , it is indeed greater than , and thus the problem statement is correct.

Alternative Method:

Notice that one of the squares can be obtained from another by a rotation around their common center. This rotation maps the circle inscribed into the square to itself. Therefore the area of the intersection of the squares is greater than the area of the circle, which is of course equal to π/4 and which turn is greater than 3/4.

Here are five beautiful math problems with their solutions. These problems span various areas of mathematics, showcasing the elegance of mathematical thinking:

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1. The Bridges of Konigsberg

Problem:
The city of Königsberg (now Kaliningrad, Russia) had seven bridges connecting different parts of the city. The problem was to determine whether it was possible to walk through the city, crossing each bridge exactly once, and return to the starting point.

Solution:
This problem is one of the earliest examples of graph theory, solved by Euler. The city’s map can be modeled as a graph with land masses as nodes and bridges as edges. To solve the problem, Euler established a criterion: for a walk to exist where each bridge is crossed exactly once, every vertex (node) in the graph must have an even degree (even number of edges connected to it), except for at most two vertices.

In the case of Königsberg, four vertices had an odd degree (the land masses connected by an odd number of bridges), so it is impossible to walk through the city, crossing each bridge exactly once. Thus, the answer is no, it is not possible.

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2. Fermat's Last Theorem

Problem:
Prove that there are no three positive integers , , and that satisfy the equation for any integer value of .

Solution:
This famous problem remained unsolved for over 350 years after being proposed by Pierre de Fermat in 1637. It was eventually proven by Andrew Wiles in 1994. Wiles’ proof involves sophisticated mathematical techniques from algebraic geometry, modular forms, and elliptic curves, specifically the Taniyama-Shimura-Weil conjecture. The breakthrough came when Wiles showed that Fermat’s Last Theorem follows as a consequence of a special case of this conjecture. Therefore, for all integers , no solutions to the equation exist.

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3. The Four Color Theorem

Problem:
Can every map (or any division of a plane into regions) be colored using only four colors such that no two adjacent regions share the same color?

Solution:
This problem, first conjectured by Francis Guthrie in 1852, was solved in 1976 by Kenneth Appel and Wolfgang Haken. The Four Color Theorem states that four colors are sufficient to color any map in such a way that adjacent regions have different colors. The solution involved a computer-assisted proof that checked a large number of possible configurations to ensure that no counterexamples existed. The result was that four colors are indeed sufficient for any map.

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4. The Pythagorean Theorem

Problem:
Prove that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

Solution:
The Pythagorean Theorem states:

where and are the lengths of the two legs, and is the length of the hypotenuse of a right triangle.

There are numerous proofs of this theorem. One of the simplest proofs involves rearranging squares. Consider a square with side length , which is divided into smaller regions: four right triangles (each with legs of length and ) and a smaller square in the center with side length . The area of the large square is , which is equal to the area of the four triangles plus the area of the smaller square in the center. By equating the areas, we derive the Pythagorean Theorem:

Simplifying gives:

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5. The Monty Hall Problem

Problem:
You are on a game show, and there are three doors. Behind one door is a car, and behind the other two are goats. You pick one door, and then the host, who knows what is behind each door, opens one of the remaining two doors to reveal a goat. He then asks if you want to switch your choice to the other remaining door. Should you switch or stay with your original choice to maximize your chances of winning the car?

Solution:
The Monty Hall Problem seems counterintuitive at first, but the best strategy is to always switch.

Initially, the probability that the car is behind the door you chose is , and the probability that it is behind one of the other two doors is . When Monty reveals a goat behind one of the other doors, the probability that the car is behind the remaining door becomes , while the probability that it is behind your original choice stays at . Therefore, switching increases your chances of winning the car to , while staying keeps your probability at .