You may have seen the pigeonhole principle before, sometimes called Dirichlet’s box principle. It says that if you have more pigeons than pigeonholes, and you put all of the pigeons into some pigeonhole, then there exists at least one pigeonhole with at least two pigeons. While it sounds quite simple, it’s a powerful technique. The difficult thing is often choosing the appropriate pigeons and pigeonholes.
It has multiple applications in various situations.
Today we will see how to use it in geometric problems.
Show that given any nine points on a sphere, there is a closed hemisphere that contains six of them. A closed hemiphere is one that contains the equator with respect to the division.
I’m thinking of a positive number less than \(100\). This number has remainder \(1\) when divided by \(3\), it has remainder \(2\) when divided by \(4\), and finally, it leaves remainder \(3\) when divided by \(5\). What number am I thinking of?
I’m thinking of two prime numbers. The first prime number squared is thirty-six more than the second prime number. What’s the second prime number?
How many integers less than \(2025\) are divisible by \(18\) or \(21\), but not both?
Determine all prime numbers \(p\) such that \(p^2-6\) and \(p^2+6\) are both prime numbers.
Let \(ABCD\) be a square and let \(X\) be any point on side \(BC\) between \(B\) and \(C\). Let \(Y\) be the point on line \(CD\) such that \(BX=YD\) and \(D\) is between \(C\) and \(Y\). Prove that the midpoint of \(XY\) lies on diagonal \(BD\).
Let \(ABCD\) be a trapezium such that \(AB\) is parallel to \(CD\). Let \(E\) be the intersection of diagonals \(AC\) and \(BD\). Suppose that \(AB=BE\) and \(AC=DE\). Prove that the internal angle bisector of \(\angle BAC\) is perpendicular to \(AD\).
Let \(ABC\) be an isosceles triangle with \(AB=AC\). Point \(D\) lies on side \(AC\) such that \(BD\) is the angle bisector of \(\angle ABC\). Point \(E\) lies on side \(BC\) between \(B\) and \(C\) such that \(BE=CD\). Prove that \(DE\) is parallel to \(AB\).
Is it possible to place a positive integer in every cell of a \(10\times10\) array in such a way that both the following conditions are satisfied?
Each number (not in the bottom row) is a proper divisor of the number immediately below.
The numbers in each row, rearrange if necessary, form a sequence of 10 consecutive numbers.