鸽巢原理

2012 USAMO Problems/Problem 2

Problem
A circle is divided into 432 congruent arcs by 432 points. The points are colored in four colors such that some 108 points are colored Red, some 108 points are colored Green, some 108 points are colored Blue, and the remaining 108 points are colored Yellow. Prove that one can choose three points of each color in such a way that the four triangles formed by the chosen points of the same color are congruent.

Solution
If you rotate the red points 431 times, they will overlap with blue points $108\times 108$ times, for an average of $\frac{108\times 108}{431}$ per rotation. Note that this average is slightly greater than 27. Therefore at some point 28 red points overlap with blue points. In other words, there exist 28 red and blue points such that the convex 28-gons formed by them are congruent.

Rotate these 28 red points 431 times. They will overlap with green points $108\times 28$ times, for an average of $\frac{108\times 28}{431}$ per rotation. This average is slightly greater than 7, so at some point 8 of those red points overlap with green points. In other words, there exist 8 red points, 8 blue points, and 8 green points such that the convex octagons formed by them are congruent.

Rotate these 8 red points 431 times. They will overlap with yellow points $108\times 8$ times, for an average of $\frac{108\cdot 8}{431}$ per rotation. This average is slightly greater than 2, so at some point 3 of those red points will overlap with yellow points. In other words, there exist 3 red points, 3 blue points, 3 green points, and 3 yellow points such that the triangles formed by them are congruent.

Note
Huge bash of Pigeonhole Principle:

$$108\longrightarrow 28\longrightarrow 8\longrightarrow 3$$
$\left \lceil{\left \lceil{\left \lceil{108\cdot \frac{108}{431}}\right \rceil \cdot \frac{108}{431}}\right \rceil \cdot \frac{108}{431}}\right \rceil =\left \lceil{\left \lceil{28\cdot \frac{108}{431}}\right \rceil \cdot \frac{108}{431}}\right \rceil =\left \lceil{8\cdot \frac{108}{431}}\right \rceil =3$

https://www.docin.com/p-1900556536.html

https://wenku.baidu.com/view/22965a4b5e0e7cd184254b35eefdc8d376ee14b2.html

https://wenku.baidu.com/view/6dffcd2acfc789eb162dc805.html?sxts=1570529223251

https://math.stackexchange.com/questions/2816894/circles-inside-a-square/2816917

There are several circles inside a square of side length $1$. The sum of the circumferences of the circles is $10$. Prove that there exists a line that intersects at least $4$ of the circles.

We need to solve this using expected value, I tried making some diagrams, but gained nothing out of it. I really need some help.

Since we need to solve this using expected value, it is more "useful" to think of probability. And we are lucky that the square is a unit square:

[![enter image description here][1]][1]

Therefore, the probability of a line intersecting a circle, is simply *the diameter of that circle* (think of vertical movement of the green line!). Let's move on: We know from the question $$\sum_i \pi d_i=10\implies \sum_i d_i=\dfrac{10}{\pi},$$ where $d_i$ are diameter of each circle $i$. Then $$\mathbb{E}[\text{line intersects with circle}]=\mathbb{E}[C_1 \text{ intersect}]+\mathbb{E}[C_2 \text{ intersect}]+\cdots+\mathbb{E}[C_n \text{ intersect}]=\sum_i \mathbb{E}[C_i \text{ intersect}]=\sum_i d_i=\dfrac{10}{\pi}\approx \dfrac{10}{3.14}>3$$
This reads: it is expected that there are more than 3 circles that intersect with the line. We can then conclude that there exists a line that intersects at least 4 of the circles.
[1]: https://i.stack.imgur.com/LtMAW.jpg

 

$$\begin{array}{l} \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{\sin \ln \left( {n + 1} \right)}}{{\sin \ln n}} - 1} \right] = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{\sin \ln \left( {n + 1} \right) - \sin \ln n}}{{\sin \ln n}}} \right]\\ = \mathop {\lim }\limits_{n \to \infty } \frac{2}{{\sin \ln n}}\cos \frac{{\ln \left( {{n^2} + n} \right)}}{2}\sin \frac{{\ln \left( {1 + \frac{1}{n}} \right)}}{2} = 0 \end{array}$$