[ Dirichlet's Principle (area and volume).] $[$ Geometry on graph paper $]$

A corner square has been cut out of a chessboard. What is the minimum number of equal-area triangles the resulting figure can be cut into?
Evaluation. Let's take the area of one cell as a unit. The given figure is a non-convex hexagon $A B C D E F$ with an area of 63 and an angle of $270^{\circ}$ at vertex $D$ (see the figure on the right). If the figure is divided into triangles, it is clear that point $D$ must belong to at least two triangles, one of which has a side on the line $D E$, and the other on $D C$. Moreover, for at least one of them, it lies on the corresponding segment. For definiteness, let's assume this is triangle $D K L$, where $K$ lies on the segment $D C$. Then the base $D K$ of this triangle is no more than

$D C=1$, and the height is no more than $B C=7$. Therefore, $S_{D K L} \leq{ }^{7} / 2$. If all triangles are equal in area, then there are no fewer than $63:{ }^{7} / 2=18$.

An example of cutting into 18 equal-area triangles is shown in the figure on the left.
![](https://cdn.mathpix.com/cropped/2024_05_06_9e7ee8b57660b9721937g-22.jpg?height=480&width=1062&top_left_y=2384&top_left_x=502)

## [  Coloring  Problem $\underline{32099}$ Topics: [Equivalence relation. Equivalence classes] [ Equilateral (equilateral) triangle ]

On a plane, a certain number of equilateral triangles are drawn. They do not intersect, but they can share parts of their sides. We want to color each triangle in some color so that those which touch each other are painted in different colors (triangles that have one common point can be painted in the same color). Will two colors be enough for such a coloring?

## Solution

We will call two triangles parallel if some side of one of them is parallel to some side of the other. It is clear that then each side of one of them is parallel to some side of the other. Therefore, parallelism on the set of equilateral triangles on the plane is an equivalence relation.

Since each triangle can share a part of a side only with a triangle parallel to it, the coloring can be done independently for each "parallel class". We will show how to color in two colors, satisfying the conditions of the problem, the triangles of one "parallel class". For this, we will introduce the concept of orientation on the "parallel class": two triangles in the figure on the left are oriented the same way, and two triangles in the figure on the right are oriented oppositely.
![](https://cdn.mathpix.com/cropped/2024_05_06_9e7ee8b57660b9721937g-23.jpg?height=368&width=1236&top_left_y=1138&top_left_x=412)

Obviously, two triangles can share a part of a side only if they are oppositely oriented. Therefore, if we paint all triangles of one orientation in one color, and all triangles of the opposite orientation in another color, we will get the required coloring in two colors.

## Answer

It will be enough.