Problem 1277

In order to reuse the principle of drawers as in the previous exercise, we attempt to cut the $4 \times 3$ rectangle into five zones with a diameter of at most $\sqrt{5}$. Instead of abruptly presenting such a partition, let's build it step by step.

Moreover, since replacing a zone with its convex hull does not change its diameter, we can always assume our zones are convex. Similarly, if a zone contains the interior of a segment $[X, Y]$, we observe that all its points are at a distance $d \leqslant \sqrt{5}$ from $X$ and $Y$, so we will assume that $X$ and $Y$ also belong to our zone.

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The five points $A, E, P, T$, and any point inside the segment $[M H]$ represented above are at a distance $d>\sqrt{5}$ from each other, so they belong to five different zones. We then draw these zones with different colors, considering points at the boundary of several zones as multicolored. In particular, since the black polygon contains the interior of the segment $[M H]$, it contains the entire segment $[M H]$.

Next, any point inside the segment $[A B]$ is at a distance $d>\sqrt{5}$ from the points $E, M$, $P$, and $T$, so it belongs to the blue zone. Thus, the segment $[A B]$ is in the blue zone. But then any point inside the segment $[K P]$ is at a distance $d>\sqrt{5}$ from the points $B, E, H$, and $Y$, so it belongs to the purple zone. Thus, the segment $[P K]$ is in the purple zone. We similarly determine the colors of the points $I, J, K, L, N$, and $P$.

Finally, any point at a distance $d \leqslant \sqrt{5}$ from $A$ and $B$ is also at a distance $d^{\prime} \leqslant \sqrt{5}$ from $H$. We therefore assume $H$ is blue. Similarly, we assume $L$ is blue, then $H$ and $N$ are red, $I$ and $M$ are green, and $G$ and $M$ are purple.

Furthermore, the opposite sides of the rectangle are at least 3 units apart, so the black zone cannot contain points from two opposite sides. Without loss of generality, we assume it contains no points from the sides $[A E]$ and $[A P]$. Then, any point inside the segment $[B C]$ is at a distance $d>\sqrt{5}$ from the points $J, P$, and $T$, so it belongs to the blue zone. Thus, the segment $[B C]$ is in the blue zone, and similarly, the segment $[C D]$ is in the red zone.

But then any point inside the segment $[F K]$ is at a distance $d>\sqrt{5}$ from the points $C$ and $T$, so it belongs to the purple zone. Thus, the segment $[F K]$ is in the purple zone. We similarly prove that $[Q R]$ is in the black zone, that $[J O]$ is in the green zone, and finally that $[R S]$ is in the black zone. This gives us the partition below.

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We are almost done, but the two small gray triangles have not yet been included in any zone. We could include some points of the left triangle in the blue or purple zones, but others are too far from the points $A$ and $P$, and must necessarily belong to the black zone. However, we have a lot of room to expand it, and we include our two gray triangles in it. We can even push the perfectionism to include the points $L$ and $N$ in the black zone, so that our $3 \times 4$ rectangle will be covered by five monochromatic zones in the shape of houses, as below.

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House

In particular, among the six points that survived Gabrielle's intervention, two points belong to the same zone, and they are therefore at a distance of at most $\sqrt{5}$ from each other.

## Attention!

1. As this exercise clearly illustrates, it is sometimes (and, in reality, often) more difficult to find the right drawers than to apply the principle of drawers afterward.
2. Nothing excludes that a point belongs to several houses or, more generally, that a shirt is stored in several drawers: if this is the case, it increases our chances of finding a really full drawer!

In any case, in such a situation, we can always choose the drawer to which each element will belong in priority and exclude this element from the other drawers to which it could have belonged a priori. Here, for example, this means replacing our house-shaped zones with the colored zones in the left drawing, where we have truncated the green and purple houses.

## 2 Angle Hunting (Mathieu Ba.)

Angle hunting is a tool that proves extremely useful in solving geometry problems. Many seemingly complex problems can be reformulated in terms