There are $43$ points in the space: $3$ yellow and $40$ red. Any four of them are not coplanar. May the number of triangles with red vertices hooked with the triangle with yellow vertices be equal to $2023$? Yellow triangle is hooked with the red one if the boundary of the red triangle meet the part of the plane bounded by the yellow triangle at the unique point. The triangles obtained by the transpositions of vertices are identical.
Problem
Source: Sharygin Finals 2023 10.7
Tags: geometry, Sharygin Geometry Olympiad, Sharygin 2023
i3435
03.08.2023 17:32
I can't find problems with this but it doesn't seem right. The answer is no. Label the red vertices $1,2,\dots ,40$ and for each $i$ let $d_i$ be the number of red points $j$ for which $ij$ passes through the yellow triangle. $\sum d_i$ is even because it counts each pair of vertices whose segment passes through the yellow triangle twice. In each triangle of red points, there are at most $2$ segments which pass through the yellow triangle because at least one pair of red points in the triangle will be on the same side of the plane formed by the yellow triangle. $38\frac{\sum d_i}{2}$ counts the pairs (line segment passing through the yellow triangle, point not forming the line segment), so it counts the number of hooked red triangles plus twice the number of red triangles which have two segments passing through the yellow triangle. The number of red triangles which have two segments passing through the yellow triangle is $\sum \binom{d_i}{2}$, so the number of hooked red triangles is $19\sum d_i-2\sum \binom{d_i}{2}=\sum 20d_i-\sum d_i^2$. Since $\sum d_i$ is even, $\sum d_i^2$ is even, so the number of hooked red triangles is even and can't be $2023$.
N4palm
21.11.2023 19:50
I have solved this problem during the final round and I came up with a solution that I think is a little bit more elegant than the official solution using bipartite graphs and complex calculations.
The answer is no. Let the yellow triangle be $T$ and its plane be $\alpha$. 'Let's look at a random tetrahedron with red vertices. Notice that if we start from any vertice of $T$ and walk along his edges, when we come back to the initial point, we will cross the faces of the tetrahedron an even amount of times, since each time we cross a side we "switch" our position relative to the tetrahedron: from inside to outside and vice versa, and when we come back, we are in the same state as when we began. Then, notice that the edges of $T$ cannot intersect any face $R$ of the tetrahedron more than 2 times: no edge of $T$ intersects any face more than once and all 3 edges of $T$ can't intersect $R$ in inner points, since that would contradict Menelaus theorem in vector form for $T$ and line $r \cap \alpha$, where $r$ is the plane that contains $R$. Thus, the parity of intersections is equal to parity of 1-point intersections of $T$ and the tetrahedron. But note that this is exactly the amount of hooked triangles. The only thing left to do is to sum this equality over all tetrahedrons and notice that we counted every hooked triangle 37 times, 1 time in each tetrahedron that has this triangle as a face. Thus, the amount of hooked triangles is even and can't be equal to 2023.
Amy_Chen
18.10.2024 10:28
Can somebody post the diagram.