IMO 1967 LL CZS10

Origin: CZS

Problem

The square ABCD is to be decomposed into n triangles (nonoverlapping) all of whose angles are acute. Find the smallest inte- ger n for which there exists a solution to this problem and construct at least one decomposition for this n. Answer whether it is possible to ask additionally that (at least) one of these triangles has a perimeter less than an arbitrarily given positive number.

Solution

Let n be the number of triangles and let b and i be the numbers of vertices on the boundary and in the interior of the square, respectively. Since all the triangles are acute, each of the vertices of the square belongs to at least two triangles. Additionally, every vertex on the boundary be- longs to at least three, and every vertex in the interior belongs to at least five triangles. Therefore 3n \geq8 + 3b + 5i. (1) Moreover, the sum of angles at any vertex that lies in the interior, on the boundary, or at a vertex of the square is equal to 2\pi, \pi, \pi/2 respec- tively. The sum of all angles of the triangles equals n\pi, which gives us n\pi = 4 \cdot \pi/2 + b\pi + 2i\pi, i.e., n = 2 + b + 2i. This relation together with (1) easily yields that i \geq2. Since each of the vertices inside the square belongs to at least five trian- gles, and at most two contain both, it follows that n \geq8. A B C D K L

It is shown in the figure that the square can be decomposed into eight acute triangles. Obviously one of them can have an arbitrarily small perimeter.