IMO 1968 SL 13

Origin: POL

Problem

Given two congruent triangles $A_1A_2A_3$ and $B_1B_2B_3$ $(A_iA_k = B_iB_k)$, prove that there exists a plane such that the orthogonal projections of these triangles onto it are congruent and equally oriented.

Solution

Translating one of the triangles if necessary, we may assume w.l.o.g. that $B_1 \equiv A_1$. We also assume that $B_2 \not\equiv A_2$ and $B_3 \not\equiv A_3$, since the result is obvious otherwise.

There exists a plane $\pi$ through $A_1$ that is parallel to both $A_2B_2$ and $A_3B_3$. Let $A'_2$, $A'_3$, $B'_2$, $B'_3$ denote the orthogonal projections of $A_2$, $A_3$, $B_2$, $B_3$ onto $\pi$, and let $h_2$, $h_3$ denote the distances of $A_2$, $B_2$ and of $A_3$, $B_3$ from $\pi$. By the Pythagorean theorem,

$$A'_2A'_3{}^2 = A_2A_3{}^2 - (h_2 + h_3)^2 = B_2B_3{}^2 - (h_2 + h_3)^2 = B'_2B'_3{}^2,$$

and similarly $A_1A'_2 = A_1B'_2$ and $A_1A'_3 = A_1B'_3$; hence $\triangle A_1A'_2A'_3$ and $\triangle A_1B'_2B'_3$ are congruent. If these two triangles are equally oriented, then we have finished. Otherwise, they are symmetric with respect to some line $a$ passing through $A_1$, and consequently the projections of the triangles $A_1A_2A_3$ and $A_1B_2B_3$ onto the plane through $a$ perpendicular to $\pi$ coincide.