Competition problem — Geometry — Maths Olympiad Prep

Library / /748 of 860

Geometry Difficulty 5.5 AIME, harder Find the answer

Regular tetrahedron $A B C D$ is projected onto a plane sending $A, B, C$ , and $D$ to $A^{\prime}, B^{\prime}, C^{\prime}$ , and $D^{\prime}$ respectively. Suppose $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is a convex quadrilateral with $A^{\prime} B^{\prime}=A^{\prime} D^{\prime}$ and $C^{\prime} B^{\prime}=C^{\prime} D^{\prime}$ , and suppose that the area of $A^{\prime} B^{\prime} C^{\prime} D^{\prime}=4$ . Given these conditions, the set of possible lengths of $A B$ consists of all real numbers in the interval $[a, b)$ . Compute $b$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

The value of $b$ occurs when the quadrilateral $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ degenerates to an isosceles triangle. This occurs when the altitude from $A$ to $B C D$ is parallel to the plane. Let $s=A B$ . Then the altitude from $A$ intersects the center $E$ of face $B C D$ . Since $E B=\frac{s}{\sqrt{3}}$ , it follows that $A^{\prime} C^{\prime}=A E=\sqrt{s^{2}-\frac{s^{2}}{3}}=\frac{s \sqrt{6}}{3}$ . Then since $B D$ is parallel to the plane, $B^{\prime} D^{\prime}=s$ . Then the area of $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is $4=\frac{1}{2} \cdot \frac{s^{2} \sqrt{6}}{3}$ , implying $s^{2}=4 \sqrt{6}$ , or $s=2 \sqrt[4]{6}$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.