Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.2 AIME, harder Find the answer

8. As shown in Figure 3, given a regular tetrahedron $O-ABC$ with three lateral edges $OA, OB, OC$ mutually perpendicular, and each of length $2, E, F$ are the midpoints of edges $AB, AC$ respectively, $H$ is the midpoint of line segment $EF$ , and a plane is constructed through $EF$ intersecting the lateral edges $OA, OB, OC$ or their extensions at points $A_{1}, B_{1}, C_{1}$ . If $OA_{1}=\frac{3}{2}$ , then the tangent value of the dihedral angle $O-A_{1}B_{1}-C_{1}$ is $\qquad$

Your answer

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

Solution

8. $\sqrt{5}$ .

As shown in Figure 5, draw $O N \perp A_{1} B_{1}$ at point $N$ , and connect $C_{1} N$ .
Since $O C_{1} \perp$ plane $O A_{1} B_{1}$ , by the theorem of three perpendiculars, we know $C_{1} N \perp A_{1} B_{1}$ :
Therefore, $\angle O N C_{1}$ is the plane angle of the dihedral angle $O-A_{1} B_{1}-C_{1}$ . Draw $E M \perp O B_{1}$ at point $M$ . Then $E M / / O A$ .
Thus, $M$ is the midpoint of $O B$ . Therefore,
$E M = O M = 1$ .
Let $O B_{1} = x$ .
From $\frac{O B_{1}}{M B_{1}} = \frac{O A_{1}}{E M} \Rightarrow \frac{x}{x-1} = \frac{3}{2} \Rightarrow x = 3$ .
In the right triangle $\triangle O A_{1} B_{1}$ , we have
$A_{1} B_{1} = \sqrt{O A_{1}^{2} + O B_{1}^{2}} = \frac{3 \sqrt{5}}{2}$ .
Thus, $O N = \frac{O A_{1} \cdot O B_{1}}{A_{1} B_{1}} = \frac{3}{\sqrt{5}}$ .
Therefore, $\tan \angle O N C_{1} = \frac{O C_{1}}{O N} = \sqrt{5}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.