Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.3 AIME, harder Prove it China

Suppose two regular triangular pyramids $P$ - $ABC$ and $Q$ - $ABC$ sharing the same base are inscribed in the same sphere. If the angle between the side-face and the base of $P$ - $ABC$ is $45^\circ$ , then the tangent value of the angle between the side-face and the base of $\triangle ABC$ is ______.

Solution

As seen in Fig. 5.1, connecting $PQ$ , then $PQ$ is perpendicular to plane $ABC$ with the foot point $H$ being the center of $\triangle ABC$ . The center of the sphere $O$ is also on $PQ$ . Connect and extend $CH$ to let it intersect with $AB$ at point $M$ . $M$ is then the midpoint of $AB$ , and $CM \perp AB$ . It is easy to see that $\angle PMH$ and $\angle QMH$ are the plane angles formed by the sides-faces and the bases of the regular triangular pyramids $P$ - $ABC$ and $Q$ - $ABC$ , respectively. Then $\angle PMH = 45^\circ$ , so $PH = MH = \frac{1}{2}AH$ .

Since $\angle PAQ = 90^\circ$ , $AH \perp PQ$ , then $AH^2 = PH \cdot QH$ . We then have $AH^2 = \frac{1}{2}AH \cdot QH$ .
Therefore, $QH = 2AH = 4MH$ .
Finally, $\tan \angle QMH = \frac{QH}{MH} = 4$ .

The answer is 4.

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