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Problem 256

AMC 10/12, early questions

Geometry Difficulty 4.0 Find the answer China Mathematical Competition · China

Suppose the side of the base and the height of regular triangular pyramid $P$ - $ABC$ are $1$ and $\sqrt{2}$ , respectively. Then the radius of the inscribed sphere of the pyramid is ______.

Your answer

A number or a short expression. Spacing, $ signs and \frac vs / are all fine.

Official solution

As seen in Fig. 4.1, suppose the projections of the inscribed sphere's center $O$ on faces $ABC$ and $ABP$ are $H$ , $K$ , respectively, the midpoint of $AB$ is $M$ , and the radius of the sphere is $r$ . Then $P$ , $K$ , $M$ are collinear, $\angle PHM = \angle PKO = \frac{\pi}{2}$ , and

Fig. 4.1
$OH = OK = r, \quad PO = PH - OH = \sqrt{2} - r, \\ MH = \frac{\sqrt{3}}{6} AB = \frac{\sqrt{3}}{6}, \quad PM = \sqrt{MH^2 + PH^2} = \sqrt{\frac{1}{12} + 2} = \frac{5\sqrt{3}}{6}.$
Then we have
$\frac{r}{\sqrt{2} - r} = \frac{OK}{PO} = \sin \angle KPO = \frac{MH}{PM} = \frac{1}{5}.$
Therefore, $r = \frac{\sqrt{2}}{6}$ .
The answer is $\frac{\sqrt{2}}{6}$ .

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