Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.4 AIME Find the answer China

The vertical cross-section of a circular cone with vertex $P$ is an isosceles right-angled triangle. Point $A$ is on the circumference of the base circle, point $B$ is interior to the base circle, $O$ is the center of the base circle, $AB \perp OB$ and intersecting at $B$ , $OH \perp PB$ and intersecting at $H$ , $PA = 4$ , and $C$ is the midpoint of $PA$ . When the tetrahedron $O-HPC$ has the maximum volume, the length of $OB$ is:

Pick one

Solution

Since $AB \perp OB$ , and $AB \perp OP$ , we have $AB \perp PB$ , and $PAB \perp POB$ . Moreover, from $OH \perp PB$ we obtain that $OH \perp HC$ and $OH \perp PA$ . Since $C$ is the midpoint of $PA$ , $OC \perp PA$ . Thus, $PC$ is the altitude of the

tetrahedron $O-HPC$ and $PC = 2$ .

In $\triangle OHC$ , $OC = 2$ . Therefore when $HO = HC$ , $S_{\triangle ABC}$ reaches its maximum, that is, $V_{O-HPC} = V_{P-HCO}$ reaches its maximum. In this case, $HO = \sqrt{2}$ , and $HO = \frac{1}{2} OP$ . Hence, $\angle HPO = 30^\circ$ , and $OB = OP \cdot \tan 30^\circ = \frac{2\sqrt{6}}{3}$ .

Answer: D.

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