China Mathematical Competition (Hainan)

Question

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,

A B ⊥ O B

and intersecting at

B

,

O H ⊥ P B

and intersecting at

H

,

P A = 4

, and

C

is the midpoint of

P A

. When the tetrahedron

O - H P C

has the maximum volume, the length of

O B

is ( ).

(A)

\frac{5}{3}

(B)

\frac{2 5}{3}

(C)

\frac{6}{3}

(D)

\frac{2 6}{3}

Accepted Answer

Since AB OB, and AB OP, we have AB PB, and PAB POB. Moreover, from OH PB we obtain that OH HC and OH PA. Since C is the midpoint of PA, OC PA. Thus, PC is the altitude of the FIGURE_0 tetrahedron O-HPC and PC = 2. In OHC, OC = 2. Therefore when HO = HC, S_ ABC reaches its maximum, that is, V_O-HPC = V_P-HCO reaches its maximum. In this case, HO = 2, and HO = 12 OP. Hence, HPO = 30^, and OB = OP 30^ = 263. Answer: D.