Track / Stage 4 / 73 of 340 #333 of 2000

Problem 333

AMC 12 late, AIME early

Geometry Difficulty 4.6 Find the answer

$24 \cdot 8$ When a lake froze, a ball was floating on it. After the ball was removed (without breaking the ice), a hole with a top diameter of $24 \mathrm{~cm}$ and a depth of $8 \mathrm{~cm}$ was left on the ice surface. What is the radius of the ball?
(A) $8 \mathrm{~cm}$ .
(B) $12 \mathrm{~cm}$ .
(C) $13 \mathrm{~cm}$ .
(D) $8 \sqrt{3} \mathrm{~cm}$ .
(E) $6 \sqrt{6} \mathrm{~cm}$ .
(38th American High School Mathematics Examination, 1987)

Your answer

Official solution

[Solution] Let the radius of the sphere be $R$ . On the great circle of the sphere, using the intersecting chords theorem, we have $(2 R-8) \cdot 8=12 \cdot 12$ , solving for $R$ gives $R=13$ . Therefore, the answer is $(C)$ .

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