1987 AHSME Problems/Problem 22

Revision as of 17:56, 15 January 2024 by Mathchampion 1234 (talk | contribs) (Solution)

Problem

A ball was floating in a lake when the lake froze. The ball was removed (without breaking the ice), leaving a hole $24$ cm across as the top and $8$ cm deep. What was the radius of the ball (in centimeters)?

$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 13 \qquad \textbf{(D)}\ 8\sqrt{3} \qquad \textbf{(E)}\ 6\sqrt{6}$

Solution

Consider a cross-section of this problem in which a circle lies with its center somewhere above a line. A line segment of $8$ cm can be drawn from the line to the bottom of the ball. Denote the distance between the center of the circle and the line as $x$. We can construct a right triangle by dragging the center of the circle to the intersection of the circle and the line. We then have the equation $x^2+(12)^2=(x+8)^2$, $x^2+144=x^2+16x+64$. Solving, the answer is (C) $13$

See also

1987 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png