Difference between revisions of "1987 AHSME Problems/Problem 22"

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==Solution==
 
==Solution==
Consider a cross-section of this problem in which a circle lies with it's center somewhere above a line. A line segment of <math>8</math> 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 <math>x</math>. 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 <math>x^2+(12)^2=(x+8)^2</math>, <math>x^2+144=x^2+16x+64</math>. Solving, the answer is <math>\textbf{(C)}\ 13 \qquad</math>
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Consider a cross-section of this problem in which a circle lies with its center somewhere above a line. A line segment of <math>8</math> 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 <math>x</math>. 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 <math>x^2+(12)^2=(x+8)^2</math>, <math>x^2+144=x^2+16x+64</math>. Solving, the answer is <math>{\textbf{(C)}\ 13}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 17:59, 15 January 2024

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 ${\textbf{(C)}\ 13}$.

See also

1987 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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