1952 AHSME Problems/Problem 29

Problem

In a circle of radius $5$ units, $CD$ and $AB$ are perpendicular diameters. A chord $CH$ cutting $AB$ at $K$ is $8$ units long. The diameter $AB$ is divided into two segments whose dimensions are:

$\textbf{(A)}\ 1.25, 8.75 \qquad \textbf{(B)}\ 2.75,7.25 \qquad \textbf{(C)}\ 2,8 \qquad \textbf{(D)}\ 4,6 \qquad \textbf{(E)}\ \text{none of these}$

Solution

$\fbox{}$

Let the intersection of $CH$ and $AB$ be $N$ , $O$ be the center of the circle, $ON = a$ and $CN = x$ . By power of a point on $N$ , we have $(5+a)(5-a) = 25-a^2 = x(8-x).$ $\triangle CON$ is a right triangle, so we also know that $x^2 = a^2 + 25$ , thus $25 - a^2 = 25-(x^2-25) = 50-x^2 = 8x-x^2 \Rightarrow x = \frac{50}{8} = \frac{25}{4}.$ It follows that $a = \sqrt{x^2 - 25} = 5\sqrt{\frac{25}{16}-1} = \frac{15}{4}$ .

1952 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 28	Followed by Problem 30
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1952 AHSME Problems/Problem 29

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