1961 AHSME Problems/Problem 38

Problem

$\triangle ABC$ is inscribed in a semicircle of radius $r$ so that its base $AB$ coincides with diameter $AB$ . Point $C$ does not coincide with either $A$ or $B$ . Let $s=AC+BC$ . Then, for all permissible positions of $C$ :

$\textbf{(A)}\ s^2\le8r^2\qquad \textbf{(B)}\ s^2=8r^2 \qquad \textbf{(C)}\ s^2 \ge 8r^2 \qquad\\ \textbf{(D)}\ s^2\le4r^2 \qquad \textbf{(E)}\ s^2=4r^2$

Solution

$[asy] draw((-50,0)--(-30,40)--(50,0)--(-50,0)); draw(Arc((0,0),50,0,180)); draw(rightanglemark((-50,0),(-30,40),(50,0),200)); dot((-50,0)); label("A",(-50,0),SW); dot((-30,40)); label("C",(-30,40),NW); dot((50,0)); label("B",(50,0),SE); [/asy]$ Since $s=AC+BC$ , $s^2 = AC^2 + 2 \cdot AC \cdot BC + BC^2$ . Since $\triangle ABC$ is inscribed and $AB$ is the diameter, $\triangle ABC$ is a right triangle, and by the Pythagorean Theorem, $AC^2 + BC^2 = AC^2 = (2r)^2$ . Thus, $s^2 = 4r^2 + 2 \cdot AC \cdot BC$ .

The area of $\triangle ABC$ is $\frac{AC \cdot BC}{2}$ , so $2 \cdot [ABC] = AC \cdot BC$ . That means $s^2 = 4r^2 + 4 \cdot [ABC]$ . The area of $\triangle ABC$ can also be calculated by using base $AB$ and the altitude from $C$ . The maximum possible value of the altitude is $r$ , so the maximum area of $\triangle ABC$ is $r^2$ .

Therefore, $s^2 \le 8r^2$ , so the answer is $\boxed{\textbf{(A)}}$ .

1961 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 37	Followed by Problem 39
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
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1961 AHSME Problems/Problem 38

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