1985 AHSME Problems/Problem 3

Problem

In right $\triangle ABC$ with legs $5$ and $12$ , arcs of circles are drawn, one with center $A$ and radius $12$ , the other with center $B$ and radius $5$ . They intersect the hypotenuse in $M$ and $N$ . Then $MN$ has length

$[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(12,7), C=(12,0), M=12*dir(A--B), N=B+B.y*dir(B--A); real r=degrees(B); draw(A--B--C--cycle^^Arc(A,12,0,r)^^Arc(B,B.y,180+r,270)); pair point=incenter(A,B,C); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$M$", M, dir(point--M)); label("$N$", N, dir(point--N)); label("$12$", (6,0), S); label("$5$", (12,3.5), E);[/asy]$

$\mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }\frac{13}{5} \qquad \mathrm{(C) \ } 3 \qquad \mathrm{(D) \ } 4 \qquad \mathrm{(E) \ }\frac{24}{5}$

Solution

Firstly, the Pythagorean theorem gives $\begin{align*}AB &=\sqrt{AC^2+BC^2} \\ &= \sqrt{12^2+5^2} \\ & =\sqrt{144+25} \\ &=\sqrt{169} \\ &= 13.\end{align*}$ Also, $AM = AC = 12$ and $BN = BC = 5$ since they are both radii of the respective circles. Thus $MB = AB-AM = 13-12 = 1$ , and so $MN = BN-BM = 5-1 = \boxed{\text{(D)} \ 4}$ .

1985 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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
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1985 AHSME Problems/Problem 3

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