1985 AHSME Problems/Problem 3

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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 at $M$ 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

First of all, from the Pythagorean Theorem, $AB=\sqrt{AC^2+BC^2}=\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}=13$. Also, since $AM$ and $AC$ are radii of the same circle, $AM=AC=12$. Therefore, $MB=AB-AM=13-12=1$. Also, since $BN$ and $BC$ are radii of the same circle, $BN=BC=5$. We therefore have $MN=BN-BM=5-1=4, \boxed{\text{D}}$.

See Also

1985 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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