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3

Part of 1985 AMC 12/AHSME

Problems(1)

Finding the Length between Two Circular Arcs

Source:

3/2/2009
In right △ABC \triangle ABC with legs 5 5 and 12 12, arcs of circles are drawn, one with center A A and radius 12 12, the other with center B B and radius 5 5. They intersect the hypotenuse at M M and N N. Then, MN MN has length:
[asy]size(200); 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("AA", A, dir(point--A)); label("BB", B, dir(point--B)); label("CC", C, dir(point--C)); label("MM", M, dir(point--M)); label("NN", N, dir(point--N));
label("1212", (6,0), S); label("55", (12,3.5), E);[/asy]
<spanclass=′latex−bold′>(A)</span> 2<spanclass=′latex−bold′>(B)</span> 135<spanclass=′latex−bold′>(C)</span> 3<spanclass=′latex−bold′>(D)</span> 4<spanclass=′latex−bold′>(E)</span> 245 <span class='latex-bold'>(A)</span>\ 2 \qquad <span class='latex-bold'>(B)</span>\ \frac {13}{5} \qquad <span class='latex-bold'>(C)</span>\ 3 \qquad <span class='latex-bold'>(D)</span>\ 4 \qquad <span class='latex-bold'>(E)</span>\ \frac {24}{5}
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