Problem

2004 AMC 12B Problem 14

In $\triangle ABC$ , $AB=13$ , $AC=5$ , and $BC=12$ . Points $M$ and $N$ lie on $AC$ and $BC$ , respectively, with $CM=CN=4$ . Points $J$ and $K$ are on $AB$ so that $MJ$ and $NK$ are perpendicular to $AB$ . What is the area of pentagon $CMJKN$ ?

[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,5), B=(12,0), M=(0,4), N=(4,0); pair J=intersectionpoint(A--B, M--(M+rotate(90)*(B-A)) ); pair K=intersectionpoint(A--B, N--(N+rotate(90)*(B-A)) ); draw( A--B--C--cycle ); draw( M--J ); draw( N--K ); label("$A$",A,NW); label("$B$",B,SE); label("$C$",C,SW); label("$M$",M,SW); label("$N$",N,S); label("$J$",J,NE); label("$K$",K,NE); [/asy]

$\mathrm{(A)}\ 15 \qquad \mathrm{(B)}\ \frac{81}{5} \qquad \mathrm{(C)}\ \frac{205}{12} \qquad \mathrm{(D)}\ \frac{240}{13} \qquad \mathrm{(E)}\ 20$

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.

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Category: AMC 12B
Points: 3
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