# Difference between revisions of "2004 AMC 12B Problems/Problem 14"

## Problem

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 $CMJKN$? $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,5), B=(12,0), M=(0,4), N=(4,0); pair J=intersectionpoint(A--B, M--(M+rotate(90)*(B-A)) ); pair K=intersectionpoint(A--B, N--(N+rotate(90)*(B-A)) ); draw( A--B--C--cycle ); draw( M--J ); draw( N--K ); label("A",A,NW); label("B",B,SE); label("C",C,SW); label("M",M,SW); label("N",N,S); label("J",J,NE); label("K",K,NE); [/asy]$ $\mathrm{(A)}\ 15 \qquad \mathrm{(B)}\ \frac{81}{5} \qquad \mathrm{(C)}\ \frac{205}{12} \qquad \mathrm{(D)}\ \frac{240}{13} \qquad \mathrm{(E)}\ 20$

## Solution

### Solution 1

The triangle $ABC$ is clearly a right triangle, its area is $\frac{5\cdot 12}2 = 30$. If we knew the areas of triangles $AMJ$ and $BNK$, we could subtract them to get the area of the pentagon.

Draw the height $CL$ from $C$ onto $AB$. As $AB=13$ and the area is $30$, we get $CL=\frac{60}{13}$. The situation is shown in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,5), B=(12,0), M=(0,4), N=(4,0); pair J=intersectionpoint(A--B, M--(M+rotate(90)*(B-A)) ); pair K=intersectionpoint(A--B, N--(N+rotate(90)*(B-A)) ); pair L=intersectionpoint(A--B, C--(C+rotate(90)*(B-A)) ); draw( A--B--C--cycle ); draw( M--J ); draw( N--K ); draw( C--L, dashed ); label("A",A,NW); label("B",B,SE); label("C",C,SW); label("M",M,SW); label("N",N,S); label("J",J,NE); label("K",K,NE); label("L",L,NE); [/asy]$

Now note that the triangles $ABC$, $AMJ$, $ACL$, $CBL$ and $NBK$ all have the same angles and therefore they are similar. We already know some of their sides, and we will use this information to compute their areas. Note that if two polygons are similar with ratio $k$, their areas have ratio $k^2$. We will use this fact repeatedly. Below we will use $[XYZ]$ to denote the area of the triangle $XYZ$.

We have $\frac{CL}{BC} = \frac{60/13}{12} = \frac 5{13}$, hence $[ACL] = \frac{ 25[ABC] }{169} = \frac{750}{169}$.

Also, $\frac{CL}{AC} = \frac{60/13}5 = \frac{12}{13}$, hence $[CBL] = \frac{ 144[ABC] }{169} = \frac{4320}{169}$.

Now for the smaller triangles:

We know that $\frac{AM}{AC}=\frac 15$, hence $[AMJ] = \frac{[ACL]}{25} = \frac{30}{169}$.

Similarly, $\frac{BN}{BC}=\frac 8{12} = \frac 23$, hence $[NBK] = \frac{4[CBL]}9 = \frac{1920}{169}$.

Finally, the area of the pentagon is $30 - \frac{30}{169} - \frac{1920}{169} = \boxed{\frac{240}{13}}$.

### Solution 2

Split the pentagon along a different diagonal as follows:

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 );
draw( M--N, dashed );
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);
label("$L$",L,NE);
(Error compiling LaTeX. label("$L$",L,NE);
^
0125ab1cc6f4632b43bedf8b6e324d871fc40be7.asy: 19.6: no matching function 'label(string, path(int k=<default>, pair A, pair B, real a=<default>, real b=<default>), pair)')

The area of the pentagon is then the sum of the areas of the resulting right triangle and trapezoid. As before, triangles $ABC$, $AMJ$, and $NBK$ are all similar.

Since $BN=12-4=8$, $NK=\frac{5}{13}(8)=\frac{40}{13}$ and $BK=\frac{12}{13}(8)=\frac{96}{13}$. Since $AM=5-4=1$, $JM=\frac{12}{13}$ and $AJ=\frac{5}{13}$.

The trapezoid's height is therefore $13-\frac{5}{13}-\frac{96}{13}=\frac{68}{13}$, and its area is $\frac{1}{2}\left(\frac{68}{13}\right)\left(\frac{12}{13}+\frac{40}{13}\right)=\frac{34}{13}(4)=\frac{136}{13}$.

Triangle $MCN$ has area $\frac{1}{2}(4)(4)=8$, and the total area is $\frac{104+136}{13}=\boxed{\frac{240}{13}}$.