# 1999 AMC 8 Problems/Problem 23

## Problem

Square $ABCD$ has sides of length 3. Segments $CM$ and $CN$ divide the square's area into three equal parts. How long is segment $CM$?

$[asy] pair A,B,C,D,M,N; A = (0,0); B = (0,3); C = (3,3); D = (3,0); M = (0,1); N = (1,0); draw(A--B--C--D--cycle); draw(M--C--N); label("A",A,SW); label("M",M,W); label("B",B,NW); label("C",C,NE); label("D",D,SE); label("N",N,S); [/asy]$

$\text{(A)}\ \sqrt{10} \qquad \text{(B)}\ \sqrt{12} \qquad \text{(C)}\ \sqrt{13} \qquad \text{(D)}\ \sqrt{14} \qquad \text{(E)}\ \sqrt{15}$

## Solution

Since the square has side length $3$, the area of the entire square is $9$.

The segments divide the square into 3 equal parts, so the area of each part is $9 \div 3 = 3$.

Since $\triangle CBM$ has area $3$ and base $CB = 3$, using the area formula for a triangle:

$A_{tri} = \frac{1}{2}bh$

$3 = \frac{1}{2}3h$

$h = 2$

Thus, height $BM = 2$.

Since $\triangle CBM$ is a right triangle, $CM = \sqrt{BM^2 + BC^2} = \sqrt{2^2 + 3^2} = \boxed{\text{(C)}\ \sqrt{13}}$.

## Solution 2

Connect $AC$, $S_\triangle AMC=S_\triangle ANC$. To satisfied the three area is equal, we have $2S_\triangle AMC=S_\triangle BMC$, $2S_\triangle ANC=S_\triangle DNC$. Thus, $AM=AN=\frac{1}{2}BM=\frac{1}{2}AB=1$. $BM=2,BC=3,MC=\boxed{\text{(C)}\ \sqrt{13}}$.