# 2009 AMC 12A Problems/Problem 10

The following problem is from both the 2009 AMC 12A #10 and 2009 AMC 10A #12, so both problems redirect to this page.

## Problem

In quadrilateral $ABCD$, $AB = 5$, $BC = 17$, $CD = 5$, $DA = 9$, and $BD$ is an integer. What is $BD$? $[asy] unitsize(4mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=4; pair C=(0,0), B=(17,0); pair D=intersectionpoints(Circle(C,5),Circle(B,13)); pair A=intersectionpoints(Circle(D,9),Circle(B,5)); pair[] dotted={A,B,C,D}; draw(D--A--B--C--D--B); dot(dotted); label("D",D,NW); label("C",C,W); label("B",B,E); label("A",A,NE); [/asy]$ $\textbf{(A)}\ 11 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 13 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 15$

## Solution

By the triangle inequality we have $BD < DA + AB = 9 + 5 = 14$, and also $BD + CD > BC$, hence $BD > BC - CD = 17 - 5 = 12$.

We get that $12 < BD < 14$, and as we know that $BD$ is an integer, we must have $BD=\boxed{13}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 