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2009 AMC 12A Problems/Problem 10

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(Redirected from 2009 AMC 10A Problems/Problem 12)

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 $BD$ ?

unitsize(4mm);defaultpen(linewidth(.8pt)+fontsize(8pt));dotfactor=4;pair C=(0,0), B=(17,0);pair D=intersectionpoints(Circle(C...

$\textbf{(A)}\ 11 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 13 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 15$

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

See Also

2009 AMC 12A (Problems • Resources)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

2009 AMC 10A (Problems • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2009_AMC_12A_Problems/Problem_10"

Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.

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