Difference between revisions of "2009 AMC 12A Problems/Problem 10"
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label("$A$",A,NE); | label("$A$",A,NE); | ||
</asy></center><math>\textbf{(A)}\ 11 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 13 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 15</math> | </asy></center><math>\textbf{(A)}\ 11 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 13 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 15</math> | ||
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== Solution == | == Solution == | ||
Revision as of 10:40, 13 August 2014
- 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 
, 
, 
, 
, 
, and 
is an integer. What is ?
![[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))[0]; pair A=intersectionpoints(Circle(D,9),Circle(B,5))[0]; 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]](http://latex.artofproblemsolving.com/5/7/0/570b698fea7c6d66beab04dd6aee44dd24f2b6c1.png)
Solution
By the triangle inequality we have 
, and also 
, hence 
.
We got that 
, and as we know that is an integer, we must have 
.
See Also
| 2009 AMC 12A (Problems • Answer Key • 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 | |
| All AMC 12 Problems and Solutions | |
| 2009 AMC 10A (Problems • Answer Key • 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 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
