Difference between revisions of "2010 AMC 10A Problems/Problem 16"
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== See also == | == See also == | ||
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| + | {{AMC12 box|year=2010|num-b=13|num-a=15|ab=A}} | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 21:08, 10 May 2020
Problem
Nondegenerate 
has integer side lengths, 
is an angle bisector, 
, and 
. What is the smallest possible value of the perimeter?
Solution
![[asy] pair A,B,C,D; C=(0,0); B=(4,0); A=(3,1); D=(2,0.666); draw(A--B--C--cycle); draw(B--D); label("$A$",A,N); label("$B$",B,S); label("$C$",C,S); label("$D$",D,NW); label("$3$",A--D,N); label("$8$",C--D,N); [/asy]](http://latex.artofproblemsolving.com/4/6/6/4661024a902bb549ef50c3798e944b3d178a7c4c.png)
By the Angle Bisector Theorem, we know that 
. If we use the lowest possible integer values for AB and BC (the measures of AD and DC, respectively), then 
, contradicting the Triangle Inequality. If we use the next lowest values (
and 
), the Triangle Inequality is satisfied. Therefore, our answer is 
, or choice 
.
See also
| 2010 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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 | ||
| 2010 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 13 |
Followed by Problem 15 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
