Difference between revisions of "2013 AIME II Problems/Problem 13"
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so the height of this <math>\triangle ABC</math> is <math>\sqrt{4^2-(\sqrt{7})^2}=3</math> | so the height of this <math>\triangle ABC</math> is <math>\sqrt{4^2-(\sqrt{7})^2}=3</math> | ||
| − | Then the area of <math>\triangle ABC</math> is <math>3\sqrt{7}</math>, so the answer is <math>\boxed{010}</math> | + | Then the area of <math>\triangle ABC</math> is <math>3\sqrt{7}</math>, so the answer is <math>\boxed{010}</math>. |
==See Also== | ==See Also== | ||
{{AIME box|year=2013|n=II|num-b=12|num-a=14}} | {{AIME box|year=2013|n=II|num-b=12|num-a=14}} | ||
Revision as of 15:44, 6 April 2013
Problem 13
In 
, 
, and point 
is on 
so that 
. Let 
be the midpoint of 
. Given that 
and 
, the area of can be expressed in the form 
, where 
and 
are positive integers and is not divisible by the square of any prime. Find 
.
Solution
After drawing the figure, we suppose 
, so that
,
, and 
.
Using cosine law for 
and 
,we get
... 
...
So, 
, we get
...
Using cosine law in 
,we get
So, 
...
Using cosine law in 
and 
, we get
...
...
, and according to , we can get 
...
Using and , we can solve 
and 
Finally, we use cosine law for 
,
$4(\frac{\sqrt{22}}{2})^2+1+2\cdot\2(\frac{\sqrt{22}}{2})\cdot cos(ADC)=AB^2$ (Error compiling LaTeX. Unknown error_msg)
then 
so the height of this is 
Then the area of is 
, so the answer is 
.
See Also
| 2013 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
