Difference between revisions of "2019 AIME II Problems/Problem 15"
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==Solution== | ==Solution== | ||
| + | Let <math>AP=a, AQ=b, cos\angle A = k</math> | ||
| + | |||
| + | Therefore <math>AB= \frac{b}{k} , AC= \frac{a}{k} | ||
| + | |||
| + | By power of point, we have | ||
| + | </math>AP*BP=XP*YP , AQ*CQ=YQ*XQ<math> | ||
| + | Which are simplified to | ||
| + | </math>400= \frac{ab}{k} - a^2<math> | ||
| + | </math>525= \frac{ab}{k} - b^2<math> | ||
| + | Or | ||
| + | </math>a^2= \frac{ab}{k} - 400<math> | ||
| + | </math>b^2= \frac{ab}{k} - 525<math> | ||
| + | (1) | ||
| + | |||
| + | Or | ||
| + | </math>k= \frac{ab}{a^2+400} = \frac{ab}{b^2+525} | ||
| + | |||
| + | Let <math>u=a^2+400=b^2+525</math> | ||
| + | Then, <math>a=\sqrt{u-400},b=\sqrt{u-525},k=\frac{\sqrt{(u-400)(u-525)}}{u} | ||
| + | |||
| + | |||
| + | In triangle </math>APQ<math>, by law of cosine | ||
| + | |||
| + | </math>25^2= a^2 + b^2 - 2abk<math> | ||
| + | |||
| + | Pluging (1) | ||
| + | |||
| + | </math>625= \frac{ab}{k} - 400 + \frac{ab}{k} - 525 -2abk<math> | ||
| + | |||
| + | Or | ||
| + | |||
| + | </math> \frac{ab}{k} - abk =775<math> | ||
| + | |||
| + | Substitute everything by </math>u<math> | ||
| + | |||
| + | </math>u- \frac{(u-400)(u-525)}{u} =775<math> | ||
| + | |||
| + | The quadratic term is cancelled out after simplified | ||
| + | |||
| + | Which gives </math>u=1400<math> | ||
| + | |||
| + | Plug back in, </math>a= \sqrt{1000} , b=\sqrt{775}<math> | ||
| + | |||
| + | Then | ||
| + | |||
| + | </math>AB*AC= \frac{a}{k} \frac{b}{k} = \frac{ab}{\frac{ab}{u} *\frac{ab}{u} } = \frac{u^2}{ab} | ||
| + | = \frac{1400 * 1400}{ \sqrt{ 1000* 875 }} = 560 \sqrt{14} | ||
| + | |||
| + | So the final answer is 560 + 14 = 574 | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2019|n=II|num-b=14|after=Last Question}} | {{AIME box|year=2019|n=II|num-b=14|after=Last Question}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 20:14, 22 March 2019
Problem
In acute triangle 
points 
and 
are the feet of the perpendiculars from 
to 
and from 
to 
, respectively. Line 
intersects the circumcircle of 
in two distinct points, 
and 
. Suppose 
, 
, and 
. The value of 
can be written in the form 
where 
and 
are positive relatively prime integers. Find 
.
Solution
Let 
Therefore $AB= \frac{b}{k} , AC= \frac{a}{k}
By power of point, we have$ (Error compiling LaTeX. Unknown error_msg)AP*BP=XP*YP , AQ*CQ=YQ*XQ
400= \frac{ab}{k} - a^2$$ (Error compiling LaTeX. Unknown error_msg)525= \frac{ab}{k} - b^2
a^2= \frac{ab}{k} - 400$$ (Error compiling LaTeX. Unknown error_msg)b^2= \frac{ab}{k} - 525$(1)
Or$ (Error compiling LaTeX. Unknown error_msg)k= \frac{ab}{a^2+400} = \frac{ab}{b^2+525}
Let 
Then, $a=\sqrt{u-400},b=\sqrt{u-525},k=\frac{\sqrt{(u-400)(u-525)}}{u}
In triangle$ (Error compiling LaTeX. Unknown error_msg)APQ
25^2= a^2 + b^2 - 2abk
625= \frac{ab}{k} - 400 + \frac{ab}{k} - 525 -2abk \frac{ab}{k} - abk =775
u$$ (Error compiling LaTeX. Unknown error_msg)u- \frac{(u-400)(u-525)}{u} =775$The quadratic term is cancelled out after simplified
Which gives$ (Error compiling LaTeX. Unknown error_msg)u=1400
a= \sqrt{1000} , b=\sqrt{775}
AB*AC= \frac{a}{k} \frac{b}{k} = \frac{ab}{\frac{ab}{u} *\frac{ab}{u} } = \frac{u^2}{ab}
= \frac{1400 * 1400}{ \sqrt{ 1000* 875 }} = 560 \sqrt{14}
So the final answer is 560 + 14 = 574
See Also
| 2019 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Last Question | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
