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2019 AIME II Problems/Problem 15: Difference between revisions

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Let <math>AP=a, AQ=b, cos\angle A = k</math>
Let <math>AP=a, AQ=b, cos\angle A = k</math>


Therefore <math>AB= \frac{b}{k} , AC= \frac{a}{k}
Therefore <math>AB= \frac{b}{k} , AC= \frac{a}{k}</math>


By power of point, we have  
By power of point, we have  
</math>AP*BP=XP*YP , AQ*CQ=YQ*XQ<math>
<math>AP*BP=XP*YP , AQ*CQ=YQ*XQ</math>
Which are simplified to  
Which are simplified to  
</math>400= \frac{ab}{k} - a^2<math>
<math>400= \frac{ab}{k} - a^2</math>
</math>525= \frac{ab}{k} - b^2<math>
<math>525= \frac{ab}{k} - b^2</math>
Or
Or
</math>a^2= \frac{ab}{k} - 400<math>
<math>a^2= \frac{ab}{k} - 400</math>
</math>b^2= \frac{ab}{k} - 525<math>
<math>b^2= \frac{ab}{k} - 525</math>
(1)
(1)


Or
Or
</math>k= \frac{ab}{a^2+400} = \frac{ab}{b^2+525}
<math>k= \frac{ab}{a^2+400} = \frac{ab}{b^2+525}</math>


Let <math>u=a^2+400=b^2+525</math>
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}
Then, <math>a=\sqrt{u-400},b=\sqrt{u-525},k=\frac{\sqrt{(u-400)(u-525)}}{u} </math>




In triangle </math>APQ<math>, by law of cosine
In triangle <math>APQ</math>, by law of cosine


</math>25^2= a^2 + b^2 - 2abk<math>
<math>25^2= a^2 + b^2 - 2abk</math>


Pluging (1)
Pluging (1)


</math>625=  \frac{ab}{k} - 400 + \frac{ab}{k} - 525 -2abk<math>
<math>625=  \frac{ab}{k} - 400 + \frac{ab}{k} - 525 -2abk</math>


Or
Or


</math> \frac{ab}{k} - abk =775<math>
<math> \frac{ab}{k} - abk =775</math>


Substitute everything by </math>u<math>
Substitute everything by <math>u</math>


</math>u- \frac{(u-400)(u-525)}{u} =775<math>
<math>u- \frac{(u-400)(u-525)}{u} =775</math>


The quadratic term is cancelled out after simplified
The quadratic term is cancelled out after simplified


Which gives </math>u=1400<math>
Which gives <math>u=1400</math>


Plug back in, </math>a= \sqrt{1000} , b=\sqrt{775}<math>
Plug back in, <math>a= \sqrt{1000} , b=\sqrt{775}</math>


Then
Then


</math>AB*AC= \frac{a}{k} \frac{b}{k} = \frac{ab}{\frac{ab}{u} *\frac{ab}{u} } = \frac{u^2}{ab}
<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}
= \frac{1400 * 1400}{ \sqrt{ 1000* 875 }} = 560 \sqrt{14}</math>


So the final answer is 560 + 14 = 574
So the final answer is 560 + 14 = 574
By SpecialBeing2017


==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 19:15, 22 March 2019

Problem

In acute triangle $ABC$ points $P$ and $Q$ are the feet of the perpendiculars from $C$ to $\overline{AB}$ and from $B$ to $\overline{AC}$, respectively. Line $PQ$ intersects the circumcircle of $\triangle ABC$ in two distinct points, $X$ and $Y$. Suppose $XP=10$, $PQ=25$, and $QY=15$. The value of $AB\cdot AC$ can be written in the form $m\sqrt n$ where $m$ and $n$ are positive relatively prime integers. Find $m+n$.

Solution

Let $AP=a, AQ=b, cos\angle A = k$

Therefore $AB= \frac{b}{k} , AC= \frac{a}{k}$

By power of point, we have $AP*BP=XP*YP , AQ*CQ=YQ*XQ$ Which are simplified to $400= \frac{ab}{k} - a^2$ $525= \frac{ab}{k} - b^2$ Or $a^2= \frac{ab}{k} - 400$ $b^2= \frac{ab}{k} - 525$ (1)

Or $k= \frac{ab}{a^2+400} = \frac{ab}{b^2+525}$

Let $u=a^2+400=b^2+525$ Then, $a=\sqrt{u-400},b=\sqrt{u-525},k=\frac{\sqrt{(u-400)(u-525)}}{u}$


In triangle $APQ$, by law of cosine

$25^2= a^2 + b^2 - 2abk$

Pluging (1)

$625= \frac{ab}{k} - 400 + \frac{ab}{k} - 525 -2abk$

Or

$\frac{ab}{k} - abk =775$

Substitute everything by $u$

$u- \frac{(u-400)(u-525)}{u} =775$

The quadratic term is cancelled out after simplified

Which gives $u=1400$

Plug back in, $a= \sqrt{1000} , b=\sqrt{775}$

Then

$AB*AC= \frac{a}{k} \frac{b}{k} = \frac{ab}{\frac{ab}{u} *\frac{ab}{u} } = \frac{u^2}{ab} = \frac{1400 * 1400}{ \sqrt{ 1000* 875 }} = 560 \sqrt{14}$

So the final answer is 560 + 14 = 574

By SpecialBeing2017

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

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