Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2019 AIME II Problems/Problem 15"

m (→‎Solution)
Line 61: Line 61:
  
 
By SpecialBeing2017
 
By SpecialBeing2017
 +
 +
==Solution 2==
 +
 +
Let <math>\overline{AP}=a, \overline{PB} = b, \overline{AQ} = c</math> and <math>\overline{QC} = d</math>
 +
 +
By power of point, we have
 +
<math>\overline{AP}\cdot \overline{PB}=\overline{XP}\cdot \overline{YP}</math> and <math>\overline{AQ}\cdot \overline{QC}=\overline{YQ}\cdot \overline{XQ}</math>
 +
 +
Therefore, substituting in the values:
 +
 +
<math>ab = 400</math>
 +
 +
<math>cd = 525</math>
 +
 +
Notice than quadrilateral <math>BPQC</math> is cyclic.
 +
 +
From this fact, we can deduce that <math>\angle PQA= \angle B</math> and <math>\angle QPA = \angle C</math>
 +
 +
Therefore <math>\triangle ABC</math> is similar to <math>\triangle AQP</math>.
 +
 +
Therefore:
 +
<math>\frac{a}{c+d}=\frac{c}{a+b} \implies a^2 + ab = c^2 +cd \implies a^2 + 400 = c^2 + 525 \implies \bf{a^2 = c^2 + 125}</math>
 +
 +
Now using Law of Cosines on <math>\triangle AQP</math> we get:
 +
 +
<math>625 = a^2 + c^2 - 2ac\cos{A}</math>
 +
 +
Notice <math>\cos{A} = \frac{c}{a+b}</math>
 +
 +
Substituting and Simplifying:
 +
 +
<math>625 = a^2 + c^2 - 2ac\frac{c}{a+b}</math>
 +
 +
<math>625 = a^2 + c^2 - 2ac\frac{c}{a+\frac{400}{a}}</math>
 +
 +
<math>625 = c^2 + 125 + c^2 - 2\frac{(ac)^2}{a^2+400}</math>
 +
 +
<math>625 = c^2 + 125 + c^2 - 2\frac{c^2(c^2+125)}{c^2+125+400}</math>
 +
 +
Now we solve for <math>c</math> using regular algebra which actually turns out to be very easy.
 +
 +
We get <math>c = 5\sqrt{35}</math> and from the above relations between the variables we quickly determine <math>d = 3\sqrt{35}</math>,  <math>a = 10\sqrt{10}</math> and <math>b = 4\sqrt{10}</math>
 +
 +
Therefore <math>AB\cdot AC = (a+b)\cdot(c+d) = 560\sqrt{14}</math>
 +
 +
So the answer is <math>560 + 14 = \boxed{574} </math>
 +
 +
  
 
==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 16:23, 27 June 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 integers, and $n$ is not divisible by the square of any prime. 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 $AP\cdot BP=XP\cdot YP , AQ\cdot CQ=YQ\cdot XQ$ 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

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

So the final answer is $560 + 14 = \boxed{574}$

By SpecialBeing2017

Solution 2

Let $\overline{AP}=a, \overline{PB} = b, \overline{AQ} = c$ and $\overline{QC} = d$

By power of point, we have $\overline{AP}\cdot \overline{PB}=\overline{XP}\cdot \overline{YP}$ and $\overline{AQ}\cdot \overline{QC}=\overline{YQ}\cdot \overline{XQ}$

Therefore, substituting in the values:

$ab = 400$

$cd = 525$

Notice than quadrilateral $BPQC$ is cyclic.

From this fact, we can deduce that $\angle PQA= \angle B$ and $\angle QPA = \angle C$

Therefore $\triangle ABC$ is similar to $\triangle AQP$.

Therefore: $\frac{a}{c+d}=\frac{c}{a+b} \implies a^2 + ab = c^2 +cd \implies a^2 + 400 = c^2 + 525 \implies \bf{a^2 = c^2 + 125}$

Now using Law of Cosines on $\triangle AQP$ we get:

$625 = a^2 + c^2 - 2ac\cos{A}$

Notice $\cos{A} = \frac{c}{a+b}$

Substituting and Simplifying:

$625 = a^2 + c^2 - 2ac\frac{c}{a+b}$

$625 = a^2 + c^2 - 2ac\frac{c}{a+\frac{400}{a}}$

$625 = c^2 + 125 + c^2 - 2\frac{(ac)^2}{a^2+400}$

$625 = c^2 + 125 + c^2 - 2\frac{c^2(c^2+125)}{c^2+125+400}$

Now we solve for $c$ using regular algebra which actually turns out to be very easy.

We get $c = 5\sqrt{35}$ and from the above relations between the variables we quickly determine $d = 3\sqrt{35}$, $a = 10\sqrt{10}$ and $b = 4\sqrt{10}$

Therefore $AB\cdot AC = (a+b)\cdot(c+d) = 560\sqrt{14}$

So the answer is $560 + 14 = \boxed{574}$


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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_15&oldid=107028"