2019 AIME II Problems/Problem 15

Revision as of 17:34, 3 March 2021 by Crazyeyemoody907 (talk | contribs) (Solution)


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 $m+n$.


First we have $a\cos A=PQ=25$, and $(a\cos A)(c\cos C)=(a\cos C)(c\cos A)=AP\cdot PB=10(25+15)=400.$ Similarly, $(a\cos A)(b\cos B)=15(10+25)=525,$ and dividing these each by $a\cos A$ gives $b\cos B=21,c\cos C=16$.

It is known that the sides of the orthic triangle are $a\cos A,b\cos B,c\cos C$, and its angles are $\pi-2A$,$\pi-2B$, and $\pi-2C$. We thus have the three sides of the orthic triangle now. Letting $D$ be the foot of the altitude from $A$, we have, in $\triangle DPQ$, \[\cos P,\cos Q=\frac{21^2+25^2-16^2}{2\cdot 21\cdot 25},\frac{16^2+25^2-21^2}{2\cdot 16\cdot 25}=27/35,11/20.\] \[\Rightarrow \cos B=\cos\biggl(\frac{\pi-P}{2}\biggr)=\sin\frac{P}{2}=\sqrt{4/35},\] similarly, we get \[\cos C=\cos\biggl(\frac{\pi-Q}{2}\biggr)=\sin\frac{Q}{2}=\sqrt{9/40}.\] Our final answer is then \[bc= \frac{(b\cos B)(c\cos C)}{\cos B\cos C}=\frac{16\cdot 21}{6\sqrt{1400}}\] \[=56\cdot\sqrt{1400}=560\sqrt{14}.\]

The requested sum is $\boxed{574}$.


Solution 1

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

$400= \frac{ab}{k} - a^2$

$525= \frac{ab}{k} - b^2$


$a^2= \frac{ab}{k} - 400$

$b^2= \frac{ab}{k} - 525$



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

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)

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


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

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, $a= \sqrt{1000} , b=\sqrt{875}$


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

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 that 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 $560 + 14 = \boxed{574}$

By asr41

See Also

2019 AIME II (ProblemsAnswer KeyResources)
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