2019 AIME II Problems/Problem 15

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 $m+n$ .

Diagram

$[asy] size(200); defaultpen(linewidth(0.4)+fontsize(10)); pen s = linewidth(0.8)+fontsize(8); pair A,B,C,P,Q,X,Y,O; O = origin; real theta = 32; A = dir(180+theta); B = dir(-theta); C = dir(75); Q = foot(B,A,C); P = foot(C,A,B); path c = circumcircle(A,B,C); X = IP(c, Q--(2*P-Q)); Y = IP(c, P--(2*Q-P)); draw(A--B--C--A, black+0.8); draw(c^^X--Y^^B--Q^^C--P); dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, N); dot("$P$", P, SW); dot("$Q$", Q, W); dot("$X$", X, SE); dot("$Y$", Y, NW); label("$25$", P--Q, SW); label("$15$", Q--Y, SW); label("$10$", X--P, SW); [/asy]$

Solution 1

First we have $a\cos A=PQ=25$ , and $(a\cos A)(c\cos C)=(a\cos C)(c\cos A)=AP\cdot PB=10(25+15)=400$ by PoP. 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 $\triangle DPQ$ , $\cos P,\cos Q=\frac{21^2+25^2-16^2}{2\cdot 21\cdot 25},\frac{16^2+25^2-21^2}{2\cdot 16\cdot 25}= \frac{27}{35}, \frac{11}{20}.$ $\Rightarrow \cos B=\cos\biggl(\frac{\pi-P}{2}\biggr)=\sin\frac{P}{2}=\sqrt{\frac{4}{35}},$ similarly, we get $\cos C=\cos\biggl(\frac{\pi-Q}{2}\biggr)=\sin\frac{Q}{2}=\sqrt{\frac{9}{40}}.$ To finish, $bc= \frac{(b\cos B)(c\cos C)}{\cos B\cos C}=\frac{16\cdot 21}{(2/\sqrt{35})(3/\sqrt{40})}$ $=56\sqrt{1400}=560\sqrt{14}.$

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

༺\\ crazyeyemoody9❂7 //༻

Solution 2

Let $BC=a$ , $AC=b$ , and $AB=c$ . Let $\cos\angle A=k$ . Then $AP=bk$ and $AQ=ck$ .

By Power of a Point theorem, $\begin{align} AP\cdot BP=XP\cdot YP \quad &\Longrightarrow \quad b^2k^2-bck+400=0\\ AQ\cdot CQ=YQ\cdot XQ \quad &\Longrightarrow \quad c^2k^2-bck+525=0 \end{align}$ Thus $bck = (bk)^2+400=(ck)^2+525 = u$ . Then $bk=\sqrt{u-400}$ , $ck=\sqrt{u-525}$ , and $k=\sqrt{\frac{(u-400)(u-525)}{u^2}}$ Use the Law of Cosines in $\triangle APQ$ to get $25^2=b^2k^2+c^2k^2-2bck^3 = 2bck-925-2bck^3$ , which rearranges to $775=bck - k^2\cdot bck = u-\frac{(u-400)(u-525)}{u}$ Upon simplification, this reduces to a linear equation in $u$ , with solution $u=1400$ . Then $AB\cdot AC = bc = \frac 1{k}\cdot bck = \frac{u^2}{\sqrt{(u-400)(u-525)}}=560 \sqrt{14}$ So the final answer is $560 + 14 = \boxed{574}$

By SpecialBeing2017

Solution 3

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

Solution 4 (Clean)

This solution is directly based of @CantonMathGuy's solution. We start off with a key claim.

Claim. $XB \parallel AC$ and $YC \parallel AB$ .

Proof. Let $H_p$ and $H_q$ denote the reflections of the orthocenter over points $P$ and $Q$ , respectively. Since $H_p H_q \parallel XY$ and $H_p H_q = 2 PQ = XP + PQ + QY = XY$ , we have that $H_p X Y H_q$ is a rectangle. Then, since $\angle XYH_q = 90^\circ$ we obtain $\angle XBH_q = 90^\circ$ (which directly follows from $XBYH_q$ being cyclic); hence $\angle XBQ = \angle AQB$ , or $XB \parallel AQ \implies XB \parallel AC$ . Similarly, we can obtain $YC \parallel AB$ .

A direct result of this claim is that $\triangle BPX \sim \triangle APQ \sim \triangle CYQ$ . Thus, we can set $AP = 5k$ and $BP = 2k$ , then applying Power of a Point on $P$ we get $10 \cdot 40 = 10k^2 \implies k = 2\sqrt{10} \implies AB = 14 \sqrt{10}$ . Also, we can set $AQ = 5l$ and $CQ = 3l$ and once again applying Power of a Point (but this time to $Q$ ) we get $15 \cdot 35 = 15l^2 \implies l = \sqrt{35} \implies AC = 8 \sqrt{35}$ . Hence, $AB \cdot AC = 112 \sqrt{350} = 112 \cdot 5 \sqrt{14} = 560 \sqrt{14}$ and the answer is $560 + 14 = \boxed{574}$ . ~rocketsri

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