AIME 2019 II · 第 15 题

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 $\triangle DPQ$,

$$\cos P,\cos Q=\frac{21^2+25^2-16^2}{2\cdot 21\cdot 25},\frac{16^2+25^2-21^2}{2\cdot 16\cdot 25}= \frac{27}{35}, \frac{11}{20}.$$ $$\Rightarrow \cos B=\cos\left(\tfrac 12 (\pi-P)\right)=\sin\tfrac 12 P =\sqrt{\frac{4}{35}},$$ similarly, we get

$$\cos C=\cos\left(\tfrac 12 (\pi-Q)\right)=\sin\tfrac 12 Q=\sqrt{\frac{9}{40}}.$$ To finish,

$$bc= \frac{(b\cos B)(c\cos C)}{\cos B\cos C}=\frac{16\cdot 21}{(2/\sqrt{35})(3/\sqrt{40})}=560\sqrt{14}.$$ The requested sum is $\boxed{574}$. - crazyeyemoody907

Remark: The proof that $a \cos A = PQ$ can be found here: http://www.irmo.ie/5.Orthic_triangle.pdf

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,

$$\begin{aligned} AP\cdot BP=XP\cdot YP \quad &\Longrightarrow \quad b^2k^2-bck+400=0\\ AQ\cdot CQ=YQ\cdot XQ \quad &\Longrightarrow \quad c^2k^2-bck+525=0 \end{aligned}$$ 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 $AP=p$, $PB=q$, $AQ=r$, and $QC=s$. By Power of a Point,

$$\begin{aligned} AP\cdot PB=XP\cdot YP \quad &\Longrightarrow \quad pq=400\\ AQ\cdot QC=YQ\cdot XQ \quad &\Longrightarrow \quad rs=525 \end{aligned}$$ Points $P$ and $Q$ lie on the circle, $\omega$, with diameter $BC$, and pow$(A,\omega) = AP\cdot AB = AQ\cdot AC$, so

$$p(p+q)=r(r+s)\quad \Longrightarrow \quad p^2-r^2=125$$ Use Law of Cosines in $\triangle APQ$ to get $25^2=p^2+r^2-2pr\cos A$; since $\cos A = \frac r{p+q}$, this simplifies as

$$500 \ =\ 2r^2-\frac{2pr^2}{p+q} \ =\ 2r^2-\frac{2p^2r^2}{p^2+400} \ =\ \frac{800r^2}{r^2+525}$$ We get $r=5\sqrt{35}$ and thus

$$r=5\sqrt{35}, \quad p = \sqrt{r^2+125} = 10\sqrt{10}, \quad q = \frac{400}{p} =4\sqrt{10}, \quad s= \frac{525}{r} = 3\sqrt{35}.$$ Therefore $AB\cdot AC = (p+q)\cdot(r+s) = 560\sqrt{14}$. 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 $E$ and $F$ denote the reflections of the orthocenter over points $P$ and $Q$, respectively. Since $EF \parallel XY$ and

$$EF = 2 PQ = XP + PQ + QY = XY,$$ we have that $E X Y F$ is a rectangle. Then, since $\angle XYF = 90^\circ$ we obtain $\angle XBF = 90^\circ$ (which directly follows from $XBYF$ being cyclic); hence $\angle XBQ = \angle AQB$, or $XB \parallel AQ \Rightarrow XB \parallel AC$.

Similarly, we can obtain $YC \parallel AB$. $\ \blacksquare$

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

$\phantom{...................}15 \cdot 35 = 15l^2 \implies l = \sqrt{35} \implies AC = 8 \sqrt{35}$.

Hence,

$\phantom{...................}AB \cdot AC = 112 \sqrt{350} = 112 \cdot 5 \sqrt{14} = 560 \sqrt{14}$

and the answer is $560 + 14 = \boxed{574}$. ~rocketsri

Solution 5

mathboy282

Solution 6

Certainly $\triangle AQP\sim\triangle ABC$. Let $k=AP$ and $\ell=AQ$, and let $c$ be the constant with $BC = PQ\cdot c = 25c$. Then $PB = \ell c-k$ and $QC = kc-\ell$. By Power of a Point at $P$ and $Q$, we obtain the equations \begin{align*} AP\cdot PB &= k(\ell c-k) = k\ell c-k^2 = 400,\\ AQ\cdot QC &= \ell(kc-\ell) = k\ell c-\ell^2 = 525. \end{align*} This implies that $k^2-\ell^2 = 125$. Next, by the Pythagorean theorem we have \begin{align*} PC^2 &= BC^2-BP^2 = 625c^2-(\ell c-k)^2,\\ PC^2 &= AC^2-AP^2 = k^2c^2 - k^2. \end{align*} Equating the two and simplifying, we arrive at the equation

$$c = \frac{2k\ell}{k^2+\ell^2-625}.$$ From here, substitute $\ell = \sqrt{k^2-125}$. Then the first Power of a Point equation becomes

$$k^2+400=k\ell\cdot\frac{2k\ell}{2k^2-750} = \frac{k^2(k^2-125)}{k^2-375}.$$ Solving, we obtain $k=\sqrt{1000}$ and $\ell = \sqrt{875}$. Then we evaluate: \begin{align*} AB\cdot AC &= (ck)(c\ell) = \frac{(k\ell)^3}{(k^2-375)^2}\\ &= \frac{(\sqrt{875000})^3}{625^2} = \frac{(250\sqrt{14})^3}{625^2} \\ &= \frac{(2\cdot 5^3)^3\cdot 14\sqrt{14}}{5^8}=560\sqrt{14}. \end{align*} The answer is $560+14 = \boxed{574}.$

~TThB0501

Video Solution

2019 AIME II #15

MathProblemSolvingSkills.com

Video Solution by MOP 2024

https://youtu.be/aYV09qIwTqs

~r00tsOfUnity

Video Solution by Mr. Math

https://www.youtube.com/watch?v=rpNnK5n0_P0