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

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In acute triangle <math>ABC</math> points <math>P</math> and <math>Q</math> are the feet of the perpendiculars from <math>C</math> to <math>\overline{AB}</math> and from <math>B</math> to <math>\overline{AC}</math>, respectively. Line <math>PQ</math> intersects the circumcircle of <math>\triangle ABC</math> in two distinct points, <math>X</math> and <math>Y</math>. Suppose <math>XP=10</math>, <math>PQ=25</math>, and <math>QY=15</math>. The value of <math>AB\cdot AC</math> can be written in the form <math>m\sqrt n</math> where <math>m</math> and <math>n</math> are positive integers, and <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>.
 
In acute triangle <math>ABC</math> points <math>P</math> and <math>Q</math> are the feet of the perpendiculars from <math>C</math> to <math>\overline{AB}</math> and from <math>B</math> to <math>\overline{AC}</math>, respectively. Line <math>PQ</math> intersects the circumcircle of <math>\triangle ABC</math> in two distinct points, <math>X</math> and <math>Y</math>. Suppose <math>XP=10</math>, <math>PQ=25</math>, and <math>QY=15</math>. The value of <math>AB\cdot AC</math> can be written in the form <math>m\sqrt n</math> where <math>m</math> and <math>n</math> are positive integers, and <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>.
  
==Solution==
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==Diagram==
Let <math>AP=a, AQ=b, cos\angle A = k</math>
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<asy>
 +
size(200);
 +
defaultpen(linewidth(0.4)+fontsize(10));
 +
pen s = linewidth(0.8)+fontsize(8);
  
Therefore <math>AB= \frac{b}{k} , AC= \frac{a}{k}</math>
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pair A,B,C,P,Q,X,Y,O;
 +
O = origin;
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real theta = 32;
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A = dir(180+theta);
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B = dir(-theta);
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C = dir(75);
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Q = foot(B,A,C);
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P = foot(C,A,B);
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path c = circumcircle(A,B,C);
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X = IP(c, Q--(2*P-Q));
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Y = IP(c, P--(2*Q-P));
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draw(A--B--C--A, black+0.8);
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draw(c^^X--Y^^B--Q^^C--P);
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dot("$A$", A, SW);
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dot("$B$", B, SE);
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dot("$C$", C, N);
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dot("$P$", P, SW);
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dot("$Q$", Q, W);
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dot("$X$", X, SE);
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dot("$Y$", Y, NW);
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label("$25$", P--Q, SW);
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label("$15$", Q--Y, SW);
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label("$10$", X--P, SW);
 +
</asy>
  
By power of point, we have
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==Solution 1==
<math>AP\cdot BP=XP\cdot YP , AQ\cdot CQ=YQ\cdot XQ</math>
 
Which are simplified to
 
  
<math>400= \frac{ab}{k} - a^2</math>
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First we have <math>a\cos A=PQ=25</math>, and <math>(a\cos A)(c\cos C)=(a\cos C)(c\cos A)=AP\cdot PB=10(25+15)=400</math> by PoP. Similarly, <math>(a\cos A)(b\cos B)=15(10+25)=525,</math> and dividing these each by <math>a\cos A</math> gives
 +
<math>b\cos B=21,c\cos C=16</math>.
  
<math>525= \frac{ab}{k} - b^2</math>
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It is known that the sides of the orthic triangle are <math>a\cos A,b\cos B,c\cos C</math>, and its angles are <math>\pi-2A</math>,<math>\pi-2B</math>, and <math>\pi-2C</math>. We thus have the three sides of the orthic triangle now.
 +
Letting <math>D</math> be the foot of the altitude from <math>A</math>, we have, in <math>\triangle DPQ</math>,
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<cmath>\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}.</cmath>
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<cmath>\Rightarrow \cos B=\cos\left(\tfrac 12 (\pi-P)\right)=\sin\tfrac 12 P =\sqrt{\frac{4}{35}},</cmath>
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similarly, we get 
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<cmath>\cos C=\cos\left(\tfrac 12 (\pi-Q)\right)=\sin\tfrac 12 Q=\sqrt{\frac{9}{40}}.</cmath>
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To finish, <cmath>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}.</cmath>
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The requested sum is <math>\boxed{574}</math>.
  
Or
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༺\\ crazyeyemoody9❂7 //༻
  
<math>a^2= \frac{ab}{k} - 400</math>
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==Solution 2==
 
 
<math>b^2= \frac{ab}{k} - 525</math>
 
 
 
(1)
 
 
 
Or
 
 
 
<math>k= \frac{ab}{a^2+400} = \frac{ab}{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} </math>
 
 
 
 
 
In triangle <math>APQ</math>, by law of cosine
 
 
 
<math>25^2= a^2 + b^2 - 2abk</math>
 
  
Pluging (1)
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Let <math>BC=a</math>, <math>AC=b</math>, and <math>AB=c</math>. Let <math>\cos\angle A=k</math>. Then <math>AP=bk</math> and <math>AQ=ck</math>.
  
<math>625=  \frac{ab}{k} - 400 + \frac{ab}{k} - 525 -2abk</math>
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By Power of a Point theorem,
 
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<cmath>\begin{align}
Or
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    AP\cdot BP=XP\cdot YP \quad &\Longrightarrow \quad b^2k^2-bck+400=0\\
 
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    AQ\cdot CQ=YQ\cdot XQ \quad &\Longrightarrow \quad c^2k^2-bck+525=0
<math> \frac{ab}{k} - abk =775</math>
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\end{align}</cmath>
 
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Thus <math>bck = (bk)^2+400=(ck)^2+525 = u</math>. Then <math>bk=\sqrt{u-400}</math>, <math>ck=\sqrt{u-525}</math>, and
Substitute everything by <math>u</math>
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<cmath>k=\sqrt{\frac{(u-400)(u-525)}{u^2}}</cmath>
 
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Use the Law of Cosines in <math>\triangle APQ</math> to get <math>25^2=b^2k^2+c^2k^2-2bck^3 = 2bck-925-2bck^3</math>, which rearranges to <cmath>775=bck - k^2\cdot bck  = u-\frac{(u-400)(u-525)}{u}</cmath>Upon simplification, this reduces to a linear equation in <math>u</math>, with solution <math>u=1400</math>. Then <cmath>AB\cdot AC = bc = \frac 1{k}\cdot bck = \frac{u^2}{\sqrt{(u-400)(u-525)}}=560 \sqrt{14}</cmath>
<math>u- \frac{(u-400)(u-525)}{u} =775</math>
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So the final answer is <math>560 + 14 = \boxed{574}</math>
 
 
The quadratic term is cancelled out after simplified
 
 
 
Which gives <math>u=1400</math>
 
 
 
Plug back in, <math>a= \sqrt{1000} , b=\sqrt{775}</math>
 
 
 
Then
 
 
 
<math>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}</math>
 
 
 
So the final answer is <math>560 + 14 = \boxed{574} </math>
 
  
 
By SpecialBeing2017
 
By SpecialBeing2017
  
==Solution 2==
+
==Solution 3==
  
 
Let <math>\overline{AP}=a, \overline{PB} = b, \overline{AQ} = c</math> and <math>\overline{QC} = d</math>
 
Let <math>\overline{AP}=a, \overline{PB} = b, \overline{AQ} = c</math> and <math>\overline{QC} = d</math>
Line 75: Line 78:
 
<math>cd = 525</math>
 
<math>cd = 525</math>
  
Notice than quadrilateral <math>BPQC</math> is cyclic.
+
Notice that 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>
 
From this fact, we can deduce that <math>\angle PQA= \angle B</math> and <math>\angle QPA = \angle C</math>
Line 108: Line 111:
 
So the answer is <math>560 + 14 = \boxed{574} </math>
 
So the answer is <math>560 + 14 = \boxed{574} </math>
  
 +
By asr41
 +
 +
==Solution 4 (Clean)==
 +
This solution is directly based of @CantonMathGuy's solution.
 +
We start off with a key claim.
 +
 +
 +
<i> Claim. </i> <math>XB \parallel AC</math> and <math>YC \parallel AB</math>.
 +
 +
<i> Proof. </i> Let <math>H_p</math> and <math>H_q</math> denote the reflections of the orthocenter over points <math>P</math> and <math>Q</math>, respectively. Since <math>H_p H_q \parallel XY</math> and <math>H_p H_q = 2 PQ = XP + PQ + QY = XY</math>, we have that <math>H_p X Y H_q</math> is a rectangle. Then, since <math>\angle XYH_q = 90^\circ</math> we obtain <math>\angle XBH_q = 90^\circ</math> (which directly follows from <math>XBYH_q</math> being cyclic); hence <math>\angle XBQ = \angle AQB</math>, or <math>XB \parallel AQ \implies XB \parallel AC</math>. Similarly, we can obtain <math>YC \parallel AB</math>.
 +
 +
 +
A direct result of this claim is that <math>\triangle BPX \sim \triangle APQ \sim \triangle CYQ</math>. Thus, we can set <math>AP = 5k</math> and <math>BP = 2k</math>, then applying Power of a Point on <math>P</math> we get <math>10 \cdot 40 = 10k^2 \implies k = 2\sqrt{10} \implies AB = 14 \sqrt{10}</math>. Also, we can set <math>AQ = 5l</math> and <math>CQ = 3l</math> and once again applying Power of a Point (but this time to <math>Q</math>) we get <math>15 \cdot 35 = 15l^2 \implies l = \sqrt{35} \implies AC = 8 \sqrt{35}</math>. Hence, <math>AB \cdot AC = 112 \sqrt{350} = 112 \cdot 5 \sqrt{14} = 560 \sqrt{14}</math> and the answer is <math>560 + 14 = \boxed{574}</math>. ~rocketsri
  
  
 
==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}}
 +
[[Category: Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 11:54, 17 January 2022

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 $\triangle DPQ$, \[\cos P,\cos Q=\frac{21^2+25^2-16^2}{2\cdot 21\cdot 25},\frac{16^2+25^2-21^2}{2\cdot 16\cdot 25}= \frac{27}{35}, \frac{11}{20}.\] \[\Rightarrow \cos B=\cos\left(\tfrac 12 (\pi-P)\right)=\sin\tfrac 12 P =\sqrt{\frac{4}{35}},\] 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 $\boxed{574}$.

༺\\ 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


See Also

2019 AIME II (ProblemsAnswer KeyResources)
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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