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Difference between revisions of "1999 AIME Problems/Problem 14"

 
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== Problem ==
 
== Problem ==
 +
[[Point]] <math>P_{}</math> is located inside [[triangle]] <math>ABC</math> so that [[angle]]s <math>PAB, PBC,</math> and <math>PCA</math> are all congruent.  The sides of the triangle have lengths <math>AB=13, BC=14,</math> and <math>CA=15,</math> and the [[tangent]] of angle <math>PAB</math> is <math>m/n,</math> where <math>m_{}</math> and <math>n_{}</math> are relatively [[prime]] positive integers.  Find <math>m+n.</math>
  
 +
__TOC__
 
== Solution ==
 
== Solution ==
 +
<center><asy>
 +
real theta = 29.66115; /* arctan(168/295) to five decimal places .. don't know other ways to construct Brocard */
 +
pathpen = black +linewidth(0.65); pointpen = black;
 +
pair A=(0,0),B=(13,0),C=IP(circle(A,15),circle(B,14));
 +
D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle);
 +
 +
/* constructing P, C is there as check */
 +
pair Aa=A+(B-A)*dir(theta),Ba=B+(C-B)*dir(theta),Ca=C+(A-C)*dir(theta), P=IP(A--Aa,B--Ba);
 +
D(A--MP("P",P,NW)--B);D(P--C);
 +
D(anglemark(B,A,P,30));D(anglemark(C,B,P,30));D(anglemark(A,C,P,30));
 +
MP("13",(A+B)/2,S);MP("15",(A+C)/2,NW);MP("14",(C+B)/2,NE);
 +
</asy></center>
 +
<!--This image does exist now: [[Image:1999_AIME-14.png]]-->
 +
 +
=== Solution 1 ===
 +
Drop [[perpendicular]]s from <math>P</math> to the three sides of <math>\triangle ABC</math> and let them meet <math>\overline{AB}, \overline{BC},</math> and <math>\overline{CA}</math> at <math>D, E,</math> and <math>F</math> respectively. 
 +
 +
<center><asy>
 +
import olympiad;
 +
real theta = 29.66115; /* arctan(168/295) to five decimal places .. don't know other ways to construct Brocard */
 +
pathpen = black +linewidth(0.65); pointpen = black;
 +
pair A=(0,0),B=(13,0),C=IP(circle(A,15),circle(B,14));
 +
D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle);
 +
 +
/* constructing P, C is there as check */
 +
pair Aa=A+(B-A)*dir(theta),Ba=B+(C-B)*dir(theta),Ca=C+(A-C)*dir(theta), P=IP(A--Aa,B--Ba);
 +
D(A--MP("P",P,SSW)--B);D(P--C);
 +
D(anglemark(B,A,P,30));D(anglemark(C,B,P,30));D(anglemark(A,C,P,30));
 +
MP("13",(A+B)/2,S);MP("15",(A+C)/2,NW);MP("14",(C+B)/2,NE);
 +
 +
/* constructing D,E,F as foot of perps from P */
 +
pair D=foot(P,A,B),E=foot(P,B,C),F=foot(P,C,A);
 +
D(MP("D",D,NE)--P--MP("E",E,SSW),dashed);D(P--MP("F",F),dashed);
 +
D(rightanglemark(P,E,C,15));D(rightanglemark(P,F,C,15));D(rightanglemark(P,D,A,15));
 +
</asy></center>
 +
 +
Let <math>BE = x, CF = y,</math> and <math>AD = z</math>.  We have that
 +
<cmath>
 +
\begin{align*}DP&=z\tan\theta\\
 +
EP&=x\tan\theta\\
 +
FP&=y\tan\theta\end{align*}
 +
</cmath>
 +
We can then use the tool of calculating area in two ways
 +
<cmath>
 +
\begin{align*}[ABC]&=[PAB]+[PBC]+[PCA]\\
 +
&=\frac{1}{2}(13)(z\tan\theta)+\frac{1}{2}(14)(x\tan\theta)+\frac{1}{2}(15)(y\tan\theta)\\
 +
&=\frac{1}{2}\tan\theta(13z+14x+15y)\end{align*}
 +
</cmath>
 +
On the other hand,
 +
<cmath>
 +
\begin{align*}[ABC]&=\sqrt{s(s-a)(s-b)(s-c)}\\
 +
&=\sqrt{21\cdot6\cdot7\cdot8}\\
 +
&=84\end{align*}
 +
</cmath>
 +
We still need <math>13z+14x+15y</math> though. We have all these [[right triangle]]s and we haven't even touched [[Pythagorean theorem|Pythagoras]]. So we give it a shot:
 +
<cmath>
 +
\begin{align}x^2+x^2\tan^2\theta&=z^2\tan^2\theta+(13-z)^2\\
 +
z^2+z^2\tan^2\theta&=y^2\tan^2\theta+(15-y)^2\\
 +
y^2+y^2\tan^2\theta&=x^2\tan^2\theta+(14-x)^2\end{align}
 +
</cmath>
 +
Adding <math>(1) + (2) + (3)</math> gives
 +
<cmath>
 +
\begin{align*}x^2+y^2+z^2&=(14-x)^2+(15-y)^2+(13-z)^2\\
 +
\Rightarrow13z+14x+15y&=295\end{align*}
 +
</cmath>
 +
Recall that we found that <math>[ABC]=\frac{1}{2}\tan\theta(13z+14x+15y)=84</math>. Plugging in <math>13z+14x+15y=295</math>, we get <math>\tan\theta=\frac{168}{295}</math>, giving us <math>\boxed{463}</math> for an answer.
 +
 +
=== Solution 2 ===
 +
Let <math>AB=c</math>, <math>BC=a</math>, <math>AC=b</math>, <math>PA=x</math>, <math>PB=y</math>, and <math>PC=z</math>.
 +
 +
So by the [[Law of Cosines]], we have:
 +
<cmath>
 +
\begin{align*}x^2 &= z^2 + b^2 - 2bz\cos{\theta}\\
 +
y^2 &= x^2 + c^2 - 2cx\cos{\theta}\\
 +
z^2 &= y^2 + a^2 - 2ay\cos{\theta}\end{align*}
 +
</cmath>
 +
Adding these equations and rearranging, we have:
 +
<cmath>
 +
a^2 + b^2 + c^2 = (2bz + 2cx + 2ay)\cos{\theta}\qquad(1)
 +
</cmath>
 +
Now <math>[CAP] + [ABP] + [BCP] = [ABC] = \sqrt {(21)(8)(7)(6)} = 84</math>, by [[Heron's formula]].
 +
 +
Now the area of a triangle, <math>[A] = \frac {mn\sin{\beta}}{2}</math>, where <math>m</math> and <math>n</math> are sides on either side of an angle, <math>\beta</math>. So,
 +
<cmath>
 +
\begin{align*}[CAP] &= \frac {bz\sin{\theta}}{2}\\
 +
[ABP] &= \frac {cx\sin{\theta}}{2}\\
 +
[BCP] &= \frac {ay\sin{\theta}}{2}\end{align*}
 +
</cmath>
 +
Adding these equations yields:
 +
<cmath>\begin{align*}[ABC]= 84 &= \frac {(bz + cx + ay)\sin{\theta}}{2}\\
 +
\Rightarrow 168&= (bz + cx + ay)\sin{\theta}\qquad (2)\end{align*}
 +
</cmath>
 +
Dividing <math>(2)</math> by <math>(1)</math>, we have:
 +
<cmath>
 +
\begin{align*}\frac {168}{a^2 + b^2 + c^2} &= \frac {(bz + cx + ay)\sin{\theta}}{(2bz + 2cx + 2ay)\cos{\theta}}\\
 +
\Rightarrow \tan{\theta} = \frac {336}{a^2 + b^2 + c^2} &= \frac {336}{14^2 + 15^2 + 13^2} = \frac {336}{590} = \frac {168}{295}\end{align*}
 +
</cmath>
 +
Thus, <math>m + n = 168 + 295 = \boxed{463}</math>.
 +
 +
Note: In fact, this problem is unfairly easy to those who happen to have learned about Brocard point. The Brocard Angle is given by
 +
<cmath>cot(\theta)=\frac{a^2+b^2+c^2}{4\Delta}</cmath>
 +
 +
=== Solution 3 ===
 +
 +
Let <math>\angle{PAB} = \angle{PBC} = \angle{PCA} = x.</math> Then, using Law of Cosines on the three triangles containing vertex <math>P,</math> we have
 +
<cmath>
 +
\begin{align*}
 +
b^2 &= a^2 + 169 - 26a \cos x \\
 +
c^2 &= b^2 + 196 - 28b \cos x \\
 +
a^2 &= c^2 + 225 - 30c \cos x.
 +
\end{align*}
 +
</cmath>
 +
Add the three equations up and rearrange to obtain <cmath>(13a + 14b + 15c) \cos x = 295.</cmath> Also, using <math>[ABC] = \frac{1}{2}ab \sin \angle C</math> we have <cmath>[ABC] = [APB] + [BPC] + [CPA] = \dfrac{\sin x}{2}(13a + 14b + 15c) = 84 \iff (13a + 14b + 15c) \sin x = 168.</cmath> Divide the two equations to obtain <math>\tan x = \frac{168}{295} \iff \boxed{463}.~\square</math>
 +
 +
 +
=== Solution 4 (Law of sines) ===
 +
Firstly, denote angles <math>ABC</math>, <math>BCA</math>, and <math>CAB</math> as <math>B</math>, <math>A</math>, and <math>C</math> respectively. Let <math>\angle{PAB}=x</math>.
 +
Notice that by angle chasing that <math>\angle{BPC}=180-C</math> and <math>\angle{BPA}=180-B</math>.
 +
Using the nice properties of the 13-14-15 triangle, we have <math>\sin B = \frac{12}{13}</math> and <math>\sin C = \frac{4}{5}</math>. <math>\cos C</math> is easily computed, so we have <math>\cos C=\frac{3}{5}</math>.
 +
 +
Using Law of Sines,
 +
<cmath>
 +
\begin{align*}
 +
\frac{BP}{\sin x} &= \frac{13}{\sin (180 - B)} \\
 +
\frac{BP}{\sin (C - x)} &= \frac{14}{\sin (180 - C)}
 +
\end{align*}
 +
</cmath>
 +
hence,
 +
<cmath>
 +
\begin{align*}
 +
\frac{BP}{\sin x} &= \frac{13}{\sin B} \\
 +
\frac{BP}{\sin (C - x)} &= \frac{14}{\sin C}
 +
\end{align*}
 +
</cmath>
 +
Now, computation carries the rest.
 +
<cmath>
 +
\begin{align*}
 +
\frac{13 \sin x}{\sin B} &= \frac{14 \sin (C-x)}{\sin C}  \\
 +
\frac{169 \sin x}{12} &= \frac{210 \sin (C-x)}{12} \\
 +
169 \sin x &= 210 (\sin C \cos x - \cos C \sin x) \\
 +
169 \sin x &= 210 (\frac{4}{5} \cos x - \frac{3}{5} \sin x) \\
 +
169 \sin x &= 168 \cos x - 126 \sin x \\
 +
295 \sin x &= 168 \cos x \\
 +
\tan x &= \frac{168}{295}
 +
\end{align*}
 +
</cmath>
 +
Extracting yields <math>168 + 295 = \boxed{463}</math>.
  
 
== See also ==
 
== See also ==
−
* [[1999 AIME Problems]]
+
*[[Brocard point]]
 +
{{AIME box|year=1999|num-b=13|num-a=15}}
 +
 
 +
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 12:25, 22 December 2022

Problem

Point $P_{}$ is located inside triangle $ABC$ so that angles $PAB, PBC,$ and $PCA$ are all congruent. The sides of the triangle have lengths $AB=13, BC=14,$ and $CA=15,$ and the tangent of angle $PAB$ is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $m+n.$

Solution

[asy] real theta = 29.66115; /* arctan(168/295) to five decimal places .. don't know other ways to construct Brocard */ pathpen = black +linewidth(0.65); pointpen = black; pair A=(0,0),B=(13,0),C=IP(circle(A,15),circle(B,14)); D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); /* constructing P, C is there as check */ pair Aa=A+(B-A)*dir(theta),Ba=B+(C-B)*dir(theta),Ca=C+(A-C)*dir(theta), P=IP(A--Aa,B--Ba); D(A--MP("P",P,NW)--B);D(P--C); D(anglemark(B,A,P,30));D(anglemark(C,B,P,30));D(anglemark(A,C,P,30)); MP("13",(A+B)/2,S);MP("15",(A+C)/2,NW);MP("14",(C+B)/2,NE); [/asy]

Solution 1

Drop perpendiculars from $P$ to the three sides of $\triangle ABC$ and let them meet $\overline{AB}, \overline{BC},$ and $\overline{CA}$ at $D, E,$ and $F$ respectively.

[asy] import olympiad; real theta = 29.66115; /* arctan(168/295) to five decimal places .. don't know other ways to construct Brocard */ pathpen = black +linewidth(0.65); pointpen = black; pair A=(0,0),B=(13,0),C=IP(circle(A,15),circle(B,14)); D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); /* constructing P, C is there as check */ pair Aa=A+(B-A)*dir(theta),Ba=B+(C-B)*dir(theta),Ca=C+(A-C)*dir(theta), P=IP(A--Aa,B--Ba); D(A--MP("P",P,SSW)--B);D(P--C); D(anglemark(B,A,P,30));D(anglemark(C,B,P,30));D(anglemark(A,C,P,30)); MP("13",(A+B)/2,S);MP("15",(A+C)/2,NW);MP("14",(C+B)/2,NE); /* constructing D,E,F as foot of perps from P */ pair D=foot(P,A,B),E=foot(P,B,C),F=foot(P,C,A); D(MP("D",D,NE)--P--MP("E",E,SSW),dashed);D(P--MP("F",F),dashed); D(rightanglemark(P,E,C,15));D(rightanglemark(P,F,C,15));D(rightanglemark(P,D,A,15)); [/asy]

Let $BE = x, CF = y,$ and $AD = z$. We have that \begin{align*}DP&=z\tan\theta\\ EP&=x\tan\theta\\ FP&=y\tan\theta\end{align*} We can then use the tool of calculating area in two ways \begin{align*}[ABC]&=[PAB]+[PBC]+[PCA]\\ &=\frac{1}{2}(13)(z\tan\theta)+\frac{1}{2}(14)(x\tan\theta)+\frac{1}{2}(15)(y\tan\theta)\\ &=\frac{1}{2}\tan\theta(13z+14x+15y)\end{align*} On the other hand, \begin{align*}[ABC]&=\sqrt{s(s-a)(s-b)(s-c)}\\ &=\sqrt{21\cdot6\cdot7\cdot8}\\ &=84\end{align*} We still need $13z+14x+15y$ though. We have all these right triangles and we haven't even touched Pythagoras. So we give it a shot: \begin{align}x^2+x^2\tan^2\theta&=z^2\tan^2\theta+(13-z)^2\\ z^2+z^2\tan^2\theta&=y^2\tan^2\theta+(15-y)^2\\ y^2+y^2\tan^2\theta&=x^2\tan^2\theta+(14-x)^2\end{align} Adding $(1) + (2) + (3)$ gives \begin{align*}x^2+y^2+z^2&=(14-x)^2+(15-y)^2+(13-z)^2\\ \Rightarrow13z+14x+15y&=295\end{align*} Recall that we found that $[ABC]=\frac{1}{2}\tan\theta(13z+14x+15y)=84$. Plugging in $13z+14x+15y=295$, we get $\tan\theta=\frac{168}{295}$, giving us $\boxed{463}$ for an answer.

Solution 2

Let $AB=c$, $BC=a$, $AC=b$, $PA=x$, $PB=y$, and $PC=z$.

So by the Law of Cosines, we have: \begin{align*}x^2 &= z^2 + b^2 - 2bz\cos{\theta}\\ y^2 &= x^2 + c^2 - 2cx\cos{\theta}\\ z^2 &= y^2 + a^2 - 2ay\cos{\theta}\end{align*} Adding these equations and rearranging, we have: \[a^2 + b^2 + c^2 = (2bz + 2cx + 2ay)\cos{\theta}\qquad(1)\] Now $[CAP] + [ABP] + [BCP] = [ABC] = \sqrt {(21)(8)(7)(6)} = 84$, by Heron's formula.

Now the area of a triangle, $[A] = \frac {mn\sin{\beta}}{2}$, where $m$ and $n$ are sides on either side of an angle, $\beta$. So, \begin{align*}[CAP] &= \frac {bz\sin{\theta}}{2}\\ [ABP] &= \frac {cx\sin{\theta}}{2}\\ [BCP] &= \frac {ay\sin{\theta}}{2}\end{align*} Adding these equations yields: \begin{align*}[ABC]= 84 &= \frac {(bz + cx + ay)\sin{\theta}}{2}\\ \Rightarrow 168&= (bz + cx + ay)\sin{\theta}\qquad (2)\end{align*} Dividing $(2)$ by $(1)$, we have: \begin{align*}\frac {168}{a^2 + b^2 + c^2} &= \frac {(bz + cx + ay)\sin{\theta}}{(2bz + 2cx + 2ay)\cos{\theta}}\\ \Rightarrow \tan{\theta} = \frac {336}{a^2 + b^2 + c^2} &= \frac {336}{14^2 + 15^2 + 13^2} = \frac {336}{590} = \frac {168}{295}\end{align*} Thus, $m + n = 168 + 295 = \boxed{463}$.

Note: In fact, this problem is unfairly easy to those who happen to have learned about Brocard point. The Brocard Angle is given by \[cot(\theta)=\frac{a^2+b^2+c^2}{4\Delta}\]

Solution 3

Let $\angle{PAB} = \angle{PBC} = \angle{PCA} = x.$ Then, using Law of Cosines on the three triangles containing vertex $P,$ we have \begin{align*} b^2 &= a^2 + 169 - 26a \cos x \\ c^2 &= b^2 + 196 - 28b \cos x \\ a^2 &= c^2 + 225 - 30c \cos x. \end{align*} Add the three equations up and rearrange to obtain \[(13a + 14b + 15c) \cos x = 295.\] Also, using $[ABC] = \frac{1}{2}ab \sin \angle C$ we have \[[ABC] = [APB] + [BPC] + [CPA] = \dfrac{\sin x}{2}(13a + 14b + 15c) = 84 \iff (13a + 14b + 15c) \sin x = 168.\] Divide the two equations to obtain $\tan x = \frac{168}{295} \iff \boxed{463}.~\square$


Solution 4 (Law of sines)

Firstly, denote angles $ABC$, $BCA$, and $CAB$ as $B$, $A$, and $C$ respectively. Let $\angle{PAB}=x$. Notice that by angle chasing that $\angle{BPC}=180-C$ and $\angle{BPA}=180-B$. Using the nice properties of the 13-14-15 triangle, we have $\sin B = \frac{12}{13}$ and $\sin C = \frac{4}{5}$. $\cos C$ is easily computed, so we have $\cos C=\frac{3}{5}$.

Using Law of Sines, \begin{align*} \frac{BP}{\sin x} &= \frac{13}{\sin (180 - B)} \\ \frac{BP}{\sin (C - x)} &= \frac{14}{\sin (180 - C)} \end{align*} hence, \begin{align*} \frac{BP}{\sin x} &= \frac{13}{\sin B} \\ \frac{BP}{\sin (C - x)} &= \frac{14}{\sin C} \end{align*} Now, computation carries the rest. \begin{align*} \frac{13 \sin x}{\sin B} &= \frac{14 \sin (C-x)}{\sin C} \\ \frac{169 \sin x}{12} &= \frac{210 \sin (C-x)}{12} \\ 169 \sin x &= 210 (\sin C \cos x - \cos C \sin x) \\ 169 \sin x &= 210 (\frac{4}{5} \cos x - \frac{3}{5} \sin x) \\ 169 \sin x &= 168 \cos x - 126 \sin x \\ 295 \sin x &= 168 \cos x \\ \tan x &= \frac{168}{295} \end{align*} Extracting yields $168 + 295 = \boxed{463}$.

See also

1999 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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All AIME Problems and Solutions

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