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1999 AIME Problems/Problem 14

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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 $m+n.$

Contents

1 Problem
2 Solution
- 2.1 Solution 1
- 2.2 Solution 2
3 See also

Solution

real theta = 29.66115; /* arctan(168/295) to five decimal places .. don't know other ways to construct Brocard */ pathpen = b...

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.

import olympiad;real theta = 29.66115; /* arctan(168/295) to five decimal places .. don't know other ways to construct Brocar...

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 $\begin{align*}[ABC]&=[PAB]+[PBC]+[PCA]\\&=\frac{1}{2}(13)(z\tan\theta)+\frac{1}{2}(14)(x\tan\theta)+\frac{1}{2}(15)(y...$ 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: $\begin{align}x^2+x^2\tan^2\theta&=z^2\tan^2\theta+(13-z)^2\\z^2+z^2\tan^2\theta&=y^2\tan^2\theta+(15-y)^2\\y^2+y^2\ta...$ 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: $\begin{align*}x^2 &= z^2 + b^2 - 2bz\cos{\theta}\\y^2 &= x^2 + c^2 - 2cx\cos{\theta}\\z^2 &= y^2 + a^2 - 2ay\cos{...$ 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, $\begin{align*}[CAP] &= \frac {bz\sin{\theta}}{2}\\[ABP] &= \frac {cx\sin{\theta}}{2}\\[BCP] &= \frac {ay\sin{\the...$ Adding these equations yields: $\begin{align*}[ABC]= 84 &= \frac {(bz + cx + ay)\sin{\theta}}{2}\\\Rightarrow 168&= (bz + cx + ay)\sin{\theta}\qquad ...$ Dividing $(2)$ by $(1)$ , we have: $\begin{align*}\frac {168}{a^2 + b^2 + c^2} &= \frac {(bz + cx + ay)\sin{\theta}}{(2bz + 2cx + 2ay)\cos{\theta}}\\\Rightar...$ Thus, $m + n = 168 + 295 = \boxed{463}$ .

See also

Brocard point

1999 AIME (Problems • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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Category: Intermediate Geometry Problems

Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.

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