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1999 AIME Problems/Problem 14

Revision as of 21:51, 18 October 2007 by Azjps (talk | contribs) (wik)

Problem

Point $P_{}$ is located inside traingle $ABC$ 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

File:1999 AIME-14.png

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. Let $BE = x, CF = y,$ and $AD = z$. We have that \begin{eqnarray*} DP & = & z\tan \theta \\ EP & = & x\tan \theta \\ FP & = & y\tan \theta \end{eqnarray*} We can then use the tool of calculating area in two ways \begin{eqnarray*} [ABC] & = & [PAB] + [PBC] + [PCA] \\ & = & \frac 12 (13)(z\tan \theta) + \frac 12 (14)(x\tan\theta) + \frac 12 (15)(y\tan\theta) \\ & = & \frac 12 \tan\theta(13z + 14x + 15y) \end{eqnarray*} On the other hand \begin{eqnarray*} [ABC] & = & \sqrt {s(s - a)(s - b)(s - c)} \\ & = & \sqrt {21\cdot 6\cdot 7\cdot 8} \\ & = & 84 \end{eqnarray*} 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{eqnarray} 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{eqnarray*} (Error compiling LaTeX. Unknown error_msg)

But then $(1) + (2) + (3)$ gives \begin{eqnarray*} x^2 + y^2 + z^2 & = & (14 - x)^2 + (15 - y)^2 + (13 - z)^2 \\ \Rightarrow 13z + 14x + 15y & = & 295 \end{eqnarray*} Recall that we found that $[ABC] = \frac 12 \tan\theta(13z + 14x + 15y) = 84$. Plugging in $13z + 14x + 15y = 295$ we get $\tan \theta = \frac {168}{295}$ giving us $\boxed{463}$ for an answer.

Solution 2

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

So by the Law of Cosines, we have: $x^2 = z^2 + b^2 - 2bz\cos{\theta}$ $y^2 = x^2 + c^2 - 2cx\cos{\theta}$ $z^2 = y^2 + a^2 - 2ay\cos{\theta}$

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, $[CAP] = \frac {bz\sin{\theta}}{2}$. $[ABP] = \frac {cx\sin{\theta}}{2}$. $[BCP] = \frac {ay\sin{\theta}}{2}$

So adding these equations yields: $[ABC] = 84 = \frac {(bz + cx + ay)\sin{\theta}}{2}$ $\Rightarrow 168 = (bz + cx + ay)\sin{\theta}\qquad (2)$

Dividing (2) by (1), we have: $\frac {168}{a^2 + b^2 + c^2} = \frac {(bz + cx + ay)\sin{\theta}}{(2bz + 2cx + 2ay)\cos{\theta}}$ $\Rightarrow \tan{\theta} = \frac {336}{a^2 + b^2 + c^2} = \frac {336}{14^2 + 15^2 + 13^2} = \frac {336}{590} = \frac {168}{295}$

So $m + n = 168 + 295 = \boxed{463}$.

See also

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