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

Revision as of 11:59, 26 April 2008

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

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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 $\begin{align*}DP&=z\tan\theta\\ EP&=x\tan\theta\\ FP&=y\tan\theta\end{align*}$ 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, $m + n = 168 + 295 = \boxed{463}$ .

@@ Line 4: / Line 4: @@
 __TOC__
 == Solution ==
-[[Image:1999_AIME-14.png]]
+{{image}}
+<!--This image does not exist: [[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.  Let <math>BE = x, CF = y,</math> and <math>AD = z</math>.  We have that
 <cmath>
-\begin{eqnarray*} DP & = & z\tan \theta \\
+\begin{align*}DP&=z\tan\theta\\
-EP & = & x\tan \theta \\
+EP&=x\tan\theta\\
-FP & = & y\tan \theta \end{eqnarray*}
+FP&=y\tan\theta\end{align*}
 </cmath>
 We can then use the tool of calculating area in two ways
 <cmath>
-\begin{eqnarray*} [ABC] & = & [PAB] + [PBC] + [PCA] \\
+\begin{align*}[ABC]&=[PAB]+[PBC]+[PCA]\\
-& = & \frac 12 (13)(z\tan \theta) + \frac 12 (14)(x\tan\theta) + \frac 12 (15)(y\tan\theta) \\
+&=\frac{1}{2}(13)(z\tan\theta)+\frac{1}{2}(14)(x\tan\theta)+\frac{1}{2}(15)(y\tan\theta)\\
-& = & \frac 12 \tan\theta(13z + 14x + 15y) \end{eqnarray*}
+&=\frac{1}{2}\tan\theta(13z+14x+15y)\end{align*}
 </cmath>
-On the other hand
+On the other hand,
 <cmath>
-\begin{eqnarray*} [ABC] & = & \sqrt {s(s - a)(s - b)(s - c)} \\
+\begin{align*}[ABC]&=\sqrt{s(s-a)(s-b)(s-c)}\\
-& = & \sqrt {21\cdot 6\cdot 7\cdot 8} \\
+&=\sqrt{21\cdot6\cdot7\cdot8}\\
-& = & 84 \end{eqnarray*}
+&=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
+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{eqnarray} x^2 + x^2\tan^2\theta & = & z^2\tan^2\theta + (13 - z)^2 \\
+\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 \\
+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 \\
+y^2+y^2\tan^2\theta&=x^2\tan^2\theta+(14-x)^2\end{align}
-\end{eqnarray*}
 </cmath>
-But then <math>(1) + (2) + (3)</math> gives
+Adding <math>(1) + (2) + (3)</math> gives
 <cmath>
-\begin{eqnarray*} x^2 + y^2 + z^2 & = & (14 - x)^2 + (15 - y)^2 + (13 - z)^2 \\
+\begin{align*}x^2+y^2+z^2&=(14-x)^2+(15-y)^2+(13-z)^2\\
-\Rightarrow 13z + 14x + 15y & = & 295 \end{eqnarray*}
+\Rightarrow13z+14x+15y&=295\end{align*}
 </cmath>
-Recall that we found that <math>[ABC] = \frac 12 \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.
+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, BC = a, AC = b, PA = x, PB = y, PC = z</math>.
+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:
-<math>x^2 = z^2 + b^2 - 2bz\cos{\theta}</math>
+<cmath>
-<math>y^2 = x^2 + c^2 - 2cx\cos{\theta}</math>
+\begin{align*}x^2 &= z^2 + b^2 - 2bz\cos{\theta}\\
-<math>z^2 = y^2 + a^2 - 2ay\cos{\theta}</math>
+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:
-<math>a^2 + b^2 + c^2 = (2bz + 2cx + 2ay)\cos{\theta}\qquad (1)</math>
+<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,
-<math>[CAP] = \frac {bz\sin{\theta}}{2}</math>.
+<cmath>
-<math>[ABP] = \frac {cx\sin{\theta}}{2}</math>.
+\begin{align*}[CAP] &= \frac {bz\sin{\theta}}{2}\\
-<math>[BCP] = \frac {ay\sin{\theta}}{2}</math>
+[ABP] &= \frac {cx\sin{\theta}}{2}\\
+[BCP] &= \frac {ay\sin{\theta}}{2}\end{align*}
-So adding these equations yields:
+</cmath>
-<math>[ABC] = 84 = \frac {(bz + cx + ay)\sin{\theta}}{2}</math>
+Adding these equations yields:
-<math>\Rightarrow 168 = (bz + cx + ay)\sin{\theta}\qquad (2)</math>
+<cmath>\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:
+</cmath>
-<math>\frac {168}{a^2 + b^2 + c^2} = \frac {(bz + cx + ay)\sin{\theta}}{(2bz + 2cx + 2ay)\cos{\theta}}</math>
+Dividing <math>(2)</math> by <math>(1)</math>, we have:
-<math>\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}</math>
+<cmath>
+\begin{align*}\frac {168}{a^2 + b^2 + c^2} &= \frac {(bz + cx + ay)\sin{\theta}}{(2bz + 2cx + 2ay)\cos{\theta}}\\
-So <math>m + n = 168 + 295 = \boxed{463}</math>.
+\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>.
 == 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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