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

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

Revision as of 10:59, 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, $m + n = 168 + 295 = \boxed{463}$.

See also

1999 AIME (ProblemsAnswer KeyResources)
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