Difference between revisions of "1999 AIME Problems/Problem 14"
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== Solution == | == Solution == | ||
− | [[Image:1999_AIME-14.png]] | + | {{image}} |
+ | <!--This image does not exist: [[Image:1999_AIME-14.png]]--> | ||
+ | |||
=== 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{ | + | \begin{align*}DP&=z\tan\theta\ |
− | EP & = | + | EP&=x\tan\theta\ |
− | FP & = | + | 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{ | + | \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> | </cmath> | ||
− | On the other hand | + | On the other hand, |
<cmath> | <cmath> | ||
− | \begin{ | + | \begin{align*}[ABC]&=\sqrt{s(s-a)(s-b)(s-c)}\ |
− | & = | + | &=\sqrt{21\cdot6\cdot7\cdot8}\ |
− | & = | + | &=84\end{align*} |
</cmath> | </cmath> | ||
− | We still need <math>13z + 14x + 15y</math> though. | + | 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{ | + | \begin{align}x^2+x^2\tan^2\theta&=z^2\tan^2\theta+(13-z)^2\ |
− | z^2 + z^2\tan^2\theta & = | + | z^2+z^2\tan^2\theta&=y^2\tan^2\theta+(15-y)^2\ |
− | y^2 + y^2\tan^2\theta & = | + | y^2+y^2\tan^2\theta&=x^2\tan^2\theta+(14-x)^2\end{align} |
− | \end{ | ||
</cmath> | </cmath> | ||
− | + | Adding <math>(1) + (2) + (3)</math> gives | |
<cmath> | <cmath> | ||
− | \begin{ | + | \begin{align*}x^2+y^2+z^2&=(14-x)^2+(15-y)^2+(13-z)^2\ |
− | \ | + | \Rightarrow13z+14x+15y&=295\end{align*} |
</cmath> | </cmath> | ||
− | Recall that we found that <math>[ABC] = \frac | + | 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>. | + | 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: | ||
− | < | + | <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: | 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 <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, | ||
− | < | + | <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*} | |
− | Dividing (2) by (1), we have: | + | </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>. | ||
== See also == | == See also == |
Revision as of 10:59, 26 April 2008
Problem
Point is located inside traingle so that angles and are all congruent. The sides of the triangle have lengths and and the tangent of angle is where and are relatively prime positive integers. Find
Contents
[hide]Solution
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Solution 1
Drop perpendiculars from to the three sides of and let them meet and at and respectively. Let and . We have that We can then use the tool of calculating area in two ways On the other hand, We still need though. We have all these right triangles and we haven't even touched Pythagoras. So we give it a shot: Adding gives Recall that we found that . Plugging in , we get , giving us for an answer.
Solution 2
Let , , , , , and .
So by the Law of Cosines, we have: Adding these equations and rearranging, we have: Now , by Heron's formula.
Now the area of a triangle, , where and are sides on either side of an angle, . So, Adding these equations yields: Dividing by , we have: Thus, .
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 |