Difference between revisions of "1999 AIME Problems/Problem 14"
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== Problem == | == Problem == | ||
− | Point <math> | + | Point <math>P_{}</math> is located inside traingle <math>ABC</math> so that angles <math>PAB, PBC,</math> and <math>PCA</math> are all congruent. The sides of the triangle have lengths <math>AB=13, BC=14,</math> and <math>CA=15,</math> and the tangent of angle <math>PAB</math> is <math>m/n,</math> where <math>m_{}</math> and <math>n_{}</math> are relatively prime positive integers. Find <math>m+n.</math> |
+ | __TOC__ | ||
== Solution == | == Solution == | ||
− | {{ | + | [[Image:1999_AIME-14.png]] |
+ | === Solution 1 === | ||
+ | Drop perpendiculars 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 \ | ||
+ | EP & = & x\tan \theta \ | ||
+ | FP & = & y\tan \theta \end{eqnarray*} | ||
+ | </cmath> | ||
+ | We can then use the tool of calculating area in two ways | ||
+ | <cmath> | ||
+ | \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*} | ||
+ | </cmath> | ||
+ | On the other hand | ||
+ | <cmath> | ||
+ | \begin{eqnarray*} [ABC] & = & \sqrt {s(s - a)(s - b)(s - c)} \ | ||
+ | & = & \sqrt {21\cdot 6\cdot 7\cdot 8} \ | ||
+ | & = & 84 \end{eqnarray*} | ||
+ | </cmath> | ||
+ | We still need <math>13z + 14x + 15y</math> though. We have all these right triangles and we haven't even touched 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 \ | ||
+ | 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*} | ||
+ | </cmath> | ||
+ | But then <math>(1) + (2) + (3)</math> gives | ||
+ | <cmath> | ||
+ | \begin{eqnarray*} x^2 + y^2 + z^2 & = & (14 - x)^2 + (15 - y)^2 + (13 - z)^2 \ | ||
+ | \Rightarrow 13z + 14x + 15y & = & 295 \end{eqnarray*} | ||
+ | </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. | ||
+ | |||
+ | === Solution 2 === | ||
+ | Let <math>AB = c, BC = a, AC = b, PA = x, PB = y, PC = z</math>. | ||
+ | |||
+ | So by the Law of Cosines, we have: | ||
+ | <math>x^2 = z^2 + b^2 - 2bz\cos{\theta}</math> | ||
+ | <math>y^2 = x^2 + c^2 - 2cx\cos{\theta}</math> | ||
+ | <math>z^2 = y^2 + a^2 - 2ay\cos{\theta}</math> | ||
+ | |||
+ | Adding these equations and rearranging, we have: | ||
+ | <math>a^2 + b^2 + c^2 = (2bz + 2cx + 2ay)\cos{\theta}</math> (1) | ||
+ | |||
+ | 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>. | ||
+ | <math>[ABP] = \frac {cx\sin{\theta}}{2}</math>. | ||
+ | <math>[BCP] = \frac {ay\sin{\theta}}{2}</math> | ||
+ | |||
+ | So adding these equations yields: | ||
+ | <math>[ABC] = 84 = \frac {(bz + cx + ay)\sin{\theta}}{2}</math> | ||
+ | <math>\Rightarrow 168 = (bz + cx + ay)\sin{\theta}</math> (2) | ||
+ | |||
+ | Dividing (2) by (1), we have: | ||
+ | <math>\frac {168}{a^2 + b^2 + c^2} = \frac {(bz + cx + ay)\sin{\theta}}{(2bz + 2cx + 2ay)\cos{\theta}}</math> | ||
+ | <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> | ||
+ | |||
+ | So <math>m + n = 168 + 295 = \boxed{463}</math>. | ||
== See also == | == See also == | ||
+ | *[[Brocard point]] | ||
{{AIME box|year=1999|num-b=13|num-a=15}} | {{AIME box|year=1999|num-b=13|num-a=15}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] |
Revision as of 20:46, 18 October 2007
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
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
\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 gives Recall that we found that . Plugging in we get giving us for an answer.
Solution 2
Let .
So by the Law of Cosines, we have:
Adding these equations and rearranging, we have: (1)
Now , by Heron's Formula.
Now the area of a triangle, , where and are sides on either side of an angle, . So, . .
So adding these equations yields: (2)
Dividing (2) by (1), we have:
So .
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 |