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 |