Difference between revisions of "1999 AIME Problems/Problem 14"
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== Problem == | == Problem == | ||
− | Point <math>P_{}</math> is located inside traingle <math>ABC</math> so that | + | [[Point]] <math>P_{}</math> is located inside [[traingle]] <math>ABC</math> so that [[angle]]s <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__ | __TOC__ | ||
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[[Image:1999_AIME-14.png]] | [[Image:1999_AIME-14.png]] | ||
=== Solution 1 === | === Solution 1 === | ||
− | Drop | + | 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{eqnarray*} DP & = & z\tan \theta \ | ||
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& = & 84 \end{eqnarray*} | & = & 84 \end{eqnarray*} | ||
</cmath> | </cmath> | ||
− | We still need <math>13z + 14x + 15y</math> though. We have all these right | + | 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 \ | \begin{eqnarray} x^2 + x^2\tan^2\theta & = & z^2\tan^2\theta + (13 - z)^2 \ | ||
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Let <math>AB = c, BC = a, AC = b, PA = x, PB = y, PC = z</math>. | Let <math>AB = c, BC = a, AC = b, PA = x, PB = y, 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> | <math>x^2 = z^2 + b^2 - 2bz\cos{\theta}</math> | ||
<math>y^2 = x^2 + c^2 - 2cx\cos{\theta}</math> | <math>y^2 = x^2 + c^2 - 2cx\cos{\theta}</math> | ||
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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}</math> | + | <math>a^2 + b^2 + c^2 = (2bz + 2cx + 2ay)\cos{\theta}\qquad (1)</math> |
− | Now <math>[CAP] + [ABP] + [BCP] = [ABC] = \sqrt {(21)(8)(7)(6)} = 84</math>, by Heron's | + | 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, | ||
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So adding these equations yields: | So adding these equations yields: | ||
<math>[ABC] = 84 = \frac {(bz + cx + ay)\sin{\theta}}{2}</math> | <math>[ABC] = 84 = \frac {(bz + cx + ay)\sin{\theta}}{2}</math> | ||
− | <math>\Rightarrow 168 = (bz + cx + ay)\sin{\theta}</math> | + | <math>\Rightarrow 168 = (bz + cx + ay)\sin{\theta}\qquad (2)</math> |
Dividing (2) by (1), we have: | Dividing (2) by (1), we have: |
Revision as of 20:51, 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:
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:
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 |