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Difference between revisions of "1990 AIME Problems/Problem 7"

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Use the [[angle bisector theorem]] to find that the angle bisector of <math>\angle P</math> divides <math>QR</math> into segments of length <math>\frac{25}{x} = \frac{15}{20 -x} \Longrightarrow x = \frac{500}{40} = \frac{25}{2},\ \frac{15}{2}</math>. If we draw a triangle using the points <math>Q</math>, the point by which the angle bisector touches <math>QR</math>, and the point directly to the right of <math>Q</math> and the bottom of the aforementioned point, we get another <math>3-4-5 \triangle</math> (this can be shown by proving its [[similar triangle|similarity]] to the triangle drawn using the side of length <math>20</math> as the [[hypotenuse]]). Using this, the lengths of the triangle are <math>\frac{15}{2}, 10, \frac{25}{2}</math>.
 
Use the [[angle bisector theorem]] to find that the angle bisector of <math>\angle P</math> divides <math>QR</math> into segments of length <math>\frac{25}{x} = \frac{15}{20 -x} \Longrightarrow x = \frac{500}{40} = \frac{25}{2},\ \frac{15}{2}</math>. If we draw a triangle using the points <math>Q</math>, the point by which the angle bisector touches <math>QR</math>, and the point directly to the right of <math>Q</math> and the bottom of the aforementioned point, we get another <math>3-4-5 \triangle</math> (this can be shown by proving its [[similar triangle|similarity]] to the triangle drawn using the side of length <math>20</math> as the [[hypotenuse]]). Using this, the lengths of the triangle are <math>\frac{15}{2}, 10, \frac{25}{2}</math>.
  
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Thus, the angle bisector touches <math>QR</math> at the point <math>(-15 + 10, -19 + \frac{15}{2}) \Rightarrow (-5,-\frac{23}{2})</math>. The [[slope]] of these two points is <math>\frac{5 - (-\frac{23}{2})}{-8 - (-5)} = \frac{-11}{2}</math>. Setting the slope equal (we could also write out the [[x-intercept]] form of the equation and substitute) to <math>\frac{-11}{2} = \frac{y + 8}{x - 5} \Longrightarrow -11x + 55 = 2y + 16 \Longrightarrow 11x + 2y + 78 = 0</math>. Thus, the solution is <math>11 + 78 = 089</math>.
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Thus, the angle bisector touches <math>QR</math> at the point <math>\left(-15 + 10, -19 + \frac{15}{2}\right) \Rightarrow \left(-5,-\frac{23}{2}\right) = \frac{y + 8}{x - 5}</math> <math>\Longrightarrow -11x + 55 = 2y + 16</math> <math>\Longrightarrow 11x + 2y + 78 = 0</math>. Thus, the solution is <math>11 + 78 = 089</math>.
  
 
=== Solution 2 ===
 
=== Solution 2 ===
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Extend <math>PR</math> to a point <math>S</math> such that <math>PS = 25</math>. This forms an [[isosceles triangle]] <math>PQS</math>. The [[coordinate]]s of <math>S</math>, using the slope of <math>PR</math> (which is <math>-\frac{4}{3}</math>), can be determined to be <math>(7,-15)</math>. Since the [[angle bisector]] of <math>\angle P</math> must touch the midpoint of <math>\displaystyle QS \Rightarrow (-4,-17)</math>, we have found our two points.
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Extend <math>PR</math> to a point <math>S</math> such that <math>PS = 25</math>. This forms an [[isosceles triangle]] <math>PQS</math>. The [[coordinate]]s of <math>S</math>, using the slope of <math>PR</math> (which is <math>-\frac{4}{3}</math>), can be determined to be <math>(7,-15)</math>. Since the [[angle bisector]] of <math>\angle P</math> must touch the midpoint of <math>QS \Rightarrow (-4,-17)</math>, we have found our two points.
  
 
The [[slope]] and equation of the line in general form will remain the same, yielding the same answer of <math>11x + 2y + 78 = 0</math>.
 
The [[slope]] and equation of the line in general form will remain the same, yielding the same answer of <math>11x + 2y + 78 = 0</math>.
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== See also ==
 
== See also ==
 
{{AIME box|year=1990|num-b=6|num-a=8}}
 
{{AIME box|year=1990|num-b=6|num-a=8}}
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[[Category:Intermediate Geometry Problems]]

Revision as of 21:04, 24 November 2007

Problem

A triangle has vertices $P_{}^{}=(-8,5)$, $Q_{}^{}=(-15,-19)$, and $R_{}^{}=(1,-7)$. The equation of the bisector of $\angle P$ can be written in the form $ax+2y+c=0_{}^{}$. Find $a+c_{}^{}$.

1990 AIME Problem 7.png

Solution

Use the distance formula to determine the lengths of each of the sides of the triangle. We find that it has lengths of side $15,\ 20,\ 25$, indicating that it is a $3-4-5$ right triangle. At this point, we just need to find another point that lies on the bisector of $\angle P$.

Solution 1

Use the angle bisector theorem to find that the angle bisector of $\angle P$ divides $QR$ into segments of length $\frac{25}{x} = \frac{15}{20 -x} \Longrightarrow x = \frac{500}{40} = \frac{25}{2},\ \frac{15}{2}$. If we draw a triangle using the points $Q$, the point by which the angle bisector touches $QR$, and the point directly to the right of $Q$ and the bottom of the aforementioned point, we get another $3-4-5 \triangle$ (this can be shown by proving its similarity to the triangle drawn using the side of length $20$ as the hypotenuse). Using this, the lengths of the triangle are $\frac{15}{2}, 10, \frac{25}{2}$.

Thus, the angle bisector touches $QR$ at the point $\left(-15 + 10, -19 + \frac{15}{2}\right) \Rightarrow \left(-5,-\frac{23}{2}\right) = \frac{y + 8}{x - 5}$ $\Longrightarrow -11x + 55 = 2y + 16$ $\Longrightarrow 11x + 2y + 78 = 0$. Thus, the solution is $11 + 78 = 089$.

Solution 2

Extend $PR$ to a point $S$ such that $PS = 25$. This forms an isosceles triangle $PQS$. The coordinates of $S$, using the slope of $PR$ (which is $-\frac{4}{3}$), can be determined to be $(7,-15)$. Since the angle bisector of $\angle P$ must touch the midpoint of $QS \Rightarrow (-4,-17)$, we have found our two points.

The slope and equation of the line in general form will remain the same, yielding the same answer of $11x + 2y + 78 = 0$.

See also

1990 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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