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1990 AIME Problems/Problem 7

Revision as of 09:34, 3 March 2007 by Azjps (talk | contribs) (shift TOC downwards)

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 $a+c_{}^{}$.

Solution

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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 $(-15 + 10, -19 + \frac{15}{2}) \Rightarrow (-5,-\frac{23}{2})$. The slope of these two points is $\frac{5 - (-\frac{23}{2})}{-8 - (-5)} = \frac{-11}{2}$. Setting the slope equal (we could also write out the x-intercept form of the equation and substitute) to $\frac{-11}{2} = \frac{y + 8}{x - 5} \Longrightarrow -11x + 55 = 2y + 16 \Longrightarrow 11x + 2y + 78 = 0$. 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 $\displaystyle QS \Rightarrow (-4,-17)$, 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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