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

Revision as of 20:59, 2 March 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 $a+c_{}^{}$ .

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 $11x + 2y + 78 = 0$ .

@@ Line 1: / Line 1: @@
 == Problem ==
-A triangle has vertices <math>P_{}^{}=(-8,5)</math>, <math>Q_{}^{}=(-15,-19)</math>, and <math>R_{}^{}=(1,-7)</math>. The equation of the bisector of <math>\angle P</math> can be written in the form <math>ax+2y+c=0_{}^{}</math>. Find <math>a+c_{}^{}</math>.
+A [[triangle]] has [[vertex|vertices]] <math>P_{}^{}=(-8,5)</math>, <math>Q_{}^{}=(-15,-19)</math>, and <math>R_{}^{}=(1,-7)</math>. The [[equation]] of the [[bisector]] of <math>\angle P</math> can be written in the form <math>ax+2y+c=0_{}^{}</math>. Find <math>a+c_{}^{}</math>.
 == Solution ==
-{{solution}}
+{{image}}
+Use the [[distance formula]] to determine the lengths of each of the sides of the triangle. We find that it has lengths of side <math>15,\ 20,\ 25</math>, indicating that it is a <math>3-4-5</math> [[right triangle]]. At this point, we just need to find another [[point]] that lies on the bisector of <math>\angle P</math>.
+=== Solution 1 ===
+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>.
+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>.
+=== Solution 2 ===
+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.
+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>.
 == See also ==
 {{AIME box|year=1990|num-b=6|num-a=8}}

1990 AIME (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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