Difference between revisions of "1990 AIME Problems/Problem 7"
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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>. | 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>. | ||
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== Solution == | == Solution == | ||
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Revision as of 09:34, 3 March 2007
Problem
A triangle has vertices 
, 
, and 
. The equation of the bisector of 
can be written in the form 
. Find 
.
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 
, indicating that it is a 
right triangle. At this point, we just need to find another point that lies on the bisector of .
Solution 1
Use the angle bisector theorem to find that the angle bisector of divides 
into segments of length 
. If we draw a triangle using the points 
, the point by which the angle bisector touches , and the point directly to the right of and the bottom of the aforementioned point, we get another 
(this can be shown by proving its similarity to the triangle drawn using the side of length 
as the hypotenuse). Using this, the lengths of the triangle are 
.
Thus, the angle bisector touches at the point 
. The slope of these two points is 
. Setting the slope equal (we could also write out the x-intercept form of the equation and substitute) to 
. Thus, the solution is 
.
Solution 2
Extend 
to a point 
such that 
. This forms an isosceles triangle 
. The coordinates of , using the slope of (which is 
), can be determined to be 
. Since the angle bisector of must touch the midpoint of 
, 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 
.
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 | ||
