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>. | ||
− | Thus, the angle bisector touches <math>QR</math> at the point <math>(-15 + 10, -19 + \frac{15}{2}) \Rightarrow (-5,-\frac{23}{2}) | + | 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 === | ||
− | 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> | + | 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}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] |
Revision as of 21:04, 24 November 2007
Problem
A triangle has vertices , , and . The equation of the bisector of can be written in the form . Find .
Contents
[hide]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 , 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 . 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 |