Difference between revisions of "1990 AIME Problems/Problem 7"
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By the angle bisector theorem as in solution 1, we find that <math>QP' = 25/2</math>. If we draw the right triangle formed by <math>Q, P',</math> and the point directly to the right of <math>Q</math> and below <math>P'</math>, we get another <math>3-4-5 \triangle</math> (since the slope of <math>QR</math> is <math>3/4</math>). Using this, we find that the horizontal projection of <math>QP'</math> is <math>10</math> and the vertical projection of <math>QP'</math> is <math>15/2</math>. | By the angle bisector theorem as in solution 1, we find that <math>QP' = 25/2</math>. If we draw the right triangle formed by <math>Q, P',</math> and the point directly to the right of <math>Q</math> and below <math>P'</math>, we get another <math>3-4-5 \triangle</math> (since the slope of <math>QR</math> is <math>3/4</math>). Using this, we find that the horizontal projection of <math>QP'</math> is <math>10</math> and the vertical projection of <math>QP'</math> is <math>15/2</math>. | ||
− | Thus, the angle bisector touches <math>QR</math> at the point <math>\left(-15 + 10, -19 + \frac{15}{2}\right) = \left(-5,-\frac{23}{2}\right)</math>, from where we continue with the first solution. | + | Thus, the angle bisector touches <math>QR</math> at the point <math>\left(-15 + 10, -19 + \frac{15}{2}\right) = \left(-5,-\frac{23}{2}\right)</math>, from where we continue with the first solution. |
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+ | === Solution 4 === | ||
== See also == | == See also == |
Revision as of 20:34, 2 September 2019
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 . It follows that , and so .
The desired answer is the equation of the line . has slope , from which we find the equation to be . Therefore, .
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. We reach the same answer of .
Solution 3
By the angle bisector theorem as in solution 1, we find that . If we draw the right triangle formed by and the point directly to the right of and below , we get another (since the slope of is ). Using this, we find that the horizontal projection of is and the vertical projection of is .
Thus, the angle bisector touches at the point , from where we continue with the first solution.
Solution 4
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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