1990 AIME Problems/Problem 7
Problem
A triangle has vertices , , and . The equation of the bisector of can be written in the form . Find .
Contents
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
This solution uses terminology from the other solutions. The incenter is a much easier point to find on the line . Note that the inradius of is . If you do not understand this, substitute values into the equation. If lines are drawn from the incenter perpendicular to and , then a square with side length will be created. Call the point opposite in this square . Since has coordinates , and the sides of the squares are on a ratio, the coordinates of are . This is because the x-coordinate is moving to the left units and the y-coordinate is moving up units. The line through and is .
See also
1990 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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