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 .
Solution 5 (Trigonometry)
Transform triangle so that is at the origin. Note that the slopes do not change when we transform the triangle.
We know that the slope of is and the slope of is . Thus, in the complex plane, they are equivalent to and , respectively. Here is the angle formed by the -axis and , and is the angle formed by the -axis and . The equation of the angle bisector is .
As the tangents are in very neat Pythagorean triples, we can easily calculate and .
Angle is in the third quadrant, so is negative. Thus .
Angle is in the fourth quadrant, so is positive. Thus .
By the Half-Angle Identities, and .
Since and must be in the second quadrant, their tangent values are both negative. Thus and .
By the sum of tangents formula, , which is the slope of the angle bisector.
Finally, the equation of the angle bisector is or . Rearranging, we get , so our sum is . ~eevee9406
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.