1990 AIME Problems/Problem 7
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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.
See also
| 1990 AIME (Problems • Resources) | ||
| Preceded by Problem 6 | Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
