Difference between revisions of "1990 AIME Problems/Problem 7"
(→Solution 4) |
(→Solution 4) |
||
Line 67: | Line 67: | ||
</asy></center> | </asy></center> | ||
− | Let <math>X_1</math>, <math>X_2</math> be the | + | Let <math>X_1</math>, <math>X_2</math> be the x-intercepts of <math>PQ</math>, <math>PR</math>, respectively. Calculating the slopes and plugging in the coordinates, we have the equations of <math>PQ</math> and <math>PR</math> |
+ | <cmath>y_{PQ}=\frac{24}{7}x_{PQ}+\frac{227}{7}</cmath> | ||
+ | <cmath>y_{PQ}=-\frac{4}{3}x_{PQ}-\frac{17}{3}</cmath> | ||
== See also == | == See also == |
Revision as of 21:03, 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
Let , be the x-intercepts of , , respectively. Calculating the slopes and plugging in the coordinates, we have the equations of and
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.