Difference between revisions of "1990 AIME Problems/Problem 7"
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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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== See also == | == See also == |
Revision as of 22:12, 2 September 2019
Problem
A triangle has vertices ,
, and
. The equation of the bisector of
can be written in the form
. Find
.
![[asy] import graph; pointpen=black;pathpen=black+linewidth(0.7);pen f = fontsize(10); pair P=(-8,5),Q=(-15,-19),R=(1,-7),S=(7,-15),T=(-4,-17); MP("P",P,N,f);MP("Q",Q,W,f);MP("R",R,E,f); D(P--Q--R--cycle);D(P--T,EndArrow(2mm)); D((-17,0)--(4,0),Arrows(2mm));D((0,-21)--(0,7),Arrows(2mm)); [/asy]](http://latex.artofproblemsolving.com/b/7/6/b7650dc9ace1a94c8dc1634c4bfa218183f6602c.png)
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
![[asy] import graph; pointpen=black;pathpen=black+linewidth(0.7);pen f = fontsize(10); pair P=(-8,5),Q=(-15,-19),R=(1,-7),S=(7,-15),T=(-4,-17),U=IP(P--T,Q--R); MP("P",P,N,f);MP("Q",Q,W,f);MP("R",R,E,f);MP("P'",U,SE,f); D(P--Q--R--cycle);D(U);D(P--U); D((-17,0)--(4,0),Arrows(2mm));D((0,-21)--(0,7),Arrows(2mm)); [/asy]](http://latex.artofproblemsolving.com/e/5/1/e51a70bc1f21cb48775b66de3054d6f7622a3627.png)
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
![[asy] import graph; pointpen=black;pathpen=black+linewidth(0.7);pen f = fontsize(10); pair P=(-8,5),Q=(-15,-19),R=(1,-7),S=(7,-15),T=(-4,-17); MP("P",P,N,f);MP("Q",Q,W,f);MP("R",R,NE,f);MP("S",S,E,f); D(P--Q--R--cycle);D(R--S--Q,dashed);D(T);D(P--T); D((-17,0)--(4,0),Arrows(2mm));D((0,-21)--(0,7),Arrows(2mm)); [/asy]](http://latex.artofproblemsolving.com/7/8/6/786d51d5c72e94f0b159e6406b8cc29bbc70f311.png)
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
![[asy] import graph; pointpen=black;pathpen=black+linewidth(0.7);pen f = fontsize(10); pair P=(-8,5),Q=(-15,-19),R=(1,-7),S=(7,-15),T=(-4,-17),U=IP(P--T,Q--R); MP("P",P,N,f);MP("Q",Q,W,f);MP("R",R,E,f);MP("P'",U,SE,f); D(P--Q--R--cycle);D(U);D(P--U); D((-17,0)--(4,0),Arrows(2mm));D((0,-21)--(0,7),Arrows(2mm)); D(Q--(U.x,Q.y)--U,dashed);D(rightanglemark(Q,(U.x,Q.y),U,20),dashed); [/asy]](http://latex.artofproblemsolving.com/3/b/e/3be1357e127e0b9b6aeea9267697d70227e84c44.png)
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 • 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.