Difference between revisions of "1990 AIME Problems/Problem 7"
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import graph; | import graph; | ||
pointpen=black;pathpen=black+linewidth(0.7);pen f = fontsize(10); | pointpen=black;pathpen=black+linewidth(0.7);pen f = fontsize(10); | ||
− | pair P=(-8,5),Q=(-15,-19),R=(1,-7),S=(7,-15 | + | 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); | 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(P--Q--R--cycle);D(U);D(P--U); | ||
D((-17,0)--(4,0),Arrows(2mm));D((0,-21)--(0,7),Arrows(2mm)); | D((-17,0)--(4,0),Arrows(2mm));D((0,-21)--(0,7),Arrows(2mm)); | ||
+ | label("$x_1$,(-10,3),W); | ||
</asy></center> | </asy></center> | ||
Revision as of 20:40, 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
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)); label("$x_1$,(-10,3),W); (Error making remote request. Unknown error_msg)
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.