Difference between revisions of "2001 AIME II Problems/Problem 4"
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Latest revision as of 19:34, 4 July 2013
Problem
Let 
. The lines whose equations are 
and 
contain points 
and 
, respectively, such that 
is the midpoint of 
. The length of 
equals 
, where 
and 
are relatively prime positive integers. Find 
.
Solution
![[asy] pointpen = black; pathpen = black+linewidth(0.7); pair R = (8,6), P = (32,60)/7, Q= (80,24)/7; D((0,0)--MP("x",(13,0),E),EndArrow(6)); D((0,0)--MP("y",(0,10),N),EndArrow(6)); D((0,0)--(10/(15/8),10),EndArrow(6)); D((0,0)--(13,13 * 3/10),EndArrow(6)); D(D(MP("P",P,NW))--D(MP("Q",Q),SE),linetype("4 4")); D(MP("R",R,NE)); [/asy]](http://latex.artofproblemsolving.com/9/7/f/97f7b283e5b032415f83e3964b09dce0c599a9bc.png)
The coordinates of can be written as 
and the coordinates of point can be written as 
. By the midpoint formula, we have 
and 
. Solving for 
gives 
, so the point is 
. The answer is twice the distance from to 
, which by the distance formula is 
. Thus, the answer is 
.
See also
| 2001 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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. 
