2009 AIME I Problems/Problem 4
Problem 4
In parallelogram 
, point 
is on 
so that 
and point 
is on 
so that 
. Let 
be the point of intersection of 
and 
. Find 
.
Solution
Solution 1
One of the ways to solve this problem is to make this parallelogram a straight line.
So the whole length of the line is 
(
or 
), and 
is 
(
or 
) is 
So the answer is 
Solution 2
Draw a diagram with all the given points and lines involved. Construct parallel lines 
and 
to , where for the lines the endpoints are on 
and 
, respectively, and each point refers to an intersection. Also, draw the median of quadrilateral 
where the points are in order from top to bottom. Clearly, by similar triangles, 
and 
. It is not difficult to see that 
is the center of quadrilateral and thus the midpoint of as well as the midpoint of 
(all of this is easily proven with symmetry). From more triangle similarity, 
.
Solution 3
Using vectors, note that 
and 
. Note that 
for some positive x and y, but at the same time is a scalar multiple of 
. So, writing the equation in terms of 
and 
, we have 
. But the coefficients of the two vectors must be equal because, as already stated, 
is a scalar multiple of . We then see that 
and 
. Finally, we have 
and, simplifying, $\overrightarrow{AB}+\overrightarrow{AD}}=177\overrightarrow{AP}$ (Error compiling LaTeX. Unknown error_msg) and the desired quantity is 
.
See also
| 2009 AIME I (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. 
