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 |