1986 AIME Problems/Problem 9
Problem
In ,
,
, and
. An interior point
is then drawn, and segments are drawn through
parallel to the sides of the triangle. If these three segments are of an equal length
, find
.
Contents
[hide]Solution
Solution 1
Let the points at which the segments hit the triangle be called as shown above. All three smaller triangles and the larger triangle are similar (
). This is easy to find using repeated alternate interior angles. The remaining three sections are parallelograms, which is also simple to see by the parallel lines.
Since is a parallelogram, we find
, and similarly
. So
. Thus
. By the same logic,
.
Since , we have the proportion:


Doing the same with , we find that
. Now,
.
Solution 2
Define the points the same as above.
Let ,
,
,
,
and
Key theorem: the ratio of the areas of 2 similar triangles is the ratio of a pair of corresponding sides squared.
Let the length of the segment be and the area of the triangle be
, using the theorem, we get:
,
,
adding all these together and using
we get
Using corresponding angles from parallel lines, it is easy to show that , since
and
are parallelograms, it is easy to show that
Now we have the side length ratio, so we have the area ratio
, by symmetry, we have
and
Substituting these into our initial equation, we have
answer follows after some hideous computation
See also
1986 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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