2016 AMC 10A Problems/Problem 19
Problem
In rectangle
and
. Point
between
and
, and point
between
and
are such that
. Segments
and
intersect
at
and
, respectively. The ratio
can be written as
where the greatest common factor of
and
is
What is
?
Solution 1
Since
Similarly,
. This means that
. As
and
are similar, we see that
. Thus
. Therefore,
so
Solution 2
Coordinate Bash:
We can set coordinates for the points. and
. The line
's equation is
, line
's equation is
, and line
's equation is
. Adding the equations of lines
and
, we find that the coordinates of
is
. Furthermore we find that the coordinates
is
. Using the Pythagorean Theorem, the length of
is
, and the length of
=
The length of
. Then
The ratio
Then
and
is
and
, respectively. The problem tells us to find
, so
An alternate solution is to perform the same operations, but only solve for the x-coordinates. By similar triangles, the ratios will be the same.
Solution 3
Extend to meet
at point
. Since
and
,
by similar triangles
and
. It follows that
. Now, using similar triangles
and
,
. WLOG let
. Solving for
gives
and
. So our desired ratio is
and
.
See Also
2016 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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All AMC 10 Problems and Solutions |
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