Difference between revisions of "2016 AMC 10B Problems/Problem 19"
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==Solution 1 (Coordinate Geometry)== | ==Solution 1 (Coordinate Geometry)== |
Revision as of 16:40, 24 December 2020
Contents
Problem
Rectangle has
and
. Point
lies on
so that
, point
lies on
so that
, and point
lies on
so that
. Segments
and
intersect
at
and
, respectively. What is the value of
?
Solution 1 (Coordinate Geometry)
First, we will define point as the origin. Then, we will find the equations of the following three lines:
,
, and
. The slopes of these lines are
,
, and
, respectively. Next, we will find the equations of
,
, and
. They are as follows:
After drawing in altitudes to
from
,
, and
, we see that
because of similar triangles, and so we only need to find the x-coordinates of
and
.
Finding the intersections of
and
, and
and
gives the x-coordinates of
and
to be
and
. This means that
. Now we can find
Solution 2 (Similar Triangles)
Extend to intersect
at
. Letting
, we have that
Then, notice that and
. Thus, we see that
and
Thus, we see that
Solution 3 (Answer Choices)
Since the opposite sides of a rectangle are parallel and
due to vertical angles,
. Furthermore, the ratio between the side lengths of the two triangles is
. Labeling
and
, we see that
turns out to be equal to
. Since the denominator of
must now be a multiple of 7, the only possible solution in the answer choices is
.
See Also
2016 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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