Difference between revisions of "2016 AMC 10B Problems/Problem 19"
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Then, notice that <math>\triangle{AEQ}\sim\triangle{HFQ}</math> and <math>\triangle{AEP}\sim\triangle{CFP}</math>. Thus, we see that <cmath>\dfrac{AE}{HF}=\dfrac{EQ}{QF}\implies \dfrac{EQ}{QF} = \dfrac{4}{3+\frac{5}{3}} = \dfrac{12}{14}=\dfrac{6}{7}\implies \dfrac{EQ}{EF}=\dfrac{6}{13}</cmath> | Then, notice that <math>\triangle{AEQ}\sim\triangle{HFQ}</math> and <math>\triangle{AEP}\sim\triangle{CFP}</math>. Thus, we see that <cmath>\dfrac{AE}{HF}=\dfrac{EQ}{QF}\implies \dfrac{EQ}{QF} = \dfrac{4}{3+\frac{5}{3}} = \dfrac{12}{14}=\dfrac{6}{7}\implies \dfrac{EQ}{EF}=\dfrac{6}{13}</cmath> | ||
and <cmath>\dfrac{AE}{CF}=\dfrac{EP}{FP} \implies \dfrac{4}{3}=\dfrac{EP}{FP}\implies \dfrac{FP}{FE} = \dfrac{3}{7}.</cmath> | and <cmath>\dfrac{AE}{CF}=\dfrac{EP}{FP} \implies \dfrac{4}{3}=\dfrac{EP}{FP}\implies \dfrac{FP}{FE} = \dfrac{3}{7}.</cmath> | ||
− | Thus, we see that <cmath>\dfrac{PQ}{EF} = 1-\left(\dfrac{6}{13}+\dfrac{3}{7}\right) = 1-\left(\dfrac{42+39}{91}\right) = 1-\left(\dfrac{81}{91}\right) = \boxed{\dfrac{10}{91}}.</cmath> | + | Thus, we see that <cmath>\dfrac{PQ}{EF} = 1-\left(\dfrac{6}{13}+\dfrac{3}{7}\right) = 1-\left(\dfrac{42+39}{91}\right) = 1-\left(\dfrac{81}{91}\right) = \boxed{\textbf{(D)} \dfrac{10}{91}}.</cmath> |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=B|num-b=18|num-a=20}} | {{AMC10 box|year=2016|ab=B|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:48, 21 February 2016
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 (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
.
Solution 2 (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 3 (Similar Triangles)
Extend to intersect
at
. Letting
, we have that
Then, notice that and
. Thus, we see that
and
Thus, we see that
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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