Difference between revisions of "2016 AMC 10B Problems/Problem 19"
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<cmath>=\sqrt{\frac{400}{8281}+\frac{1600}{8281}}=\sqrt{\frac{2000}{8281}}=\frac{20\sqrt{5}}{91}</cmath> Also using the distance formula, <cmath>EF=\sqrt{\left(4-2\right)^{2}+\left(4-0\right)^{2}}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}</cmath> Finally, <cmath>\frac{PQ}{EF}=\frac{\frac{20\sqrt{5}}{91}}{2\sqrt{5}}=\frac{10}{91} \Longrightarrow \boxed{\textbf{(D)}~\frac{10}{91}}</cmath> | <cmath>=\sqrt{\frac{400}{8281}+\frac{1600}{8281}}=\sqrt{\frac{2000}{8281}}=\frac{20\sqrt{5}}{91}</cmath> Also using the distance formula, <cmath>EF=\sqrt{\left(4-2\right)^{2}+\left(4-0\right)^{2}}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}</cmath> Finally, <cmath>\frac{PQ}{EF}=\frac{\frac{20\sqrt{5}}{91}}{2\sqrt{5}}=\frac{10}{91} \Longrightarrow \boxed{\textbf{(D)}~\frac{10}{91}}</cmath> | ||
~JH. L | ~JH. L | ||
+ | |||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/GrCtzL0S-Uo?t=911 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==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}} |
Latest revision as of 03:18, 23 January 2023
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
.
Solution 4 (Area)
I will calculate using similar triangle, and
using ratio of area of
to
.
Because
and
share the same base
, the ratio
is equal to the ratio of the altitude of
to
to that of
to
, which is equal to
:
Solution 5 (Area)
I will calculate using the ratio of area of
to that of
.
Because
and
share the same base
, the ratio
is equal to the ratio of altitude of
to
to that of
to
, which is equal to
:
Solution 6 (Coordinate Bash, not as efficient as Solution 1 but it works)
We set the points
,
,
,
,
,
, and
.
The equation of
is
, the equation of
is
, and the equation of
is
. Solving the system of equations for
and
to find point
,
and
. So the coordinate of point P is
. Next find point Q by solving the system of equations for
and
to get
. Using the distance formula,
Also using the distance formula,
Finally,
~JH. L
Video Solution by OmegaLearn
https://youtu.be/GrCtzL0S-Uo?t=911
~ pi_is_3.14
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.