Difference between revisions of "2016 AMC 10A Problems/Problem 19"
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We set coordinates for the points. Let <math>A=(0,3), B=(6,3), C=(6,0)</math> and <math>D=(0,0)</math>. Then the equation of line <math>AE</math> is <math>y = -\frac{1}{6}x + 3,</math> the equation of line <math>AF</math> is <math>y = -\frac{1}{3}x + 3,</math> and the equation of line <math>BD</math> is <math>y = \frac{1}{2}x</math>. We find that the x-coordinate of point <math>P</math> is <math>\frac 9 2</math> by solving <math> -\frac{1}{6}x + 3=\frac{1}{2}x.</math> Similarly we find that the x-coordinate of point <math>Q</math> is <math>\frac {18} 5</math> by solving <math>-\frac{1}{3}x + 3=\frac{1}{2}x.</math> It follows that <math>BP:PQ:QD=6-\frac 9 2 : \frac 9 2 - \frac {18} 5 : \frac {18} 5= \frac 3 2 : \frac 9 {10} : \frac {18} 5 = 5:3:12.</math> Hence <math>r,s,t=5,3,12</math> and <math>r+s+t=5+3+12=\boxed{\textbf{(E) } 20}.</math> ~ Solution by dolphin7 | We set coordinates for the points. Let <math>A=(0,3), B=(6,3), C=(6,0)</math> and <math>D=(0,0)</math>. Then the equation of line <math>AE</math> is <math>y = -\frac{1}{6}x + 3,</math> the equation of line <math>AF</math> is <math>y = -\frac{1}{3}x + 3,</math> and the equation of line <math>BD</math> is <math>y = \frac{1}{2}x</math>. We find that the x-coordinate of point <math>P</math> is <math>\frac 9 2</math> by solving <math> -\frac{1}{6}x + 3=\frac{1}{2}x.</math> Similarly we find that the x-coordinate of point <math>Q</math> is <math>\frac {18} 5</math> by solving <math>-\frac{1}{3}x + 3=\frac{1}{2}x.</math> It follows that <math>BP:PQ:QD=6-\frac 9 2 : \frac 9 2 - \frac {18} 5 : \frac {18} 5= \frac 3 2 : \frac 9 {10} : \frac {18} 5 = 5:3:12.</math> Hence <math>r,s,t=5,3,12</math> and <math>r+s+t=5+3+12=\boxed{\textbf{(E) } 20}.</math> ~ Solution by dolphin7 | ||
− | == | + | ==Video Solution== |
− | |||
https://www.youtube.com/watch?v=aG9JiBMd0ag | https://www.youtube.com/watch?v=aG9JiBMd0ag | ||
+ | |||
+ | ==Video Solution 2== | ||
+ | https://youtu.be/us3e2jMjWnw | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | ==See Also== | ||
{{AMC10 box|year=2016|ab=A|num-b=18|num-a=20}} | {{AMC10 box|year=2016|ab=A|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 03:50, 19 May 2020
Contents
[hide]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
Use similar triangles. Our goal is to put the ratio in terms of . 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
are
. Furthermore we find that the coordinates of
are
. Using the Pythagorean Theorem, we get that the length of
is
, and the length of
is
The length of
. Then
The ratio
Then
and
is
and
, respectively. The problem tells us to find
, so
~ minor LaTeX edits by dolphin7
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
.
Solution 4 (Mass Points)
Draw line segment , and call the intersection between
and
point
. In
, observe that
and
. Using mass points, find that
. Again utilizing
, observe that
and
. Use mass points to find that
. Now, draw a line segment with points
,
,
, and
ordered from left to right. Set the values
,
,
and
. Setting both sides segment
equal, we get
. Plugging in and solving gives
,
,
. The question asks for
, so we add
to
and multiply the ratio by
to create integers. This creates
. This sums up to
Solution 5 (Easy Coord Bash)
We set coordinates for the points. Let and
. Then the equation of line
is
the equation of line
is
and the equation of line
is
. We find that the x-coordinate of point
is
by solving
Similarly we find that the x-coordinate of point
is
by solving
It follows that
Hence
and
~ Solution by dolphin7
Video Solution
https://www.youtube.com/watch?v=aG9JiBMd0ag
Video Solution 2
~IceMatrix
See Also
2016 AMC 10A (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.