Difference between revisions of "2016 AMC 10A Problems/Problem 19"
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Use similar triangles. Our goal is to put the ratio in terms of <math>{BD}</math>. Since <math>\triangle APD \sim \triangle EPB,</math> <math>\frac{DP}{PB}=\frac{AD}{BE}=3.</math> Therefore, <math>PB=\frac{BD}{4}</math>. Similarly, <math>\frac{DQ}{QB}=\frac{3}{2}</math>. This means that <math>{DQ}=\frac{3\cdot BD}{5}</math>. Therefore, <math>r:s:t=\frac{1}{4}:\frac{2}{5}-\frac{1}{4}:\frac{3}{5}=5:3:12,</math> so <math>r+s+t=\boxed{\textbf{(E) }20.}</math> | Use similar triangles. Our goal is to put the ratio in terms of <math>{BD}</math>. Since <math>\triangle APD \sim \triangle EPB,</math> <math>\frac{DP}{PB}=\frac{AD}{BE}=3.</math> Therefore, <math>PB=\frac{BD}{4}</math>. Similarly, <math>\frac{DQ}{QB}=\frac{3}{2}</math>. This means that <math>{DQ}=\frac{3\cdot BD}{5}</math>. Therefore, <math>r:s:t=\frac{1}{4}:\frac{2}{5}-\frac{1}{4}:\frac{3}{5}=5:3:12,</math> so <math>r+s+t=\boxed{\textbf{(E) }20.}</math> | ||
− | ==Solution 2 (Mass points and Similar Triangles)== | + | ==Solution 2 (Mass points and Similar Triangles - Easy)== |
+ | <asy> | ||
+ | size(9cm); | ||
+ | pair D=(0,0), C=(6,0), B=(6,3), A=(0,3), G=(2, 1), H=(9, 0); | ||
+ | draw(A--B--C--D--cycle); | ||
+ | draw(B--D); | ||
+ | draw(A--(6,2)); | ||
+ | draw(A--(6,1)); | ||
+ | draw(A--H); | ||
+ | draw((6,1)--G); | ||
+ | draw(D--H); | ||
+ | label("$A$", A, dir(135)); | ||
+ | label("$B$", B, dir(45)); | ||
+ | label("$C$", C, dir(-45)); | ||
+ | label("$D$", D, dir(-135)); | ||
+ | label("$Q$", extension(A,(6,1),B,D),dir(-90)); | ||
+ | label("$P$", extension(A,(6,2),B,D), dir(90)); | ||
+ | label("$F$", (6,1), dir(45)); | ||
+ | label("$E$", (6,2), dir(45)); | ||
+ | label("$H$", H, dir(0)); | ||
+ | label("$G$", G, dir(135)); | ||
+ | </asy> | ||
+ | |||
This problem breaks down into finding <math>QP:PB</math> and <math>DQ:QB</math>. We can find the first using Mass Points, and the second using similar triangles. | This problem breaks down into finding <math>QP:PB</math> and <math>DQ:QB</math>. We can find the first using Mass Points, and the second using similar triangles. | ||
Revision as of 15:47, 30 December 2021
Contents
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 (Similar Triangles)
Use similar triangles. Our goal is to put the ratio in terms of . Since
Therefore,
. Similarly,
. This means that
. Therefore,
so
Solution 2 (Mass points and Similar Triangles - Easy)
This problem breaks down into finding and
. We can find the first using Mass Points, and the second using similar triangles.
Draw point on
such that
. Then, by similar triangles
. Again, by similar triangles
and
,
. Now we begin Mass Points.
We will consider the triangle with center
, so that
balances
and
, and
balances
and
. Assign a mass of
to
. Then,
so
. By mass points addition,
since
balances
and
.
Also,
so
so
. Then,
.
To calculate , extend
past
to point
such that
lies on
. Then
is similar to
so
. Also,
is similar to
so
Now, we wish to get . Observe thaat
. So,
so (since
has sum
),
. now, we may combine the two and get
so
.
~Firebolt360
Solution 3(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 4
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 5 (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 6 (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
Video Solution 3
https://youtu.be/4_x1sgcQCp4?t=3406
~ pi_is_3.14
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.