Difference between revisions of "2016 AMC 10A Problems/Problem 19"
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== Problem == | == Problem == | ||
− | In rectangle <math> | + | In rectangle <math>ABCD,</math> <math>AB=6</math> and <math>BC=3</math>. Point <math>E</math> between <math>B</math> and <math>C</math>, and point <math>F</math> between <math>E</math> and <math>C</math> are such that <math>BE=EF=FC</math>. Segments <math>\overline{AE}</math> and <math>\overline{AF}</math> intersect <math>\overline{BD}</math> at <math>P</math> and <math>Q</math>, respectively. The ratio <math>BP:PQ:QD</math> can be written as <math>r:s:t</math> where the greatest common factor of <math>r,s,</math> and <math>t</math> is <math>1.</math> What is <math>r+s+t</math>? |
<math>\textbf{(A) } 7 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 20</math> | <math>\textbf{(A) } 7 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 20</math> |
Revision as of 21:39, 3 April 2021
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 (Similar Triangles)
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
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.