Difference between revisions of "2016 AMC 10A Problems/Problem 19"
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<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> | ||
− | == Solution 1== | + | == Solution 1 (Similar Triangles)== |
<asy> | <asy> | ||
Line 24: | Line 24: | ||
</asy> | </asy> | ||
− | Since <math>\triangle APD \sim \triangle EPB,</math> <math>\frac{DP}{PB}=\frac{AD}{BE}=3.</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== | + | ==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. | |
− | |||
− | ==Solution 3== | + | Draw point <math>G</math> on <math>DB</math> such that <math>FG\parallel CD</math>. Then, by similar triangles <math>FG=BF\cdot 2=4</math>. Again, by similar triangles <math>AQB</math> and <math>FQG</math>, <math>AQ:FQ=AB:FG=6:4=3:2</math>. Now we begin Mass Points. |
+ | |||
+ | We will consider the triangle <math>ABF</math> with center <math>P</math>, so that <math>E</math> balances <math>B</math> and <math>F</math>, and <math>Q</math> balances <math>A</math> and <math>F</math>. Assign a mass of <math>1</math> to <math>B</math>. Then, <math>BE:FE=1:1</math> so <math>F=B=1</math>. By mass points addition, <math>E=B+F=2</math> since <math>E</math> balances <math>B</math> and <math>F</math>. | ||
+ | Also, <math>AQ:QF=3:2</math> so <math>A=\frac{2}{3}F=\frac{2}{3}</math> so <math>Q=\frac{5}{3}</math>. Then, <math>QP:PB=1:\frac{5}{3}=3:5</math>. | ||
+ | |||
+ | To calculate <math>DQ:BQ</math>, extend <math>AF</math> past <math>F</math> to point <math>H</math> such that <math>H</math> lies on <math>BC</math>. Then <math>AFB</math> is similar to <math>HAD</math> so <math>DH=3\cdot AD=9</math>. Also, <math>AQB</math> is similar to <math>HQD</math> so <math>DQ:BQ=9:6=3:2</math> | ||
+ | |||
+ | Now, we wish to get <math>DQ:QP:PB</math>. Observe that <math>BQ=QP+PB</math>. So, <math>DQ:BQ=DQ:QP+PB=3:2</math> so (since <math>QP:PB=3:5</math> has sum <math>3+5=8</math>), <math>DQ:QP+PB=3:2=12:8</math>. now, we may combine the two and get <math>DQ:QP:PB=12:3:5</math> so <math>r+s+t=12+3+5=\boxed{\textbf{(E) }20}</math>. | ||
+ | |||
+ | ~Firebolt360(minor edits by vadava_lx) | ||
+ | |||
+ | ==Solution 3(Coordinate Bash)== | ||
+ | |||
+ | We can set coordinates for the points. <math>D=(0,0), C=(6,0), B=(6,3),</math> and <math>A=(0,3)</math>. The line <math>BD</math>'s equation is <math>y = \frac{1}{2}x</math>, line <math>AE</math>'s equation is <math>y = -\frac{1}{6}x + 3</math>, and line <math>AF</math>'s equation is <math>y = -\frac{1}{3}x + 3</math>. Adding the equations of lines <math>BD</math> and <math>AE</math>, we find that the coordinates of <math>P</math> are <math>\left(\frac{9}{2},\frac{9}{4}\right)</math>. Furthermore we find that the coordinates of <math>Q</math> are <math>\left(\frac{18}{5}, \frac{9}{5}\right)</math>. Using the [[Pythagorean Theorem]], we get that the length of <math>QD</math> is <math>\sqrt{\left(\frac{18}{5}\right)^2+\left(\frac{9}{5}\right)^2} = \sqrt{\frac{405}{25}} = \frac{\sqrt{405}}{5} = \frac{9\sqrt{5}}{5}</math>, and the length of <math>DP</math> is <math>\sqrt{\left(\frac{9}{2}\right)^2+\left(\frac{9}{4}\right)^2} = \sqrt{\frac{81}{4} + \frac{81}{16}} = \sqrt{\frac{405}{16}} = \frac{\sqrt{405}}{4} = \frac{9\sqrt{5}}{4}.</math> <math>PQ = DP - DQ = \frac{9\sqrt{5}}{4} - \frac{9\sqrt{5}}{5} = \frac{9\sqrt{5}}{20}.</math> The length of <math>DB = \sqrt{6^2 + 3^2} = \sqrt{45} = 3\sqrt{5}</math>. Then <math>BP= 3\sqrt{5} - \frac{9\sqrt{5}}{4} = \frac{3\sqrt{5}}{4}.</math> The ratio <math>BP : PQ : QD = \frac{3\sqrt{5}}{4} : \frac{9\sqrt{5}}{20} : \frac{9\sqrt{5}}{5} = 15\sqrt{5} : 9\sqrt{5} : 36\sqrt{5} = 15 : 9 : 36 = 5 : 3 : 12.</math> Then <math>r, s,</math> and <math>t</math> is <math>5, 3,</math> and <math>12</math>, respectively. The problem tells us to find <math>r + s + t</math>, so <math>5 + 3 + 12 = \boxed{\textbf{(E) }20}</math> ~ minor LaTeX edits by dolphin7 | ||
+ | |||
+ | ==Solution 4== | ||
Extend <math>AF</math> to meet <math>CD</math> at point <math>T</math>. Since <math>FC=1</math> and <math>BF=2</math>, <math>TC=3</math> by similar triangles <math>\triangle TFC</math> and <math>\triangle AFB</math>. It follows that <math>\frac{BQ}{QD}=\frac{BP+PQ}{QD}=\frac{2}{3}</math>. Now, using similar triangles <math>\triangle BEP</math> and <math>\triangle DAP</math>, <math>\frac{BP}{PD}=\frac{BP}{PQ+QD}=\frac{1}{3}</math>. WLOG let <math>BP=1</math>. Solving for <math>PQ, QD</math> gives <math>PQ=\frac{3}{5}</math> and <math>QD=\frac{12}{5}</math>. So our desired ratio is <math>5:3:12</math> and <math>5+3+12=\boxed{\textbf{(E) } 20}</math>. | Extend <math>AF</math> to meet <math>CD</math> at point <math>T</math>. Since <math>FC=1</math> and <math>BF=2</math>, <math>TC=3</math> by similar triangles <math>\triangle TFC</math> and <math>\triangle AFB</math>. It follows that <math>\frac{BQ}{QD}=\frac{BP+PQ}{QD}=\frac{2}{3}</math>. Now, using similar triangles <math>\triangle BEP</math> and <math>\triangle DAP</math>, <math>\frac{BP}{PD}=\frac{BP}{PQ+QD}=\frac{1}{3}</math>. WLOG let <math>BP=1</math>. Solving for <math>PQ, QD</math> gives <math>PQ=\frac{3}{5}</math> and <math>QD=\frac{12}{5}</math>. So our desired ratio is <math>5:3:12</math> and <math>5+3+12=\boxed{\textbf{(E) } 20}</math>. | ||
+ | |||
+ | ==Solution 5 (Mass Points)== | ||
+ | |||
+ | Draw line segment <math>AC</math>, and call the intersection between <math>AC</math> and <math>BD</math> point <math>K</math>. In <math>\delta ABC</math>, observe that <math>BE:EC=1:2</math> and <math>AK:KC=1:1</math>. Using mass points, find that <math>BP:PK=1:1</math>. Again utilizing <math>\delta ABC</math>, observe that <math>BF:FC=2:1</math> and <math>AK:KC=1:1</math>. Use mass points to find that <math>BQ:QK=4:1</math>. Now, draw a line segment with points <math>B</math>,<math>P</math>,<math>Q</math>, and <math>K</math> ordered from left to right. Set the values <math>BP=x</math>,<math>PK=x</math>,<math>BQ=4y</math> and <math>QK=y</math>. Setting both sides segment <math>BK</math> equal, we get <math>y= \frac{2}{5}x</math>. Plugging in and solving gives <math>QK= \frac{2}{5}x</math>, <math>PQ=\frac{3}{5}x</math>,<math>BP=x</math>. The question asks for <math>BP:PQ:QD</math>, so we add <math>2x</math> to <math>QK</math> and multiply the ratio by <math>5</math> to create integers. This creates <math>5(1:\frac{3}{5}:\frac{12}{5})= 5:3:12</math>. This sums up to <math>3+5+12=\boxed{\textbf{(E) }20}</math> | ||
+ | |||
+ | ==Solution 6 (Easy Coord Bash)== | ||
+ | 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 | ||
+ | |||
+ | ==Video Solution 2== | ||
+ | https://youtu.be/us3e2jMjWnw | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | == Video Solution 3 by OmegaLearn == | ||
+ | https://youtu.be/4_x1sgcQCp4?t=3406 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==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}} |
Latest revision as of 15:10, 8 October 2023
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 that
. So,
so (since
has sum
),
. now, we may combine the two and get
so
.
~Firebolt360(minor edits by vadava_lx)
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 by OmegaLearn
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.