Difference between revisions of "2016 AMC 10B Problems/Problem 19"
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==Solution 4 (Area)== | ==Solution 4 (Area)== | ||
− | I will calculate <math>\frac{EP}{EF}</math> using similar triangle, and <math>\frac{EQ}{EF}</math> using area of <math>\triangle AEG</math> | + | I will calculate <math>\frac{EP}{EF}</math> using similar triangle, and <math>\frac{EQ}{EF}</math> using ratio of area of <math>\triangle AEG</math> to <math>\triangle AFG</math>. |
<asy>pair A1=(2,0),A2=(4,4); | <asy>pair A1=(2,0),A2=(4,4); | ||
Line 139: | Line 139: | ||
<cmath>\triangle AEP \sim \triangle CFP, \frac{AE}{CF}=\frac{EP}{FP}, \frac{EP}{FP}=\frac{4}{3}, \frac{EP}{EF}=\frac{4}{7}</cmath> | <cmath>\triangle AEP \sim \triangle CFP, \frac{AE}{CF}=\frac{EP}{FP}, \frac{EP}{FP}=\frac{4}{3}, \frac{EP}{EF}=\frac{4}{7}</cmath> | ||
− | <cmath>[AEG]= | + | <cmath>[AEG]=\frac{1}{2} \cdot 4\cdot 3=6</cmath> |
<cmath>[AFG]=[ABCD]-[ADF]-[CFG]-[ABG]=20-4-\frac{3}{2}-\frac{15}{2}=7</cmath> | <cmath>[AFG]=[ABCD]-[ADF]-[CFG]-[ABG]=20-4-\frac{3}{2}-\frac{15}{2}=7</cmath> | ||
− | Because <math>\triangle AEG</math> and <math>\triangle AFG</math> share the same base <math>AG</math>, the ratio <math>\frac{[AEG]}{[AFG]}</math> is equal to the ratio of the | + | Because <math>\triangle AEG</math> and <math>\triangle AFG</math> share the same base <math>AG</math>, the ratio <math>\frac{[AEG]}{[AFG]}</math> is equal to the ratio of the altitude of <math>\triangle AEG</math> to <math>AG</math> to that of <math>\triangle AFG</math> to <math>AG</math>, which is equal to <math>\frac{EQ}{QF}</math>: |
<cmath>\frac{[AEG]}{[AFG]}=\frac{EQ}{QF}=\frac{6}{7}</cmath> | <cmath>\frac{[AEG]}{[AFG]}=\frac{EQ}{QF}=\frac{6}{7}</cmath> | ||
<cmath>\frac{EQ}{EF}=\frac{6}{13}</cmath> | <cmath>\frac{EQ}{EF}=\frac{6}{13}</cmath> | ||
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<cmath>\frac{PQ}{EF}=\boxed{\textbf{(D)}~\frac{10}{91}}</cmath> | <cmath>\frac{PQ}{EF}=\boxed{\textbf{(D)}~\frac{10}{91}}</cmath> | ||
− | ~isabelchen | + | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] |
+ | |||
+ | ==Solution 5 (Area)== | ||
+ | |||
+ | I will calculate <math>\frac{PQ}{QE}</math> using the ratio of area of <math>\triangle APG</math> to that of <math>\triangle AEG</math>. | ||
+ | |||
+ | <asy>pair A1=(2,0),A2=(4,4); | ||
+ | pair B1=(0,4),B2=(5,1),B3=(20/7,12/7); | ||
+ | pair C1=(5,0),C2=(0,4); | ||
+ | draw(A1--A2); | ||
+ | draw(B1--B2); | ||
+ | draw(C1--C2); | ||
+ | draw(A2--B2); | ||
+ | draw(B2--B3); | ||
+ | draw((0,0)--B1--(5,4)--C1--cycle); | ||
+ | dot((20/7,12/7)); | ||
+ | dot((3.07692307692,2.15384615384)); | ||
+ | label("$Q$",(3.07692307692,2.15384615384),N); | ||
+ | label("$P$",(20/7,12/7),W); | ||
+ | label("$A$",(0,4), NW); | ||
+ | label("$B$",(5,4), NE); | ||
+ | label("$C$",(5,0),SE); | ||
+ | label("$D$",(0,0),SW); | ||
+ | label("$F$",(2,0),S); label("$G$",(5,1),E); | ||
+ | label("$E$",(4,4),N);</asy> | ||
+ | |||
+ | <cmath>[ACG]=\frac{1}{2} \cdot 5 \cdot 1 = \frac{5}{2}</cmath> | ||
+ | <cmath>[CAB]=\frac{1}{2} \cdot 5 \cdot 4=10</cmath> | ||
+ | <cmath>\triangle AEP \sim \triangle CFP</cmath> | ||
+ | <cmath>\frac{CP}{AP}=\frac{CF}{AE}=\frac{3}{4}</cmath> | ||
+ | <cmath>\frac{CP}{AC}=\frac{3}{7}</cmath> | ||
+ | <cmath>\frac{[CPG]}{[CAB]}=\frac{CP}{CA} \cdot \frac{CG}{CB}=\frac{3 \cdot 1}{7 \cdot 4}=\frac{3}{28}</cmath> | ||
+ | <cmath>[CPG]=\frac{3}{28} \cdot [CAB]=\frac{3}{28} \cdot 10=\frac{15}{14}</cmath> | ||
+ | <cmath>[APG]=[ACG]-[CPG]=\frac{5}{2}-\frac{15}{14}=\frac{35-15}{14}=\frac{20}{14}=\frac{10}{7}</cmath> | ||
+ | <cmath>[AEG]=\frac{1}{2} \cdot 4 \cdot 3=6</cmath> | ||
+ | Because <math>\triangle APG</math> and <math>\triangle AEG</math> share the same base <math>AG</math>, the ratio <math>\frac{[APG]}{[AEG]}</math> is equal to the ratio of altitude of <math>\triangle APG</math> to <math>AG</math> to that of <math>\triangle AEG</math> to <math>AG</math>, which is equal to <math>\frac{PQ}{QE}</math>: | ||
+ | <cmath>\frac{PQ}{QE}=\frac{[APG]}{[AEG]}=\frac{\frac{10}{7}}{6}=\frac{10}{42}=\frac{5}{21}</cmath> | ||
+ | <cmath>\frac{PQ}{PE}=\frac{5}{21+5}=\frac{5}{26}</cmath> | ||
+ | <cmath>\frac{PE}{PF}=\frac{AE}{CF}=\frac{4}{3}</cmath> | ||
+ | <cmath>\frac{PE}{EF}=\frac{4}{7}</cmath> | ||
+ | <cmath>\frac{PQ}{PE} \cdot \frac{PE}{EF} = \frac{5}{26} \cdot \frac{4}{7} = \frac{10}{91}</cmath> | ||
+ | <cmath>\frac{PQ}{EF}=\boxed{\textbf{(D)}~\frac{10}{91}}</cmath> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | ==Solution 6 (Coordinate Bash, not as efficient as Solution 1 but it works)== | ||
+ | We set the points | ||
+ | <math>D(0, 0)</math>, <math>A(0, 4)</math>, <math>E(4, 4)</math>, <math>B(5, 4)</math>, <math>G(5, 1)</math>, <math>C(5, 0)</math>, and <math>F(2, 0)</math>. | ||
+ | The equation of <math>\overline{AC}</math> is <math>y=-\frac{4}{5}x+4</math>, the equation of <math>\overline{AG}</math> is <math>y=-\frac{3}{5}x+4</math>, and the equation of <math>\overline{EF}</math> is <math>y=2x-4</math>. Solving the system of equations for <math>\overline{AC}</math> and <math>\overline{EF}</math> to find point <math>P</math>, <math>y=-\frac{4}{5}x+4=2x-4 \longrightarrow \frac{14}{5}x=8 \longrightarrow x=\frac{20}{7}</math> and <math>y=2x-4=\frac{12}{7}</math>. So the coordinate of point P is <math>P(\frac{20}{7}, \frac{12}{7})</math>. Next find point Q by solving the system of equations for <math>\overline{AG}</math> and <math>\overline{EF}</math> to get <math>Q(\frac{40}{13}, \frac{28}{13})</math>. Using the distance formula, <cmath>PQ=\sqrt{\left(\frac{40}{13}-\frac{20}{7}\right)^{2}+\left(\frac{28}{13}-\frac{12}{7}\right)^{2}}=\sqrt{\left(\frac{20}{91}\right)^{2}+\left(\frac{40}{91}\right)^{2}}</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 | ||
+ | |||
+ | |||
+ | == 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.