Difference between revisions of "2016 AMC 10B Problems/Problem 19"
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==Problem== | ==Problem== | ||
− | Rectangle <math>ABCD</math> has <math>AB=5</math> and <math>BC=4</math>. Point <math>E</math> lies on <math>\overline{AB}</math> so that <math>EB=1</math>, point <math>G</math> lies on <math>\overline{BC}</math> so that <math>CG=1</math> | + | Rectangle <math>ABCD</math> has <math>AB=5</math> and <math>BC=4</math>. Point <math>E</math> lies on <math>\overline{AB}</math> so that <math>EB=1</math>, point <math>G</math> lies on <math>\overline{BC}</math> so that <math>CG=1</math>, and point <math>F</math> lies on <math>\overline{CD}</math> so that <math>DF=2</math>. Segments <math>\overline{AG}</math> and <math>\overline{AC}</math> intersect <math>\overline{EF}</math> at <math>Q</math> and <math>P</math>, respectively. What is the value of <math>\frac{PQ}{EF}</math>? |
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− | |||
− | ==Solution | + | |
+ | ==Solution 1 (Coordinate Geometry)== | ||
First, we will define point <math>D</math> as the origin. Then, we will find the equations of the following three lines: <math>AG</math>, <math>AC</math>, and <math>EF</math>. The slopes of these lines are <math>-\frac{3}{5}</math>, <math>-\frac{4}{5}</math>, and <math>2</math>, respectively. Next, we will find the equations of <math>AG</math>, <math>AC</math>, and <math>EF</math>. They are as follows: | First, we will define point <math>D</math> as the origin. Then, we will find the equations of the following three lines: <math>AG</math>, <math>AC</math>, and <math>EF</math>. The slopes of these lines are <math>-\frac{3}{5}</math>, <math>-\frac{4}{5}</math>, and <math>2</math>, respectively. Next, we will find the equations of <math>AG</math>, <math>AC</math>, and <math>EF</math>. They are as follows: | ||
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dot(C1); dot(C2); | dot(C1); dot(C2); | ||
dot((0,0)); dot((5,4));</asy> | dot((0,0)); dot((5,4));</asy> | ||
− | Finding the intersections of <math>AC</math> and <math>EF</math>, and <math>AG</math> and <math>EF</math> gives the x-coordinates of <math>P</math> and <math>Q</math> to be <math>\frac{20}{7}</math> and <math>\frac{40}{13}</math>. This means that <math>P'Q' = DQ' - DP' = \frac{ | + | Finding the intersections of <math>AC</math> and <math>EF</math>, and <math>AG</math> and <math>EF</math> gives the x-coordinates of <math>P</math> and <math>Q</math> to be <math>\frac{20}{7}</math> and <math>\frac{40}{13}</math>. This means that <math>P'Q' = DQ' - DP' = \frac{40}{13} - \frac{20}{7} = \frac{20}{91}</math>. Now we can find <math>\frac{PQ}{EF} = \frac{P'Q'}{E'F} = \frac{\frac{20}{91}}{2} = \boxed{\textbf{(D)}~\frac{10}{91}}</math> |
+ | |||
+ | ==Solution 2 (Similar Triangles)== | ||
+ | <asy> | ||
+ | pair A1=(2,0),A2=(4,4); | ||
+ | pair B1=(0,4),B2=(5,1); | ||
+ | pair C1=(5,0),C2=(0,4); | ||
+ | pair H = (20/3,0); | ||
+ | draw(A1--A2); | ||
+ | draw(B1--B2); | ||
+ | draw(C1--C2); | ||
+ | draw(B1--H); | ||
+ | draw((0,0)--H); | ||
+ | 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); | ||
+ | label("$H$",H,E); | ||
+ | |||
+ | |||
+ | </asy> | ||
+ | |||
+ | Extend <math>AG</math> to intersect <math>CD</math> at <math>H</math>. Letting <math>x=\overline{HC}</math>, we have that <cmath>\triangle{HCG}\sim\triangle{HDA}\implies \dfrac{\overline{HC}}{\overline{CG}}=\dfrac{\overline{HD}}{\overline{AD}}\implies \dfrac{x}{1}=\dfrac{x+5}{4}\implies x=\dfrac{5}{3}.</cmath> | ||
+ | |||
+ | Then, notice that <math>\triangle{AEQ}\sim\triangle{HFQ}</math> and <math>\triangle{AEP}\sim\triangle{CFP}</math>. Thus, we see that <cmath>\dfrac{AE}{HF}=\dfrac{EQ}{QF}\implies \dfrac{AE}{HF} = \dfrac{4}{3+\frac{5}{3}} = \dfrac{12}{14}=\dfrac{6}{7}\implies \dfrac{EQ}{EF}=\dfrac{6}{13}</cmath> | ||
+ | and <cmath>\dfrac{AE}{CF}=\dfrac{EP}{FP} \implies \dfrac{4}{3}=\dfrac{EP}{FP}\implies \dfrac{FP}{FE} = \dfrac{3}{7}.</cmath> | ||
+ | Thus, we see that <cmath>\dfrac{PQ}{EF} = 1-\left(\dfrac{6}{13}+\dfrac{3}{7}\right) = 1-\left(\dfrac{42+39}{91}\right) = 1-\left(\dfrac{81}{91}\right) = \boxed{\textbf{(D)}~ \dfrac{10}{91}}.</cmath> | ||
+ | |||
+ | ==Solution 3 (Answer Choices)== | ||
+ | |||
+ | Since the opposite sides of a rectangle are parallel and <math>\angle{APE}</math> <math>=</math> <math>\angle{CPF}</math> due to vertical angles, <math>\triangle{APE}</math> <math>\sim</math> <math>\triangle{CPF}</math>. Furthermore, the ratio between the side lengths of the two triangles is <math>\frac{AE}{FC}</math> <math>=</math> <math>\frac{4}{3}</math>. Labeling <math>EP</math> <math>=</math> <math>4x</math> and <math>FP</math> <math>=</math> <math>3x</math>, we see that <math>EF</math> turns out to be equal to <math>7x</math>. Since the denominator of <math>\frac{PQ}{EF}</math> must now be a multiple of 7, the only possible solution in the answer choices is <math>\boxed{\textbf{(D)}~\frac{10}{91}}</math>. | ||
==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}} |
Revision as of 16:40, 24 December 2020
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 .
See Also
2016 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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