Difference between revisions of "2016 AMC 10B Problems/Problem 19"
Isabelchen (talk | contribs) |
Isabelchen (talk | contribs) |
||
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 altitudes of <math>\triangle AEG</math> and <math>\triangle AFG</math> to <math>AG</math>, which is equal to <math>\frac{EQ}{QF}</math>: | 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 altitudes of <math>\triangle AEG</math> and <math>\triangle AFG</math> to <math>AG</math>, which is equal to <math>\frac{EQ}{QF}</math>: | ||
Line 151: | Line 151: | ||
==Solution 5== | ==Solution 5== | ||
+ | |||
+ | I will calculate <math>\frac{PQ}{QE}</math> using the ratio of the area of <math>\triangle APG</math> and <math>\triangle AEG</math>. | ||
<asy>pair A1=(2,0),A2=(4,4); | <asy>pair A1=(2,0),A2=(4,4); | ||
Line 171: | Line 173: | ||
label("$F$",(2,0),S); label("$G$",(5,1),E); | label("$F$",(2,0),S); label("$G$",(5,1),E); | ||
label("$E$",(4,4),N);</asy> | 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 AEG</math> and <math>\triangle APG</math> share the same base <math>AG</math>, the ratio <math>\frac{[AEG]}{[APG]}</math> is equal to the ratio of the altitudes of <math>\triangle APG</math> and <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}</cmath> | ||
+ | <cmath>\frac{PQ}{PE}=\frac{10}{42+10}=\frac{10}{52}</cmath> | ||
+ | <cmath>\frac{FP}{PE}=\frac{CF}{AE}=\frac{3}{4}</cmath> | ||
+ | <cmath>\frac{FP}{EF}=\frac{3}{7}</cmath> | ||
~isabelchen | ~isabelchen |
Revision as of 11:09, 8 October 2021
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 area of and .
Because and share the same base , the ratio is equal to the ratio of the altitudes of and to , which is equal to :
~isabelchen
Solution 5
I will calculate using the ratio of the area of and .
Because and share the same base , the ratio is equal to the ratio of the altitudes of and to , which is equal to :
~isabelchen
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.