Difference between revisions of "2018 AMC 10A Problems/Problem 24"
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+ | {{duplicate|[[2018 AMC 10A Problems/Problem 24|2018 AMC 10A #24]] and [[2018 AMC 12A Problems/Problem 18|2018 AMC 12A #18]]}} | ||
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== Problem == | == Problem == | ||
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== Solution 1 == | == Solution 1 == | ||
− | Let <math>BC = a</math>, <math>BG = x</math>, <math>GC = y</math>, and the length of the perpendicular to <math>BC</math> through <math>A</math> be <math>h</math>. By angle bisector theorem, we have that <cmath>\frac{50}{x} = \frac{10}{y},</cmath> where <math>y = -x+a</math>. Therefore substituting we have that <math>BG=\frac{5a}{6}</math>. By similar triangles, we have that <math>DF=\frac{5a}{12}</math>, and the height of this trapezoid is <math>\frac{h}{2}</math>. Then, we have that <math>\frac{ah}{2}=120</math>. We wish to compute <math>\frac{5a}{8}\cdot\frac{h}{2}</math>, and we have that it is <math>\boxed{75}</math> by substituting. | + | Let <math>BC = a</math>, <math>BG = x</math>, <math>GC = y</math>, and the length of the perpendicular to <math>BC</math> through <math>A</math> be <math>h</math>. By angle bisector theorem, we have that <cmath>\frac{50}{x} = \frac{10}{y},</cmath> where <math>y = -x+a</math>. Therefore substituting we have that <math>BG=\frac{5a}{6}</math>. By similar triangles, we have that <math>DF=\frac{5a}{12}</math>, and the height of this trapezoid is <math>\frac{h}{2}</math>. Then, we have that <math>\frac{ah}{2}=120</math>. We wish to compute <math>\frac{5a}{8}\cdot\frac{h}{2}</math>, and we have that it is <math>\boxed{75}</math> by substituting. |
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== Solution 2 == | == Solution 2 == | ||
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== Solution 3 == | == Solution 3 == | ||
− | The area of <math>\bigtriangleup ABG</math> to the area of <math>\bigtriangleup ACG</math> is <math>5:1</math> | + | The ratio of the <math>\overline{BG}</math> to <math>\overline{GC}</math> is <math>5:1</math> by the Angle Bisector Theorem, so area of <math>\bigtriangleup ABG</math> to the area of <math>\bigtriangleup ACG</math> is also <math>5:1</math> (They have the same height). Therefore, the area of <math>\bigtriangleup ABG</math> is <math>\frac{5}{5+1}\times120=100</math>. Since <math>\overline{DE}</math> is the midsegment of <math>\bigtriangleup ABC</math>, so <math>\overline{DF}</math> is the midsegment of <math>\bigtriangleup ABG</math> . Thus, the ratio of the area of <math>\bigtriangleup ADF</math> to the area of <math>\bigtriangleup ABG</math> is <math>1:4</math>, so the area of <math>\bigtriangleup ACG</math> is <math>\frac{1}{4}\times100=25</math>. Therefore, the area of quadrilateral <math>FDBG</math> is <math>[ABG]-[ADF]=100-25=\boxed{75}</math> |
− | |||
==Solution 4 == | ==Solution 4 == | ||
The area of quadrilateral <math>FDBG</math> is the area of <math>\bigtriangleup ABG</math> minus the area of <math>\bigtriangleup ADF</math>. Notice, <math>\overline{DE} || \overline{BC}</math>, so <math>\bigtriangleup ABG \sim \bigtriangleup ADF</math>, and since <math>\overline{AD}:\overline{AB}=1:2</math>, the area of <math>\bigtriangleup ADF:\bigtriangleup ABG=(1:2)^2=1:4</math>. Given that the area of <math>\bigtriangleup ABC</math> is <math>120</math>, using <math>\frac{bh}{2}</math> on side <math>AB</math> yields <math>\frac{50h}{2}=120\implies h=\frac{240}{50}=\frac{24}{5}</math>. Using the Angle Bisector Theorem, <math>\overline{BG}:\overline{BC}=50:(10+50)=5:6</math>, so the height of <math>\bigtriangleup ABG: \bigtriangleup ACB=5:6</math>. Therefore our answer is <math>\big[ FDBG\big] = \big[ABG\big]-\big[ ADF\big] = \big[ ABG\big]\big(1-\frac{1}{4}\big)=\frac{3}{4}\cdot \frac{bh}{2}=\frac{3}{8}\cdot 50\cdot \frac{5}{6}\cdot \frac{24}{5}=\frac{3}{8}\cdot 200=\boxed{75}</math> | The area of quadrilateral <math>FDBG</math> is the area of <math>\bigtriangleup ABG</math> minus the area of <math>\bigtriangleup ADF</math>. Notice, <math>\overline{DE} || \overline{BC}</math>, so <math>\bigtriangleup ABG \sim \bigtriangleup ADF</math>, and since <math>\overline{AD}:\overline{AB}=1:2</math>, the area of <math>\bigtriangleup ADF:\bigtriangleup ABG=(1:2)^2=1:4</math>. Given that the area of <math>\bigtriangleup ABC</math> is <math>120</math>, using <math>\frac{bh}{2}</math> on side <math>AB</math> yields <math>\frac{50h}{2}=120\implies h=\frac{240}{50}=\frac{24}{5}</math>. Using the Angle Bisector Theorem, <math>\overline{BG}:\overline{BC}=50:(10+50)=5:6</math>, so the height of <math>\bigtriangleup ABG: \bigtriangleup ACB=5:6</math>. Therefore our answer is <math>\big[ FDBG\big] = \big[ABG\big]-\big[ ADF\big] = \big[ ABG\big]\big(1-\frac{1}{4}\big)=\frac{3}{4}\cdot \frac{bh}{2}=\frac{3}{8}\cdot 50\cdot \frac{5}{6}\cdot \frac{24}{5}=\frac{3}{8}\cdot 200=\boxed{75}</math> | ||
− | -Solution by | + | |
+ | ==Solution 5: Trig == | ||
+ | We try to find the area of quadrilateral <math>FDBG</math> by subtracting the area outside the quadrilateral but inside triangle <math>ABC</math>. Note that the area of <math>\triangle ADE</math> is equal to <math>\frac{1}{2} \cdot 25 \cdot 5 \cdot \sin{A}</math> and the area of triangle <math>ABC</math> is equal to <math>\frac{1}{2} \cdot 50 \cdot 10 \cdot \sin A</math>. The ratio <math>\frac{[ADE]}{[ABC]}</math> is thus equal to <math>\frac{1}{4}</math> and the area of triangle <math>ADE</math> is <math>\frac{1}{4} \cdot 120 = 30</math>. Let side <math>BC</math> be equal to <math>6x</math>, then <math>BG = 5x, GC = x</math> by the angle bisector theorem. Similarly, we find the area of triangle <math>AGC</math> to be <math>\frac{1}{2} \cdot 10 \cdot x \cdot \sin C</math> and the area of triangle <math>ABC</math> to be <math>\frac{1}{2} \cdot 6x \cdot 10 \cdot \sin C</math>. A ratio between these two triangles yields <math>\frac{[ACG]}{[ABC]} = \frac{x}{6x} = \frac{1}{6}</math>, so <math>[AGC] = 20</math>. Now we just need to find the area of triangle <math>AFE</math> and subtract it from the combined areas of <math>[ADE]</math> and <math>[ACG]</math>, since we count it twice. Note that the angle bisector theorem also applies for <math>\triangle ADE</math> and <math>\frac{AE}{AD} = \frac{1}{5}</math>, so thus <math>\frac{EF}{ED} = \frac{1}{6}</math> and we find <math>[AFE] = \frac{1}{6} \cdot 30 = 5</math>, and the area outside <math>FDBG</math> must be <math> [ADE] + [AGC] - [AFE] = 30 + 20 - 5 = 45</math>, and we finally find <math>[FDBG] = [ABC] - 45 = 120 -45 = \boxed{75}</math>, and we are done. | ||
+ | |||
+ | ==Solution 6: Areas == | ||
+ | <asy> | ||
+ | draw((0,0)--(1,3)--(5,0)--cycle); | ||
+ | draw((0,0)--(2,2.25)); | ||
+ | draw((0.5,1.5)--(2.5,0)); | ||
+ | label("A",(0,0),SW); | ||
+ | label("B",(5,0),SE); | ||
+ | label("C",(1,3),N); | ||
+ | label("G",(2,2.25),NE); | ||
+ | label("D",(2.5,0),S); | ||
+ | label("E",(0.5,1.5),NW); | ||
+ | label("3Y",(2.5,0.75),N); | ||
+ | label("Y",(1,0.2),N); | ||
+ | label("X",(0.5,0.5),N); | ||
+ | label("3X",(1.25,1.75),N); | ||
+ | </asy> | ||
+ | Give triangle <math>AEF</math> area X. Then, by similarity, since <math>\frac{AC}{AE} = \frac{2}{1}</math>, <math>ACG</math> has area 4X. Thus, <math>FGCE</math> has area 3X. | ||
+ | Doing the same for triangle <math>AGB</math>, we get that triangle <math>AFD</math> has area Y and quadrilateral <math>GFDB</math> has area 3Y. Since <math>AEF</math> has the same height as <math>AFD</math>, the ratios of the areas is equal to the ratios of the bases. Because of the Angle Bisector Theorem, <math>\frac{CG}{GB} = \frac{1}{5}</math>. So, <math>\frac{[AEF]}{[AFD]} = \frac{1}{5}</math>. Since <math>AEF</math> has area X, we can write the equation 5X = Y and substitute 5X for Y. | ||
+ | <asy> | ||
+ | draw((0,0)--(1,3)--(5,0)--cycle); | ||
+ | draw((0,0)--(2,2.25)); | ||
+ | draw((0.5,1.5)--(2.5,0)); | ||
+ | label("A",(0,0),SW); | ||
+ | label("B",(5,0),SE); | ||
+ | label("C",(1,3),N); | ||
+ | label("G",(2,2.25),NE); | ||
+ | label("D",(2.5,0),S); | ||
+ | label("E",(0.5,1.5),NW); | ||
+ | label("",(2.5,0.75),N); | ||
+ | label("",(1,0.2),N); | ||
+ | label("F", (1, 1.5), N); | ||
+ | label("",(0.5,1.5),N); | ||
+ | label("",(1.25,1.75),N); | ||
+ | </asy> | ||
+ | Now we can solve for X by adding up all the sums. X + 3X + 5X + 15X = 120, so X = 5. Since we want to find <math>GFDB</math>, we substitute 5 for 15X to get <math>\boxed{75}</math>. | ||
+ | <math>\sim</math>krishkhushi09 | ||
+ | |||
+ | == Video Solution by Richard Rusczyk == | ||
+ | |||
+ | https://artofproblemsolving.com/videos/amc/2018amc10a/469 | ||
+ | |||
+ | ~ dolphin7 | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/4_x1sgcQCp4?t=4898 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==See Also== |
Latest revision as of 03:01, 5 January 2022
- The following problem is from both the 2018 AMC 10A #24 and 2018 AMC 12A #18, so both problems redirect to this page.
Contents
Problem
Triangle with and has area . Let be the midpoint of , and let be the midpoint of . The angle bisector of intersects and at and , respectively. What is the area of quadrilateral ?
Solution 1
Let , , , and the length of the perpendicular to through be . By angle bisector theorem, we have that where . Therefore substituting we have that . By similar triangles, we have that , and the height of this trapezoid is . Then, we have that . We wish to compute , and we have that it is by substituting.
Solution 2
For this problem, we have because of SAS and . Therefore, is a quarter of the area of , which is . Subsequently, we can compute the area of quadrilateral to be . Using the angle bisector theorem in the same fashion as the previous problem, we get that is times the length of . We want the larger piece, as described by the problem. Because the heights are identical, one area is times the other, and .
Solution 3
The ratio of the to is by the Angle Bisector Theorem, so area of to the area of is also (They have the same height). Therefore, the area of is . Since is the midsegment of , so is the midsegment of . Thus, the ratio of the area of to the area of is , so the area of is . Therefore, the area of quadrilateral is
Solution 4
The area of quadrilateral is the area of minus the area of . Notice, , so , and since , the area of . Given that the area of is , using on side yields . Using the Angle Bisector Theorem, , so the height of . Therefore our answer is
Solution 5: Trig
We try to find the area of quadrilateral by subtracting the area outside the quadrilateral but inside triangle . Note that the area of is equal to and the area of triangle is equal to . The ratio is thus equal to and the area of triangle is . Let side be equal to , then by the angle bisector theorem. Similarly, we find the area of triangle to be and the area of triangle to be . A ratio between these two triangles yields , so . Now we just need to find the area of triangle and subtract it from the combined areas of and , since we count it twice. Note that the angle bisector theorem also applies for and , so thus and we find , and the area outside must be , and we finally find , and we are done.
Solution 6: Areas
Give triangle area X. Then, by similarity, since , has area 4X. Thus, has area 3X. Doing the same for triangle , we get that triangle has area Y and quadrilateral has area 3Y. Since has the same height as , the ratios of the areas is equal to the ratios of the bases. Because of the Angle Bisector Theorem, . So, . Since has area X, we can write the equation 5X = Y and substitute 5X for Y. Now we can solve for X by adding up all the sums. X + 3X + 5X + 15X = 120, so X = 5. Since we want to find , we substitute 5 for 15X to get . krishkhushi09
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2018amc10a/469
~ dolphin7
Video Solution
https://youtu.be/4_x1sgcQCp4?t=4898
~ pi_is_3.14
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 |
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.