Difference between revisions of "2006 AMC 10B Problems/Problem 23"
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== Problem == | == Problem == | ||
− | == Solution == | + | A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7, as shown. What is the area of the shaded quadrilateral? |
− | == See | + | |
− | + | <asy> | |
+ | unitsize(1.5cm); | ||
+ | defaultpen(.8); | ||
+ | |||
+ | pair A = (0,0), B = (3,0), C = (1.4, 2), D = B + 0.4*(C-B), Ep = A + 0.3*(C-A); | ||
+ | pair F = intersectionpoint( A--D, B--Ep ); | ||
+ | |||
+ | draw( A -- B -- C -- cycle ); | ||
+ | draw( A -- D ); | ||
+ | draw( B -- Ep ); | ||
+ | filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black ); | ||
+ | |||
+ | label("$7$",(1.25,0.2)); | ||
+ | label("$7$",(2.2,0.45)); | ||
+ | label("$3$",(0.45,0.35)); | ||
+ | </asy> | ||
+ | |||
+ | <math> \mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 17\qquad \mathrm{(C) \ } \frac{35}{2}\qquad \mathrm{(D) \ } 18\qquad \mathrm{(E) \ } \frac{55}{3} </math> | ||
+ | |||
+ | == Solution 1== | ||
+ | |||
+ | Label the points in the figure as shown below, and draw the segment <math>CF</math>. This segment divides the quadrilateral into two triangles, let their areas be <math>x</math> and <math>y</math>. | ||
+ | |||
+ | <asy> | ||
+ | unitsize(2cm); | ||
+ | defaultpen(.8); | ||
+ | |||
+ | pair A = (0,0), B = (3,0), C = (1.4, 2), D = B + 0.4*(C-B), Ep = A + 0.3*(C-A); | ||
+ | pair F = intersectionpoint( A--D, B--Ep ); | ||
+ | |||
+ | draw( A -- B -- C -- cycle ); | ||
+ | draw( A -- D ); | ||
+ | draw( B -- Ep ); | ||
+ | filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black ); | ||
+ | |||
+ | label("$7$",(1.45,0.15)); | ||
+ | label("$7$",(2.2,0.45)); | ||
+ | label("$3$",(0.45,0.35)); | ||
+ | |||
+ | draw( C -- F, dashed ); | ||
+ | |||
+ | label("$A$",A,SW); | ||
+ | label("$B$",B,SE); | ||
+ | label("$C$",C,N); | ||
+ | label("$D$",D,NE); | ||
+ | label("$E$",Ep,NW); | ||
+ | label("$F$",F,S); | ||
+ | |||
+ | label("$x$",(1,1)); | ||
+ | label("$y$",(1.6,1)); | ||
+ | </asy> | ||
+ | |||
+ | Since [[triangle]]s <math>AFB</math> and <math>DFB</math> share an [[altitude]] from <math>B</math> and have equal area, their bases must be equal, hence <math>AF=DF</math>. | ||
+ | |||
+ | Since triangles <math>AFC</math> and <math>DFC</math> share an altitude from <math>C</math> and their respective bases are equal, their areas must be equal, hence <math>x+3=y</math>. | ||
+ | |||
+ | Since triangles <math>EFA</math> and <math>BFA</math> share an altitude from <math>A</math> and their respective areas are in the ratio <math>3:7</math>, their bases must be in the same ratio, hence <math>EF:FB = 3:7</math>. | ||
+ | |||
+ | Since triangles <math>EFC</math> and <math>BFC</math> share an altitude from <math>C</math> and their respective bases are in the ratio <math>3:7</math>, their areas must be in the same ratio, hence <math>x:(y+7) = 3:7</math>, which gives us <math>7x = 3(y+7)</math>. | ||
+ | |||
+ | Substituting <math>y=x+3</math> into the second equation we get <math>7x = 3(x+10)</math>, which solves to <math>x=\frac{15}{2}</math>. Then <math>y=x+3 = \frac{15}{2}+3 = \frac{21}{2}</math>, and the total area of the quadrilateral is <math>x+y = \frac{15}{2}+\frac{21}{2} = \boxed{\textbf{(D) }18}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Connect points <math>E</math> and <math>D</math>. Triangles <math>EFA</math> and <math>FAB</math> share an altitude and their areas are in the ration <math>3:7</math>. Their bases, <math>EF</math> and <math>FB</math>, must be in the same <math>3:7</math> ratio. | ||
+ | |||
+ | Triangles <math>EFD</math> and <math>FBD</math> share an altitude and their bases are in a <math>3:7</math> ratio. Therefore, their areas are in a <math>3:7</math> ratio and the area of triangle <math>EFD</math> is <math>3</math>. | ||
+ | |||
+ | Triangle <math>CED</math> and <math>DEA</math> share an altitude. Therefore, the ratio of their areas is equal to the ratio of bases <math>CE</math> and <math>EA</math>. The ratio is <math>A:(3+3) \Rightarrow A:6</math> where <math>A</math> is the area of triangle <math>CED</math> | ||
+ | |||
+ | Triangles <math>CEB</math> and <math>EAB</math> also share an altitude. The ratio of their areas is also equal to the ratio of bases <math>CE</math> and <math>EA</math>. The ratio is <math>(A+3+7):(3+7) \Rightarrow (A+10):10</math> | ||
+ | |||
+ | Because the two ratios are equal, we get the equation <math>\frac{A}{6} = \frac {A+10}{10} \Rightarrow 10A = 6A+60 \Rightarrow A = 15</math>. We add the area of triangle <math>EDF</math> to get that the total area of the quadrilateral is <math>\boxed{\textbf{(D) }18}</math>. | ||
+ | |||
+ | ~Zeric Hang | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | We use mass points (similar to above). Let the triangle be <math>ABC</math> with cevians (lines to opposite side) from <math>B</math> and <math>C</math>. Let the points opposite <math>B</math> and <math>C</math> be <math>D</math> and <math>F</math> respectively and the intersection as <math>P</math>. | ||
+ | |||
+ | Assign masses of 1 at <math>B</math> and <math>D</math> since <math>[BPC] = [DPC]</math>. Then the mass at <math>P</math> is 2. To find masses at <math>F</math> and <math>C</math>, we let the mass at <math>F</math> be x and the mass at <math>C</math> be y. Then <math>3x = 7y</math> and <math>y = \frac{3}{7}x</math>. Then <math>\frac{10}{7}x = 2</math> since we add the masses for the fulcrum mass, and <math>x = \frac{7}{5}</math> and <math>y = \frac{3}{5}</math>. | ||
+ | |||
+ | To calculate the mass at a, it is merely <math>\frac{7}{5} - 1 = \frac{2}{5}</math> which means <math>\frac{[BCF]}{[ACF]} = \frac{2}{5}</math> or <math>[ACF] = 25</math>. It is easy to see the answer is <math>\boxed{\textbf{(D) }18}</math>. | ||
+ | |||
+ | == See also == | ||
+ | {{AMC10 box|year=2006|ab=B|num-b=22|num-a=24}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
+ | [[Category:Area Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 20:00, 1 November 2020
Problem
A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7, as shown. What is the area of the shaded quadrilateral?
Solution 1
Label the points in the figure as shown below, and draw the segment . This segment divides the quadrilateral into two triangles, let their areas be and .
Since triangles and share an altitude from and have equal area, their bases must be equal, hence .
Since triangles and share an altitude from and their respective bases are equal, their areas must be equal, hence .
Since triangles and share an altitude from and their respective areas are in the ratio , their bases must be in the same ratio, hence .
Since triangles and share an altitude from and their respective bases are in the ratio , their areas must be in the same ratio, hence , which gives us .
Substituting into the second equation we get , which solves to . Then , and the total area of the quadrilateral is .
Solution 2
Connect points and . Triangles and share an altitude and their areas are in the ration . Their bases, and , must be in the same ratio.
Triangles and share an altitude and their bases are in a ratio. Therefore, their areas are in a ratio and the area of triangle is .
Triangle and share an altitude. Therefore, the ratio of their areas is equal to the ratio of bases and . The ratio is where is the area of triangle
Triangles and also share an altitude. The ratio of their areas is also equal to the ratio of bases and . The ratio is
Because the two ratios are equal, we get the equation . We add the area of triangle to get that the total area of the quadrilateral is .
~Zeric Hang
Solution 3
We use mass points (similar to above). Let the triangle be with cevians (lines to opposite side) from and . Let the points opposite and be and respectively and the intersection as .
Assign masses of 1 at and since . Then the mass at is 2. To find masses at and , we let the mass at be x and the mass at be y. Then and . Then since we add the masses for the fulcrum mass, and and .
To calculate the mass at a, it is merely which means or . It is easy to see the answer is .
See also
2006 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 |
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