Difference between revisions of "2003 AMC 10B Problems/Problem 20"

m (Problem)
(Solution 3)
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Then calculate the area of trapezoid <math>FGBA</math> and triangle <math>EGF</math> separately and add them. The area of the trapezoid is <math>\frac {2+5}{2}\cdot 3 = \frac {21}{2}</math> and the area of the triangle is <math>\frac{1}{2}\cdot 2 \cdot 2 = 2</math>. <math>\frac{21}{2}+2=\boxed{\textbf{(D)}\ \frac{25}{2}}</math>
 
Then calculate the area of trapezoid <math>FGBA</math> and triangle <math>EGF</math> separately and add them. The area of the trapezoid is <math>\frac {2+5}{2}\cdot 3 = \frac {21}{2}</math> and the area of the triangle is <math>\frac{1}{2}\cdot 2 \cdot 2 = 2</math>. <math>\frac{21}{2}+2=\boxed{\textbf{(D)}\ \frac{25}{2}}</math>
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==Solution 4==
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Since <math>\Delta{ABE}\sim{\Delta{FGE}}</math> then <math>[AFGB]\sim{[FXYG]}</math>, where <math>X</math> and <math>Y</math> are ponts on <math>EF</math> and <math>EG</math> respectivley which make the areas similar. This process can be done over and over again multiple times by the ratio of <math>\frac{FG}{AB}=\frac{2}{5}</math>, or something like this
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<cmath>[AEB]=[AFGB]+[FXYZ]+...</cmath><cmath>[AEB]=[AFGB]+\frac{2}{5}[AFGB]+(\frac{2}{5})^2[AFGB]+...</cmath>we have to find the ratio of the areas when the sides have shrunk by length <math>\frac{2}{5}l</math>
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[asy]
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unitsize(0.6 cm);
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pair A, B, C, D, E, F, G;
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 +
A = (0,0);
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B = (5,0);
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C = (5,3);
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D = (0,3);
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F = (1,3);
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G = (3,3);
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E = extension(A,F,B,G);
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draw(A--B--C--D--cycle);
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draw(A--E--B);
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label("<math>A</math>", A, SW);
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label("<math>B</math>", B, SE);
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label("<math>C</math>", C, NE);
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label("<math>D</math>", D, NW);
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label("<math>E</math>", E, N);
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label("<math>F</math>", F, SE);
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label("<math>G</math>", G, SW);
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label("<math>2/5</math>", (D + F)/2, N);
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label("<math>4/5</math>", (G + C)/2, N);
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label("<math>6/5</math>", (B + C)/2, dir(0));
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label("<math>6/5</math>", (A + D)/2, W);
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label("<math>2</math>", (A + B)/2, S);
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[/asy]
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Let <math>[AFGB]'</math> be the area of the shape whose length is <math>\frac{2}{5}l</math>
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<cmath>[AFGB]'=[ADCB]-[ADF]-[BCG]</cmath><cmath>[AFGB]'=12/5-6/25-12/25</cmath><cmath>[AFGB]'=42/25</cmath>Now comparing the ratios of <math>[AFGB]'</math> to <math>[AFGB]</math> we get
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<cmath>\frac{[AFGB]'}{[AFGB]}=\frac{42}{25}/\frac{21}{2}\implies \frac{[AFGB]'}{[AFGB]}=\frac{4}{25}</cmath>By applying an infinite summation
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<cmath>[AEB]=\sum_{n=0}^{\infty} \frac{21}{2}\cdot{(\frac{4}{25})^n}</cmath><cmath>S=\frac{a_1}{1-r}</cmath><cmath>\boxed{[AEB]=\frac{25}{2}}</cmath>
  
 
==See Also==
 
==See Also==

Revision as of 01:18, 15 December 2019

The following problem is from both the 2003 AMC 12B #14 and 2003 AMC 10B #20, so both problems redirect to this page.

Problem

In rectangle $ABCD, AB=5$ and $BC=3$. Points $F$ and $G$ are on $\overline{CD}$ so that $DF=1$ and $GC=2$. Lines $AF$ and $BG$ intersect at $E$. Find the area of $\triangle AEB$.

[asy] unitsize(8mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=0;  pair A=(0,0), B=(5,0), C=(5,3), D=(0,3); pair F=(1,3), G=(3,3); pair E=(5/3,5);  draw(A--B--C--D--cycle); draw(A--E); draw(B--E);  pair[] ps={A,B,C,D,E,F,G}; dot(ps); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); label("$E$",E,N); label("$F$",F,SE); label("$G$",G,SW); label("$1$",midpoint(D--F),N); label("$2$",midpoint(G--C),N); label("$5$",midpoint(A--B),S); label("$3$",midpoint(A--D),W); [/asy]

$\textbf{(A) } 10 \qquad\textbf{(B) } \frac{21}{2} \qquad\textbf{(C) } 12 \qquad\textbf{(D) } \frac{25}{2} \qquad\textbf{(E) } 15$

Solution 1

$\triangle EFG \sim \triangle EAB$ because $FG \parallel AB.$ The ratio of $\triangle EFG$ to $\triangle EAB$ is $2:5$ since $AB=5$ and $FG=2$ from subtraction. If we let $h$ be the height of $\triangle EAB,$

\[\frac{2}{5} = \frac{h-3}{h}\] \[2h = 5h-15\] \[3h = 15\] \[h = 5\]

The height is $5$ so the area of $\triangle EAB$ is $\frac{1}{2}(5)(5) = \boxed{\textbf{(D)}\ \frac{25}{2}}$.

Solution 2

We can look at this diagram as if it were a coordinate plane with point $A$ being $(0,0)$. This means that the equation of the line $AE$ is $y=3x$ and the equation of the line $EB$ is $y=\frac{-3}{2}x+\frac{15}{2}$. From this we can set of the follow equation to find the $x$ coordinate of point $E$:

\[3x=\frac{-3}{2}x+\frac{15}{2}\] \[6x=-3x+15\] \[9x=15\] \[x=\frac{5}{3}\]

We can plug this into one of our original equations to find that the $y$ coordinate is $5$, meaning the area of $\triangle EAB$ is $\frac{1}{2}(5)(5) = \boxed{\textbf{(D)}\ \frac{25}{2}}$

Solution 3

At points $A$ and $B$, segment $AE$ is 5 units from segment $BE$. At points $F$ and $G$, the segments are 2 units from each other. This means that collectively, the two lines closed the distance between them by 3 units over a height of 3 units. Therefore, to close the next two units of distance, they will have to travel a height of 2 units.

Then calculate the area of trapezoid $FGBA$ and triangle $EGF$ separately and add them. The area of the trapezoid is $\frac {2+5}{2}\cdot 3 = \frac {21}{2}$ and the area of the triangle is $\frac{1}{2}\cdot 2 \cdot 2 = 2$. $\frac{21}{2}+2=\boxed{\textbf{(D)}\ \frac{25}{2}}$

Solution 4

Since $\Delta{ABE}\sim{\Delta{FGE}}$ then $[AFGB]\sim{[FXYG]}$, where $X$ and $Y$ are ponts on $EF$ and $EG$ respectivley which make the areas similar. This process can be done over and over again multiple times by the ratio of $\frac{FG}{AB}=\frac{2}{5}$, or something like this \[[AEB]=[AFGB]+[FXYZ]+...\]\[[AEB]=[AFGB]+\frac{2}{5}[AFGB]+(\frac{2}{5})^2[AFGB]+...\]we have to find the ratio of the areas when the sides have shrunk by length $\frac{2}{5}l$ [asy] unitsize(0.6 cm);

pair A, B, C, D, E, F, G;

A = (0,0); B = (5,0); C = (5,3); D = (0,3); F = (1,3); G = (3,3); E = extension(A,F,B,G);

draw(A--B--C--D--cycle); draw(A--E--B);

label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, NW); label("$E$", E, N); label("$F$", F, SE); label("$G$", G, SW); label("$2/5$", (D + F)/2, N); label("$4/5$", (G + C)/2, N); label("$6/5$", (B + C)/2, dir(0)); label("$6/5$", (A + D)/2, W); label("$2$", (A + B)/2, S); [/asy] Let $[AFGB]'$ be the area of the shape whose length is $\frac{2}{5}l$ \[[AFGB]'=[ADCB]-[ADF]-[BCG]\]\[[AFGB]'=12/5-6/25-12/25\]\[[AFGB]'=42/25\]Now comparing the ratios of $[AFGB]'$ to $[AFGB]$ we get \[\frac{[AFGB]'}{[AFGB]}=\frac{42}{25}/\frac{21}{2}\implies \frac{[AFGB]'}{[AFGB]}=\frac{4}{25}\]By applying an infinite summation \[[AEB]=\sum_{n=0}^{\infty} \frac{21}{2}\cdot{(\frac{4}{25})^n}\]\[S=\frac{a_1}{1-r}\]\[\boxed{[AEB]=\frac{25}{2}}\]

See Also

2003 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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
2003 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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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