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2003 AMC 10B Problems/Problem 20

Revision as of 16:26, 3 July 2023 by Jeffrey16 (talk | contribs) (→‎Solution 1)
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 2

$\triangle EFG \sim \triangle EAB$ 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 \[[AEB]=\sum_{n=0}^{\infty} \frac{21}{2}\cdot{(\frac{4}{25})^n}\]\[S=\frac{a_1}{1-r}\]\[\boxed{[AEB]=\frac{25}{2}}\]

Solution 5

Drop a perpendicular from $E$ to $AB$ and call the intersection point $X.$ $\triangle EXA$ and $\triangle DFA$ are similar and so are $\triangle BCG$ and $\triangle EXB$ (You can prove this with $AA$ by observing that $\angle DFA$ is congruent to $\angle EAB$ and so on). This means that $\dfrac{AX}{BX}=\dfrac{1}{2}$ so $AX=\dfrac{5}{3}.$ $\dfrac{EX}{AX}=3,$ which means that $EX=5.$ Then, $[AEB]=\dfrac{5\cdot 5}{2}=\boxed{\textbf{(D) } \dfrac{25}{2}}.$

See Also

2003 AMC 12B (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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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