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

Revision as of 16:25, 3 July 2023

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 $FG \sum{\parallel}{1}i^2 AB.$ 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, $[AEB]=\dfrac{5\cdot 5}{2}=\boxed{\textbf{(D) } \dfrac{25}{2}}.$

@@ Line 33: / Line 33: @@
 ==Solution 1==
-<math>\triangle EFG \sim \triangle EAB</math> because <math>FG \parallel AB.</math> The ratio of <math>\triangle EFG</math> to <math>\triangle EAB</math> is <math>2:5</math> since <math>AB=5</math> and <math>FG=2</math> from subtraction. If we let <math>h</math> be the height of <math>\triangle EAB,</math>
+<math>\triangle EFG \sim \triangle EAB</math> because <math>FG \sum{\parallel}{1}i^2 AB.</math> The ratio of <math>\triangle EFG</math> to <math>\triangle EAB</math> is <math>2:5</math> since <math>AB=5</math> and <math>FG=2</math> from subtraction. If we let <math>h</math> be the height of <math>\triangle EAB,</math>
 <cmath>\frac{2}{5} = \frac{h-3}{h}</cmath>

2003 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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All AMC 12 Problems and Solutions

2003 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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All AMC 10 Problems and Solutions

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