Difference between revisions of "2010 AMC 12A Problems/Problem 3"

Revision as of 09:34, 16 February 2010

Problem 3

Rectangle $ABCD$ , pictured below, shares $50\%$ of its area with square $EFGH$ . Square $EFGH$ shares $20\%$ of its area with rectangle $ABCD$ . What is $\frac{AB}{AD}$ ?

$[asy] unitsize(1mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((0,0)--(0,25)--(25,25)--(25,0)--cycle); fill((0,20)--(0,15)--(25,15)--(25,20)--cycle,gray); draw((0,15)--(0,20)--(25,20)--(25,15)--cycle); draw((25,15)--(25,20)--(50,20)--(50,15)--cycle); label("$A$",(0,20),W); label("$B$",(50,20),E); label("$C$",(50,15),E); label("$D$",(0,15),W); label("$E$",(0,25),NW); label("$F$",(25,25),NE); label("$G$",(25,0),SE); label("$H$",(0,0),SW); [/asy]$

$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10$

Solution

Solution 1

Let $EF = FG = GF = HE = s$ , let $AD = BC = h$ , and let $AB = CD = x$ .

$0.2 \cdot s^2 = hs$

$s = 5h$

$0.5 \cdot hx = hs$

$x = 2s = 10h$

$\frac{AB}{AD} = \frac{x}{h} = \boxed{10\ \textbf{(E)}}$

Solution 2

The answer does not change if we shift $A$ to coincide with $E$ , and add new horizontal lines to divide $EFGH$ into five equal parts:

$[asy] unitsize(1mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((0,0)--(0,25)--(25,25)--(25,0)--cycle); fill((0,25)--(0,20)--(25,20)--(25,25)--cycle,gray); draw((25,20)--(25,25)--(50,25)--(50,20)--cycle); draw((0,5)--(25,5)); draw((0,10)--(25,10)); draw((0,15)--(25,15)); label("$A=E$",(0,25),W); label("$B$",(50,25),E); label("$C$",(50,20),E); label("$D$",(0,20),W); label("$F$",(25,25),NE); label("$G$",(25,0),SE); label("$H$",(0,0),SW); [/asy]$

This helps us to see that $AD=a/5$ and $AB=2a$ , where $a=EF$ . Hence $\dfrac{AB}{AD}=\dfrac{2a}{a/5}=10$ .

@@ Line 25: / Line 25: @@
 == Solution ==
+=== Solution 1 ===
 Let <math>EF = FG = GF = HE = s</math>, let <math>AD = BC = h</math>, and let <math>AB = CD = x</math>.
-<math>.2s^2 = hs</math>
+<math>0.2 \cdot s^2 = hs</math>
 <math>s = 5h</math>
-<math>.5hx = hs</math>
+<math>0.5 \cdot hx = hs</math>
 <math>x = 2s = 10h</math>
+<math>\frac{AB}{AD} = \frac{x}{h} = \boxed{10\ \textbf{(E)}}</math>
-<math>\frac{AB}{AD} = \frac{x}{h} = \boxed{10\ \textbf{(E)}}</math>
+=== Solution 2 ===
+The answer does not change if we shift <math>A</math> to coincide with <math>E</math>, and add new horizontal lines to divide <math>EFGH</math> into five equal parts:
+<center><asy>
+unitsize(1mm);
+defaultpen(linewidth(.8pt)+fontsize(8pt));
+draw((0,0)--(0,25)--(25,25)--(25,0)--cycle);
+fill((0,25)--(0,20)--(25,20)--(25,25)--cycle,gray);
+draw((25,20)--(25,25)--(50,25)--(50,20)--cycle);
+draw((0,5)--(25,5));
+draw((0,10)--(25,10));
+draw((0,15)--(25,15));
+label("$A=E$",(0,25),W);
+label("$B$",(50,25),E);
+label("$C$",(50,20),E);
+label("$D$",(0,20),W);
+label("$F$",(25,25),NE);
+label("$G$",(25,0),SE);
+label("$H$",(0,0),SW);
+</asy></center>
+This helps us to see that <math>AD=a/5</math> and <math>AB=2a</math>, where <math>a=EF</math>.
+Hence <math>\dfrac{AB}{AD}=\dfrac{2a}{a/5}=10</math>.

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