Difference between revisions of "2009 AMC 10A Problems/Problem 17"

Revision as of 23:25, 31 January 2012

Problem

Rectangle $ABCD$ has $AB=4$ and $BC=3$ . Segment $EF$ is constructed through $B$ so that $EF$ is perpendicular to $DB$ , and $A$ and $C$ lie on $DE$ and $DF$ , respectively. What is $EF$ ?

$\mathrm{(A)}\ 9 \qquad \mathrm{(B)}\ 10 \qquad \mathrm{(C)}\ \frac {125}{12} \qquad \mathrm{(D)}\ \frac {103}{9} \qquad \mathrm{(E)}\ 12$

Solution

Solution 1

The situation is shown in the picture below.

$[asy] unitsize(0.6cm); defaultpen(0.8); pair A=(0,0), B=(4,0), C=(4,3), D=(0,3); pair EF=rotate(90)*(D-B); pair E=intersectionpoint( (0,-100)--(0,100), (B-100*EF)--(B+100*EF) ); pair F=intersectionpoint( (-100,3)--(100,3), (B-100*EF)--(B+100*EF) ); draw(A--B--C--D--cycle); draw(B--D, dashed); draw(E--F); draw(A--E, dashed); draw(C--F, dashed); label("$A$",A,W); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,NW); label("$E$",E,SW); label("$F$",F,NE); label("$3$",A--D,W); label("$4$",C--D,N); [/asy]$

Obviously, from the Pythagorean theorem we have $BD=5$ .

Triangle $EAB$ is similar to $ABD$ , as they have the same angles. Hence $BE/AB = DB/AD$ , and therefore $BE = AB\cdot DB/AD = 20/3$ .

Also triangle $CBF$ is similar to $ABD$ . Hence $BF/BC = DB/AB$ , and therefore $BF=BC\cdot DB / AB = 15/4$ .

We then have $EF = EB+BF = \frac{20}3 + \frac{15}4 = \frac{80 + 45}{12} = \boxed{\frac{125}{12}}$ .

Solution 2

Since $BD$ is the altitude from $B$ to $EF$ , we can use the equation $BD^2 = EB\cdot BF$ .

Looking at the angles, we see that triangle $EAB$ is similar to $DCB$ . Because of this, $\frac{AB}{CB} = \frac{EB}{DB}$ . From the given information and the Pythagorean theorem, $AB=4$ , $CB=3$ , and $DB=5$ . Solving gives $EB=20/3$ .

We can use the above formula to solve for $BF$ . $BD^2 = 20/3\cdot BF$ . Solve to obtain $BF=15/4$ .

We now know $EB$ and $BF$ . $EF = EB+BF = \frac{20}3 + \frac{15}4 = \frac{80 + 45}{12} = \boxed{\frac{125}{12}}$ .

@@ Line 15: / Line 15: @@
 == Solution ==
+=== Solution 1 ===
 The situation is shown in the picture below.
@@ Line 47: / Line 49: @@
 We then have <math>EF = EB+BF = \frac{20}3 + \frac{15}4 = \frac{80 + 45}{12} = \boxed{\frac{125}{12}}</math>.
+=== Solution 2 ===
+Since <math>BD</math> is the altitude from <math>B</math> to <math>EF</math>, we can use the equation <math>BD^2 = EB\cdot BF</math>.
+Looking at the angles, we see that triangle <math>EAB</math> is similar to <math>DCB</math>. Because of this, <math>\frac{AB}{CB} = \frac{EB}{DB}</math>. From the given information and the [[Pythagorean theorem]], <math>AB=4</math>, <math>CB=3</math>, and <math>DB=5</math>. Solving gives <math>EB=20/3</math>.
+We can use the above formula to solve for <math>BF</math>. <math>BD^2 = 20/3\cdot BF</math>. Solve to obtain <math>BF=15/4</math>.
+We now know <math>EB</math> and <math>BF</math>. <math>EF = EB+BF = \frac{20}3 + \frac{15}4 = \frac{80 + 45}{12} = \boxed{\frac{125}{12}}</math>.
 == See Also ==
 {{AMC10 box|year=2009|ab=A|num-b=16|num-a=18}}

2009 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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