2009 AMC 10A Problems/Problem 17

Revision as of 01:02, 10 June 2020 by Binderclips1 (talk | contribs) (Added solution)

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 $EF$?

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

Solutions

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]

From the Pythagorean theorem we have $BD=5$.

Triangle $EAB$ is similar to $DAB$, as they have the same angles. Segment $BA$ is perpendicular to $DA$, meaning that angle $DAB$ and $BAE$ are right angles and congruent. Also, angle $DBE$ is a right angle. Because it is a rectangle, angle $BDC$ is congruent to $DBA$ and angle $ADC$ is also a right angle. By the transitive property:

$mADB + mBDC = mDBA + mABE$

$mBDC = mDBA$

$mADB + mBDC = mBDC + mABE$

$mADB = mABE$

Next, because every triangle has a degree measure of 180, angle $E$ and angle $DBA$ are congruent.


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

Solution 3 There is a better solution where we find EF directly instead of in parts (use similarity). The strategy is similar.

Solution 4(Coordinate Bash)

To keep things simple, we will use coordinates in only the first quadrant. The picture will look like the diagram above reflected over the $x$-axis.It is also worth noting the $F$ will lie on the $x$ axis and $E$ on the $y$. Let $D$ be the origin, $A(3,0)$, $C(4,0)$, and $B(4,3)$. We can express segment $DB$ as the line $y=\frac{3x}{4}$. Since $EF$ is perpendicular to $DB$, and we know that $(4,3)$ lies on it, we can use this information to find that segment $EF$ is on the line $y=\frac{-4x}{3}+\frac{25}{3}$. Since $E$ and $F$ are on the $y$ and $x$ axis, respectively, we plug in $0$ for $x$ and $y$, we find that point $E$ is at $(0,\frac{25}{3})$, and point $F$ is at $(\frac{25}{4},0)$. Applying the distance formula, we obtain that $EF$= $\boxed{\frac{125}{12}}$.

See Also

2009 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AMC 10 Problems and Solutions

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