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

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\mathrm{(E)}\ 12
 
\mathrm{(E)}\ 12
 
</math>
 
</math>
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[[Category: Introductory Geometry Problems]]
  
−
== Solution ==
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== Solutions ==
  
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=== [[Solution 1]] ===
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=== Solution 1 ===
  
 
The situation is shown in the picture below.
 
The situation is shown in the picture below.
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From the [[Pythagorean theorem]] we have <math>BD=5</math>.
 
From the [[Pythagorean theorem]] we have <math>BD=5</math>.
  
−
Triangle <math>EAB</math> is similar to <math>DAB</math>, as they have the same angles. Segment <math>BA</math> is perpendicular to <math>DA</math>, meaning that angle <math>DAB</math> and <math>BAE</math> are right angles and congruent. Also, angle <math>DBE</math> is a right angle. Because it is a rectangle, angle <math>BDC</math> is congruent to <math>DBA</math> and angle <math>ADC</math> is also a right angle. By the transitive property:
+
Triangle <math>EAB</math> is similar to <math>BAD</math>, as they have the same angles. Segment <math>BA</math> is perpendicular to <math>DA</math>, meaning that angle <math>DAB</math> and <math>BAE</math> are right angles and congruent. Also, angle <math>DBE</math> is a right angle. Because it is a rectangle, angle <math>BDC</math> is congruent to <math>DBA</math> and angle <math>ADC</math> is also a right angle. By the transitive property:
  
 
<math>mADB + mBDC = mDBA + mABE</math>
 
<math>mADB + mBDC = mDBA + mABE</math>
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<math>mADB = mABE</math>
 
<math>mADB = mABE</math>
  
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Next, because every triangle has a degree measure of 180, angle <math>E</math> and angle <math>DBA</math> are congruent.  
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Next, because every triangle has a degree measure of <math>180</math>, angle <math>BEA</math> and angle <math>DBA</math> are similar.  
  
  
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=== Solution 2 ===
 
=== Solution 2 ===
  
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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>.  
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Since <math>BD</math> is the altitude from <math>D</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>.  
+
Looking at the angles, we see that triangle <math>BDE</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 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>.
 
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>.
 +
 +
===Solution 3(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 <math>x</math>-axis.It is also worth noting the <math>F</math> will lie on the <math>x</math> axis and <math>E</math> on the <math>y</math>. Let <math>D</math> be the origin, <math>A(3,0)</math>, <math>C(4,0)</math>, and  <math>B(4,3)</math>. We can express segment <math>DB</math> as the line <math>y=\frac{3x}{4}</math>.
 +
Since <math>EF</math> is perpendicular to <math>DB</math>, and we know that <math>(4,3)</math> lies on it, we can use this information to find that segment <math>EF</math>
 +
is on the line <math>y=\frac{-4x}{3}+\frac{25}{3}</math>. Since <math>E</math> and <math>F</math> are on the <math>y</math> and <math>x</math> axis, respectively, we plug in <math>0</math> for <math>x</math>
 +
and <math>y</math>, we find that point <math>E</math> is at <math>(0,\frac{25}{3})</math>, and point <math>F</math> is at <math>(\frac{25}{4},0)</math>. Applying the distance formula,
 +
we obtain that <math>EF</math>= <math>\boxed{\frac{125}{12}}</math>.
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC10 box|year=2009|ab=A|num-b=16|num-a=18}}
 
{{AMC10 box|year=2009|ab=A|num-b=16|num-a=18}}
 +
{{MAA Notice}}

Latest revision as of 20:14, 23 October 2023

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 $BAD$, 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 $BEA$ and angle $DBA$ are similar.


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 $D$ 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(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 (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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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