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Difference between revisions of "2003 AMC 10A Problems/Problem 22"

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Plugging in <math> y=2x+8 </math> for <math>6y </math>, we get <math> 4x-24 = 0 </math>, so <math> x=6 </math>.
 
Plugging in <math> y=2x+8 </math> for <math>6y </math>, we get <math> 4x-24 = 0 </math>, so <math> x=6 </math>.
 
Lastly, to find the value of y, which is the value of Line <math> FG </math>, our desired value, we plug in <math> 6 </math> for <math> x </math> in the equation <math> y=2x+8 </math>, we get <math> 2(6)+8 </math>, which, finally, we get our <math>y</math> value of <math>20</math>, so thus, our answer is <math>GF= \boxed{\mathrm{(B)}\ 20}</math>
 
Lastly, to find the value of y, which is the value of Line <math> FG </math>, our desired value, we plug in <math> 6 </math> for <math> x </math> in the equation <math> y=2x+8 </math>, we get <math> 2(6)+8 </math>, which, finally, we get our <math>y</math> value of <math>20</math>, so thus, our answer is <math>GF= \boxed{\mathrm{(B)}\ 20}</math>
 +
 +
~Darth_Cadet
  
 
== See Also ==
 
== See Also ==

Revision as of 19:04, 1 March 2023

Problem

In rectangle $ABCD$, we have $AB=8$, $BC=9$, $H$ is on $BC$ with $BH=6$, $E$ is on $AD$ with $DE=4$, line $EC$ intersects line $AH$ at $G$, and $F$ is on line $AD$ with $GF \perp AF$. Find the length of $GF$.

[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair D=(0,0), Ep=(4,0), A=(9,0), B=(9,8), H=(3,8), C=(0,8), G=(-6,20), F=(-6,0); draw(D--A--B--C--D--F--G--Ep); draw(A--G); label("$F$",F,W); label("$G$",G,W); label("$C$",C,WSW); label("$H$",H,NNE); label("$6$",(6,8),N); label("$B$",B,NE); label("$A$",A,SW); label("$E$",Ep,S); label("$4$",(2,0),S); label("$D$",D,S);[/asy]

$\mathrm{(A) \ } 16\qquad \mathrm{(B) \ } 20\qquad \mathrm{(C) \ } 24\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 30$

Solutions

Solution 1

$\angle GHC = \angle AHB$ (Vertical angles are equal).

$\angle F = \angle B$ (Both are 90 degrees).

$\angle BHA = \angle HAD$ (Alt. Interior Angles are congruent).

Therefore $\triangle GFA$ and $\triangle ABH$ are similar. $\triangle GCH$ and $\triangle GEA$ are also similar.

$DA$ is 9, therefore $EA$ must equal 5. Similarly, $CH$ must equal 3.

Because $GCH$ and $GEA$ are similar, the ratio of $CH\; =\; 3$ and $EA\; =\; 5$, must also hold true for $GH$ and $HA$. $\frac{GH}{GA} = \frac{3}{5}$, so $HA$ is $\frac{2}{5}$ of $GA$. By Pythagorean theorem, $(HA)^2\; =\; (HB)^2\; +\; (BA)^2\;...\;HA=10$.

$HA\: =\: 10 =\: \frac{2}{5}*(GA)$.

$GA\: =\: 25.$

So $\frac{GA}{HA}\: =\: \frac{GF}{BA}$.

$\frac{25}{10}\: =\: \frac{GF}{8}$.

Therefore $GF= \boxed{\mathrm{(B)}\ 20}$.

Solution 2

Since $ABCD$ is a rectangle, $CD=AB=8$.

Since $ABCD$ is a rectangle and $GF \perp AF$, $\angle GFE = \angle CDE = \angle ABC = 90^\circ$.

Since $ABCD$ is a rectangle, $AD || BC$.

So, $AH$ is a transversal, and $\angle GAF = \angle AHB$.

This is sufficient to prove that $GFE \approx CDE$ and $GFA \approx ABH$.

Using ratios:

$\frac{GF}{FE}=\frac{CD}{DE}$

$\frac{GF}{FD+4}=\frac{8}{4}=2$

$GF=2 \cdot (FD+4)=2 \cdot FD+8$

$\frac{GF}{FA}=\frac{AB}{BH}$

$\frac{GF}{FD+9}=\frac{8}{6}=\frac{4}{3}$

$GF=\frac{4}{3} \cdot (FD+9)=\frac{4}{3} \cdot FD+12$

Since $GF$ can't have 2 different lengths, both expressions for $GF$ must be equal.

$2 \cdot FD+8=\frac{4}{3} \cdot FD+12$

$\frac{2}{3} \cdot FD=4$

$FD=6$

$GF=2 \cdot FD+8=2\cdot6+8=\boxed{\mathrm{(B)}\ 20}$

Solution 3 (fastest)

We extend $BC$ such that it intersects $GF$ at $X$. Since $ABCD$ is a rectangle, it follows that $CD=8$, therefore, $XF=8$. Let $GX=y$. From the similarity of triangles $GCH$ and $GEA$, we have the ratio $3:5$ (as $CH=9-6=3$, and $EA=9-4=5$). $GX$ and $GF$ are the altitudes of $GCH$ and $GEA$, respectively. Thus, $y:y+8 = 3:5$, from which we have $y=12$, thus $GF=y+8=12+8=\boxed{\mathrm{(B)}\ 20}$

Solution 4

Since $GF\perp AF$ and $AF\perp CD,$ we have $GF\parallel CD\parallel AB.$ Thus, $\triangle CDE\sim GFE.$ Suppose $GF=x$ and $FD=y.$ Thus, we have $\dfrac{x}{8}=\dfrac{y+4}{4}.$ Additionally, now note that $\triangle GAF\sim AHB,$ which is pretty obvious from insight, but can be proven by AA with extending $BH$ to meet $GF.$ From this new pair of similar triangles, we have $\dfrac{x}{8}=\dfrac{y+9}{6}.$ Therefore, we have by combining those two equations, \[\dfrac{y+9}{6}=\dfrac{y+4}{4}.\] Solving, we have $y=6,$ and therefore $x=\boxed{\mathrm{(B)}\ 20}$

Solution 5

Since there are only lines, you can resort to coordinate bashing. Let $FD=k$. Three lines, line $GF$, line $GE$, and line $GA$, intersect at $G$. Our goal is to find the y-coordinate of that intersection point.

Line $GF$ is $x=0$

Line $GE$ passes through $(k+4, 0)$ and $(k, 8)$. Therefore the slope is $-2$ and the line is $y-0=-2(x-k-4)$ which is $y=-2x+2k+8$

Line $GA$ passes through $(k+9, 0)$ and $(k+3, 8)$. Therefore the slope is $\frac{-4}{3}$ and the line is $y-0=-\frac{4}{3}(x-k-9)$ which simplifies to $y=-\frac{4}{3}x+\frac{4}{3}k+12$

We solve the system of equations with these three lines. First we plug in $x=0$

$y=2k+8$

$y=\frac{4}{3}k+12$

Next, we solve for k. $k=6$ Therefore $y=20$. The y-coordinate of this intersection point is indeed our answer. $\boxed{\mathrm{(B)}\ 20}$ ~superagh

Solution 6 (simple coordinates)

Let $A$ be the origin of our coordinate system. Now line $GA$ has equation $-\frac{4}{3}x$. We can use point-slope form to find the equation for line $GE$. First, we know that its slope is $-2$, and we know that it passes through $E=(-5,0)$, so line $GE$ has equation $-2(x+5)$. Solving for the intersection by letting $-\frac{4}{3}x=-2(x+5)$, we get $x=-15$. Plugging this into our equation for line $GA$ gives us $G=(-15,20)$, so $GF= \boxed{\mathrm{(B)}\ 20}$ ~chrisdiamond10

Solution 7 (system of equations through angle similarity)

First, using given information, we can find the values of some line segments in the figure. We find that $HA = 10$ (through Pythagorean Theorem), $CH = 3$, and $EA = 5$. Let Line $FD = x$ and let Line $FG = y$. We find that $\triangle FGE \sim \triangle CDE$ through some angle chasing (they both have a right angle, and they both share angle $\angle CED$. Using this information, we can write the equation $\frac{4}{8} = \frac{4+x}{y}$. Through simplifying this equation, we get that $y=2x+8$. Let point $I$ be the point on line $FG$ so that lines $CI$ and $FG$ are perpendicular, and we get that $GI = 2x$ and $FI =8$. Doing some more angle chasing, we can find that $\triangle GIH \sim \triangle GFA$, as they both share $\angle FGH$ and they both have a right angle.

With this information, we can write the equation $\frac{x+3}{y-8} = \frac{x+9}{y}.$ Simplifying this equation we get the equation $-8x+6y-72 = 0$. Plugging in $y=2x+8$ for $6y$, we get $4x-24 = 0$, so $x=6$. Lastly, to find the value of y, which is the value of Line $FG$, our desired value, we plug in $6$ for $x$ in the equation $y=2x+8$, we get $2(6)+8$, which, finally, we get our $y$ value of $20$, so thus, our answer is $GF= \boxed{\mathrm{(B)}\ 20}$

~Darth_Cadet

See Also

2003 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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All AMC 10 Problems and Solutions

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