2003 AMC 10A Problems/Problem 22: Difference between revisions

Latest revision as of 15:07, 4 July 2025

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 and easiest)

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 therefore, our answer is $GF= \boxed{\mathrm{(B)}\ 20}$

~Darth_Cadet

Solution 8 Line and Slope

Draw a coordinate plane with the y-axis centered on $CD$ and the x-axis centered on $AD$ . From there, call the line passing through $CE$ $a$ and the line passing through $HA$ $b$ . From there, you can find that the equations for these lines are $y=-2x+8$ and $y=-4/3 x+12$ respectively. We need to find the length of $GF$ , so we are finding the y-value of the point $G$ . Solving for this point, we get $-4/3 x+12=-2x+8$ ; $2/3x=-4$ ; $x=-6$ . Now, plugging in the values, we find that $y=-2(-6)+8=20$ , so our answer is $GF= \boxed{\mathrm{(B)}\ 20}$ .

~iamcalifornia'sresidentidiot

@@ Line 2: / Line 2: @@
 In rectangle <math>ABCD</math>, we have <math>AB=8</math>, <math>BC=9</math>, <math>H</math> is on <math>BC</math> with <math>BH=6</math>, <math>E</math> is on <math>AD</math> with <math>DE=4</math>, line <math>EC</math> intersects line <math>AH</math> at <math>G</math>, and <math>F</math> is on line <math>AD</math> with <math>GF \perp AF</math>. Find the length of <math>GF</math>.
-[[Image:2003amc10a22.gif]]
+<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>
 <math> \mathrm{(A) \ } 16\qquad \mathrm{(B) \ } 20\qquad \mathrm{(C) \ } 24\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 30 </math>
-== Solution ==
+== Solutions ==
 === Solution 1 ===
-<math>\angle GCH = \angle ABH</math> (Opposite angles are equal).
+<math>\angle GHC = \angle AHB</math> (Vertical angles are equal).
 <math>\angle F = \angle B</math> (Both are 90 degrees).
@@ Line 29: / Line 44: @@
 <math>\frac{25}{10}\: =\: \frac{GF}{8}</math>.
-Therefore <math>GF= \boxed{\mathrm{(B)}\ 20}</math>.
+Therefore   <math>GF= \boxed{\mathrm{(B)}\ 20}</math>.
 === Solution 2 ===
@@ Line 66: / Line 81: @@
 <math>GF=2 \cdot FD+8=2\cdot6+8=\boxed{\mathrm{(B)}\ 20}</math>
-=== Solution 3 ===
+=== Solution 3 (fastest and easiest)===
-Since <math>ABCD</math> is a rectangle, <math>CD=3</math>, <math>EA=5</math>, and <math>CD=8</math>. From the [[Pythagorean Theorem]], <math>CE^2=CD^2+DE^2=80\Rightarrow CE=4\sqrt{5}</math>.
+We extend <math>BC</math> such that it intersects <math>GF</math> at <math>X</math>. Since <math>ABCD</math> is a rectangle, it follows that <math>CD=8</math>, therefore, <math>XF=8</math>. Let <math>GX=y</math>. From the similarity of triangles <math>GCH</math> and <math>GEA</math>, we have the ratio <math>3:5</math> (as <math>CH=9-6=3</math>, and <math>EA=9-4=5</math>). <math>GX</math> and <math>GF</math> are the altitudes of <math>GCH</math> and <math>GEA</math>, respectively. Thus, <math>y:y+8 = 3:5</math>, from which we have <math>y=12</math>, thus <math>GF=y+8=12+8=\boxed{\mathrm{(B)}\ 20}</math>
-==== Lemma ====
-Statement: <math>GCH \approx GEA</math>
+=== Solution 4 ===
-Proof: <math>\angle CGH=\angle EGA</math>, obviously.
+Since <math>GF\perp AF</math> and <math>AF\perp CD, </math> we have <math>GF\parallel CD\parallel AB.</math> Thus, <math>\triangle CDE\sim GFE.</math> Suppose <math>GF=x</math> and <math>FD=y.</math> Thus, we have <math>\dfrac{x}{8}=\dfrac{y+4}{4}.</math> Additionally, now note that <math>\triangle GAF\sim AHB,</math> which is pretty obvious from insight, but can be proven by AA with extending <math>BH</math> to meet <math>GF.</math> From this new pair of similar triangles, we have <math>\dfrac{x}{8}=\dfrac{y+9}{6}.</math> Therefore, we have by combining those two equations, <cmath>\dfrac{y+9}{6}=\dfrac{y+4}{4}.</cmath> Solving, we have <math>y=6,</math> and therefore <math>x=\boxed{\mathrm{(B)}\ 20}</math>
-<math>\begin{eqnarray}
+=== Solution 5 ===
-\angle HCE=180^{\circ}-\angle CHG\\
-\angle DCE=\angle CHG-90^{\circ}\\
-\angle CEED=180-\angle CHG\\
-\angle GEA=\angle GCH
-\end{eqnarray}</math>
-Since two angles of the triangles are equal, the third angles must equal each other. Therefore, the triangles are similar.
+Since there are only lines, you can resort to coordinate bashing. Let <math>FD=k</math>. Three lines, line <math>GF</math>, line <math>GE</math>, and line <math>GA</math>, intersect at <math>G</math>. Our goal is to find the y-coordinate of that intersection point.
+Line <math>GF</math> is <math>x=0</math>
+Line <math>GE</math> passes through <math>(k+4, 0)</math> and <math>(k, 8)</math>. Therefore the slope is <math>-2</math> and the line is
+<math>y-0=-2(x-k-4)</math> which is <math>y=-2x+2k+8</math>
-Let <math>GC=x</math>.
+Line <math>GA</math> passes through <math>(k+9, 0)</math> and <math>(k+3, 8)</math>. Therefore the slope is <math>\frac{-4}{3}</math> and the line is
+<math>y-0=-\frac{4}{3}(x-k-9)</math> which simplifies to <math>y=-\frac{4}{3}x+\frac{4}{3}k+12</math>
-<cmath>\begin{eqnarray}
+We solve the system of equations with these three lines. First we plug in <math>x=0</math>
-\dfrac{x}{3}=\dfrac{x+4\sqrt{5}}{5}\\
-x=3x+12\sqrt{5}\\
-x=12\sqrt{5}\\
-x=6\sqrt{5}
-\end{eqnarray}</cmath>
-Also, <math>\triangle GFE\approx \triangle CDE</math>, therefore
+<math>y=2k+8</math>
-<cmath>\dfrac{8}{4\sqrt{5}}=\dfrac{GF}{10\sqrt{5}}</cmath>
+<math>y=\frac{4}{3}k+12</math>
-We can multiply both sides by <math>\sqrt{5}</math> to get that <math>GF</math> is twice of 10, or <math>\boxed{\mathrm{(B)}\ 20}</math>
+Next, we solve for k. <math>k=6</math> Therefore <math>y=20</math>. The y-coordinate of this intersection point is indeed our answer. <math>\boxed{\mathrm{(B)}\ 20}</math>
+~superagh
-=== Solution 4 ===
+=== Solution 6 (simple coordinates)===
-We extend <math>BC</math> such that it intersects <math>GF</math> at <math>X</math>. Since <math>ABCD</math> is a rectangle, it follows that <math>CD=8</math>, therefore, <math>XF=8</math>. Let <math>GX=y</math>. From the similarity of triangles <math>GCH</math> and <math>GEA</math>, we have the ratio <math>3:5</math> (as <math>CH=9-6=3</math>, and <math>EA=9-4=5</math>). <math>GX</math> and <math>GF</math> are the altitudes of <math>GCH</math> and <math>GEA</math>, respectively. Thus, <math>y:y+8 = 3:5</math>, from which we have <math>y=12</math>, thus <math>GF=y+8=12+8=\boxed{\mathrm{(B)}\ 20}</math>
+Let <math>A</math> be the origin of our coordinate system. Now line <math>GA</math> has equation <math>-\frac{4}{3}x</math>. We can use point-slope form to find the equation for line <math>GE</math>. First, we know that its slope is <math>-2</math>, and we know that it passes through <math>E=(-5,0)</math>, so line <math>GE</math> has equation <math>-2(x+5)</math>. Solving for the intersection by letting <math>-\frac{4}{3}x=-2(x+5)</math>, we get <math>x=-15</math>. Plugging this into our equation for line <math>GA</math> gives us <math>G=(-15,20)</math>, so <math>GF= \boxed{\mathrm{(B)}\ 20}</math>
+~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 <math> HA = 10 </math> (through Pythagorean Theorem), <math> CH = 3 </math>, and <math> EA = 5 </math>.
+Let Line <math>FD = x</math> and let Line <math>FG = y</math>.
+We find that <math>\triangle FGE \sim \triangle CDE </math> through some angle chasing (they both have a right angle, and they both share angle <math>\angle CED </math>. Using this information, we can write the equation <math>\frac{4}{8} = \frac{4+x}{y} </math>. Through simplifying this equation, we get that <math> y=2x+8 </math>. Let point <math> I </math> be the point on line <math> FG </math> so that lines <math> CI </math> and <math> FG </math> are perpendicular, and we get that <math> GI = 2x </math> and <math> FI =8 </math>. Doing some more angle chasing, we can find that <math>\triangle GIH \sim \triangle GFA </math>, as they both share <math>\angle FGH </math> and they both have a right angle.
+With this information, we can write the equation <math>\frac{x+3}{y-8} = \frac{x+9}{y}. </math>
+Simplifying this equation we get the equation <math> -8x+6y-72 = 0 </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 therefore, our answer is <math>GF= \boxed{\mathrm{(B)}\ 20}</math>
+~Darth_Cadet
+===Solution 8 Line and Slope===
+Draw a coordinate plane with the y-axis centered on <math>CD</math> and the x-axis centered on <math>AD</math>. From there, call the line passing through <math>CE</math> <math>a</math> and the line passing through <math>HA</math> <math>b</math>. From there, you can find that the equations for these lines are <math>y=-2x+8</math> and <math>y=-4/3 x+12</math> respectively. We need to find the length of <math>GF</math>, so we are finding the y-value of the point <math>G</math>. Solving for this point, we get <math>-4/3 x+12=-2x+8</math>;<math>2/3x=-4</math>;<math>x=-6</math>. Now, plugging in the values, we find that <math>y=-2(-6)+8=20</math>, so our answer is <math>GF= \boxed{\mathrm{(B)}\ 20}</math>.
+~iamcalifornia'sresidentidiot
 == See Also ==
@@ Line 106: / Line 137: @@
 [[Category:Introductory Geometry Problems]]
+{{MAA Notice}}

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