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2014 AMC 10A Problems/Problem 16

Problem

In rectangle $ABCD$, $AB=1$, $BC=2$, and points $E$, $F$, and $G$ are midpoints of $\overline{BC}$, $\overline{CD}$, and $\overline{AD}$, respectively. Point $H$ is the midpoint of $\overline{GE}$. What is the area of the shaded region?

[asy] import graph; size(9cm); pen dps = fontsize(10); defaultpen(dps); pair D = (0,0); pair F = (1/2,0); pair C = (1,0); pair G = (0,1); pair E = (1,1); pair A = (0,2); pair B = (1,2); pair H = (1/2,1); // do not look pair X = (1/3,2/3); pair Y = (2/3,2/3); draw(A--B--C--D--cycle); draw(G--E); draw(A--F--B); draw(D--H--C); filldraw(H--X--F--Y--cycle,grey); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$E$",E,E); label("$F$",F,S); label("$G$",G,W); label("$H$",H,N); label("$\frac12$",(0.25,0),S); label("$\frac12$",(0.75,0),S); label("$1$",(1,0.5),E); label("$1$",(1,1.5),E); [/asy]

$\textbf{(A)}\ \dfrac1{12}\qquad\textbf{(B)}\ \dfrac{\sqrt3}{18}\qquad\textbf{(C)}\ \dfrac{\sqrt2}{12}\qquad\textbf{(D)}\ \dfrac{\sqrt3}{12}\qquad\textbf{(E)}\ \dfrac16$

Solution 1

Denote $D=(0,0)$. Then $A= (0,2), F = \left(\frac12,0\right), H = \left(\frac12,1\right)$. Let the intersection of $AF$ and $DH$ be $X$, and the intersection of $BF$ and $CH$ be $Y$. Then we want to find the coordinates of $X$ so we can find $XY$. From our points, the slope of $AF$ is $\bigg(\dfrac{-2}{\tfrac12}\bigg) = -4$, and its $y$-intercept is just $2$. Thus the equation for $AF$ is $y = -4x + 2$. We can also quickly find that the equation of $DH$ is $y = 2x$. Setting the equations equal, we have $2x = -4x +2 \implies x = \frac13$. Because of symmetry, we can see that the distance from $Y$ to $BC$ is also $\frac13$, so $XY = 1 - 2 \cdot \frac13 = \frac13$. Now the area of the kite is simply the product of the two diagonals over $2$. Since the length $HF = 1$, our answer is $\dfrac{\dfrac{1}{3} \cdot 1}{2} = \boxed{\textbf{(E)} \: \dfrac16}$.

[asy] import graph; size(9cm); pen dps = fontsize(10); defaultpen(dps); pair D = (0,0); pair F = (1/2,0); pair C = (1,0); pair G = (0,1); pair E = (1,1); pair A = (0,2); pair B = (1,2); pair H = (1/2,1); // do not look pair X = (1/3,2/3); pair Y = (2/3,2/3); draw(A--B--C--D--cycle); draw(G--E); draw(A--F--B); draw(D--H--C); filldraw(H--X--F--Y--cycle,grey); draw(X--Y,dashed); label("$A\: (0,2)$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D \: (0,0)$",D,SW); label("$E$",E,E); label("$F\: (\frac12,0)$",F,S); label("$G$",G,W); label("$H \: (\frac12,1)$",H,N); label("$Y$",Y,E); label("$X$",X,W); label("$\frac12$",(0.25,0),S); label("$\frac12$",(0.75,0),S); label("$1$",(1,0.5),E); label("$1$",(1,1.5),E); [/asy]

Solution 2

Let the area of the shaded region be $x$. Let the other two vertices of the kite be $I$ and $J$ with $I$ closer to $AD$ than $J$. Note that $[ABCD] = [ABF] + [DCH] - x + [ADI] + [BCJ]$. The area of $ABF$ is $1$ and the area of $DCH$ is $\dfrac{1}{2}$. We will solve for the areas of $ADI$ and $BCJ$ in terms of x by noting that the area of each triangle is the length of the perpendicular from $I$ to $AD$ and $J$ to $BC$ respectively. Because the area of $x$ = $\dfrac{1}{2} \cdot IJ$ based on the area of a kite formula, $\dfrac{ab}{2}$ for diagonals of length $a$ and $b$, $IJ = 2x$. So each perpendicular is length $\dfrac{1-2x}{2}$. So taking our numbers and plugging them into $[ABCD] =[ABF] + [DCH] - x + [ADI] + [BCJ]$ gives us $2 = \dfrac{5}{2} - 3x$ Solving this equation for $x$ gives us $x = \boxed{\textbf{(E)} \: \frac{1}{6}}$

Solution 3

From the diagram in Solution 1, let $e$ be the height of $XHY$ and $f$ be the height of $XFY$. It is clear that their sum is $1$ as they are parallel to $GD$. Let $k$ be the ratio of the sides of the similar triangles $XFY$ and $AFB$, which are similar because $XY$ is parallel to $AB$ and the triangles share angle $F$. Then $k = f/2$, as 2 is the height of $AFB$. Since $XHY$ and $DHC$ are similar for the same reasons as $XFY$ and $AFB$, the height of $XHY$ will be equal to the base, like in $DHC$, making $XY = e$. However, $XY$ is also the base of $XFY$, so $k = e / AB$ where $AB = 1$ so $k = e$. Subbing into $k = f/2$ gives a system of linear equations, $e + f = 1$ and $e = f/2$. Solving yields $e = XY = 1/3$ and $f = \frac{2}{3}$, and since the area of the kite is simply the product of the two diagonals over $2$ and $HF = 1$, our answer is $\frac{\frac{1}{3} \cdot 1}{2} = \boxed{\textbf{(E)} \: \dfrac16}$.

Solution 4

Let the unmarked vertices of the shaded area be labeled $I$ and $J$, with $I$ being closer to $GD$ than $J$. Noting that kite $HJFI$ can be split into triangles $HJI$ and $JIF$.

Lemma: The distance from line segment $JI$ to $H$ is half the distance from $JI$ to $F$

Proof: Drop perpendiculars of triangles $HJI$ and $JIF$ to line $JI$, and let the point of intersection be $Q$. Note that $HJI$ and $JIF$ are similar to $HDC$ and $ABF$, respectively. Now, the ratio of $DC$ to $HF$ is $1:1$, which shows that the ratio of $JI$ to $HQ$ is $1:1$, because of similar triangles as described above. Similarly, the ratio of $JI$ to $FQ$ is $1:2$. Since these two triangles contain the same base, $JI$, the ratio of $HQ:FQ = 1:2$.

Because kite $HJFI$ is orthodiagonal, we multiply $\frac{1\cdot\tfrac{1}{3}}{2} = \boxed{\textbf{(E)} \: \frac{1}{6}}$

~Lemma proof by sakshamsethi

Solution 5 (Similarity)

[asy] import graph; size(9cm); pen dps = fontsize(10); defaultpen(dps); pair D = (0,0); pair F = (1/2,0); pair C = (1,0); pair G = (0,1); pair E = (1,1); pair A = (0,2); pair B = (1,2); pair H = (1/2,1); // do not look pair X = (1/3,2/3); pair Y = (2/3,2/3); draw(A--B--C--D--cycle); draw(G--E); draw(A--F--B); draw(D--H--C); filldraw(H--X--F--Y--cycle,grey); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$E$",E,E); label("$F$",F,S); label("$G$",G,W); label("$H$",H,N); label("$\frac12$",(0.25,0),S); label("$\frac12$",(0.75,0),S); label("$1$",(1,0.5),E); label("$1$",(1,1.5),E); [/asy]

The area of the shaded area is the area of $\triangle DHC$ minus the two triangles on the side. Extend $\overline{DH}$ so that it hits point $B$. Call the intersection of $\overline{AF}$ and $\overline{DB}$ point $P$. \[\triangle APB \sim \triangle FPD\] Drop altitudes from $P$ down to $\overline{DF}$ and $\overline{AB}$; call the intersection points $L$ and $M$ respectively. \[\frac{DF}{AB}=\frac{LP}{PM}=\frac{\frac{1}{2}}{1}=\frac{1}{2}\] \[LP=\frac{1}{3}\cdot LM=\frac{2}{3}\] Thus the two triangles on the side have area $\frac{1}{2} \cdot \frac{2}{2} \cdot \frac {1}{3} = \frac{1}{6}$. Since there are two, their total area is $2 \cdot \frac{1}{6} = \frac{1}{3}$. The area of $\triangle DHC$ is $\frac{1}{2} \cdot 1 \cdot 1=\frac{1}{2}$. The shaded region is $\frac{1}{2}-\frac{1}{3} = \frac{1}{6}$ which is $\boxed{\textbf{(E)} \frac{1}{6}}$.

~JH. L :)

See Also

2014 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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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