Difference between revisions of "2014 AMC 10A Problems/Problem 16"

Revision as of 15:17, 10 April 2025

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 :)

Solution 6 (Elementary Construction)

Call the unmarked vertices of the quadrilateral P, and Q. draw $\overline{HF}$ . $\linebreak$ $\overline{HF}$ connects midpoints of opposite sides of a square, so $\overline{HF}$ \parallel \overline{AD} $and,$ HF = 1 $Then, by AA similarity we get that$ \deltaHFP \sim \deltaDAP $, with a scale factor of$ \frac{1}{2} $(look at the sides).$ \linebreak $Dropping altitudes from P and Q, to points P' and Q', and P'' and Q'' we get: PP'' + PP' = QQ'' + QQ' =$ \frac{1}{2} $.$ \linebreak $Then, as$ \deltaHFP \sim \deltaDAP $, as well as$ \deltaHFQ \sim \deltaCBQ $, we get that 3PP'= 3QQ' =$ \frac{1}{2} $So, PP`=QQ` = 1/6$ \linebreak $Then as [HFPQ] = [$ \deltaHFP$]+[\deltaHFQ], we get that [HFPQ] = 1*$ (Error compiling LaTeX. Unknown error_msg)\frac{1}{6} $*\frac{1}{2}$ + 1* $\frac{1}{6}$ * $\frac{1}{2}$ or [HFPQ] = $\boxed{\textbf{(E)} \: \frac{1}{6}}$ Feel free to correct any Errors! ~StereoTypicalMathNerd.

@@ Line 157: / Line 157: @@
 ~JH. L :)
-==Solution 6(Elementary Construction)==
+==Solution 6 (Elementary Construction)==
-draw <math>\overline{HF}</math>.
-<math>\overline{HF}</math> connects midpoints of opposite sides of a square, so
+Call the unmarked vertices of the quadrilateral P, and Q. draw <math>\overline{HF}</math>.
+<math>\linebreak</math>
+<math>\overline{HF}</math> connects midpoints of opposite sides of a square, so <math>\overline{HF}</math> \parallel \overline{AD}<math> and, </math>HF = 1 <math>
+Then, by AA similarity we get that </math>\deltaHFP \sim \deltaDAP<math>, with a scale factor of </math>\frac{1}{2}<math> (look at the sides).
+</math>\linebreak<math>
+Dropping altitudes from P and Q, to points P' and Q', and P'' and Q'' we get:  PP'' + PP' = QQ'' + QQ' =</math>\frac{1}{2}<math>.
+</math>\linebreak<math>
+Then, as </math>\deltaHFP \sim \deltaDAP<math>, as well as </math>\deltaHFQ \sim \deltaCBQ<math>, we get that 3PP'= 3QQ' = </math>\frac{1}{2}<math>
+So, PP`=QQ` = 1/6
+</math>\linebreak<math>
+Then as [HFPQ] = [</math>\deltaHFP<math>]+[\deltaHFQ], we get that [HFPQ] = 1*</math>\frac{1}{6}<math>*\frac{1}{2}</math> + 1*<math>\frac{1}{6}</math>*<math>\frac{1}{2}</math> or [HFPQ] = <math>\boxed{\textbf{(E)} \: \frac{1}{6}}</math>
+Feel free to correct any Errors!
+~StereoTypicalMathNerd.
 ==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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