AMC 10 · 2006 · #17

Grade 8 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Line Intersection Vertices then Area

coordinate-geometrysimilar-trianglesarea-triangles symmetry-argument ↑ Prerequisites: coordinate-geometry

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

In rectangle $ADEH$ , points $B$ and $C$ trisect $\overline{AD}$ , and points $G$ and $F$ trisect $\overline{HE}$ . In addition, $AH=AC=2$ , and $AD=3$ . What is the area of quadrilateral $WXYZ$ shown in the figure?

Pick an answer.

(A)

$\frac{1}{2}$

(B)

$\frac{\sqrt{2}}{2}$

(C)

$\frac{\sqrt{3}}{2}$

(D)

$\sqrt{2}$

(E)

$\frac{2\sqrt{3}}{3}$

AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: Rectangle $ADEH$ has a top side $AD$ of length $3$ and a height $AH$ of length $2$. Points $B$ and $C$ cut the top side $AD$ into three equal parts, and points $G$ and $F$ cut the bottom side $HE$ into three equal parts. Four slanted segments are drawn — $A$ to $F$, $B$ to $E$, $C$ to $H$, and $D$ to $G$ — and they cross each other to carve out a small tilted quadrilateral $WXYZ$ in the middle. Find the area of $WXYZ$.

Givens: $ADEH$ is a rectangle with $AD=3$ (the width) and $AH=2$ (the height); $B$ and $C$ trisect $\overline{AD}$, so $AB=BC=CD=1$; note $AC=2$ as stated; $G$ and $F$ trisect $\overline{HE}$, so $HG=GF=FE=1$; The four drawn segments are $A\!-\!F$, $B\!-\!E$, $C\!-\!H$, and $D\!-\!G$; $W=B\!-\!E\cap C\!-\!H$, $X=B\!-\!E\cap D\!-\!G$, $Y=A\!-\!F\cap D\!-\!G$, $Z=A\!-\!F\cap C\!-\!H$; Answer choices: (A) $\frac{1}{2}$, (B) $\frac{\sqrt{2}}{2}$, (C) $\frac{\sqrt{3}}{2}$, (D) $\sqrt{2}$, (E) $\frac{2\sqrt{3}}{3}$

Unknowns: The four corner points $W$, $X$, $Y$, $Z$ where the slanted segments cross; The area of quadrilateral $WXYZ$

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #4 Introduce a Variable, #7 Identify Subproblems

The picture is a tangle of four crossing segments, so the safest move is to pin the whole thing onto a coordinate grid (Tool #1, Draw a Diagram, made exact with coordinates). Once every labeled point has an $(x,y)$ address, each drawn segment becomes a simple line equation (Tool #4, Introduce a Variable — here the variables are $x$ and $y$). Two of the lines slant one way and two slant the other, and the two slants turn out to be perpendicular, which already tells us $WXYZ$ is a rectangle. Then finding each corner is just a small subproblem: solve where two lines cross (Tool #7, Identify Subproblems). Finally the area comes straight from the corner coordinates. Coordinates trade a confusing figure for four tidy pieces of algebra.

Execute — Answer: A

#1 Draw a Diagram 6.G.A.3 Step 1

Put the rectangle on a grid

Place corner $H$ at the origin and let the width run right, the height run up.
Then $H=(0,0)$, $E=(3,0)$, $A=(0,2)$, $D=(3,2)$.
The trisection points follow: on the top, $B=(1,2)$ and $C=(2,2)$; on the bottom, $G=(1,0)$ and $F=(2,0)$.
This matches the givens — $AC$ goes from $(0,2)$ to $(2,2)$, length $2$, and $AD$ has length $3$.

$$H(0,0),\;E(3,0),\;A(0,2),\;D(3,2),\;B(1,2),\;C(2,2),\;G(1,0),\;F(2,0)$$

💡 Giving every point an address turns a messy drawing into numbers you can compute with.

#4 Introduce a Variable 8.EE.B.6 Step 2

Write the four slanted lines

Each drawn segment is a straight line; find its slope and equation.
$A(0,2)\!\to\!F(2,0)$ drops $2$ over a run of $2$, slope $-1$: $y=2-x$.
$B(1,2)\!\to\!E(3,0)$ also has slope $-1$: $y=3-x$.
$C(2,2)\!\to\!H(0,0)$ rises $2$ over $2$, slope $+1$: $y=x$.
$D(3,2)\!\to\!G(1,0)$ also has slope $+1$: $y=x-1$.
So $\{A F,\,B E\}$ are a parallel pair (slope $-1$) and $\{C H,\,D G\}$ are a parallel pair (slope $+1$).
Slopes $-1$ and $+1$ multiply to $-1$, so the two families are perpendicular — meaning $WXYZ$ has four right angles and is a rectangle.

$$AF:\,y=2-x,\quad BE:\,y=3-x,\quad CH:\,y=x,\quad DG:\,y=x-1$$

💡 Two slants of $-1$ and $+1$ meet at a right angle, so the middle piece is a true rectangle.

#7 Identify Subproblems 8.EE.C.8 Step 3

Solve for the four corners

Each corner is where two of these lines cross, so set the two equations equal and solve.
$Z=AF\cap CH$: $2-x=x\Rightarrow x=1$, so $Z=(1,1)$.
$W=BE\cap CH$: $3-x=x\Rightarrow x=\tfrac32$, so $W=(\tfrac32,\tfrac32)$.
$Y=AF\cap DG$: $2-x=x-1\Rightarrow x=\tfrac32$, so $Y=(\tfrac32,\tfrac12)$.
$X=BE\cap DG$: $3-x=x-1\Rightarrow x=2$, so $X=(2,1)$.

$$Z(1,1),\quad W\!\left(\tfrac32,\tfrac32\right),\quad X(2,1),\quad Y\!\left(\tfrac32,\tfrac12\right)$$

💡 Where two lines meet is exactly where their two equations give the same point, so just set them equal.

#7 Identify Subproblems 6.G.A.1 Step 4

Take the area from the diagonals

The two diagonals of $WXYZ$ are $WY$ and $XZ$.
$W(\tfrac32,\tfrac32)$ to $Y(\tfrac32,\tfrac12)$ is vertical with length $\tfrac32-\tfrac12=1$.
$X(2,1)$ to $Z(1,1)$ is horizontal with length $2-1=1$.
The diagonals are perpendicular (one vertical, one horizontal), cross at the center $(\tfrac32,1)$, and are equal in length, confirming a square.
For a quadrilateral with perpendicular diagonals the area is half their product: $\tfrac12\cdot d_1\cdot d_2=\tfrac12\cdot 1\cdot 1=\tfrac12$.
So the area of $WXYZ$ is $\tfrac12$, which is choice (A).

$$WY=1,\;\;XZ=1,\;\;[WXYZ]=\tfrac12\cdot WY\cdot XZ=\tfrac12\cdot 1\cdot 1=\tfrac12\;\Rightarrow\;\textbf{(A)}$$

💡 When the diagonals cross at right angles, half of their product is the whole area.

[1] #1 6.G.A.3 Place corner $H$ at the origin and let the width run right, the height run up. T

[2] #4 8.EE.B.6 Each drawn segment is a straight line; find its slope and equation. $A(0,2)\!\to

[3] #7 8.EE.C.8 Each corner is where two of these lines cross, so set the two equations equal an

[4] #7 6.G.A.1 The two diagonals of $WXYZ$ are $WY$ and $XZ$. $W(\tfrac32,\tfrac32)$ to $Y(\tfr

Review

Reasonableness: A quick side-length check agrees: one side runs from $Z(1,1)$ to $W(\tfrac32,\tfrac32)$, length $\sqrt{(\tfrac12)^2+(\tfrac12)^2}=\tfrac{\sqrt2}{2}$, and $\left(\tfrac{\sqrt2}{2}\right)^2=\tfrac12$ — the same area. The value is also sensible in scale: the whole rectangle has area $6$, and the little middle square taking up just $\tfrac12$ of that (one-twelfth) looks right for the tiny region in the picture. The distractors are the traps of stopping too early — $\tfrac{\sqrt2}{2}$ is the side length, not the area, and the $\sqrt3$ choices would need a $30$-$60$-$90$ angle that never appears here. Only (A) matches.

Alternative: Skip coordinates and argue by symmetry. Because slopes $\pm1$ meet at $45^\circ$, each crossing builds a $45$-$45$-$90$ right triangle, so $WXYZ$ is a square tilted $45^\circ$. Its diagonal $XZ$ is horizontal and stretches from line $CH$ to line $DG$ at height $y=1$: line $CH$ gives $x=1$ and line $DG$ gives $x=2$, so the diagonal has length $1$. A square with diagonal $1$ has area $\tfrac12\cdot 1^2=\tfrac12$. Same answer, using only the $45^\circ$ slopes and one horizontal cut.

CCSS standards used (min grade 8)

6.G.A.3 Draw polygons in the coordinate plane given coordinates for the vertices (Placing the rectangle and every trisection point on a coordinate grid so each has an exact $(x,y)$ address.)
8.EE.B.6 Use similar triangles to explain why the slope is the same between any two points (Finding the slope and equation of each of the four drawn segments and recognizing the two perpendicular families of lines.)
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations (Solving each pair of line equations to locate the four crossing points $W$, $X$, $Y$, $Z$.)
6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing (Computing the area of $WXYZ$ as half the product of its perpendicular diagonals, $\frac12\cdot1\cdot1=\frac12$.)

⭐ When a figure is a mess of crossing lines, drop it onto a coordinate grid — then every line is an equation and every crossing point is just where two equations agree.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Put the rectangle on a grid

Write the four slanted lines

Solve for the four corners

Take the area from the diagonals

Review

CCSS standards used (min grade 8)

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