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2014 AMC 10A Problem 16

2014 AMC 10A Problem 16AMC 10 Step by Step

Name: Date: September 13, 2026

Quadrilaterals & Polygon Areas · difficulty 3 of 5 · about 4 min · position P16-20

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?

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

Answer

Key insight

The shaded region is a kite with diagonal HF of length 1 along the vertical midline and a horizontal diagonal of length 1/3, so its area is (1/2)(1)(1/3).

Solution

Place $D = (0,0)$ , $C = (1,0)$ , $B = (1,2)$ , $A = (0,2)$ . Then $F = \left(\tfrac12, 0\right)$ , $G = (0,1)$ , $E = (1,1)$ and $H = \left(\tfrac12, 1\right)$ .

The shaded region is bounded by segments of $\overline{AF}$ , $\overline{DH}$ , $\overline{CH}$ , $\overline{BF}$ , with vertices $H$ (top), $F$ (bottom), and the two side corners $X = AF \cap DH$ and $Y = BF \cap CH$ .

Line $AF$ through $(0,2)$ and $\left(\tfrac12,0\right)$ : $y = 2 - 4x$ . Line $DH$ through $(0,0)$ and $\left(\tfrac12,1\right)$ : $y = 2x$ . Solving $2x = 2 - 4x$ gives $x = \tfrac13$ , $y = \tfrac23$ , so $X = \left(\tfrac13, \tfrac23\right)$ . By the left-right symmetry of the figure, $Y = \left(\tfrac23, \tfrac23\right)$ .

The quadrilateral $HXFY$ has perpendicular diagonals: $\overline{HF}$ is vertical with length $1$ , and $\overline{XY}$ is horizontal with length $\tfrac13$ . Its area is half their product:
$\frac12 \cdot 1 \cdot \frac13 = \frac16.$

The answer is $\boxed{\textbf{(E)}\ \frac16}$ .

Why this works

Midpoint-heavy rectangle figures are ideal for coordinates: every relevant point has small rational coordinates and the bounding lines are easy to write. Symmetry about the vertical midline halves the work, and recognizing the region as a kite (perpendicular diagonals) turns the area into one multiplication instead of a shoelace computation.

Alternative approach

Synthetically: let $P$ be the foot of the perpendicular from $X$ to side $\overline{AD}$ . Right triangles $DPX$ and $DGH$ share the angle at $D$ , so $PX = \tfrac12 DP$ ; right triangles $APX$ and $ADF$ share the angle at $A$ , so $PX = \tfrac14 AP$ . With $DP + AP = 2$ this gives $DP = \tfrac23$ . Then the shaded kite is $\triangle DHC$ (area $\tfrac12$ ) minus $\triangle DXF$ and $\triangle CYF$ (each $\tfrac12 \cdot \tfrac12 \cdot \tfrac23 = \tfrac16$ ), leaving $\tfrac12 - \tfrac13 = \tfrac16$ .

The trap

Assuming the side corners X and Y lie on the midline GE (height 1) instead of at height 2/3, which gives the wrong kite dimensions.

Common mistakes

Assuming the side corners X and Y lie on the midline GE (height 1) instead of at height 2/3, which gives the wrong kite dimensions.
Computing the kite's area as the full product of the diagonals, $\tfrac13$ , or treating $HXFY$ as a square.

Techniques

Place the figure on coordinates and compute · Exploit symmetry to reduce work or pair up objects