AMC 10 · 2014 · #16

Grade 8 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Line Intersection Vertices then Area
coordinate-geometrysystems-of-equationsarea-triangles convert-to-algebraidentify-subproblems ↑ Prerequisites: coordinate-geometryarea-triangles
📏 Medium solution 💡 2 insights 📊 Diagram
Problem
A rectangle ABCD has short side AB=1 and long side BC=2. Points E, F, G are the midpoints of BC, CD, AD, and H is the midpoint of GE. Segments GE, the two-part path A–F–B, and the two-part path D–H–C are drawn. Find the area of the shaded kite bounded by those two paths, with H as its top corner and F as its bottom corner.

Pick an answer.

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

AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

The figure pins down every point by exact midpoints, so the cleanest move is Tool #1 (Draw a Diagram) upgraded to a coordinate grid: drop the rectangle onto axes and every corner becomes a simple fraction. Once coordinates exist, Tool #4 (Introduce a Variable) turns each border of the kite into a line equation y=mx+b, and the two side corners are just where pairs of those lines meet. Tool #7 (Identify Subproblems) then splits the finish into three small steps — find the crossing points, read the two diagonals, multiply — because the shaded shape is a kite whose diagonals happen to be one vertical and one horizontal segment. No circle, no trig, no similar-triangle chase is needed; the grid does the heavy lifting.

1STEP 1

Put the figure on coordinates

Set D=(0,0) with the long side up, so A=(0,2), F=(1/2,0), G=(0,1), E=(1,1), and H=(1/2,1).

D=(0,0), C=(1,0), B=(1,2), A=(0,2), E=(1,1), F=(1/2,0), G=(0,1), H=(1/2,1)
2STEP 2

Write the four border lines

Two points fix each wall of the kite: AF has slope -4, DH slope 2, HC slope -2, FB slope 4.

AF: y=2-4x, DH: y=2x, HC: y=2-2x, FB: y=4x-2
3STEP 3

Find the two side corners

Setting AF=DH gives the left corner X=(1/3,2/3), and HC=FB gives the right corner Y=(2/3,2/3).

X: 2-4x=2x→ x=1/3, X=(1/3,2/3); Y: 2-2x=4x-2→ x=2/3, Y=(2/3,2/3)
4STEP 4

Read off the two diagonals

H and F share x=1/2, so diagonal HF is vertical of length 1; X and Y share y=2/3, so diagonal XY is horizontal of length 1/3.

HF: x=1/2, length=1-0=1; XY: y=2/3, length=2/3-1/3=1/3
5STEP 5

Multiply the diagonals

The diagonals cross at a right angle, so the area is half their product: 1/2·1·1/3=1/6, choice (E).

Area=1/2 (HF)(XY)=1/2 · 1·1/3=1/6 → (E)
Answer
1/6
The shoelace formula on the four corners H(1/2,1), X(1/3,2/3), F(1/2,0), Y(2/3,2/3) gives 1/2|{-1/3+1/3+1/3}|=1/6, matching the diagonal method. The whole rectangle has area 1×2=2, so the kite is 1/12 of it — a believable little sliver squeezed near the center. The kite is also symmetric about the vertical line x=1/2 (with X and Y mirror images), exactly as the picture looks.
💡Key takeaway

Drop a right-angled figure onto a coordinate grid, turn each segment into a line equation, and a messy shaded region becomes a kite whose area is just half the product of its two diagonals.

  • Put the figure on coordinates
  • Write the four border lines
  • Find the two side corners
  • Read off the two diagonals
  • Multiply the diagonals

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