AMC 8 · 2010 · #17

Grade 6 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation

area-rectanglesarea-trianglescoordinate-geometryratio-proportion area-differenceidentify-subproblemscoordinate-geometry ↑ Prerequisites: area-rectanglesarea-triangles

📏 Long solution 💡 4 insights 📊 Diagram

Problem

An octagon is made of 10 unit squares, so its total area is 10. A segment PQ cuts the octagon into two pieces of equal area. The piece below PQ is described as a unit square plus a triangle with base 5. Point Q sits on the vertical segment XY, where X=(5,2) is the top corner and Y=(5,1) is the bottom corner. Find the ratio

\frac{XQ}{QY}

Pick an answer.

(A)

$\frac{2}{5}$

(B)

$\frac{1}{2}$

(C)

$\frac{3}{5}$

(D)

$\frac{2}{3}$

(E)

$\frac{3}{4}$

AMC 8 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The problem hands us a clean decomposition of the lower region: one unit square plus a triangle with base 5. Tool #7 (Identify Subproblems) lets us treat "area of the lower region" as "area of square + area of triangle" and solve for the triangle's height — which is exactly the y-coordinate of Q. Tool #1 (Draw a Diagram) is the supporting move: marking X=(5,2), Y=(5,1), and Q=(5,h) on the figure makes it visually obvious that XQ = 2-h and QY = h-1, so the final ratio is just two subtractions.

1STEP 1

The octagon is 10 unit squares, so its area is 10; a bisector makes each half area 5.

Area below PQ =

\frac{10}{2}

= 5

2STEP 2

Write the lower piece as a unit square of area 1 plus a triangle with base 5 and height h, then set the sum equal to 5.

1 +

\frac{1}{2}

(5)(h) = 5

3STEP 3

Subtract 1 from both sides and divide by $\frac{5}{2}$ to get the height h = $\frac{8}{5}$ .

\frac{5}{2}

h = 4 → h =

\frac{8}{5}

4STEP 4

Q is (5, $\frac{8}{5}$ ), so XQ = 2 - $\frac{8}{5}$ = $\frac{2}{5}$ and QY = $\frac{8}{5}$ - 1 = $\frac{3}{5}$ .

XQ = 2 -

\frac{8}{5}

\frac{2}{5}

, QY =

\frac{8}{5}

- 1 =

\frac{3}{5}

5STEP 5

Both lengths share denominator 5, so $\frac{XQ}{QY}$ = ( $\frac{2}{5}$ )/( $\frac{3}{5}$ ) = $\frac{2}{3}$ , which is (D).

\frac{XQ}{QY}

= (

\frac{2}{5}

)/(

\frac{3}{5}

) =

\frac{2}{3}

→ (D)

Answer

\frac{2}{3}

Check the height h =

\frac{8}{5}

= 1.6 — it lies between 1 and 2, exactly where Q must sit on XY. Verify the area: square gives 1 and the triangle gives

\frac{1}{2}

(5)(1.6) = 4, totaling 5, which is half of 10. Finally, XQ + QY =

\frac{2}{5}

\frac{3}{5}

= 1, matching the length of XY from y=1 to y=2. All three checks pass, and the answer (D)

\frac{2}{3}

is one of the listed options.

💡Key takeaway

Once you split the lower region into a square plus a triangle, this AMC 8 problem only needs Grade 6 area and ratio skills you already have.

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