Competition · AMC preparation · step 4 of 4

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

Find the area on each side

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

Area below PQ = 10/2 = 5

2STEP 2

Write the area equation

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 + 1/2(5)(h) = 5

Splitting a compound region into a square and a triangle, then adding their areas, is the standard Grade 6 strategy for polygon areas.

The area of the region below PQ can be written as the unit square's area of 1 plus the base-5 triangle's area, and this measured area must equal half of the whole octagon.

▸ Why?

The region below PQ is a unit square set next to a triangle of base 5, with no gap and no overlap, so its area is the square's area added to the triangle's area.

▸ Why?

When a shape is cut into pieces that do not overlap and leave no gap, the pieces' areas add back to the area of the whole shape.

Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.

▸ Why?

That triangle has a right angle where its base meets the vertical side, so it is exactly half of the 5-by-h rectangle built on its base and height, giving area one half of 5 times h.

▸ Why?

The rectangle on the base is 5 rows of h unit-square strips, so its area is 5 times h.

Multiplication as equal groupsa times b means a equal groups of b — it is just repeated adding of the same amount.

▸ Why?

The diagonal of that rectangle splits it into two triangles that turn exactly onto each other, so they have equal area and each is half the rectangle.

Rigid motion preserves length and angleSliding, turning, or flipping a shape lays it exactly onto a copy, so every length and angle stays the same (folding is the flip case).

▸ Why?

The segment PQ bisects the octagon, so the region below it is exactly half of the octagon's total area.

▸ Why?

The octagon is made of 10 unit squares fitted together with no overlap, so its whole area is 10.

Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.

▸ Why?

Two equal halves rejoin to make the whole of 10, so each half is the number that doubles to 10.

Inverse operationsSubtraction undoes addition and division undoes multiplication — they reverse each other.

6.G.A.1Identify Subproblems

3STEP 3

Solve for the height

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

5/2h = 4 → h = 8/5

4STEP 4

Locate Q on the segment

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 - 8/5 = 2/5, QY = 8/5 - 1 = 3/5

5STEP 5

Form the requested ratio

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

XQ/QY = 2/5/3/5 = 2/3 → (D)

Answer

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.

Find the area on each side
Write the area equation
Solve for the height
Locate Q on the segment
Form the requested ratio

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Find the area on each side

Write the area equation

Solve for the height

Locate Q on the segment

Form the requested ratio

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