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AMC 8 · 2010 · #17

Grade 6 geometry-2d
Archetype Compound Area by Decomposition / Difference
area-rectanglesarea-trianglescoordinate-geometryratio-proportion area-differenceidentify-subproblemscoordinate-geometry ↑ Prerequisites: area-rectanglesarea-triangles
📏 Long solution 💡 4 insights 📊 Diagram

Problem

The diagram shows an octagon consisting of 1010 unit squares. The portion below PQ‾\overline{PQ} is a unit square and a triangle with base 55. If PQ‾\overline{PQ} bisects the area of the octagon, what is the ratio XQQY\dfrac{XQ}{QY}?

Pick an answer.

(A)
$\frac{2}{5}$
(B)
$\frac{1}{2}$
(C)
$\frac{3}{5}$
(D)
$\frac{2}{3}$
(E)
$\frac{3}{4}$
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Toolkit + CCSS Solution

Understand

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

Givens: Octagon area $= 10$ (it is made of $10$ unit squares); $\overline{PQ}$ bisects the octagon's area; Region below $\overline{PQ}$ $=$ one unit square $+$ a triangle with base $5$; $X=(5,2)$ and $Y=(5,1)$, and $Q$ lies on $\overline{XY}$; Answer choices: (A) $\tfrac{2}{5}$, (B) $\tfrac{1}{2}$, (C) $\tfrac{3}{5}$, (D) $\tfrac{2}{3}$, (E) $\tfrac{3}{4}$

Unknowns: The ratio $\dfrac{XQ}{QY}$

Understand

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

Givens: Octagon area $= 10$ (it is made of $10$ unit squares); $\overline{PQ}$ bisects the octagon's area; Region below $\overline{PQ}$ $=$ one unit square $+$ a triangle with base $5$; $X=(5,2)$ and $Y=(5,1)$, and $Q$ lies on $\overline{XY}$; Answer choices: (A) $\tfrac{2}{5}$, (B) $\tfrac{1}{2}$, (C) $\tfrac{3}{5}$, (D) $\tfrac{2}{3}$, (E) $\tfrac{3}{4}$

Plan

Primary tool: #7 Identify Subproblems

Secondary: #1 Draw a Diagram

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.

Execute — Answer: D

#7 Identify Subproblems 3.MD.C.7 Step 1
  • Find the area each side of $\overline{PQ}$ must have.
  • The octagon is made of $10$ unit squares, so its total area is $10$.
  • A bisector splits this evenly.
$$\text{Area below } \overline{PQ} = \dfrac{10}{2} = 5$$

💡 Counting unit squares to get area, then halving, is a Grade 3 area skill.

#7 Identify Subproblems 6.G.A.1 Step 2
  • Use the given decomposition.
  • The region below $\overline{PQ}$ is a unit square (area $1$) plus a triangle with base $5$ and some height $h$.
  • Set the total equal to $5$.
$$1 + \tfrac{1}{2}(5)(h) = 5$$

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

#7 Identify Subproblems 6.EE.B.7 Step 3
  • Solve the equation for the triangle's height $h$.
  • Subtract $1$ from both sides, then divide by $\tfrac{5}{2}$.
$$\tfrac{5}{2}h = 4 \;\Rightarrow\; h = \tfrac{8}{5}$$

💡 Solving a one-step equation of the form $ax = b$ is Grade 6 equation-solving.

#1 Draw a Diagram 5.NF.A.1 Step 4
  • Place $Q$ on the segment $\overline{XY}$.
  • Since $X=(5,2)$ and $Y=(5,1)$ share the x-coordinate $5$, the point $Q$ on the same vertical line has coordinates $(5, h) = (5, \tfrac{8}{5})$.
  • The lengths $XQ$ and $QY$ are just differences of y-coordinates.
$$XQ = 2 - \tfrac{8}{5} = \tfrac{2}{5},\quad QY = \tfrac{8}{5} - 1 = \tfrac{3}{5}$$

💡 Subtracting fractions with a common denominator to get a vertical distance is a Grade 5 fraction skill.

#7 Identify Subproblems 6.RP.A.1 Step 5
  • Form the requested ratio.
  • Both lengths have the same denominator, so it simplifies to a ratio of numerators.
$$\dfrac{XQ}{QY} = \dfrac{2/5}{3/5} = \dfrac{2}{3} \;\Rightarrow\; \textbf{(D)}$$

💡 Writing a ratio of two lengths is the basic Grade 6 ratio-reasoning move.

[1] #7 3.MD.C.7 Find the area each side of $\overline{PQ}$ must have. The octagon is made of $10
[2] #7 6.G.A.1 Use the given decomposition. The region below $\overline{PQ}$ is a unit square (
[3] #7 6.EE.B.7 Solve the equation for the triangle's height $h$. Subtract $1$ from both sides,
[4] #1 5.NF.A.1 Place $Q$ on the segment $\overline{XY}$. Since $X=(5,2)$ and $Y=(5,1)$ share th
[5] #7 6.RP.A.1 Form the requested ratio. Both lengths have the same denominator, so it simplifi

Review

Reasonableness: Check the height $h = \tfrac{8}{5} = 1.6$ — it lies between $1$ and $2$, exactly where $Q$ must sit on $\overline{XY}$. Verify the area: square gives $1$ and the triangle gives $\tfrac{1}{2}(5)(1.6) = 4$, totaling $5$, which is half of $10$. Finally, $XQ + QY = \tfrac{2}{5} + \tfrac{3}{5} = 1$, matching the length of $\overline{XY}$ from $y=1$ to $y=2$. All three checks pass, and the answer (D) $\tfrac{2}{3}$ is one of the listed options.

Alternative: Tool #3 (Eliminate Possibilities): the ratio $\dfrac{XQ}{QY}$ together with $XQ + QY = 1$ pins down $XQ$ for each choice — (A) $\tfrac{2}{5} \to XQ = \tfrac{2}{7}$, (B) $\tfrac{1}{2} \to XQ = \tfrac{1}{3}$, (C) $\tfrac{3}{5} \to XQ = \tfrac{3}{8}$, (D) $\tfrac{2}{3} \to XQ = \tfrac{2}{5}$, (E) $\tfrac{3}{4} \to XQ = \tfrac{3}{7}$. Only (D) gives the clean $XQ = \tfrac{2}{5}$ that matches the triangle-height calculation $h = \tfrac{8}{5}$.

CCSS standards used (min grade 6)

  • 3.MD.C.7 Relate area to multiplication and addition (Reading the octagon's area as $10$ unit squares and halving it to get the area $5$ on each side of $\overline{PQ}$.)
  • 5.NF.A.1 Add and subtract fractions with unlike denominators (Subtracting $2 - \tfrac{8}{5} = \tfrac{2}{5}$ and $\tfrac{8}{5} - 1 = \tfrac{3}{5}$ to find the vertical lengths $XQ$ and $QY$.)
  • 6.G.A.1 Find the area of polygons by composing into rectangles or decomposing into triangles (Treating the region below $\overline{PQ}$ as a unit square plus a triangle with base $5$ and writing its area as $1 + \tfrac{1}{2}(5)(h)$.)
  • 6.EE.B.7 Solve real-world and mathematical problems by writing and solving equations of the form $x+p=q$ and $px=q$ (Solving $1 + \tfrac{5}{2}h = 5$ for $h = \tfrac{8}{5}$, the y-coordinate of $Q$.)
  • 6.RP.A.1 Understand the concept of a ratio and use ratio language (Forming the final ratio $\dfrac{XQ}{QY} = \dfrac{2/5}{3/5} = \dfrac{2}{3}$.)

⭐ 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.

⭐ 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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