AMC 8 · 2022 · #1
Grade 6 geometry-2dProblem
The Math Team designed a logo shaped like a multiplication symbol, shown below on a grid of 1-inch squares. What is the area of the logo in square inches?
Pick an answer.
AMC 8 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: The Math Team's logo is a multiplication-symbol shape drawn on a $1$-inch grid. The shape is the polygon with vertices $(1,2),(2,1),(3,2),(4,1),(5,2),(4,3),(5,4),(4,5),(3,4),(2,5),(1,4),(2,3)$. We need its area in square inches and must choose from $10, 12, 13, 14, 15$.
Givens: Grid squares are $1$ inch by $1$ inch; The shaded region's $12$ corners all sit on lattice points; The shape fits inside the $4 \times 4$ square whose opposite corners are $(1,1)$ and $(5,5)$; Answer choices: (A) $10$, (B) $12$, (C) $13$, (D) $14$, (E) $15$
Unknowns: The area of the shaded logo in square inches
Understand
Restated: The Math Team's logo is a multiplication-symbol shape drawn on a $1$-inch grid. The shape is the polygon with vertices $(1,2),(2,1),(3,2),(4,1),(5,2),(4,3),(5,4),(4,5),(3,4),(2,5),(1,4),(2,3)$. We need its area in square inches and must choose from $10, 12, 13, 14, 15$.
Givens: Grid squares are $1$ inch by $1$ inch; The shaded region's $12$ corners all sit on lattice points; The shape fits inside the $4 \times 4$ square whose opposite corners are $(1,1)$ and $(5,5)$; Answer choices: (A) $10$, (B) $12$, (C) $13$, (D) $14$, (E) $15$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #16 Change Focus / Count the Complement
The ×-shape is awkward to measure directly, but it is trapped inside a clean $4 \times 4$ bounding box. Tool #7 (Identify Subproblems) lets us split the problem into two easy pieces: (1) the area of the $4 \times 4$ box and (2) the area of the white triangular cut-outs around the logo. Tool #1 (Draw a Diagram) is useful for marking those cut-out triangles on the picture, and Tool #16 (Change Focus / Count the Complement) names the strategy of finding the shaded area indirectly by subtracting the unshaded area from the whole box.
Execute — Answer: A
3.MD.C.7 Step 1 - Draw a $4 \times 4$ bounding square around the logo.
- Its opposite corners are $(1,1)$ and $(5,5)$, and the logo's tips $(1,2),(2,1),(5,2),(4,1),(5,4),(4,5),(1,4),(2,5)$ all lie on the edges of this square.
- The bounding square has area $4 \times 4 = 16$ in$^2$.
💡 Finding the area of a rectangle by side $\times$ side is the basic Grade 3 area idea.
6.G.A.1 Step 2 - Look at the four corners of the bounding box.
- At each corner (for example near $(1,1)$) the logo's outline cuts off a small right triangle with legs of length $1$.
- Each corner triangle has area $\tfrac{1}{2} \times 1 \times 1 = \tfrac{1}{2}$ in$^2$.
💡 Breaking the white border into right triangles uses the Grade 6 "area by composing/decomposing" idea.
6.G.A.1 Step 3 - Now look at the middle of each side of the bounding box.
- On each side there is a larger white triangle whose tip points inward; for the bottom side the vertices are $(2,1),(4,1),(3,2)$, so its base is $4 - 2 = 2$ and its height is $2 - 1 = 1$.
- Each side triangle has area $\tfrac{1}{2} \times 2 \times 1 = 1$ in$^2$.
💡 Same decomposition idea, applied to the bigger white triangles along each edge.
4.OA.A.3 Step 4 - Add up all the white area inside the bounding box: $4$ corner triangles plus $4$ side triangles.
- Then subtract the white area from the box to get the logo's area.
💡 Counting the complement (the white space) is easier than counting the shaded shape directly — a Grade 4 multi-step word-problem move.
4.OA.A.3 Step 5 Subtract the white area from the bounding box area to get the logo's area.
💡 Whole minus the unshaded part gives the shaded part — the complement strategy in one subtraction.
3.MD.C.7 Draw a $4 \times 4$ bounding square around the logo. Its opposite corners are $( 6.G.A.1 Look at the four corners of the bounding box. At each corner (for example near $ 6.G.A.1 Now look at the middle of each side of the bounding box. On each side there is a 4.OA.A.3 Add up all the white area inside the bounding box: $4$ corner triangles plus $4$ 4.OA.A.3 Subtract the white area from the bounding box area to get the logo's area. Review
Reasonableness: Sanity check by symmetry: the logo has four identical "arms" plus a central diamond. The central diamond has vertices $(2,3),(3,2),(4,3),(3,4)$ — a square of diagonal $2$, so its area is $\tfrac{1}{2} \times 2 \times 2 = 2$ in$^2$. Each arm (for the right arm, vertices $(3,2),(4,1),(5,2),(4,3)$) is also a unit-diagonal square of area $2$ in$^2$. Total $= 2 + 4 \times 2 = 10$ in$^2$, matching answer (A). The result also fits inside a $16$ in$^2$ box, which is the right order of magnitude.
Alternative: Tool #5 (Look for a Pattern) via Pick's Theorem on lattice points: the logo's boundary contains $B = 12$ lattice points (all $12$ listed vertices) and its interior contains $I = 5$ lattice points $(2,2),(3,2)\text{-no, that is on boundary}\ldots$, specifically $(2,2),(4,2),(2,4),(4,4),(3,3)$. Then $A = I + \tfrac{B}{2} - 1 = 5 + 6 - 1 = 10$ in$^2$, the same answer.
CCSS standards used (min grade 6)
3.MD.C.7Relate area to multiplication and addition operations (Finding the area of the $4 \times 4$ bounding square as $4 \times 4 = 16$ in$^2$.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Computing the area of each white right triangle (corner triangles with legs $1,1$ and side triangles with base $2$, height $1$) by decomposing the white border.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Combining the $4$ corner triangles and $4$ side triangles, then subtracting their total ($6$ in$^2$) from the box area ($16$ in$^2$) to get the logo's area.)
⭐ This AMC 8 problem only needs Grade 6 "area by cutting into triangles and rectangles" you already know!
⭐ This AMC 8 problem only needs Grade 6 "area by cutting into triangles and rectangles" you already know!
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