AMC 10 · 2002 · #8
Grade 6 geometry-2dBetsy designed a flag using blue triangles, small white squares, and a red center square, as shown. Let B be the total area of the blue triangles, W the total area of the white squares, and P the area of the red square. Which of the following is correct?
Pick an answer.
AMC 10 2002 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: A flag is built from an $8\times 8$ blue background, a red square in the middle, and a ring of small white squares tilted $45^\circ$. Let $B$, $W$, and $P$ be the total blue area, total white area, and red area. Decide which of the five equations relating $B$, $W$, and $P$ is true.
Givens: The whole flag is a square running from $(-4,-4)$ to $(4,4)$, so it is $8$ units on a side.; The red square runs from $(-2,-2)$ to $(2,2)$, so it is $4$ units on a side.; The white shapes are small squares tilted $45^\circ$ (diamonds); three of them sit along each edge of the flag, and the pattern repeats by quarter-turns.; Everything not red or white is blue.; Answer choices: (A) $B=W$, (B) $W=P$, (C) $B=P$, (D) $3B=2P$, (E) $2P=W$.
Unknowns: Which single equation among the choices correctly relates $B$, $W$, and $P$.
Understand
Restated: A flag is built from an $8\times 8$ blue background, a red square in the middle, and a ring of small white squares tilted $45^\circ$. Let $B$, $W$, and $P$ be the total blue area, total white area, and red area. Decide which of the five equations relating $B$, $W$, and $P$ is true.
Givens: The whole flag is a square running from $(-4,-4)$ to $(4,4)$, so it is $8$ units on a side.; The red square runs from $(-2,-2)$ to $(2,2)$, so it is $4$ units on a side.; The white shapes are small squares tilted $45^\circ$ (diamonds); three of them sit along each edge of the flag, and the pattern repeats by quarter-turns.; Everything not red or white is blue.; Answer choices: (A) $B=W$, (B) $W=P$, (C) $B=P$, (D) $3B=2P$, (E) $2P=W$.
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #16 Change Focus / Count the Complement
Comparing $B$, $W$, and $P$ turns into three separate area jobs, so Tool #7 (Identify Subproblems) is the backbone: find the red area, then the white area, then the blue area, and compare. Tool #1 (Draw a Diagram) supplies the trick that makes every piece countable — overlay a unit grid on the flag so the flag is $64$ little squares and each tilted white square is exactly half of a $2\times 2$ box. Tool #16 (Count the Complement) then gets the blue for free: instead of adding up all the blue triangles, subtract red and white from the whole flag.
Execute — Answer: A
3.MD.C.7 Step 1 Grid the whole flag
- Lay a unit grid over the flag.
- Since the flag is $8$ units on each side, it is made of $8\times 8 = 64$ unit squares.
- That total area, $64$, is the budget the three colors must share.
💡 Covering a square with unit tiles turns 'area' into 'how many little squares,' which you can just count.
3.MD.C.7 Step 2 Red area P
- The red square runs from $-2$ to $2$ in both directions, so it is $4$ units on a side.
- Its area is $4\times 4 = 16$ unit squares.
- So $P = 16$.
💡 A square's area is just one side times itself, so a $4$-wide square holds $16$ unit tiles.
6.G.A.1 Step 3 White area W
- Each white shape is a square tilted $45^\circ$.
- It fits inside a $2\times 2$ box and touches the middle of each side, so it covers exactly half of that box: area $\tfrac12\times(2\times 2)=2$.
- Three of them sit along the top edge, and the quarter-turn symmetry copies that trio to the other three edges, giving $3\times 4 = 12$ white diamonds.
- So $W = 12\times 2 = 24$.
💡 A diamond drawn through the midpoints of a box is always half the box, so each white square is worth $2$ tiles.
3.MD.C.7 Step 4 Blue area B, then compare
- Rather than chase every blue triangle, take the whole flag and remove what is not blue: $B = 64 - P - W = 64 - 16 - 24 = 24$.
- Now compare: $B = 24$ and $W = 24$, so $B = W$.
- Checking the others, $P=16$ matches nothing here, so the only true statement is $B=W$, which is choice (A).
💡 Since the colors tile the flag exactly, blue is just the whole minus the parts you already measured.
3.MD.C.7 Lay a unit grid over the flag. Since the flag is $8$ units on each side, it is m 3.MD.C.7 The red square runs from $-2$ to $2$ in both directions, so it is $4$ units on a 6.G.A.1 Each white shape is a square tilted $45^\circ$. It fits inside a $2\times 2$ box 3.MD.C.7 Rather than chase every blue triangle, take the whole flag and remove what is no Review
Reasonableness: The three areas should rebuild the whole flag, and they do: $P+W+B = 16+24+24 = 64$, the flag's area. The other choices all fail — $W=P$ needs $24=16$, $B=P$ needs $24=16$, $3B=2P$ needs $72=32$, and $2P=W$ needs $32=24$ — so $B=W$ is the only survivor, matching (A).
Alternative: Measure everything in 'diamonds,' where one white square counts as $1$ unit. There are $12$ white diamonds, so $W=12$ units. The blue also splits into $12$ diamond-sized pieces: the eight blue triangles hugging the red square pair up into $4$ diamonds, plus $4$ at the corners and $4$ along the edges, giving $12$ blue units as well. So $B=W$ directly, without ever computing $64$.
CCSS standards used (min grade 6)
3.MD.C.7Relate area to the operations of multiplication and addition (Tiling the flag and the red square with unit squares to get areas $64$ and $16$, and using area-is-additive to find blue as whole minus red minus white.)6.G.A.1Find the area of polygons by composing into rectangles or decomposing into triangles (Seeing each tilted white square as exactly half of its $2\times 2$ bounding box, so each has area $2$.)
⭐ Lay a grid over a colored picture, count each color's squares, and get the last color for free by subtracting from the whole.
⭐ Lay a grid over a colored picture, count each color's squares, and get the last color for free by subtracting from the whole.
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