AMC 8 · 2005 · #25
Grade 8 geometry-2dProblem
A square with side length 2 and a circle share the same center. The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square. What is the radius of the circle?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: A square with side $2$ and a circle share the same center. The area of the part inside the circle but outside the square equals the area of the part inside the square but outside the circle. Find the radius $r$ of the circle.
Givens: Square side length $= 2$, so square area $= 4$; Square and circle have the same center; Area(inside circle, outside square) $=$ Area(inside square, outside circle); Answer choices: (A) $\tfrac{2}{\sqrt{\pi}}$, (B) $\tfrac{1+\sqrt{2}}{2}$, (C) $\tfrac{3}{2}$, (D) $\sqrt{3}$, (E) $\sqrt{\pi}$
Unknowns: The radius $r$ of the circle
Understand
Restated: A square with side $2$ and a circle share the same center. The area of the part inside the circle but outside the square equals the area of the part inside the square but outside the circle. Find the radius $r$ of the circle.
Givens: Square side length $= 2$, so square area $= 4$; Square and circle have the same center; Area(inside circle, outside square) $=$ Area(inside square, outside circle); Answer choices: (A) $\tfrac{2}{\sqrt{\pi}}$, (B) $\tfrac{1+\sqrt{2}}{2}$, (C) $\tfrac{3}{2}$, (D) $\sqrt{3}$, (E) $\sqrt{\pi}$
Plan
Primary tool: #11 Find an Invariant
Secondary: #1 Draw a Diagram
The overlap region between the square and the circle has a messy shape, and the problem never asks for its area. That is the hint to use Tool #11 (Find an Invariant): call the overlap area $I$, then notice it appears on both sides of the equation and cancels. Tool #1 (Draw a Diagram) sets up the picture — circle and square sharing a center create three pieces (circle-only, square-only, overlap) — so that subtracting the overlap from the whole circle and from the whole square is obvious. After cancellation, circle area equals square area, and one area formula gives $r$.
Execute — Answer: A
7.G.B.4 Step 1 - Draw the concentric square and circle and label the three pieces.
- Let $I$ be the area of the overlap (inside both shapes).
- The region inside the circle but outside the square has area $\pi r^2 - I$.
- The region inside the square but outside the circle has area $4 - I$.
💡 The picture shows that each whole shape is overlap plus its own crescent piece, so the crescent equals whole minus overlap.
7.EE.B.4 Step 2 - Set the two crescent areas equal as the problem requires.
- The unknown overlap $I$ is the same on both sides, so it cancels — that is the invariant.
- What is left says the circle and the square have equal total area.
💡 Subtracting $I$ from both sides is the Grade 7 "do the same thing to both sides" move. Because $I$ is the only piece we cannot compute, removing it is exactly what we need.
8.EE.A.2 Step 3 - Solve $\pi r^2 = 4$ for $r$.
- Divide by $\pi$, then take the positive square root since a radius is positive.
💡 The Grade 8 square-root step undoes $r^2$. Splitting $\sqrt{4/\pi}$ as $\sqrt{4}/\sqrt{\pi}$ gives the clean form $2/\sqrt{\pi}$ that matches choice (A).
7.G.B.4 Draw the concentric square and circle and label the three pieces. Let $I$ be the 7.EE.B.4 Set the two crescent areas equal as the problem requires. The unknown overlap $I 8.EE.A.2 Solve $\pi r^2 = 4$ for $r$. Divide by $\pi$, then take the positive square root Review
Reasonableness: Check the size. $\pi \approx 3.14$, so $r = 2/\sqrt{3.14} \approx 2/1.77 \approx 1.13$. The square's inscribed circle has radius $1$ and its circumscribed circle has radius $\sqrt{2} \approx 1.41$, so the answer $1.13$ sits between them — exactly what equal crescent areas should give. Choice (C) $\tfrac{3}{2} = 1.5$ is too big (circle would swallow the square), and (B) $\tfrac{1+\sqrt{2}}{2} \approx 1.21$ is close but does not give equal areas: $\pi(1.21)^2 \approx 4.59 \neq 4$.
Alternative: Tool #9 (Solve an Easier Problem): drop the geometry and ask a simpler question — "For which $r$ are the circle's total area and the square's total area equal?" That gives $\pi r^2 = 4$ in one line. Then notice that whenever the two whole areas are equal, the parts they share (the overlap) automatically cancel, so the two crescent areas must be equal too. This reverses the argument but lands on the same equation and the same answer (A).
CCSS standards used (min grade 8)
7.G.B.4Know the formulas for the area and circumference of a circle and use them to solve problems (Writing the circle's area as $\pi r^2$ and the square's area as $2^2 = 4$ so the two crescent regions can be expressed as $\pi r^2 - I$ and $4 - I$.)7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities to solve problems (Setting $\pi r^2 - I = 4 - I$ from the equal-area condition and cancelling the unknown overlap $I$ to get $\pi r^2 = 4$.)8.EE.A.2Use square root and cube root symbols to represent solutions to equations of the form $x^2 = p$ (Solving $r^2 = 4/\pi$ by taking the positive square root to get $r = 2/\sqrt{\pi}$.)
⭐ The messy overlap area never has to be computed — it appears on both sides and cancels. Once it does, circle area equals square area, and one square root gives $r = \tfrac{2}{\sqrt{\pi}}$.
⭐ The messy overlap area never has to be computed — it appears on both sides and cancels. Once it does, circle area equals square area, and one square root gives $r = \tfrac{2}{\sqrt{\pi}}$.
More like this
Same archetype — closest grade level first.
