AMC 10 · 2005 · #7
Grade 8 geometry-2dA circle is inscribed in a square, then a square is inscribed in this circle, and finally, a circle is inscribed in this square. What is the ratio of the area of the smallest circle to the area of the largest square?
Pick an answer.
AMC 10 2005 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: Start with a square. Inscribe a circle in it, then inscribe a square in that circle, then inscribe a circle in that last square. That final circle is the smallest circle and the first square is the largest square. Find the ratio of the smallest circle's area to the largest square's area.
Givens: A circle is inscribed in a square (the circle touches all four sides); A square is inscribed in that circle (its four corners lie on the circle); A circle is inscribed in that inner square (touching all four of its sides); Answer choices: (A) $\frac{\pi}{16}$, (B) $\frac{\pi}{8}$, (C) $\frac{3\pi}{16}$, (D) $\frac{\pi}{4}$, (E) $\frac{\pi}{2}$
Unknowns: The ratio (area of the smallest circle) : (area of the largest square)
Understand
Restated: Start with a square. Inscribe a circle in it, then inscribe a square in that circle, then inscribe a circle in that last square. That final circle is the smallest circle and the first square is the largest square. Find the ratio of the smallest circle's area to the largest square's area.
Givens: A circle is inscribed in a square (the circle touches all four sides); A square is inscribed in that circle (its four corners lie on the circle); A circle is inscribed in that inner square (touching all four of its sides); Answer choices: (A) $\frac{\pi}{16}$, (B) $\frac{\pi}{8}$, (C) $\frac{3\pi}{16}$, (D) $\frac{\pi}{4}$, (E) $\frac{\pi}{2}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
The figure is a nest of shapes inside shapes, so Tool #1 (Draw a Diagram) is the natural way in — a clear picture shows exactly where each circle touches its square and where each square's corners land. Because nothing gives an actual size, I fix the largest square's side to a convenient number and let the shapes cascade inward. Tool #7 (Identify Subproblems) handles the cascade: peel the nest one layer at a time, carrying a single length from each shape to the next. At the end Tool #3 (Eliminate Possibilities) matches my computed ratio to one of the five answer choices.
Execute — Answer: B
7.G.B.4 Step 1 Size the outer square and its circle
- Let the largest square have side $2$, so its area is $2 \times 2 = 4$.
- The inscribed circle just fits inside, kissing the middle of each side, so it stretches exactly one square-side across.
- That means its diameter is $2$ and its radius is $1$.
💡 An inscribed circle touches each side once, so it spans exactly the width of the square — one full side.
8.G.B.7 Step 2 Middle square from the circle's diagonal
- The middle square has its four corners on that circle, so the longest line across it — its diagonal — is the circle's diameter, $2$.
- A square's diagonal cuts it into two right triangles, so if the square's side is $x$ the Pythagorean theorem gives $x^2 + x^2 = 2^2$, that is $2x^2 = 4$, so $x^2 = 2$ and $x = \sqrt{2}$.
💡 A diagonal splits a square into two right triangles, so the diagonal and the side are tied together by the Pythagorean theorem.
7.G.B.4 Step 3 Smallest circle and its area
- The smallest circle is inscribed in the middle square, so — just like before — its diameter equals that square's side, $\sqrt{2}$.
- Its radius is therefore $\frac{\sqrt{2}}{2}$.
- Using the circle-area formula, its area is $\pi r^2 = \pi\left(\frac{\sqrt{2}}{2}\right)^2 = \pi \cdot \frac{2}{4} = \frac{\pi}{2}$.
💡 The same inscribed-circle rule applies again: diameter equals the side it fits inside, then area is $\pi r^2$.
6.RP.A.3 Step 4 Form the ratio and match a choice
- The smallest circle's area is $\frac{\pi}{2}$ and the largest square's area is $4$.
- The ratio is $\frac{\pi/2}{4} = \frac{\pi}{2} \cdot \frac{1}{4} = \frac{\pi}{8}$.
- That matches answer choice (B).
💡 A ratio of two areas is just one area divided by the other.
7.G.B.4 Let the largest square have side $2$, so its area is $2 \times 2 = 4$. The inscr 8.G.B.7 The middle square has its four corners on that circle, so the longest line acros 7.G.B.4 The smallest circle is inscribed in the middle square, so — just like before — i 6.RP.A.3 The smallest circle's area is $\frac{\pi}{2}$ and the largest square's area is $ Review
Reasonableness: The smallest circle sits deep inside the largest square, so it should cover well under half of it — and $\frac{\pi}{8} \approx 0.39$, comfortably less than half, which fits the picture. The choice of side $2$ was only for convenience: with a general side $s$ the largest square has area $s^2$ and the smallest circle has area $\frac{\pi s^2}{8}$, so the $s^2$ cancels and the ratio is $\frac{\pi}{8}$ no matter the starting size. Trap answer $\frac{\pi}{16}$ comes from going one shape too far inward, and $\frac{\pi}{4}$ from stopping one shape too early.
Alternative: Work with areas directly using two shortcuts. First, a circle inscribed in a square of area $A$ has area $\frac{\pi}{4}A$, since its radius is half the side. Second, the square inscribed inside that circle has area exactly $\frac{A}{2}$ (a square of side $s$ becomes one of side $\frac{s}{\sqrt{2}}$). Start with the largest square, area $4$. The middle square is two layers in, so it has half the area: $2$. The smallest circle is inscribed in that middle square, so its area is $\frac{\pi}{4}\cdot 2 = \frac{\pi}{2}$. The ratio to the largest square is $\frac{\pi/2}{4} = \frac{\pi}{8}$, again (B).
CCSS standards used (min grade 8)
7.G.B.4Know the formulas for area and circumference of a circle (Turning an inscribed circle's diameter (equal to its square's side) into a radius and then into the circle's area with $\pi r^2$.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Getting the middle square's side from its diagonal (the circle's diameter) via $x^2 + x^2 = \text{diagonal}^2$.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Forming and simplifying the ratio of the smallest circle's area to the largest square's area.)
⭐ Carry one length from each shape to the next: an inscribed circle's diameter equals its square's side, and an inscribed square's diagonal equals its circle's diameter — chain those and the sizes fall out.
⭐ Carry one length from each shape to the next: an inscribed circle's diameter equals its square's side, and an inscribed square's diagonal equals its circle's diameter — chain those and the sizes fall out.
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