AMC 10 · 2023 · #20
Grade 8 geometry-3dProblem
Four congruent semicircles are drawn on the surface of a sphere with radius 2, as
shown, creating a close curve that divides the surface into two congruent regions.
The length of the curve is πn. What is n?
Pick an answer.
AMC 10 2023 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: Four congruent semicircles, joined end-to-end on a sphere of radius $2$, form a closed curve that splits the sphere's surface into two congruent pieces. The total length of the curve is $\pi\sqrt{n}$. Find $n$.
Givens: Sphere radius $R = 2$; Curve = $4$ congruent semicircles joined into a closed loop; The loop divides the sphere into two CONGRUENT regions; Total curve length $= \pi\sqrt{n}$; Answer choices: (A) $32$, (B) $12$, (C) $48$, (D) $36$, (E) $27$
Unknowns: $n$
Understand
Restated: Four congruent semicircles, joined end-to-end on a sphere of radius $2$, form a closed curve that splits the sphere's surface into two congruent pieces. The total length of the curve is $\pi\sqrt{n}$. Find $n$.
Givens: Sphere radius $R = 2$; Curve = $4$ congruent semicircles joined into a closed loop; The loop divides the sphere into two CONGRUENT regions; Total curve length $= \pi\sqrt{n}$; Answer choices: (A) $32$, (B) $12$, (C) $48$, (D) $36$, (E) $27$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #17 Visualize Spatial Relationships
The total length is $4$ times one semicircle's arc length, which is $\pi$ times its radius, which is half its diameter — and the diameter is a chord of the sphere joining adjacent junction points. So the problem chains: (a) pin down where the four junction points sit, (b) compute the chord between adjacent junctions, (c) convert to semicircle radius and arc length, (d) total length and solve for $n$. Tool #7 (Identify Subproblems) names each rung; Tool #1 (Diagram) of the great-circle cross-section makes the chord visible; Tool #17 (Visualize Spatial Relationships) confirms the four points sit at the corners of a square on a great circle.
Execute — Answer: A
8.G.A.5 Step 1 - Locate the four junction points.
- The closed curve splits the sphere into two congruent regions, forcing maximal symmetry.
- The natural arrangement: the four points $A, B, C, D$ sit at the vertices of a square inscribed in a great circle of the sphere (radius $R = 2$).
- Adjacent points like $A$ and $B$ are connected by one of the four semicircles.
💡 Two congruent regions means the curve must look the same after a $90^\circ$ rotation — so the four junction points must be evenly spaced on a great circle.
8.G.B.7 Step 2 - Compute the chord $AB$ between adjacent junctions using the Pythagorean theorem.
- In the great-circle plane, $O$ is the sphere's center; $OA = OB = 2$ and $\angle AOB = 90^\circ$ (since $A, B$ are adjacent vertices of an inscribed square).
- Triangle $OAB$ is right-isosceles.
💡 Drawing the great-circle cross-section turns a $3$D problem into a flat right triangle — then Pythagoras gives the chord length immediately.
7.G.B.4 Step 3 - Convert the chord $AB$ into the semicircle's radius.
- Each semicircle has $AB$ as its diameter, so its radius is half of $AB$.
💡 A semicircle's diameter is the chord between its endpoints — so the radius is half the chord.
7.G.B.4 Step 4 Length of one semicircle: half the circumference of a full circle of radius $\sqrt{2}$, which is $\pi r = \pi\sqrt{2}$.
💡 Half of $2\pi r$ is $\pi r$ — the basic arc-length formula for a semicircle.
8.EE.A.2 Step 5 Total curve length: $4$ congruent semicircles, then match to $\pi\sqrt{n}$ to read off $n$.
💡 Pull the coefficient $4$ inside the square root: $4\sqrt{2} = \sqrt{16 \cdot 2} = \sqrt{32}$, so the form $\pi\sqrt{n}$ forces $n = 32$.
8.G.A.5 Locate the four junction points. The closed curve splits the sphere into two con 8.G.B.7 Compute the chord $AB$ between adjacent junctions using the Pythagorean theorem. 7.G.B.4 Convert the chord $AB$ into the semicircle's radius. Each semicircle has $AB$ as 7.G.B.4 Length of one semicircle: half the circumference of a full circle of radius $\sq 8.EE.A.2 Total curve length: $4$ congruent semicircles, then match to $\pi\sqrt{n}$ to re Review
Reasonableness: Cross-check the magnitudes: the sphere's full great-circle circumference is $2\pi R = 4\pi$, so our curve length $4\pi\sqrt{2} \approx 5.66\pi$ is bigger than one great circle — sensible, since the four semicircles bulge around the sphere rather than tracing the great circle exactly. The $n = 32$ choice matches perfectly: $(4\sqrt{2})^2 = 32$. Also $32$ is the only answer choice whose square root produces a rational multiple of $\sqrt{2}$, which the geometry forces.
Alternative: Tool #13 (Convert to Algebra): set up $4\pi\sqrt{2} = \pi\sqrt{n}$, divide both sides by $\pi$ to get $4\sqrt{2} = \sqrt{n}$, then square to read $n = 16 \cdot 2 = 32$ — same answer (A). Tool #10 (Physical Representation): a beach ball of radius $2$ with four equally-spaced dots on its equator and four half-loops connecting adjacent dots makes the picture concrete and confirms the curve shape.
CCSS standards used (min grade 8)
8.G.A.5Use informal arguments to establish facts about angle sum and exterior angles (Symmetry argument: the four junction points must be evenly spaced ($90^\circ$ apart) on a great circle because the curve bisects the sphere into congruent halves.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Computing chord $AB = 2\sqrt{2}$ from the right-isosceles triangle $OAB$ with $OA = OB = 2$ in the great-circle plane.)7.G.B.4Know the formulas for area and circumference of a circle (Semicircle arc length $\pi r = \pi\sqrt{2}$, half the circumference of a circle of radius $\sqrt{2}$.)8.EE.A.2Use square root and cube root symbols to represent solutions (Rewriting $4\sqrt{2} = \sqrt{16 \cdot 2} = \sqrt{32}$ to match the form $\pi\sqrt{n}$ and read $n = 32$.)
⭐ Symmetry forces the four junction points to sit at the corners of a square inscribed in a great circle of the sphere. The Pythagorean theorem in the great-circle plane gives the chord between adjacent corners as $2\sqrt{2}$, so each semicircle has radius $\sqrt{2}$ and arc length $\pi\sqrt{2}$. Four of them total $4\pi\sqrt{2} = \pi\sqrt{32}$, so $n = \textbf{(A) }32$.
⭐ Symmetry forces the four junction points to sit at the corners of a square inscribed in a great circle of the sphere. The Pythagorean theorem in the great-circle plane gives the chord between adjacent corners as $2\sqrt{2}$, so each semicircle has radius $\sqrt{2}$ and arc length $\pi\sqrt{2}$. Four of them total $4\pi\sqrt{2} = \pi\sqrt{32}$, so $n = \textbf{(A) }32$.