AMC 8 · 2017 · #25
Grade 8 geometry-2dProblem
In the figure shown, US and UT are line segments each of length 2, and m∠TUS=60∘. Arcs \overarcTR and \overarcSR are each one-sixth of a circle with radius 2. What is the area of the region shown?
Pick an answer.
AMC 8 2017 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 figure is bounded by two line segments $\overline{US}$ and $\overline{UT}$ (each of length $2$, meeting at $U$ with $\angle TUS = 60^\circ$) and by two arcs $\overarc{SR}$ and $\overarc{TR}$ (each a $60^\circ$ arc of a circle of radius $2$). The two arcs bulge inward (concave to the region) and meet at point $R$ at the bottom. Find the area of this region.
Givens: $\overline{US} = \overline{UT} = 2$ and $\angle TUS = 60^\circ$, so $\triangle TUS$ is equilateral with side $2$; Arc $\overarc{SR}$ is one-sixth of a circle of radius $2$ (central angle $60^\circ$) centered at some point $X$; Arc $\overarc{TR}$ is one-sixth of a circle of radius $2$ (central angle $60^\circ$) centered at some point $Y$; From the figure the arcs are concave to the region — they curve inward toward $U$; Answer choices: (A) $3\sqrt{3}-\pi$, (B) $4\sqrt{3}-\frac{4\pi}{3}$, (C) $2\sqrt{3}$, (D) $4\sqrt{3}-\frac{2\pi}{3}$, (E) $4+\frac{4\pi}{3}$
Unknowns: The area of the region enclosed by the two segments and the two arcs
Understand
Restated: A figure is bounded by two line segments $\overline{US}$ and $\overline{UT}$ (each of length $2$, meeting at $U$ with $\angle TUS = 60^\circ$) and by two arcs $\overarc{SR}$ and $\overarc{TR}$ (each a $60^\circ$ arc of a circle of radius $2$). The two arcs bulge inward (concave to the region) and meet at point $R$ at the bottom. Find the area of this region.
Givens: $\overline{US} = \overline{UT} = 2$ and $\angle TUS = 60^\circ$, so $\triangle TUS$ is equilateral with side $2$; Arc $\overarc{SR}$ is one-sixth of a circle of radius $2$ (central angle $60^\circ$) centered at some point $X$; Arc $\overarc{TR}$ is one-sixth of a circle of radius $2$ (central angle $60^\circ$) centered at some point $Y$; From the figure the arcs are concave to the region — they curve inward toward $U$; Answer choices: (A) $3\sqrt{3}-\pi$, (B) $4\sqrt{3}-\frac{4\pi}{3}$, (C) $2\sqrt{3}$, (D) $4\sqrt{3}-\frac{2\pi}{3}$, (E) $4+\frac{4\pi}{3}$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram
The region mixes straight edges and curved edges, so no single area formula applies. Tool #7 (Identify Subproblems) breaks it into pieces with familiar formulas: first replace each arc by its chord to form a straight-sided figure (a kite made of two equilateral triangles), then subtract the two circular segments that the chords "hide" between chord and arc. Each segment is itself a subproblem: $60^\circ$ sector minus $60^\circ$ equilateral triangle. Tool #1 (Draw a Diagram) is the supporting move — sketching the chords $\overline{SR}$ and $\overline{TR}$ on top of the given figure makes the kite and the two segments visible, and shows that each chord has length $2$ (chord of a $60^\circ$ arc on a radius-$2$ circle).
Execute — Answer: B
7.G.A.2 Step 1 - Draw chords $\overline{SR}$ and $\overline{TR}$ across the two arcs.
- Each chord subtends a $60^\circ$ central angle of a radius-$2$ circle, so the radii and the chord form an equilateral triangle, giving $SR = TR = 2$.
- Together with $US = UT = 2$, the four straight segments $\overline{US}, \overline{SR}, \overline{RT}, \overline{TU}$ make a kite $USRT$ with all four sides of length $2$.
💡 An isosceles triangle with two sides $=2$ and the included angle $=60^\circ$ is forced to be equilateral, so the third side is also $2$.
6.G.A.1 Step 2 - Compute the area of the kite by splitting it along diagonal $\overline{ST}$.
- Since $US = UT = ST = 2$, $\triangle UST$ is equilateral; likewise $\triangle SRT$ has $SR = RT = ST = 2$ and is equilateral.
- The area of an equilateral triangle with side $s$ is $\frac{s^2\sqrt{3}}{4}$, which gives $\sqrt{3}$ for each, so the kite has area $2\sqrt{3}$.
💡 Splitting a quadrilateral into two triangles with a known area formula turns an unfamiliar shape into two easy pieces.
7.G.B.4 Step 3 - Find the area of one circular segment (the sliver between chord $\overline{SR}$ and arc $\overarc{SR}$).
- The $60^\circ$ sector of a radius-$2$ circle has area $\frac{60}{360}\pi r^2 = \frac{1}{6}\pi(4) = \frac{2\pi}{3}$.
- Removing the $60^\circ$-$60^\circ$-$60^\circ$ equilateral triangle (area $\sqrt{3}$) formed by the two radii and the chord leaves the segment.
💡 A circular segment is just "pie slice minus triangle" — two areas you already know how to compute.
8.NS.A.1 Step 4 - Subtract both segments from the kite.
- Because each arc is concave to the region, the kite area over-counts the region by exactly the two segments (one on chord $\overline{SR}$, one on chord $\overline{TR}$), and by symmetry both segments have the same area $\frac{2\pi}{3} - \sqrt{3}$.
💡 Combining the irrational pieces $\sqrt{3}$ and $\pi$ as separate quantities is exactly Grade 8 work with irrational numbers.
7.G.A.2 Draw chords $\overline{SR}$ and $\overline{TR}$ across the two arcs. Each chord 6.G.A.1 Compute the area of the kite by splitting it along diagonal $\overline{ST}$. Sin 7.G.B.4 Find the area of one circular segment (the sliver between chord $\overline{SR}$ 8.NS.A.1 Subtract both segments from the kite. Because each arc is concave to the region, Review
Reasonableness: Numerically, $4\sqrt{3} \approx 6.93$ and $\frac{4\pi}{3} \approx 4.19$, so the area is about $6.93 - 4.19 \approx 2.74$. As a sanity check, the kite alone has area $2\sqrt{3} \approx 3.46$, and the two concave bites should shave off a little more than $0.7$ together — which matches. The answer is positive (good), strictly smaller than the kite (good, since the arcs eat into the kite), and strictly bigger than zero (the arcs do not meet so deeply that the region disappears). Choices (C) $2\sqrt{3}$ ignores $\pi$ entirely (wrong — there are circular pieces), (E) is even larger than the kite (impossible since arcs cut area away), and (A), (D) have the wrong segment-area coefficient.
Alternative: Tool #1 (Draw a Diagram) with coordinates: place $R = (2, 0)$, sector centers at $X = (0,0)$ and $Y = (4,0)$, then $S = (1, \sqrt{3})$, $T = (3, \sqrt{3})$, $U = (2, 2\sqrt{3})$. The kite area follows from the Shoelace formula and the segments are computed exactly as above. The coordinate route gives the same $4\sqrt{3} - \frac{4\pi}{3}$ but is more setup-heavy than the symmetry-first decomposition used above.
CCSS standards used (min grade 8)
7.G.A.2Draw geometric shapes with given conditions including triangles (Recognizing that two sides of length $2$ meeting at a $60^\circ$ angle force an equilateral triangle, which then gives the chord lengths $SR = TR = 2$ and the kite $USRT$.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Splitting the kite $USRT$ into two equilateral triangles ($\triangle UST$ and $\triangle SRT$) and adding their areas to get $2\sqrt{3}$.)7.G.B.4Know the formulas for area and circumference of a circle (Using $\pi r^2$ scaled by $\frac{60}{360}$ to compute the area of each $60^\circ$ sector ($\frac{2\pi}{3}$), the first half of each circular-segment subtraction.)8.NS.A.1Know that numbers that are not rational are called irrational numbers (Treating $\sqrt{3}$ and $\pi$ as separate irrational quantities and combining them into the single expression $4\sqrt{3} - \frac{4\pi}{3}$ for the final answer.)
⭐ This AMC 8 problem only needs Grade 8 work with irrational numbers like $\sqrt{3}$ and $\pi$, plus the circle-area and triangle-area formulas you already know!
⭐ This AMC 8 problem only needs Grade 8 work with irrational numbers like $\sqrt{3}$ and $\pi$, plus the circle-area and triangle-area formulas you already know!
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