AMC 10 · 2013 · #7
Grade 8 geometry-2dSix points are equally spaced around a circle of radius 1. Three of these points are the vertices of a triangle that is neither equilateral nor isosceles. What is the area of this triangle?
Pick an answer.
AMC 10 2013 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: Six points are equally spaced around a circle of radius $1$. Choose three of them so the triangle they form is neither equilateral nor isosceles (a scalene triangle, all three sides different). Find the area of that triangle.
Givens: Six points are equally spaced around a circle of radius $1$; A triangle is formed by choosing three of these six points; The triangle is scalene — neither equilateral nor isosceles, so all three sides differ; Answer choices: (A) $\frac{\sqrt{3}}{3}$, (B) $\frac{\sqrt{3}}{2}$, (C) $1$, (D) $\sqrt{2}$, (E) $2$
Unknowns: The area of the scalene triangle
Understand
Restated: Six points are equally spaced around a circle of radius $1$. Choose three of them so the triangle they form is neither equilateral nor isosceles (a scalene triangle, all three sides different). Find the area of that triangle.
Givens: Six points are equally spaced around a circle of radius $1$; A triangle is formed by choosing three of these six points; The triangle is scalene — neither equilateral nor isosceles, so all three sides differ; Answer choices: (A) $\frac{\sqrt{3}}{3}$, (B) $\frac{\sqrt{3}}{2}$, (C) $1$, (D) $\sqrt{2}$, (E) $2$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #2 Make a Systematic List, #7 Identify Subproblems
Drawing the six points as a regular hexagon (tool #1) is what makes the answer visible: it reveals which side is a diameter and lets you read the height off an equilateral triangle. Tool #2 (Make a Systematic List) is used first to pin down the one triangle shape that is scalene by listing the ways the arc gaps can split. Tool #7 (Identify Subproblems) then breaks the area into two clean pieces — find the base, then find the height — instead of chasing all three side lengths at once.
Execute — Answer: B
8.G.A.1 Step 1 Find the scalene shape
- Three chosen points cut the circle into three arcs, each a whole number of $60^\circ$ gaps that add to $6$ gaps (the full circle).
- Rotating the circle shows a chord's length depends only on how many gaps it spans, so equal gaps make equal sides.
- Write $6$ as three positive parts: $2+2+2$, $1+1+4$, and $1+2+3$.
- The split $2+2+2$ gives three equal sides (equilateral) and $1+1+4$ gives two equal sides (isosceles).
- Only $1+2+3$ makes all three sides different, so that is the triangle we want.
💡 Turning the whole circle keeps a chord the same length, so only the arc-gap count can make two sides differ.
7.G.B.4 Step 2 Spot the diameter
- In the $1+2+3$ triangle, the arc that spans $3$ gaps is $3\times 60^\circ = 180^\circ$, exactly half the circle.
- A chord across half the circle passes through the center, so it is a diameter.
- The radius is $1$, so this diameter is $2\times 1 = 2$.
- Use this longest side as the base of the triangle.
💡 A chord that cuts off half the circle must run straight through the center, which makes it a diameter.
8.G.B.7 Step 3 Find the height
- Lay the diameter flat as the base.
- The third vertex sits one $60^\circ$ gap away from a base endpoint.
- Join both that vertex and that endpoint to the center $O$: each is a radius of length $1$, and the $60^\circ$ angle between them makes triangle equilateral with side $1$.
- Its base lies along the diameter, so the height of the third vertex above the diameter is the altitude of a unit equilateral triangle.
- By the Pythagorean theorem, $h^2 + \left(\tfrac12\right)^2 = 1^2$, so $h = \tfrac{\sqrt{3}}{2}$.
💡 Two radii with a $60^\circ$ angle always close into an equilateral triangle, whose altitude splits it into two right triangles.
6.G.A.1 Step 4 Compute the area
- The triangle has base $2$ (the diameter) and height $\tfrac{\sqrt{3}}{2}$.
- Area $=\tfrac12\times\text{base}\times\text{height} = \tfrac12\times 2\times\tfrac{\sqrt{3}}{2} = \tfrac{\sqrt{3}}{2}$.
- That is choice (B).
💡 Base times height over two gives any triangle's area once you know one side and the perpendicular drop onto it.
8.G.A.1 Three chosen points cut the circle into three arcs, each a whole number of $60^\ 7.G.B.4 In the $1+2+3$ triangle, the arc that spans $3$ gaps is $3\times 60^\circ = 180^ 8.G.B.7 Lay the diameter flat as the base. The third vertex sits one $60^\circ$ gap away 6.G.A.1 The triangle has base $2$ (the diameter) and height $\tfrac{\sqrt{3}}{2}$. Area Review
Reasonableness: The triangle fits inside a radius-$1$ circle, so its area must be well under the circle's area $\pi\approx 3.14$; $\tfrac{\sqrt{3}}{2}\approx 0.87$ is comfortably small. The three sides come out $2$, $\sqrt{3}$, and $1$ — a $30$-$60$-$90$ right triangle with all sides different, confirming it really is scalene. Choices (D) $\sqrt{2}\approx 1.41$ and (E) $2$ are too large for a triangle this size, and (A) $\tfrac{\sqrt{3}}{3}\approx 0.58$ and (C) $1$ do not match the base-height product, so only $\tfrac{\sqrt{3}}{2}$ survives.
Alternative: Use coordinates. Put the center at the origin and the six points at $(\cos\theta,\sin\theta)$ for $\theta = 0^\circ,60^\circ,\dots,300^\circ$. The scalene triangle has vertices $(1,0)$, $(-1,0)$, and $(\tfrac12,\tfrac{\sqrt{3}}{2})$. The base from $(-1,0)$ to $(1,0)$ has length $2$, and the third point's height is its $y$-coordinate $\tfrac{\sqrt{3}}{2}$, giving the same area $\tfrac12\times 2\times\tfrac{\sqrt{3}}{2}=\tfrac{\sqrt{3}}{2}$.
CCSS standards used (min grade 8)
8.G.A.1Verify experimentally the properties of rotations, reflections, and translations (Arguing that rotating the circle preserves chord lengths, so equal arc gaps give equal sides — which isolates $1+2+3$ as the only scalene split.)7.G.B.4Know the formulas for area and circumference of a circle (Recognizing the half-circle chord as a diameter and using diameter $=2\times\text{radius}=2$ as the base.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding the height $h=\tfrac{\sqrt{3}}{2}$ as the altitude of a unit equilateral triangle via $h^2+(\tfrac12)^2=1^2$.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Combining base $2$ and height $\tfrac{\sqrt{3}}{2}$ with area $=\tfrac12\,\text{base}\times\text{height}$ to get $\tfrac{\sqrt{3}}{2}$.)
⭐ For evenly spaced points on a circle, count the gaps: equal gaps make equal sides, and a chord across half the circle is a diameter you can stand the triangle on.
⭐ For evenly spaced points on a circle, count the gaps: equal gaps make equal sides, and a chord across half the circle is a diameter you can stand the triangle on.
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