AMC 10 · 2003 · #17
Grade 8 geometry-2dThe number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle?
Pick an answer.
AMC 10 2003 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: An equilateral triangle sits inside a circle so that all three corners touch the circle. The number giving the triangle's perimeter (in inches) is the same as the number giving the circle's area (in square inches). Find the circle's radius.
Givens: The triangle is equilateral and inscribed in the circle (all three vertices on the circle); Perimeter of the triangle (in inches) $=$ area of the circle (in square inches); Answer choices: (A) $\frac{3\sqrt{2}}{\pi}$, (B) $\frac{3\sqrt{3}}{\pi}$, (C) $\sqrt{3}$, (D) $\frac{6}{\pi}$, (E) $\sqrt{3}\,\pi$
Unknowns: The radius $R$ of the circle, in inches
Understand
Restated: An equilateral triangle sits inside a circle so that all three corners touch the circle. The number giving the triangle's perimeter (in inches) is the same as the number giving the circle's area (in square inches). Find the circle's radius.
Givens: The triangle is equilateral and inscribed in the circle (all three vertices on the circle); Perimeter of the triangle (in inches) $=$ area of the circle (in square inches); Answer choices: (A) $\frac{3\sqrt{2}}{\pi}$, (B) $\frac{3\sqrt{3}}{\pi}$, (C) $\sqrt{3}$, (D) $\frac{6}{\pi}$, (E) $\sqrt{3}\,\pi$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #1 Draw a Diagram, #7 Identify Subproblems, #13 Convert to Algebra
The radius $R$ is what we want, and everything else in the problem can be written in terms of it, so Tool #4 (Introduce a Variable) makes $R$ the single unknown to chase. Tool #1 (Draw a Diagram) is the key that unlocks the geometry: sketching the circle with its inscribed triangle and dropping a line from the center to one side turns the picture into a $30$-$60$-$90$ right triangle, which pins the side length to $R$. Once both the perimeter and the area are expressions in $R$, Tool #13 (Convert to Algebra) lets us set them equal and solve one clean equation. The plan is to express two things ($3\times$ side and $\pi R^2$) in the same variable, then let algebra finish.
Execute — Answer: B
6.EE.B.6 Step 1 Draw the picture and name the radius
- Sketch the circle with the equilateral triangle inside it, all three corners on the circle.
- Call the radius $R$ and the triangle's side length $s$.
- Draw a radius to each corner; because the triangle is equilateral, these three radii split the full turn at the center evenly, so the angle between two neighboring radii is $\tfrac{360^\circ}{3}=120^\circ$.
- The goal now is to trade $s$ for something written only in $R$.
💡 Putting a letter on the radius and drawing the spokes turns a word puzzle into a shape you can measure.
8.G.B.7 Step 2 Find the side from the radius with a 30-60-90 triangle
- Drop a straight line from the center to the middle of one side.
- It hits that side at a right angle and cuts both the side and the $120^\circ$ center angle exactly in half.
- That leaves a right triangle whose corner at the center is $60^\circ$, whose corner at the triangle's vertex is $30^\circ$, and whose longest side (the radius) is $R$.
- In any $30$-$60$-$90$ triangle the sides are in the ratio $1 : \sqrt{3} : 2$, and the hypotenuse $R$ matches the $2$, so half of the triangle's side (the leg opposite the $60^\circ$) is $\tfrac{\sqrt{3}}{2}R$.
- Doubling gives the whole side $s = \sqrt{3}\,R$.
💡 Splitting the corner angle in half makes a $30$-$60$-$90$ triangle, whose fixed side ratios hand you the side length for free.
7.G.B.4 Step 3 Write the perimeter and the area in terms of R
- The perimeter is three equal sides: $P = 3s = 3\sqrt{3}\,R$.
- The circle's area comes from the standard formula $A = \pi R^2$.
- The problem says these two numbers are equal, so set the perimeter expression equal to the area expression.
- Both are now written using only the single unknown $R$.
💡 Once every quantity speaks the same language ($R$), 'perimeter equals area' is just one equation.
8.EE.C.7 Step 4 Solve the equation for the radius
- Start from $3\sqrt{3}\,R = \pi R^2$.
- Since $R$ is a real circle's radius it is not zero, so divide both sides by $R$ to get $3\sqrt{3} = \pi R$.
- Then divide by $\pi$ to isolate the radius: $R = \tfrac{3\sqrt{3}}{\pi}$.
- That matches choice (B).
💡 Both sides carry a factor of $R$, so cancelling one $R$ drops the quadratic down to an easy linear equation.
6.EE.B.6 Sketch the circle with the equilateral triangle inside it, all three corners on 8.G.B.7 Drop a straight line from the center to the middle of one side. It hits that sid 7.G.B.4 The perimeter is three equal sides: $P = 3s = 3\sqrt{3}\,R$. The circle's area c 8.EE.C.7 Start from $3\sqrt{3}\,R = \pi R^2$. Since $R$ is a real circle's radius it is n Review
Reasonableness: Plug $R = \tfrac{3\sqrt{3}}{\pi}$ back in. The area is $\pi R^2 = \pi\cdot\tfrac{27}{\pi^2} = \tfrac{27}{\pi}$. The perimeter is $3\sqrt{3}\,R = 3\sqrt{3}\cdot\tfrac{3\sqrt{3}}{\pi} = \tfrac{9\cdot 3}{\pi} = \tfrac{27}{\pi}$. The two match, so the radius is correct. Numerically $R \approx \tfrac{5.196}{3.1416}\approx 1.65$ inches, a sensible size for a small circle whose perimeter number and area number happen to coincide.
Alternative: Instead of $30$-$60$-$90$, use the centroid fact: in an equilateral triangle the circumcenter is the centroid, sitting $\tfrac{2}{3}$ of the way down each median. The median (which is also the height) is $h = \tfrac{\sqrt{3}}{2}s$, so $R = \tfrac{2}{3}h = \tfrac{2}{3}\cdot\tfrac{\sqrt{3}}{2}s = \tfrac{s}{\sqrt{3}}$, giving $s = \sqrt{3}\,R$ again. From there the same equation $3\sqrt{3}\,R = \pi R^2$ yields $R = \tfrac{3\sqrt{3}}{\pi}$.
CCSS standards used (min grade 8)
6.EE.B.6Use variables to represent numbers and write expressions when solving a problem (Naming the radius $R$ and the side $s$ so the whole figure can be described with letters.)8.G.B.7Apply the Pythagorean Theorem to determine unknown side lengths in right triangles (Using the $30$-$60$-$90$ right triangle (a Pythagorean special case) to get the side $s = \sqrt{3}\,R$ from the radius.)7.G.B.4Know and use the formulas for the area and circumference of a circle (Writing the circle's area as $\pi R^2$ to equate it with the triangle's perimeter.)8.EE.C.7Solve linear equations in one variable (Dividing $3\sqrt{3}\,R = \pi R^2$ by $R$ and then by $\pi$ to solve $R = \tfrac{3\sqrt{3}}{\pi}$.)
⭐ Write both the perimeter and the area using the same radius $R$, set them equal, and cancel the shared $R$ to turn a scary squared equation into an easy one.
⭐ Write both the perimeter and the area using the same radius $R$, set them equal, and cancel the shared $R$ to turn a scary squared equation into an easy one.
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