AMC 10 · 2021 · #9

Grade 8 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length
equilateral-triangleinradiusperpendicular-bisectorarea-circles symmetry-argumentconvert-to-algebra ↑ Prerequisites: equilateral-trianglearea-circles
📏 Medium solution 💡 3 insights
Problem
An equilateral triangle has every side of length 6. Inside it sits the inscribed circle touching all three sides, and its centre is marked. A second circle is drawn through two vertices and that centre. Find the area of this second circle.

Pick an answer.

(A)
$9\pi$
(B)
$12\pi$
(C)
$18\pi$
(D)
$24\pi$
(E)
$27\pi$
How to solve
Strategy Introduce a Variable

Nothing can be measured until the picture becomes numbers, so the work opens with tool #1 (Draw a Diagram): drop the triangle onto a coordinate grid with AC flat on the x-axis and its midpoint at the origin. The picture then hands over a gift that tool #17 (Visualize Spatial Relationships) is there to notice — the whole figure is unchanged when it is reflected in the vertical line through that midpoint, and both of the mystery centers, the incenter O and the center of the new circle, are forced onto that one line. A point that could have been anywhere in the plane now needs only a single number to describe it. That is where tool #4 (Introduce a Variable) does the real work: name that number, write down what "the same distance from here as from there" says about it, and the squared terms cancel to leave one easy equation. Tool #7 (Identify Subproblems) fixes the order — locate O first, then use O to locate the new center — because each half is a two-line calculation while attacking both at once is not.

1STEP 1

Put AC flat on the x-axis

Put one side flat on an axis.

A=(-3,0), C=(3,0), 3²+h²=6² → h=3√(3), B=(0,3√(3))
2STEP 2

Symmetry traps O on the mirror line

Symmetry traps the centre on the mirror line.

O=(0,t), OA=OB=OC
3STEP 3

Pin down the height of O

Pin down the incentre's height.

9+t²=27-6√(3) t+t² → 6√(3) t=18 → t=√(3), O=(0,√(3)), OA²=12
4STEP 4

The new center sits on the same line

The new centre sits on the same line.

P=(0,k), PA²=9+k², PO²=(k-√(3))²
5STEP 5

Solve for R² and finish

Solving for the radius squared gives an area of twelve pi.

9+k²=k²-2√(3) k+3 → k=-√(3), R²=9+3=12, area=π R²=12π
Answer
12π
A and C are 6 apart and both sit on the circle, so the diameter is at least 6 and R ≥ 3, forcing the area to be at least 9π. Equality would require AC to be a full diameter, which happens only when the angle at O in triangle AOC is a right angle. But AO and CO bisect two 60° corners of the equilateral triangle, so that angle is 180°-30°-30°=120° — blunt, not square. So (A) 9π is out, and the answer should be a little above 9π: 12π fits, while (C) 18π, (D) 24π and (E) 27π need radii of roughly 4.2, 4.9 and 5.2 to carry a chord only 6 long, spreading the circle far wider than the picture allows. The tempting wrong answers are exactly the numbers that turn up in the work: 27π is what comes of mistaking the triangle's height 3√(3) for the radius, and 9π is what comes of assuming AC is a diameter. One more check that the arithmetic is not an accident: R=√(12)=2√(3) is exactly OA, the distance already found in step 3, so the new circle's radius equals the distance from O to a corner — a strong hint that something symmetric is behind it.
💡Key takeaway

When two of your points sit on a mirror line of the picture, the center of any circle through them has to sit on that mirror line too — so search along one line instead of all over the plane.

  • Put AC flat on the x-axis
  • Symmetry traps O on the mirror line
  • Pin down the height of O
  • The new center sits on the same line
  • Solve for R² and finish

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