AMC 10 · 2016 · #21

Grade 8 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length
tangent-circlescoordinate-geometrypythagorean-theorem convert-to-algebra ↑ Prerequisites: tangent-circles
📏 Long solution 💡 4 insights
Problem
Three circles with radii 1, 2, and 3 (centers P, Q, R) sit on the same side of a line l and each touch l at one point, with the middle circle's touch point between the other two. The radius-2 circle is externally tangent to both of the others. Find the area of triangle PQR.

Pick an answer.

(A)
0
(B)
$\sqrt{\frac{2}{3}}$
(C)
1
(D)
$\sqrt{6}-\sqrt{2}$
(E)
$\sqrt{\frac{3}{2}}$

AMC 10 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

The whole problem is about positions of three centers above a line, so tool #1 (Draw a Diagram) earns its keep: put l on the x-axis, and each center lands directly above its tangent point at a height equal to its radius. Tool #4 (Introduce a Variable) names the unknown horizontal positions of the centers and pins them down, because external tangency turns each center-to-center distance into a known number, and a right triangle (Pythagorean theorem) converts that into a horizontal gap. Once all three centers have coordinates, tool #7 (Identify Subproblems) computes the triangle's area cleanly by dropping the centers to the line and subtracting trapezoid areas, avoiding any messy formula.

1STEP 1

Set up a coordinate frame

On the x-axis a tangent circle sits one radius above its touch point, so heights are P=1, Q=2, R=3; only horizontal positions stay unknown.

P=(x_P, 1), Q=(x_Q, 2), R=(x_R, 3)
2STEP 2

Turn tangency into distances

External tangency makes each center distance the sum of radii, so PQ=3, QR=5, each the hypotenuse of a right triangle over known heights.

PQ=3, QR=5
3STEP 3

Find the horizontal gaps

Each height gap is 1, so Pythagoras gives horizontal gaps 2√(2) for PQ and 2√(6) for QR; place Q' at the origin with P' left and R' right.

x-gap_PQ=√(3²-1²)=2√(2), x-gap_QR=√(5²-1²)=2√(6)
4STEP 4

Write the three centers' coordinates

With Q=(0,2), step left to P=(-2√(2), 1) and right to R=(2√(6), 3), giving triangle PQR fully numeric vertices.

P=(-2√(2), 1), Q=(0, 2), R=(2√(6), 3)
5STEP 5

Get the area by subtracting trapezoids

Drop the centers to the x-axis: the triangle is the trapezoid under PR minus those under PQ and QR, giving area √(6)-√(2), answer (D).

[PQR]=(4√(2)+4√(6))-(3√(2)+5√(6))=√(6)-√(2)→(D)
Answer
√(6)-√(2)
The answer √(6)-√(2)≈ 2.449-1.414≈ 1.035 is a small positive number, which fits three centers strung along just above a line in a nearly flat sliver of a triangle. It rules out (A) 0, which would need the centers collinear, and they are not. A quick cross-check with the shoelace formula on P=(-2√(2),1), Q=(0,2), R=(2√(6),3) gives 1/2|(-2√(2))(2-3)+0(3-1)+2√(6)(1-2)|=1/2|2√(2)-2√(6)|=√(6)-√(2), matching exactly.
💡Key takeaway

Stand each circle's center straight above where it touches the line at a height equal to its radius, use the Pythagorean theorem for the sideways gaps, then carve the triangle out of trapezoids to get √(6)-√(2).

  • Set up a coordinate frame
  • Turn tangency into distances
  • Find the horizontal gaps
  • Write the three centers' coordinates
  • Get the area by subtracting trapezoids

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