AMC 10 · 2015 · #14

Grade 8 geometry-2d

Archetype Overlap / Complement Counting └ Sub-archetype Geometric Area via Inclusion-Exclusion or Area Balance

area-circlesequilateral-trianglearea-difference area-differenceidentify-subproblems ↑ Prerequisites: area-circlesequilateral-triangle

📏 Medium solution 💡 2 insights

Problem

Two overlapping shapes are compared by the parts each has outside the other. Find the difference.

Pick an answer.

(A)

$8-\pi$

(B)

$\pi+2$

(C)

$2\pi-\dfrac{\sqrt{2}}{2}$

(D)

$4(\pi-\sqrt{3})$

(E)

$2\pi-\dfrac{\sqrt{3}}{2}$

How to solve

Strategy Draw a Venn Diagram

Two shapes that overlap split the plane into exactly the pieces a Venn diagram is built for: circle-only, triangle-only, both. That is tool #12. Writing the two named regions in those terms turns the question into pure bookkeeping, and the overlap piece then drops out of the subtraction on its own — tool #16, changing focus away from the hard part nobody asked for. Tool #1 still earns its place, because a quick sketch is what convinces you the shapes really do overlap and tells you which area is the larger one, so you know the answer's sign. What is left after the cancellation splits into two independent area computations, so tool #7 handles them one at a time. Tool #3 finishes the job by matching the exact expression to the five printed choices, which matters here because two of them are numerically close. The whole plan rests on one structural fact, so that fact is what the work below has to establish honestly: the overlap enters both named areas the same way, and a difference kills it. A sum would not.

1STEP 1

Split the picture into three pieces

Each part is a whole minus the overlap.

[circle only] = [C] - S, [triangle only] = [T] - S

2STEP 2

Subtract and watch the overlap cancel

Subtracting makes the overlap cancel.

([C]-S)-([T]-S) = [C]-[T]

3STEP 3

Area of the whole circle

The circle's area is 4π.

[C] = π · 2² = 4π ≈ 12.57

4STEP 4

Area of the whole triangle

The triangle's area is 4√3.

h=√(4²-2²)=2√(3), [T]=1/2 · 4 · 2√(3)=4√(3)≈ 6.93

5STEP 5

Combine and match a choice

The difference is choice (D).

[C]-[T] = 4π-4√(3) = 4(π-√(3)) ≈ 5.64 → (D)

Answer

4(π-√(3))

Two checks. First the sign: the circle covers about 12.57 and the triangle about 6.93, so the circle-only region should be the bigger of the two named regions and the difference should come out positive. 4(π-√(3)) ≈ 5.64 is positive, as expected. Second, a stress test of the step that carries all the weight, the cancellation. Shrink the circle to radius 1 and redo the bookkeeping with the real overlap: the shared part becomes a 60° sector of radius 1, area π/6, so circle-only is π-π/6=5π/6 and triangle-only is 4√(3)-π/6. Their difference is 5π/6-4√(3)+π/6=π-4√(3), which is again the circle minus the triangle, this time negative because the triangle is now the larger shape. The cancellation is structural, not a coincidence of these numbers. One warning about checking by decimals alone: choice (C) is about 5.58 and choice (E) about 5.42, both within a quarter of a unit of 5.64, so a rough estimate cannot separate them. The exact expression is what settles the choice.

💡Key takeaway

When two shapes overlap and you subtract one "only" region from the other, the shared middle cancels, so all you need are the two whole areas.

Split the picture into three pieces
Subtract and watch the overlap cancel
Area of the whole circle
Area of the whole triangle
Combine and match a choice

Split the picture into three pieces

Subtract and watch the overlap cancel

Area of the whole circle

Area of the whole triangle

Combine and match a choice

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