AMC 10 · 2007 · #24

Grade 8 geometry-2d
Archetype Compound Area by Decomposition / Difference └ Sub-archetype Circle-Polygon Hybrid
tangent-circlesarea-trianglescircular-sectorpythagorean-theoremarea-difference identify-subproblems ↑ Prerequisites: pythagorean-theorem
📏 Long solution 💡 3 insights 📊 Diagram
Problem
Two circles of radius 2 are centered at A and B. The point O is the midpoint of AB, with OA = 2√(2). Segment OC is tangent to circle A at C, and segment OD is tangent to circle B at D. The line EF is a common tangent running above the circles, touching circle A at E and circle B at F. What is the area of the shaded region ECODF, the piece bounded by E, C, O, D, F with the two circular bites removed?

Pick an answer.

(A)
$\frac {8\sqrt {2}}{3}$
(B)
$8\sqrt {2} - 4 - \pi$
(C)
$4\sqrt {2}$
(D)
$4\sqrt {2} + \frac {\pi}{8}$
(E)
$8\sqrt {2} - 2 - \frac {\pi}{2}$

AMC 10 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Change Focus / Count the Complement

The shaded blob has a lumpy outline: slanted tangent segments OC, OD and two curved circular bites. Measuring it head-on is hard, so Tool #16 (Change Focus / Count the Complement) reframes it: trap the whole thing inside the plain rectangle ABFE, whose area is easy, then subtract every piece of that rectangle that is NOT shaded. Those leftover pieces split cleanly into shapes we know, which is Tool #7 (Identify Subproblems): two right triangles at the bottom corners and two circular sectors from the disks. Tool #1 (Draw a Diagram) keeps the labels straight so each subtracted piece is counted once. Finding the triangles first needs one tangent length, which the Pythagorean theorem supplies.

1STEP 1

Trap the region in a rectangle

Box the shaded blob inside rectangle ABFE: base AB = 4√(2), height AE = 2, so the rectangle's area is 8√(2).

[ABFE] = AB · AE = 4√(2) · 2 = 8√(2)
2STEP 2

Find the tangent length by Pythagoras

Tangency gives ∠ OCA = 90°, so Pythagoras yields OC = 2; equal legs make OCA isosceles right, hence ∠ CAO = 45°.

OC = √((2√(2))² - 2²) = √(8 - 4) = 2 → ∠ CAO = 45°
3STEP 3

Subtract the two corner triangles

Triangle OCA has legs 2 and 2, so its area is 2; by symmetry ODB matches it, and the two corners remove 4 from the rectangle.

[OCA] = 1/2 · 2 · 2 = 2, [OCA] + [ODB] = 2 + 2 = 4
4STEP 4

Subtract the two circular sectors

AE points up and AC tilts 45°, so each bite is a 45° sector, one-eighth of the 4π disk: π/2 each, and the pair removes π.

[sector CAE] = 45/360 · π (2)² = 1/8 · 4π = π/2, 2 · π/2 = π
5STEP 5

Combine the pieces

Peel the two triangles (4) and the two sectors (π) off the rectangle, leaving 8√(2) - 4 - π, which is choice (B).

[ECODF] = 8√(2) - 4 - π → (B)
Answer
8√(2) - 4 - π
The numerical value is 8√(2) - 4 - π ≈ 11.31 - 4 - 3.14 = 4.17, a positive area smaller than the enclosing rectangle's 8√(2) ≈ 11.31 and a bit under half of it, which fits the picture where the two corners and two disk-slices eat away a large chunk. Scanning the choices, (C) 4√(2)≈ 5.66 and (A) 8√(2)/3≈ 3.77 ignore the π that a curved boundary must produce, and (E) 8√(2)-2-π/2 subtracts only one triangle's worth and one sector's worth, forgetting the symmetric partner on the right. Only (B) removes both triangles (4) and both sectors (π), so (B) is correct.
💡Key takeaway

When a shaded shape has a lumpy edge, box it inside a simple rectangle and subtract every piece that is not shaded.

  • Trap the region in a rectangle
  • Find the tangent length by Pythagoras
  • Subtract the two corner triangles
  • Subtract the two circular sectors
  • Combine the pieces

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