AMC 10 · 2006 · #24

Grade 8 geometry-2d

tangent-circlespythagorean-theoremsimilar-triangles area-difference ↑ Prerequisites: tangent-circlespythagorean-theorem

📏 Long solution 💡 4 insights 📊 Diagram

Problem

A small circle with center

O

has radius

2

, a large circle with center

P

has radius

4

, and the two circles are externally tangent. Two common external tangent lines are drawn: one touches the small circle at

A

and the large circle at

D

, the other touches them at

B

and

C

. Find the area of the concave hexagon

AOBCPD

, which caves inward at

O

and

P

Pick an answer.

(A)

$18\sqrt{3}$

(B)

$24\sqrt{2}$

(C)

(D)

$24\sqrt{3}$

(E)

$32\sqrt{2}$

AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

A concave hexagon looks scary, but Tool #7 (Identify Subproblems) turns it into pieces we already know. The line OP splits the figure into two mirror-image halves, and each half is just the quadrilateral OADP — a right trapezoid, because both radii OA and PD stand perpendicular to the same tangent line AD and are therefore parallel. So the plan is: (1) draw the radii to the tangent points and mark the right angles (Tool #1, Draw a Diagram); (2) build a right triangle to find the tangent length AD with the Pythagorean theorem; (3) find the area of the right trapezoid OADP; (4) double it, since symmetry (Tool #17) makes the lower half congruent to the upper half. Two known shapes replace one strange one.

1STEP 1

Draw the radii and spot the parallel sides

External tangency gives $OP=6$ ; radii $OA$ and $PD$ both meet tangent $AD$ at right angles, so $OA \parallel PD$ and $OADP$ is a right trapezoid.

OP=2+4=6, OA ⊥ AD, PD ⊥ AD → OA ∥ PD

2STEP 2

Build a right triangle to find AD

Drop $OQ$ from $O$ perpendicular to $PD$ : $OADQ$ is a rectangle, so $PQ=4-2=2$ , and Pythagoras with $OP=6$ gives $AD=OQ=4\sqrt{2}$ .

PQ=4-2=2, AD=OQ=√(6²-2²)=√(32)=4√(2)

3STEP 3

Find the area of trapezoid OADP

Since $AD$ is perpendicular to both parallel sides, $[OADP]=\frac{1}{2}(2+4)\cdot 4\sqrt{2}$ , giving $12\sqrt{2}$ .

[OADP]=1/2(OA+PD) · AD=1/2(2+4)(4√(2))=12√(2)

4STEP 4

Double it using the symmetry

Line $OP$ mirrors the top half onto the bottom, so the hexagon is two such trapezoids: $2\cdot 12\sqrt{2}=24\sqrt{2}$ , choice (B).

[AOBCPD]=2·[OADP]=2 · 12√(2)=24√(2) → (B)

Answer

24√(2)

Turn the answer into a decimal to sanity-check: 24√(2)≈ 33.9. The tangent length came out to 4√(2)≈ 5.66, and one trapezoid to about 17, which looks right for a strip roughly 6 wide averaging 3 tall. The other choices sit noticeably apart — 18√(3)≈ 31.2, 36, 24√(3)≈ 41.6, 32√(2)≈ 45.3 — and only 24√(2) matches our computed value, so there is no rounding ambiguity. The √(2) is expected, since the tangent length √(32) produced it.

💡Key takeaway

A radius always meets its tangent at a right angle, so a tangent problem secretly hides right triangles and simple trapezoids — find them, and the Pythagorean theorem hands you the missing length.

Draw the radii and spot the parallel sides
Build a right triangle to find AD
Find the area of trapezoid OADP
Double it using the symmetry

Draw the radii and spot the parallel sides

Build a right triangle to find AD

Find the area of trapezoid OADP

Double it using the symmetry

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