AMC 10 · 2006 · #24
Grade 8 geometry-2dCircles with centers O and P have radii 2 and 4, respectively, and are externally tangent. Points A and B on the circle with center O and points C and D on the circle with center P are such that AD and BC are common external tangents to the circles. What is the area of the concave hexagon AOBCPD?
Pick an answer.
AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A small circle (center $O$, radius $2$) and a large circle (center $P$, radius $4$) touch each other on the outside. Two lines are drawn that are tangent to both circles: one touches the circles at $A$ (on the small circle) and $D$ (on the large circle), the other touches at $B$ and $C$. Find the area of the six-sided figure $AOBCPD$, which caves inward at $O$ and $P$.
Givens: Circle $O$ has radius $2$; circle $P$ has radius $4$; The two circles are externally tangent, so they touch at one point from outside; Line $AD$ is tangent to both circles ($A$ on circle $O$, $D$ on circle $P$); Line $BC$ is tangent to both circles ($B$ on circle $O$, $C$ on circle $P$); Answer choices: (A) $18\sqrt{3}$, (B) $24\sqrt{2}$, (C) $36$, (D) $24\sqrt{3}$, (E) $32\sqrt{2}$
Unknowns: The length of the tangent segment $AD$; The area of the concave hexagon $AOBCPD$
Understand
Restated: A small circle (center $O$, radius $2$) and a large circle (center $P$, radius $4$) touch each other on the outside. Two lines are drawn that are tangent to both circles: one touches the circles at $A$ (on the small circle) and $D$ (on the large circle), the other touches at $B$ and $C$. Find the area of the six-sided figure $AOBCPD$, which caves inward at $O$ and $P$.
Givens: Circle $O$ has radius $2$; circle $P$ has radius $4$; The two circles are externally tangent, so they touch at one point from outside; Line $AD$ is tangent to both circles ($A$ on circle $O$, $D$ on circle $P$); Line $BC$ is tangent to both circles ($B$ on circle $O$, $C$ on circle $P$); Answer choices: (A) $18\sqrt{3}$, (B) $24\sqrt{2}$, (C) $36$, (D) $24\sqrt{3}$, (E) $32\sqrt{2}$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #17 Visualize Spatial Relationships
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.
Execute — Answer: B
4.G.A.1 Step 1 Draw the radii and spot the parallel sides
- Because the circles are externally tangent, their centers are $OP=2+4=6$ apart.
- Draw radius $OA$ to the point where line $AD$ touches the small circle, and radius $PD$ to where it touches the large circle.
- A tangent line meets its radius at a right angle, so $OA\perp AD$ and $PD\perp AD$.
- Two segments that are both perpendicular to the same line are parallel, so $OA\parallel PD$.
- That makes quadrilateral $OADP$ a right trapezoid: its two parallel sides are $OA=2$ and $PD=4$.
💡 Both radii lean against the same tangent line at right angles, so they point the same way and are parallel.
8.G.B.7 Step 2 Build a right triangle to find AD
- In trapezoid $OADP$ the parallel sides $OA=2$ and $PD=4$ have different lengths, so slide across from $O$: drop a segment from $O$ perpendicular to line $PD$, meeting it at $Q$.
- Then $OADQ$ is a rectangle, so $OQ=AD$ and $QD=OA=2$.
- That leaves $PQ=PD-QD=4-2=2$.
- Triangle $OQP$ has a right angle at $Q$, hypotenuse $OP=6$, and leg $PQ=2$, so by the Pythagorean theorem $OQ=\sqrt{6^{2}-2^{2}}=\sqrt{32}=4\sqrt{2}$.
- Therefore $AD=4\sqrt{2}$.
💡 Chopping off a rectangle leaves one clean right triangle whose missing leg is exactly the tangent length.
6.G.A.1 Step 3 Find the area of trapezoid OADP
- The right trapezoid $OADP$ has parallel sides $OA=2$ and $PD=4$, and the distance between those parallel sides is the tangent segment $AD=4\sqrt{2}$ (it is perpendicular to both).
- The area of a trapezoid is the average of the two parallel sides times the distance between them: $\frac{1}{2}(2+4)(4\sqrt{2})=\frac{1}{2}\cdot 6\cdot 4\sqrt{2}=12\sqrt{2}$.
💡 A trapezoid's area is just the average width times how far apart the parallel edges sit.
4.G.A.3 Step 4 Double it using the symmetry
- The line $OP$ is an axis of symmetry: everything above it (the trapezoid $OADP$) is the mirror image of everything below it (the trapezoid $OBCP$), so the two halves have equal area.
- The concave hexagon $AOBCPD$ is exactly these two trapezoids joined along $OP$.
- So its area is $2\times 12\sqrt{2}=24\sqrt{2}$, which is choice (B).
💡 Mirror symmetry across $OP$ means the bottom half is a free copy of the top half, so just double.
4.G.A.1 Because the circles are externally tangent, their centers are $OP=2+4=6$ apart. 8.G.B.7 In trapezoid $OADP$ the parallel sides $OA=2$ and $PD=4$ have different lengths, 6.G.A.1 The right trapezoid $OADP$ has parallel sides $OA=2$ and $PD=4$, and the distanc 4.G.A.3 The line $OP$ is an axis of symmetry: everything above it (the trapezoid $OADP$) Review
Reasonableness: Turn the answer into a decimal to sanity-check: $24\sqrt{2}\approx 33.9$. The tangent length came out to $4\sqrt{2}\approx 5.66$, and one trapezoid to about $17$, which looks right for a strip roughly $6$ wide averaging $3$ tall. The other choices sit noticeably apart — $18\sqrt{3}\approx 31.2$, $36$, $24\sqrt{3}\approx 41.6$, $32\sqrt{2}\approx 45.3$ — and only $24\sqrt{2}$ matches our computed value, so there is no rounding ambiguity. The $\sqrt{2}$ is expected, since the tangent length $\sqrt{32}$ produced it.
Alternative: Extend the two tangent lines until they meet at a point $X$ off to the left of the small circle. By symmetry $X$, $O$, $P$ are collinear, and $X$ sees each circle inside a pair of tangent segments. The tangent length from $X$ to circle $O$ is $\sqrt{XO^{2}-2^{2}}$ and to circle $P$ is $\sqrt{XP^{2}-4^{2}}$; with $XO=6$ and $XP=12$ these are $4\sqrt{2}$ and $8\sqrt{2}$. Kite $XAOB$ has area $XA\cdot AO=4\sqrt{2}\cdot 2=8\sqrt{2}$, and kite $XDPC$ has area $XD\cdot DP=8\sqrt{2}\cdot 4=32\sqrt{2}$. The hexagon is the big kite minus the small kite: $32\sqrt{2}-8\sqrt{2}=24\sqrt{2}$ — the same answer by a completely different route.
CCSS standards used (min grade 8)
4.G.A.1Draw points, lines, line segments, rays, angles, and perpendicular and parallel lines (Recognizing that radii $OA$ and $PD$ are both perpendicular to the tangent $AD$ and therefore parallel, making $OADP$ a right trapezoid.)8.G.B.7Apply the Pythagorean Theorem to determine unknown side lengths in right triangles (Finding the tangent length $AD=\sqrt{6^{2}-2^{2}}=4\sqrt{2}$ from the right triangle $OQP$.)6.G.A.1Find the area of triangles and special quadrilaterals by composing and decomposing (Computing the area of the right trapezoid $OADP=\frac{1}{2}(2+4)(4\sqrt{2})=12\sqrt{2}$.)4.G.A.3Recognize a line of symmetry for a two-dimensional figure (Using the mirror symmetry across $OP$ to double one trapezoid into the full hexagon area $24\sqrt{2}$.)
⭐ 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.
⭐ 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.
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