AMC 10 · 2007 · #24
Grade 8 geometry-2dCircles centered at A and B each have radius 2, as shown. Point O is the midpoint of AB, and OA=22. Segments OC and OD are tangent to the circles centered at A and B, respectively, and EF is a common tangent. What is the area of the shaded region ECODF?
Pick an answer.
AMC 10 2007 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: Two circles of radius $2$ are centered at $A$ and $B$. The point $O$ is the midpoint of $\overline{AB}$, with $OA = 2\sqrt{2}$. From $O$, segment $OC$ is tangent to circle $A$ and segment $OD$ is tangent to circle $B$; the line $EF$ is tangent to both circles across the top. Find the area of the shaded region $ECODF$, which is the blob bounded by $E$, $C$, $O$, $D$, $F$ with the two circular bites removed.
Givens: Each circle has radius $2$, centered at $A$ and at $B$; $O$ is the midpoint of $\overline{AB}$ and $OA = OB = 2\sqrt{2}$; $OC$ is tangent to circle $A$ and $OD$ is tangent to circle $B$; $EF$ is a common tangent across the tops of the circles, with $E$ above $A$ and $F$ above $B$; The figure is symmetric about the perpendicular bisector of $\overline{AB}$ through $O$; Answer choices: (A) $\frac{8\sqrt{2}}{3}$, (B) $8\sqrt{2} - 4 - \pi$, (C) $4\sqrt{2}$, (D) $4\sqrt{2} + \frac{\pi}{8}$, (E) $8\sqrt{2} - 2 - \frac{\pi}{2}$
Unknowns: The area of the shaded region $ECODF$
Understand
Restated: Two circles of radius $2$ are centered at $A$ and $B$. The point $O$ is the midpoint of $\overline{AB}$, with $OA = 2\sqrt{2}$. From $O$, segment $OC$ is tangent to circle $A$ and segment $OD$ is tangent to circle $B$; the line $EF$ is tangent to both circles across the top. Find the area of the shaded region $ECODF$, which is the blob bounded by $E$, $C$, $O$, $D$, $F$ with the two circular bites removed.
Givens: Each circle has radius $2$, centered at $A$ and at $B$; $O$ is the midpoint of $\overline{AB}$ and $OA = OB = 2\sqrt{2}$; $OC$ is tangent to circle $A$ and $OD$ is tangent to circle $B$; $EF$ is a common tangent across the tops of the circles, with $E$ above $A$ and $F$ above $B$; The figure is symmetric about the perpendicular bisector of $\overline{AB}$ through $O$; Answer choices: (A) $\frac{8\sqrt{2}}{3}$, (B) $8\sqrt{2} - 4 - \pi$, (C) $4\sqrt{2}$, (D) $4\sqrt{2} + \frac{\pi}{8}$, (E) $8\sqrt{2} - 2 - \frac{\pi}{2}$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #7 Identify Subproblems, #1 Draw a Diagram
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.
Execute — Answer: B
6.G.A.1 Step 1 Trap the region in a rectangle
- Build the rectangle $ABFE$ with corners $A$, $B$, $F$, $E$.
- Its base $AB$ has length $OA + OB = 2\sqrt{2} + 2\sqrt{2} = 4\sqrt{2}$, and its height is the radius $AE = 2$.
- The shaded region sits entirely inside this rectangle, so the shaded area is the rectangle's area minus the parts of the rectangle that are not shaded.
💡 A messy shape is easier to weigh by starting from a clean box around it and taking away what does not belong.
8.G.B.7 Step 2 Find the tangent length by Pythagoras
- Look at triangle $OCA$.
- Since $OC$ is tangent to circle $A$ at $C$, the radius $AC$ meets it at a right angle, so $\angle OCA = 90^\circ$.
- The hypotenuse is $OA = 2\sqrt{2}$ and one leg is the radius $AC = 2$.
- By the Pythagorean theorem the other leg is $OC = \sqrt{OA^2 - AC^2} = \sqrt{8 - 4} = 2$.
- Both legs equal $2$, so triangle $OCA$ is an isosceles right triangle, which means each of its acute angles is $45^\circ$; in particular $\angle CAO = 45^\circ$.
💡 A tangent line makes a right angle with the radius it touches, which hands you a right triangle to measure.
6.G.A.1 Step 3 Subtract the two corner triangles
- At the bottom-left corner, the piece of the rectangle below the shaded blob is trimmed by the tangent segment $OC$.
- The right triangle $OCA$ has legs $OC = 2$ and $CA = 2$, so its area is $\frac{1}{2}\cdot 2 \cdot 2 = 2$.
- By the left-right symmetry of the figure, the triangle $ODB$ at the bottom-right corner also has area $2$.
- Together the two triangles remove $4$ from the rectangle.
💡 Half of base times height gives each right triangle, and symmetry means you only compute one.
7.G.B.4 Step 4 Subtract the two circular sectors
- The disks themselves are cut out of the shaded region.
- Near circle $A$, the removed slice is the sector $CAE$: it runs from radius $AC$ to radius $AE$.
- Since $AE$ points straight up and $AC$ points $45^\circ$ up from the base, $\angle CAE = 90^\circ - 45^\circ = 45^\circ$.
- A $45^\circ$ sector is $\frac{45}{360} = \frac{1}{8}$ of the full disk, whose area is $\pi \cdot 2^2 = 4\pi$.
- So sector $CAE = \frac{1}{8}\cdot 4\pi = \frac{\pi}{2}$, and by symmetry sector $DBF = \frac{\pi}{2}$ as well.
- The two sectors remove $\pi$ in all.
💡 A sector is just a fraction of its whole circle, and the fraction is its angle over $360^\circ$.
7.G.B.6 Step 5 Combine the pieces
- Start from the rectangle and take away everything that is not shaded: the two corner triangles ($4$ total) and the two circular sectors ($\pi$ total).
- This leaves $8\sqrt{2} - 4 - \pi$.
- That matches choice (B).
💡 The shaded area is what remains after every unshaded piece is peeled off the surrounding rectangle.
6.G.A.1 Build the rectangle $ABFE$ with corners $A$, $B$, $F$, $E$. Its base $AB$ has le 8.G.B.7 Look at triangle $OCA$. Since $OC$ is tangent to circle $A$ at $C$, the radius $ 6.G.A.1 At the bottom-left corner, the piece of the rectangle below the shaded blob is t 7.G.B.4 The disks themselves are cut out of the shaded region. Near circle $A$, the remo 7.G.B.6 Start from the rectangle and take away everything that is not shaded: the two co Review
Reasonableness: The numerical value is $8\sqrt{2} - 4 - \pi \approx 11.31 - 4 - 3.14 = 4.17$, a positive area smaller than the enclosing rectangle's $8\sqrt{2} \approx 11.31$ 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) $4\sqrt{2}\approx 5.66$ and (A) $\frac{8\sqrt{2}}{3}\approx 3.77$ ignore the $\pi$ that a curved boundary must produce, and (E) $8\sqrt{2}-2-\frac{\pi}{2}$ 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 ($\pi$), so (B) is correct.
Alternative: Instead of subtracting from a rectangle, add up the shaded region directly. Using coordinates $A=(-2\sqrt{2},0)$, $B=(2\sqrt{2},0)$, the straight-edged pentagon $O$-$C$-$E$-$F$-$D$ has area $6\sqrt{2}-4$ by the shoelace formula. The disks then bite a circular segment out of each side, each of area $\frac{\pi}{2}-\sqrt{2}$ (a $45^\circ$ sector minus its chord triangle). Removing both bites gives $(6\sqrt{2}-4) - 2\left(\frac{\pi}{2}-\sqrt{2}\right) = 8\sqrt{2}-4-\pi$, the same answer.
CCSS standards used (min grade 8)
6.G.A.1Find the area of triangles and other polygons by composing into or decomposing from familiar shapes (Getting the rectangle $ABFE$ area as base times height, and each corner right triangle as half base times height.)8.G.B.7Apply the Pythagorean Theorem to determine unknown side lengths in right triangles (Finding the tangent length $OC = \sqrt{OA^2 - AC^2} = 2$ in the right triangle $OCA$.)7.G.B.4Know and use the formulas for the area and circumference of a circle (Computing each $45^\circ$ sector as one-eighth of the disk area $\pi(2)^2$, giving $\frac{\pi}{2}$.)7.G.B.6Solve real-world and mathematical problems involving area of two-dimensional objects composed of triangles, quadrilaterals, and polygons (Combining the rectangle, triangles, and sectors into the final area $8\sqrt{2}-4-\pi$.)
⭐ When a shaded shape has a lumpy edge, box it inside a simple rectangle and subtract every piece that is not shaded.
⭐ When a shaded shape has a lumpy edge, box it inside a simple rectangle and subtract every piece that is not shaded.
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