AMC 10 · 2008 · #16
Grade 8 geometry-2dPoints A and B lie on a circle centered at O, and ∠AOB=60∘. A second circle is internally tangent to the first and tangent to both OA and OB. What is the ratio of the area of the smaller circle to that of the larger circle?
Pick an answer.
AMC 10 2008 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: Points A and B sit on a circle centered at O, with angle AOB equal to 60 degrees. A second, smaller circle lies inside the first: it touches the big circle from the inside and also touches both rays OA and OB. Find the ratio of the small circle's area to the big circle's area.
Givens: A and B are on a circle centered at O; call its radius R.; The angle AOB measures 60 degrees.; A smaller circle is tangent to both segments OA and OB.; The smaller circle is internally tangent to the larger circle.
Unknowns: The ratio of the smaller circle's area to the larger circle's area.
Understand
Restated: Points A and B sit on a circle centered at O, with angle AOB equal to 60 degrees. A second, smaller circle lies inside the first: it touches the big circle from the inside and also touches both rays OA and OB. Find the ratio of the small circle's area to the big circle's area.
Givens: A and B are on a circle centered at O; call its radius R.; The angle AOB measures 60 degrees.; A smaller circle is tangent to both segments OA and OB.; The smaller circle is internally tangent to the larger circle.
Plan
Primary tool: #1 Draw a Diagram
Secondary: #4 Introduce a Variable, #7 Identify Subproblems
The whole problem is about how one circle sits against two rays and inside another circle, so a clean picture is the key. Once it is drawn, two facts fall out: touching both rays forces the small center onto the line that bisects the 60 degree angle, and touching the big circle from inside links the two radii through the distance between centers. Name the two radii, build a right triangle to relate them, then compare their squares to get the area ratio.
Execute — Answer: B
8.G.A.5 Step 1 Place the small center on the bisector
- Draw the big circle with center O and the two rays OA and OB making a 60 degree angle.
- A circle that touches both rays must sit the same distance from each ray, so its center P lies on the line that splits the angle in half.
- That splitting line cuts the 60 degrees into two equal 30 degree angles, so the angle between OP and each ray is 30 degrees.
💡 A circle hugging both sides of an angle has to be centered right down the middle.
8.G.A.5 Step 2 Relate the radii with a right triangle
- Let the small radius be r and the big radius be R.
- Drop a line from P straight to ray OA, meeting it at T; because a radius meets a tangent at a right angle, PT equals r and angle PTO is 90 degrees.
- Triangle OTP now has a 30 degree angle at O and a right angle at T, making it a 30-60-90 triangle.
- In that shape the side facing the 30 degree angle is exactly half the longest side, so PT is half of OP.
- Since PT is r, the distance OP is 2r.
💡 In a 30-60-90 triangle the shortest side is always half the hypotenuse.
8.EE.C.7 Step 3 Use the inside-touch to link the radii
- The small circle touches the big circle from the inside.
- When one circle sits inside another and they touch, the gap between their centers equals the big radius minus the small radius.
- Here the big center is O and the small center is P, so OP equals R minus r.
- But OP is also 2r from the last step.
- Setting these equal gives 2r = R minus r, and adding r to both sides gives 3r = R, so r is one third of R.
💡 For a circle nested inside another and touching it, the centers sit apart by the difference of the radii.
7.G.B.4 Step 4 Compare the areas
- A circle's area is pi times its radius squared, so when you compare two circles the pi cancels and only the squared radii matter.
- The ratio of areas is r squared over R squared, which is the square of r over R.
- Since r over R is one third, the ratio is one third squared, which is one ninth.
- That matches choice (B).
💡 Areas of circles grow with the square of the radius, so a one-third radius means a one-ninth area.
8.G.A.5 Draw the big circle with center O and the two rays OA and OB making a 60 degree 8.G.A.5 Let the small radius be r and the big radius be R. Drop a line from P straight t 8.EE.C.7 The small circle touches the big circle from the inside. When one circle sits in 7.G.B.4 A circle's area is pi times its radius squared, so when you compare two circles Review
Reasonableness: The small radius came out to one third of the big radius, which looks right on the picture: the small circle is clearly much less than half the big one but not tiny. Squaring the one-third gives one ninth, a sensible slice of the area. A quick sanity check on the tangency: with r = R/3, the center sits at OP = 2R/3, and 2R/3 + R/3 = R reaches exactly the far edge of the big circle, confirming the inside touch. So (B) 1/9 is consistent.
Alternative: Instead of the 30-60-90 shortcut, drop the tangent line where the two circles meet; it is perpendicular to the line of centers and meets the two rays to form a triangle whose three angles are all 60 degrees, so the triangle is equilateral. The small circle is its inscribed circle, whose center is also the triangle's centroid. The centroid splits each median in a 2-to-1 ratio, which again makes the big radius three times the small one, giving the same 1/9.
CCSS standards used (min grade 8)
8.G.A.5Use informal arguments to establish facts about angles and the angle-angle criterion for similarity of triangles (Arguing the small center lies on the 30 degree bisector and that triangle OTP is a 30-60-90 triangle whose short leg is half the hypotenuse, giving OP = 2r.)8.EE.C.7Solve linear equations in one variable (Solving 2r = R - r to find that the small radius is one third of the big radius.)7.G.B.4Know and use the formulas for the area and circumference of a circle (Turning the radius ratio into an area ratio by squaring, since a circle's area is pi times the radius squared.)
⭐ Touching both sides of an angle puts a circle on the middle line, and once you know a radius is one third, the area is one ninth because area follows the radius squared.
⭐ Touching both sides of an angle puts a circle on the middle line, and once you know a radius is one third, the area is one ninth because area follows the radius squared.
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