AMC 10 · 2025 · #22
Grade 8 geometry-2dA circle of radius r is surrounded by three circles, whose radii are 1, 2, and 3, all externally tangent to the inner circle and externally tangent to each other, as shown in the diagram below.
What is r?
Pick an answer.
AMC 10 2025 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 of radius r sits in the middle, touching three larger circles of radii 1, 2, and 3. Every pair of the four circles touches from the outside, and the three big circles also touch each other. From the picture the small circle is the one nestled snugly in the gap between the three big ones. Find r.
Givens: Three circles have radii 1, 2, and 3, and they are pairwise externally tangent to each other.; A fourth circle of radius r is externally tangent to all three of them.; The diagram shows this fourth circle as a small one tucked into the central gap, not a giant circle wrapped around the outside.
Unknowns: The radius r of the small inner circle.
Understand
Restated: A small circle of radius r sits in the middle, touching three larger circles of radii 1, 2, and 3. Every pair of the four circles touches from the outside, and the three big circles also touch each other. From the picture the small circle is the one nestled snugly in the gap between the three big ones. Find r.
Givens: Three circles have radii 1, 2, and 3, and they are pairwise externally tangent to each other.; A fourth circle of radius r is externally tangent to all three of them.; The diagram shows this fourth circle as a small one tucked into the central gap, not a giant circle wrapped around the outside.
Plan
Primary tool: #4 Introduce a Variable
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities, #1 Draw a Diagram
Working with the radii directly is messy because tangency ties the centers together through square roots of sums. The clean move is to name a new variable for each circle: its curvature k = 1/r. For four circles that all touch one another, the curvatures obey Descartes' Circle Theorem, a single tidy equation. So I introduce k for every circle, split the theorem into two easy pieces (a plain sum and a sum of products under a square root), compute each, and read off the two possible circles. The picture then eliminates the wrong one, and flipping the curvature back gives r.
Execute — Answer: B
6.NS.A.1 Step 1 Switch from radius to curvature
- For circles that all touch each other, radii combine awkwardly, but their reciprocals combine neatly.
- Give each circle a new variable, its curvature k = 1/r.
- The three known circles then have curvatures k1 = 1/1 = 1, k2 = 1/2, and k3 = 1/3, and the unknown inner circle has curvature k4 = 1/r.
- Finding k4 is the same as finding r, since one is just the reciprocal of the other.
💡 Curvature turns a big circle into a small number and a small circle into a big one, and that is the language in which tangency becomes simple.
6.EE.B.5 Step 2 Write Descartes' Circle Theorem
- When four circles are all mutually tangent, their four curvatures are locked together by Descartes' Circle Theorem: the fourth curvature equals the sum of the other three, plus or minus twice the square root of (the sum of the three pairwise products of those curvatures).
- The plus-or-minus is the theorem telling us there are exactly two circles tangent to all three given ones, and we will decide the sign later.
💡 One equation captures the whole tangency picture, so the hard geometry collapses into arithmetic on four numbers.
5.NF.A.1 Step 3 Add the three curvatures
- First handle the easy part, the plain sum of the known curvatures.
- Put 1, 1/2, and 1/3 over the common denominator 6: that is 6/6 + 3/6 + 2/6 = 11/6.
- This is the part before the square root.
💡 A common denominator lets three unlike fractions merge into one clean total.
5.NF.B.4 Step 4 Add the three pairwise products
- Now the part under the square root, the sum of the three products of pairs.
- Multiply the curvatures two at a time: k1*k2 = 1 * 1/2 = 1/2, k2*k3 = 1/2 * 1/3 = 1/6, and k3*k1 = 1/3 * 1 = 1/3.
- Add them over denominator 6: 3/6 + 1/6 + 2/6 = 6/6 = 1.
- The three fractions add to exactly 1, which is what makes this problem come out clean.
💡 The pairwise products are the glue between the circles, and here they add up to a perfect 1.
8.EE.A.2 Step 5 Combine the two pieces
- Drop the two pieces into the theorem.
- The square root of 1 is 1, so the correction term is 2 * 1 = 2.
- That gives k4 = 11/6 plus or minus 2.
- Working the two signs out as fractions: 11/6 + 12/6 = 23/6, or 11/6 - 12/6 = -1/6.
- So the two circles tangent to all three have curvatures 23/6 and -1/6.
💡 Because the inside of the root is a perfect square, the correction is a clean whole number and both circles pop out at once.
6.NS.A.1 Step 6 Pick the small circle and flip back
- Two curvatures, two circles.
- A bigger curvature means a smaller, more tightly curved circle, so k4 = 23/6 is the tiny circle in the gap, while k4 = -1/6 is the big enclosing circle (negative curvature signals a circle that wraps around the others from outside, radius 6).
- The figure shows the small nestled circle, so take k4 = 23/6.
- Flip back to the radius: r = 1/k4 = 6/23.
- That is choice (B).
💡 The two signs are the two circles that fit; the picture tells you which one you were asked for.
6.NS.A.1 For circles that all touch each other, radii combine awkwardly, but their recipr 6.EE.B.5 When four circles are all mutually tangent, their four curvatures are locked tog 5.NF.A.1 First handle the easy part, the plain sum of the known curvatures. Put 1, 1/2, a 5.NF.B.4 Now the part under the square root, the sum of the three products of pairs. Mult 8.EE.A.2 Drop the two pieces into the theorem. The square root of 1 is 1, so the correcti 6.NS.A.1 Two curvatures, two circles. A bigger curvature means a smaller, more tightly cu Review
Reasonableness: The value r = 6/23 is about 0.26, a small positive length smaller than all three given radii, exactly the size of the tiny circle drawn in the central gap, so it passes the eye test. The rejected sign gave curvature -1/6, i.e. radius 6, the giant circle circling the other three, which the picture rules out. Note the answer choices 1/4 = 0.25, 6/23 approx 0.261, and 3/11 approx 0.273 are all close, so the exact fraction matters; the perfect cancellation to 6/23 confirms choice (B) rather than a near neighbor.
Alternative: You can avoid the theorem entirely with coordinates. The three big centers are 3, 4, and 5 apart, and since 3^2 + 4^2 = 5^2 they form a right triangle. Put the right-angle center at (0,0), another at (3,0), and the third at (0,4), and let the inner center be (x,y) with distances r+1, r+2, r+3 to the three. Subtracting the distance equations in pairs gives x = (3-r)/3 and y = (2-r)/2; substituting back into x^2 + y^2 = (r+1)^2 produces 23r^2 + 132r - 36 = 0, whose only positive root is 6/23. Same answer, no theorem needed.
CCSS standards used (min grade 8)
6.NS.A.1Interpret and compute quotients of fractions and solve word problems (Defining each curvature as the reciprocal k = 1/r and, at the end, flipping k4 = 23/6 back to the radius r = 6/23.)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Reading Descartes' Circle Theorem as one equation in the unknown curvature k4 and treating the plus-or-minus as its two solutions.)5.NF.A.1Add and subtract fractions with unlike denominators (Adding 1 + 1/2 + 1/3 over the common denominator 6 to get the sum 11/6.)5.NF.B.4Apply and extend understanding of multiplication to multiply a fraction by a fraction (Forming the three pairwise products of the curvatures and adding them to the clean total 1 under the square root.)8.EE.A.2Use square root and cube root symbols to represent solutions (Evaluating the square root of the perfect square 1 in the theorem to get the whole-number correction term 2.)
⭐ When circles all touch, stop juggling radii and switch to curvature (1/r); Descartes' Circle Theorem then turns the whole picture into one short fraction sum.
⭐ When circles all touch, stop juggling radii and switch to curvature (1/r); Descartes' Circle Theorem then turns the whole picture into one short fraction sum.
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