AMC 10 · 2025 · #20
Grade 8 geometry-2dFour congruent semicircles are inscribed in a square of side length 1 so that their diameters are on the sides of the square, one endpoint of each diameter is at a vertex of the square, and adjacent semicircles are tangent to each other. A small circle centered at the center of the square is tangent to each of the four semicircles, as shown below.
The diameter of the small circle can be written as (a+b)(c+d), where a, b, c, and d are integers. What is a+b+c+d?
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 unit square holds four identical semicircles. Each semicircle's diameter lies along a side of the square, one end of every diameter sits on a corner, and neighboring semicircles just touch. A small circle centered at the square's center touches all four semicircles. Its diameter can be written as $(\sqrt{a}+b)(\sqrt{c}+d)$ for integers $a$, $b$, $c$, $d$; find $a+b+c+d$.
Givens: The square has side length $1$.; Four congruent semicircles have their diameters on the sides of the square, and each diameter has one endpoint at a vertex.; Adjacent semicircles are tangent to each other.; A small circle centered at the center of the square is tangent to all four semicircles.; The small circle's diameter equals $(\sqrt{a}+b)(\sqrt{c}+d)$ for integers $a,b,c,d$.
Unknowns: The radius $r$ of the four semicircles.; The diameter of the small circle, and the value of $a+b+c+d$.
Understand
Restated: A unit square holds four identical semicircles. Each semicircle's diameter lies along a side of the square, one end of every diameter sits on a corner, and neighboring semicircles just touch. A small circle centered at the square's center touches all four semicircles. Its diameter can be written as $(\sqrt{a}+b)(\sqrt{c}+d)$ for integers $a$, $b$, $c$, $d$; find $a+b+c+d$.
Givens: The square has side length $1$.; Four congruent semicircles have their diameters on the sides of the square, and each diameter has one endpoint at a vertex.; Adjacent semicircles are tangent to each other.; A small circle centered at the center of the square is tangent to all four semicircles.; The small circle's diameter equals $(\sqrt{a}+b)(\sqrt{c}+d)$ for integers $a,b,c,d$.
Plan
Primary tool: #1 Draw a Diagram
Secondary: #4 Introduce a Variable, #7 Identify Subproblems
Everything here is about where the circles sit, so put the picture on a coordinate grid and mark each semicircle's center. Once the centers have coordinates, two facts turn the picture into equations: two circles that just touch have their centers a distance of (sum of radii) apart. Call the semicircle radius $r$, use tangency to pin down $r$, then use tangency again for the small circle. Each tangency is a right-triangle distance, so the Pythagorean theorem does all the heavy lifting.
Execute — Answer: A
6.G.A.3 Step 1 Put the picture on a grid
- Place the square with corners at $(0,0)$, $(1,0)$, $(1,1)$, $(0,1)$.
- Let the semicircle radius be $r$.
- The bottom semicircle has its diameter from the corner $(0,0)$ to $(2r,0)$, so its center is $E=(r,0)$.
- Going around, the other three centers are $F=(1,r)$, $G=(1-r,1)$, and $H=(0,1-r)$.
💡 Giving every center an address in $(x,y)$ turns a geometry picture into something you can measure with distance.
8.G.B.8 Step 2 Neighbors touch: centers are $2r$ apart
- The bottom semicircle (center $E$) and the right semicircle (center $F$) are tangent, so the distance between their centers equals the sum of their radii, $r+r=2r$.
- Measure $EF$ with the distance formula: the horizontal gap is $1-r$ and the vertical gap is $r$.
💡 When two circles kiss, the segment joining their centers is exactly both radii laid end to end.
8.EE.A.2 Step 3 Solve for the semicircle radius
- Square both sides and expand: $(1-r)^2+r^2=4r^2$ gives $1-2r+2r^2=4r^2$, so $2r^2+2r-1=0$.
- The positive root of this quadratic is $r=\dfrac{-1+\sqrt{3}}{2}=\dfrac{\sqrt{3}-1}{2}$.
💡 One clean equation from the touching condition is enough to fix the size of every semicircle.
8.G.B.8 Step 4 The small circle touches too
- Let the small circle at the center $O=(\tfrac12,\tfrac12)$ have radius $R$.
- It is tangent to the bottom semicircle, so the distance from $O$ to that semicircle's center $E=(r,0)$ equals $r+R$.
- Compute that distance: $\left(\tfrac12-r\right)^2+\left(\tfrac12\right)^2$, using $r=\tfrac{\sqrt3-1}{2}$ so $\tfrac12-r=\tfrac{2-\sqrt3}{2}$, works out to $2-\sqrt3$.
💡 Tangency is the same trick again: the center-to-center distance is just the two radii added.
8.EE.A.2 Step 5 Simplify and read off the answer
- So $r+R=\sqrt{2-\sqrt3}$.
- Rewrite $2-\sqrt3=\dfrac{4-2\sqrt3}{2}=\dfrac{(\sqrt3-1)^2}{2}$, giving $\sqrt{2-\sqrt3}=\dfrac{\sqrt3-1}{\sqrt2}=\dfrac{\sqrt6-\sqrt2}{2}$.
- The diameter is $2R=2(r+R)-2r=(\sqrt6-\sqrt2)-(\sqrt3-1)=\sqrt6-\sqrt2-\sqrt3+1$.
- Factor by grouping: $\sqrt2(\sqrt3-1)-1(\sqrt3-1)=(\sqrt2-1)(\sqrt3-1)=(\sqrt2+(-1))(\sqrt3+(-1))$.
- Matching $(\sqrt a+b)(\sqrt c+d)$ gives $a=2,\,b=-1,\,c=3,\,d=-1$, so $a+b+c+d=2-1+3-1=3$.
- The answer is $(\textbf{A})$.
💡 Turning the nested root $\sqrt{2-\sqrt3}$ into $\tfrac{\sqrt6-\sqrt2}{2}$ is what exposes the tidy factored form the problem is asking for.
6.G.A.3 Place the square with corners at $(0,0)$, $(1,0)$, $(1,1)$, $(0,1)$. Let the sem 8.G.B.8 The bottom semicircle (center $E$) and the right semicircle (center $F$) are tan 8.EE.A.2 Square both sides and expand: $(1-r)^2+r^2=4r^2$ gives $1-2r+2r^2=4r^2$, so $2r^ 8.G.B.8 Let the small circle at the center $O=(\tfrac12,\tfrac12)$ have radius $R$. It i 8.EE.A.2 So $r+R=\sqrt{2-\sqrt3}$. Rewrite $2-\sqrt3=\dfrac{4-2\sqrt3}{2}=\dfrac{(\sqrt3- Review
Reasonableness: Numerically $r\approx0.366$ (under $\tfrac12$, so the semicircles fit) and the diameter $2R=(\sqrt2-1)(\sqrt3-1)\approx0.414\times0.732\approx0.30$, a small circle radius of about $0.15$ that comfortably sits in the middle without overlapping the semicircles. Both the shape and the size look right, and $a+b+c+d=3$ is one of the offered choices, matching $(\textbf{A})$.
Alternative: You can avoid coordinates entirely. The triangle from a corner to the two nearest semicircle centers is a right triangle with legs $r$ and hypotenuse $2r$, which is a $30$-$60$-$90$ triangle, giving $r+\sqrt3\,r=1$ and the same $r=\tfrac{\sqrt3-1}{2}$. The four centers form a square of side $2r$; along its diagonal the small circle spans the gap, so $2r+2R=2r\sqrt2$, hence $2R=2r(\sqrt2-1)=(\sqrt3-1)(\sqrt2-1)$ — identical result.
CCSS standards used (min grade 8)
6.G.A.3Draw polygons in the coordinate plane given coordinates for the vertices (Placing the square and the four semicircle centers on a coordinate grid so distances can be computed.)8.G.B.8Apply the Pythagorean theorem to find distance between two points in a coordinate system (Turning each tangency (adjacent semicircles, then the small circle) into a center-to-center distance equation.)8.EE.A.2Use square root and cube root symbols to represent solutions (Solving the quadratic for $r$ and simplifying the nested radical $\sqrt{2-\sqrt3}$ into the factored diameter.)
⭐ When two circles just touch, the straight line between their centers equals their two radii added together—write that as a right-triangle distance and the algebra falls right out.
⭐ When two circles just touch, the straight line between their centers equals their two radii added together—write that as a right-triangle distance and the algebra falls right out.
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