AMC 10 · 2014 · #22
Grade 8 geometry-2dEight semicircles line the inside of a square with side length 2 as shown. What is the radius of the circle tangent to all of these semicircles?
Pick an answer.
AMC 10 2014 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 square has side length $2$. Along the inside of the square sit eight congruent semicircles, two flat against each side, their diameters covering the side. One circle is drawn in the middle so that it just touches every semicircle. Find the radius of that middle circle.
Givens: The square has side length $2$; Eight semicircles line the inside, two per side, with their diameters lying along the sides; The two semicircles on a side together span the whole side, so each diameter is $1$; A single circle in the center is tangent to all eight semicircles; Answer choices: (A) $\dfrac{1+\sqrt2}{4}$, (B) $\dfrac{\sqrt5-1}{2}$, (C) $\dfrac{\sqrt3+1}{4}$, (D) $\dfrac{2\sqrt3}{5}$, (E) $\dfrac{\sqrt5}{3}$
Unknowns: The radius of the central circle tangent to all eight semicircles
Understand
Restated: A square has side length $2$. Along the inside of the square sit eight congruent semicircles, two flat against each side, their diameters covering the side. One circle is drawn in the middle so that it just touches every semicircle. Find the radius of that middle circle.
Givens: The square has side length $2$; Eight semicircles line the inside, two per side, with their diameters lying along the sides; The two semicircles on a side together span the whole side, so each diameter is $1$; A single circle in the center is tangent to all eight semicircles; Answer choices: (A) $\dfrac{1+\sqrt2}{4}$, (B) $\dfrac{\sqrt5-1}{2}$, (C) $\dfrac{\sqrt3+1}{4}$, (D) $\dfrac{2\sqrt3}{5}$, (E) $\dfrac{\sqrt5}{3}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #4 Introduce a Variable, #7 Identify Subproblems, #17 Visualize Spatial Relationships
Tangency problems collapse to one clean fact: two circles that touch on the outside have their centers a distance apart equal to the sum of their radii. So Tool #1 (Draw a Diagram) sets up coordinates with the square centered at the origin, turning the picture into exact points. Tool #7 (Identify Subproblems) splits the work into three small pieces: find a semicircle's radius, find where its center is, and measure the distance from the square's center to that center. Tool #4 (Introduce a Variable) names the unknown radius $r$ and writes the tangency condition as an equation. Tool #17 (Visualize Spatial Relationships) is the key sighting: the touching point, the two centers all line up, so the center-to-center distance is exactly $r$ plus the semicircle radius.
Execute — Answer: B
6.G.A.3 Step 1 Put the square on coordinates
- Place the center of the square at the origin.
- Since the side length is $2$, each side is $1$ unit from the center, so the corners are at $(\pm1,\pm1)$ and the four sides lie on the lines $x=\pm1$ and $y=\pm1$.
- The central circle, by symmetry, is centered at the origin.
💡 Centering the figure at the origin makes the symmetry do the bookkeeping for you.
7.G.B.4 Step 2 Find one semicircle's radius and center
- Look at the bottom side, which lies on $y=-1$ and runs from $x=-1$ to $x=1$.
- Two semicircles share this side, so together their diameters cover a length of $2$, giving each a diameter of $1$ and therefore a radius of $\tfrac12$.
- Take the right one: its diameter runs from $(0,-1)$ to $(1,-1)$, so its center is the midpoint $\left(\tfrac12,-1\right)$.
💡 Two equal semicircles splitting a side of length $2$ each get a diameter of $1$, so radius one-half.
6.EE.B.6 Step 3 Write the tangency condition
- Let $r$ be the radius of the central circle.
- It touches the semicircle on the outside, so the two arcs meet at a single point that lies on the straight segment joining their centers.
- That means the distance between the two centers equals the sum of the two radii: $r$ from the central circle plus $\tfrac12$ from the semicircle.
💡 Externally tangent circles sit exactly a sum-of-radii apart, center to center.
8.G.B.8 Step 4 Measure the distance between centers
- Use the Pythagorean theorem on the horizontal and vertical gaps between $(0,0)$ and $\left(\tfrac12,-1\right)$.
- The horizontal gap is $\tfrac12$ and the vertical gap is $1$.
- So the distance is $\sqrt{\left(\tfrac12\right)^2+1^2}=\sqrt{\tfrac14+1}=\sqrt{\tfrac54}=\dfrac{\sqrt5}{2}$.
💡 The gap between two points is the hypotenuse of the right triangle made by their horizontal and vertical offsets.
8.EE.C.7 Step 5 Solve for the radius
- Set the measured distance equal to the sum of radii and solve for $r$.
- From $\dfrac{\sqrt5}{2}=r+\dfrac12$, subtract $\dfrac12$ from both sides to get $r=\dfrac{\sqrt5}{2}-\dfrac12=\dfrac{\sqrt5-1}{2}$.
- So the radius of the central circle is $\dfrac{\sqrt5-1}{2}$, which is choice (B).
💡 Once the equation is set, one subtraction peels off the semicircle's radius and leaves the answer.
6.G.A.3 Place the center of the square at the origin. Since the side length is $2$, each 7.G.B.4 Look at the bottom side, which lies on $y=-1$ and runs from $x=-1$ to $x=1$. Two 6.EE.B.6 Let $r$ be the radius of the central circle. It touches the semicircle on the ou 8.G.B.8 Use the Pythagorean theorem on the horizontal and vertical gaps between $(0,0)$ 8.EE.C.7 Set the measured distance equal to the sum of radii and solve for $r$. From $\df Review
Reasonableness: Numerically $r=\dfrac{\sqrt5-1}{2}\approx\dfrac{2.236-1}{2}\approx0.618$. That is larger than a semicircle's radius $0.5$ (the central circle should be the biggest disk in the picture) yet smaller than the half-side $1$ (it must fit inside the square without poking through a side). Both checks pass. Testing the near-miss choices as decimals, (A) $\approx0.604$, (C) $\approx0.683$, (D) $\approx0.693$, (E) $\approx0.745$; only the exact Pythagorean value matches (B), and (E) would already be too wide to sit clear of the semicircles.
Alternative: Skip coordinates and build the right triangle directly. Drop from the square's center a horizontal leg to below the semicircle's center and a vertical leg down to the side. The semicircle center is offset half a diameter, $\tfrac12$, sideways and a full half-side, $1$, downward, so the hypotenuse joining the two centers is $\sqrt{\left(\tfrac12\right)^2+1^2}=\dfrac{\sqrt5}{2}$. Since that hypotenuse equals $r+\tfrac12$ by tangency, $r=\dfrac{\sqrt5-1}{2}$, the same answer (B).
CCSS standards used (min grade 8)
6.G.A.3Draw polygons in the coordinate plane given coordinates for the vertices (Centering the square at the origin so its corners are $(\pm1,\pm1)$ and the central circle's center is $(0,0)$.)7.G.B.4Know the formulas for area and circumference of a circle (Using the diameter-radius relationship to get each semicircle's radius $\tfrac12$ and its center $\left(\tfrac12,-1\right)$.)6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Naming the unknown radius $r$ and writing the tangency condition as distance $=r+\tfrac12$.)8.G.B.8Apply the Pythagorean theorem to find distance between two points in a coordinate system (Computing the center-to-center distance $\sqrt{\left(\tfrac12\right)^2+1^2}=\dfrac{\sqrt5}{2}$.)8.EE.C.7Solve linear equations in one variable (Solving $\dfrac{\sqrt5}{2}=r+\dfrac12$ to get $r=\dfrac{\sqrt5-1}{2}$.)
⭐ Two circles that just touch on the outside sit a sum-of-radii apart, so measure the center-to-center distance $\tfrac{\sqrt5}{2}$ and subtract the semicircle's $\tfrac12$ to get $\tfrac{\sqrt5-1}{2}$.
⭐ Two circles that just touch on the outside sit a sum-of-radii apart, so measure the center-to-center distance $\tfrac{\sqrt5}{2}$ and subtract the semicircle's $\tfrac12$ to get $\tfrac{\sqrt5-1}{2}$.
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