AMC 10 · 2004 · #23
Grade 8 geometry-2dCircles A,B and C are externally tangent to each other, and internally tangent to circle D. Circles B and C are congruent. Circle A has radius 1 and passes through the center of D. What is the radius of circle B?
Pick an answer.
AMC 10 2004 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: Three circles $A$, $B$, $C$ each touch the other two from outside and all sit inside a big circle $D$, touching it from within. Circles $B$ and $C$ are the same size. Circle $A$ has radius $1$ and its edge passes through the center of $D$. Find the radius of circle $B$.
Givens: Circles $A$, $B$, $C$ are externally tangent to each other (each pair touches at one point, centers on opposite sides of that point).; Circles $A$, $B$, $C$ are each internally tangent to circle $D$ (they touch $D$ from the inside).; Circles $B$ and $C$ are congruent, so they have equal radii.; Circle $A$ has radius $1$ and passes through the center of circle $D$.; Answer choices: (A) $\frac23$, (B) $\frac{\sqrt3}{2}$, (C) $\frac78$, (D) $\frac89$, (E) $\frac{1+\sqrt3}{3}$.
Unknowns: The radius of circle $B$ (equal to the radius of circle $C$).
Understand
Restated: Three circles $A$, $B$, $C$ each touch the other two from outside and all sit inside a big circle $D$, touching it from within. Circles $B$ and $C$ are the same size. Circle $A$ has radius $1$ and its edge passes through the center of $D$. Find the radius of circle $B$.
Givens: Circles $A$, $B$, $C$ are externally tangent to each other (each pair touches at one point, centers on opposite sides of that point).; Circles $A$, $B$, $C$ are each internally tangent to circle $D$ (they touch $D$ from the inside).; Circles $B$ and $C$ are congruent, so they have equal radii.; Circle $A$ has radius $1$ and passes through the center of circle $D$.; Answer choices: (A) $\frac23$, (B) $\frac{\sqrt3}{2}$, (C) $\frac78$, (D) $\frac89$, (E) $\frac{1+\sqrt3}{3}$.
Plan
Primary tool: #4 Introduce a Variable
Secondary: #1 Draw a Diagram, #7 Identify Subproblems
Every fact in this problem is about distances between circle centers, so the plan is to place the centers on a coordinate grid (Tool #1) and call the unknown radius of $B$ a letter $r$ (Tool #4). Tangency turns each 'touch' into a clean distance equation: the distance between two centers equals the sum of radii when they touch outside, or the difference when one is inside the other. First a small subproblem (Tool #7) fixes the size of the big circle $D$ from the clue that $A$ passes through its center. Then two tangency equations for circle $B$ — one against $D$, one against $A$ — plus the up-down symmetry that pins $B$'s height, give enough equations to solve for $r$. Subtracting the two circle equations is the key move: the squared terms cancel and leave a simple line, which feeds straight back to a single equation in $r$.
Execute — Answer: D
7.G.B.4 Step 1 Find the big circle's radius
- Circle $A$ has radius $1$ and passes through the center of $D$.
- Start at the center of $D$: it lies on circle $A$, so it is exactly $1$ away from $A$'s center.
- Keep going the same direction to the far side of circle $A$ — that is another radius, so the far point is $2$ away from the center of $D$.
- Circle $A$ touches $D$ from the inside right at that far point, so that point is on circle $D$ too.
- Hence the radius of $D$ is $2$.
💡 A circle's diameter is two radii long, so a small circle that reaches the big circle's center stretches a full diameter across to touch the far rim.
8.G.B.8 Step 2 Set up coordinates and name the radius
- Put the center of $D$ at the origin $(0,0)$.
- Circle $A$'s center is $1$ away from it, so place it at $(-1,0)$.
- Let the radius of circle $B$ be $r$, and let its center be the point $(x,y)$.
- Circle $C$ is the mirror image of $B$ across the horizontal axis, so its center is $(x,-y)$ with the same radius $r$.
- Now every tangency becomes an equation about $x$, $y$, and $r$.
💡 Pinning the centers to a grid turns 'touches' and 'is inside' into distances you can measure with coordinates.
8.G.B.8 Step 3 Write the two tangency equations for B
- Circle $B$ sits inside $D$ and touches it, so the distance from $(0,0)$ to $(x,y)$ is the difference of the radii, $2-r$.
- Circle $B$ touches circle $A$ from outside, so the distance from $(-1,0)$ to $(x,y)$ is the sum of the radii, $1+r$.
- Using the distance formula (squared) gives two equations.
💡 Touching from inside means centers are close by the radius difference; touching from outside means they are apart by the radius sum.
8.EE.C.7 Step 4 Subtract to get x, and use symmetry for y
- Subtract the first equation from the second.
- The $y^2$ terms cancel and the $x^2$ terms cancel, leaving a straight line: $(2x+1) = (1+r)^2-(2-r)^2 = 6r-3$, so $x = 3r-2$.
- Next, circles $B$ and $C$ touch each other.
- Their centers $(x,y)$ and $(x,-y)$ are $2y$ apart, and touching from outside means that gap equals $r+r=2r$.
- So $2y=2r$, giving $y=r$.
💡 Subtracting the two circle equations erases the squared unknowns and leaves one clean straight-line relationship.
8.EE.C.7 Step 5 Substitute and solve for r
- Put $x=3r-2$ and $y=r$ into the first tangency equation $x^2+y^2=(2-r)^2$.
- Expand: $(3r-2)^2+r^2 = 9r^2-12r+4+r^2 = 10r^2-12r+4$, and $(2-r)^2 = r^2-4r+4$.
- Setting them equal and cancelling gives $10r^2-12r+4 = r^2-4r+4$, so $9r^2-8r=0$, i.e.
- $r(9r-8)=0$.
- Since $r$ is a real radius it cannot be $0$, so $9r-8=0$ and $r=\frac89$.
- That is choice (D).
💡 Feeding the two side-relations back into one circle equation leaves a single equation whose only positive solution is the radius.
7.G.B.4 Circle $A$ has radius $1$ and passes through the center of $D$. Start at the cen 8.G.B.8 Put the center of $D$ at the origin $(0,0)$. Circle $A$'s center is $1$ away fro 8.G.B.8 Circle $B$ sits inside $D$ and touches it, so the distance from $(0,0)$ to $(x,y 8.EE.C.7 Subtract the first equation from the second. The $y^2$ terms cancel and the $x^2 8.EE.C.7 Put $x=3r-2$ and $y=r$ into the first tangency equation $x^2+y^2=(2-r)^2$. Expan Review
Reasonableness: Check the radius $r=\frac89$ against the picture. Then $x=3r-2=\frac23$ and $y=r=\frac89$, so $B$'s center is $(\frac23,\frac89)$. Distance to $D$'s center: $\sqrt{(\frac23)^2+(\frac89)^2}=\sqrt{\frac{100}{81}}=\frac{10}{9}=2-\frac89$, matching the inside-touch. Distance to $A$'s center $(-1,0)$: $\sqrt{(\frac53)^2+(\frac89)^2}=\sqrt{\frac{289}{81}}=\frac{17}{9}=1+\frac89$, matching the outside-touch. Both tangency conditions hold, so $\frac89$ is right. It also passes the eye test: $B$ is a bit smaller than $A$ (radius $1$), and $\frac89$ is just under $1$. The trap answer (C) $\frac78$ is close but fails the distance checks; (B) and (E) carry a $\sqrt3$ that never appears once you use the given radius $1$.
Alternative: Skip coordinates and use right triangles directly. Let $O$, $A$, $B$ be the three centers and drop a perpendicular from $B$ to the line $OA$, meeting it at $M$. By symmetry $B$'s tangent point with $C$ is on that line, so $BM=r$. In right triangle $OMB$: $OM^2+r^2=(2-r)^2$. In right triangle $AMB$: $AM^2+r^2=(1+r)^2$, where $AM=OM+1$. Subtracting these two equations eliminates $r^2$ and the $BM$ leg, giving $OM=\frac23$; substituting back yields $9r^2-8r=0$ and $r=\frac89$ — the same answer by the Pythagorean route.
CCSS standards used (min grade 8)
7.G.B.4Know the formulas for area and circumference of a circle (Using the fact that a diameter is twice the radius to see that circle $D$ has radius $2$, since $A$ reaches its center.)8.G.B.8Apply the Pythagorean theorem to find distance between two points in a coordinate system (Turning each tangency into a distance-between-centers equation on the coordinate grid ($x^2+y^2=(2-r)^2$ and $(x+1)^2+y^2=(1+r)^2$).)8.EE.C.7Solve linear equations in one variable (Subtracting the circle equations to get $x=3r-2$ and $y=r$, then solving $9r^2-8r=0$ down to $r=\frac89$.)
⭐ When circles touch, the distance between their centers is just the sum or difference of the radii — put the centers on a grid and every 'touch' becomes an equation you can solve.
⭐ When circles touch, the distance between their centers is just the sum or difference of the radii — put the centers on a grid and every 'touch' becomes an equation you can solve.
More like this
Same archetype — closest grade level first.