AMC 10 · 2025 · #12
Grade 8 geometry-2dThe figure below shows an equilateral triangle, a rhombus with a 60∘ angle, and a regular hexagon, each of them containing some mutually tangent congruent disks. Let T,R, and H, respectively, denote the ratio in each case of the total area of the disks to the area of the enclosing polygon. Which of the following is true?
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: Three shapes each hold some equal-sized circles that touch each other snugly: an equilateral triangle holds $3$ circles, a rhombus with a $60^\circ$ angle holds $2$ circles, and a regular hexagon holds $6$ circles. For each shape, form the ratio (total area of its circles) $\div$ (area of the shape); call these $T$, $R$, and $H$. Decide how $T$, $R$, and $H$ compare in size.
Givens: An equilateral triangle contains $3$ congruent, mutually tangent disks (each disk touches two sides of the triangle); this ratio is $T$; A rhombus with a $60^\circ$ angle contains $2$ congruent tangent disks; this ratio is $R$; A regular hexagon contains $6$ congruent tangent disks; this ratio is $H$; Each ratio is (total disk area) divided by (area of the enclosing polygon); Answer choices: (A) $T=H=R$, (B) $H<R=T$, (C) $H=R<T$, (D) $H<R<T$, (E) $H<T<R$
Unknowns: The correct ordering among the three ratios $T$, $R$, and $H$
Understand
Restated: Three shapes each hold some equal-sized circles that touch each other snugly: an equilateral triangle holds $3$ circles, a rhombus with a $60^\circ$ angle holds $2$ circles, and a regular hexagon holds $6$ circles. For each shape, form the ratio (total area of its circles) $\div$ (area of the shape); call these $T$, $R$, and $H$. Decide how $T$, $R$, and $H$ compare in size.
Givens: An equilateral triangle contains $3$ congruent, mutually tangent disks (each disk touches two sides of the triangle); this ratio is $T$; A rhombus with a $60^\circ$ angle contains $2$ congruent tangent disks; this ratio is $R$; A regular hexagon contains $6$ congruent tangent disks; this ratio is $H$; Each ratio is (total disk area) divided by (area of the enclosing polygon); Answer choices: (A) $T=H=R$, (B) $H<R=T$, (C) $H=R<T$, (D) $H<R<T$, (E) $H<T<R$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #9 Solve an Easier Related Problem, #3 Eliminate Possibilities
Comparing three different shapes at once is overwhelming, so Tool #7 (Identify Subproblems) cuts each shape into equilateral triangles that each hold exactly one circle. In the rhombus and hexagon that circle is the triangle's incircle, so both reduce to copies of a single base unit: one equilateral triangle with its inscribed circle. Tool #9 (Solve an Easier Related Problem) computes that one base ratio once and reuses it for both $R$ and $H$. Tool #1 (Draw a Diagram) supplies the cutting lines and the corner geometry. Because the triangle's three disks pack differently (each hugs a corner, not the incircle), its ratio must be found separately, and Tool #3 (Eliminate Possibilities) confirms only one answer choice fits the comparison.
Execute — Answer: C
6.G.A.1 Step 1 Cut the rhombus and hexagon into triangles
- A rhombus with a $60^\circ$ angle splits along its short diagonal into two equilateral triangles.
- A regular hexagon splits from its center into six equilateral triangles.
- In the figure, each disk in these two shapes sits exactly inside one of those triangles, touching all three of its sides — that is the triangle's inscribed circle (incircle).
💡 A $60^\circ$ rhombus and a regular hexagon are both just equilateral triangles glued together.
6.RP.A.1 Step 2 Same unit means R equals H
- Both the rhombus and the hexagon are built from identical copies of one unit: an equilateral triangle holding its incircle.
- The rhombus is $2$ copies, the hexagon is $6$ copies.
- When you scale a numerator and denominator by the same count, the ratio does not change, so both $R$ and $H$ equal the disk-to-triangle ratio of that single unit.
- Therefore $H=R$.
💡 Copying the same tile many times keeps the covered-area fraction exactly the same.
8.G.B.7 Step 3 Measure the base unit's triangle and incircle
- Take an equilateral triangle of side $s$.
- Its height is $\tfrac{\sqrt3}{2}s$ (drop an altitude and use the right triangle it makes), so its area is $\tfrac{\sqrt3}{4}s^2$.
- The incircle's radius is one-third of the height, $r=\tfrac{\sqrt3}{6}s=\tfrac{s}{2\sqrt3}$, because the center is the centroid.
💡 Splitting an equilateral triangle down the middle makes a right triangle you can measure exactly.
7.G.B.4 Step 4 The base ratio (this is H = R)
- The incircle area is $\pi r^2=\pi\big(\tfrac{s}{2\sqrt3}\big)^2=\tfrac{\pi s^2}{12}$.
- Divide by the triangle area $\tfrac{\sqrt3}{4}s^2$; the $s^2$ cancels, leaving a fixed number.
- This is the common value of both $H$ and $R$, about $0.605$.
💡 One inscribed circle always fills the same slice of any equilateral triangle, whatever its size.
8.G.B.7 Step 5 Find the disk radius inside the triangle T
- The triangle's three disks are different: each hugs a $60^\circ$ corner, touching the two sides there, and neighboring disks touch each other.
- A disk of radius $\rho$ tangent to both sides of a $60^\circ$ corner has its center $2\rho$ out from the vertex along the bisector.
- Working the tangency between disks gives the side length $s=2\rho(1+\sqrt3)$, so $\rho=\dfrac{s}{2(1+\sqrt3)}$.
💡 Three circles crammed into the corners of a triangle sit differently than one circle centered inside it.
7.G.B.4 Step 6 Compute the triangle's ratio T
- The three disks have total area $3\pi\rho^2$.
- Divide by the triangle area $\tfrac{\sqrt3}{4}s^2$ and substitute $\rho=\tfrac{s}{2(1+\sqrt3)}$; again $s^2$ cancels.
- Using $(1+\sqrt3)^2=4+2\sqrt3$, this simplifies to $T=\dfrac{3\pi}{6+4\sqrt3}\approx0.729$.
💡 Packing three disks into the corners leaves less wasted space than a single central circle does.
8.NS.A.2 Step 7 Compare and pick the answer
- Now compare the numbers: $H=R\approx0.605$ while $T\approx0.729$, so $H=R<T$.
- Among the choices, only (C) says two of the ratios are equal and both below the third in this way.
- So the answer is (C).
💡 Turning the ratios into decimals makes the ordering obvious at a glance.
6.G.A.1 A rhombus with a $60^\circ$ angle splits along its short diagonal into two equil 6.RP.A.1 Both the rhombus and the hexagon are built from identical copies of one unit: an 8.G.B.7 Take an equilateral triangle of side $s$. Its height is $\tfrac{\sqrt3}{2}s$ (dr 7.G.B.4 The incircle area is $\pi r^2=\pi\big(\tfrac{s}{2\sqrt3}\big)^2=\tfrac{\pi s^2}{ 8.G.B.7 The triangle's three disks are different: each hugs a $60^\circ$ corner, touchin 7.G.B.4 The three disks have total area $3\pi\rho^2$. Divide by the triangle area $\tfra 8.NS.A.2 Now compare the numbers: $H=R\approx0.605$ while $T\approx0.729$, so $H=R<T$. Am Review
Reasonableness: Look at the pictures: the triangle looks the fullest of circle, the hexagon and rhombus look equally full but a bit emptier — matching $H=R<T$. The exact numbers confirm it: $H=R=\tfrac{\sqrt3\pi}{9}\approx0.605$ and $T=\tfrac{3\pi}{6+4\sqrt3}\approx0.729$, and $0.605<0.729$. Both stay safely below $1$, as any area-of-circles-inside-a-shape ratio must. The equality $H=R$ is exact (not a coincidence of rounding) because both shapes are literally tiled by the same triangle-plus-incircle unit.
Alternative: Skip computing $T$ exactly. Once you see that $H=R$ (same tiling unit), scan the five choices: only (C) contains "$H=R$" together with those two not equal to $T$. Since the answer must exist and the figure clearly shows the triangle is fuller than the hexagon (so $T>H$), (C) is forced by elimination — the same route the contest rewards for speed.
CCSS standards used (min grade 8)
6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Decomposing the $60^\circ$ rhombus into $2$ equilateral triangles and the regular hexagon into $6$ equilateral triangles, each holding one circle.)6.RP.A.1Understand the concept of a ratio and use ratio language (Arguing that tiling with the same triangle-plus-incircle unit leaves the area ratio unchanged, so $H=R$.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding the equilateral triangle's height and incircle radius, and locating each corner disk's center $2\rho$ from the vertex.)7.G.B.4Know the formulas for area and circumference of a circle (Computing disk areas $\pi r^2$ and $3\pi\rho^2$ to form both the base ratio $H=R$ and the triangle ratio $T$.)8.NS.A.2Use rational approximations of irrational numbers to compare their size (Approximating $\tfrac{\sqrt3\pi}{9}\approx0.605$ and $\tfrac{3\pi}{6+4\sqrt3}\approx0.729$ to order the ratios.)
⭐ A $60^\circ$ rhombus and a regular hexagon are both made of the same equilateral-triangle-with-one-circle tile, so their fill ratios match; the triangle packs three circles tighter, so it wins: $H=R<T$.
⭐ A $60^\circ$ rhombus and a regular hexagon are both made of the same equilateral-triangle-with-one-circle tile, so their fill ratios match; the triangle packs three circles tighter, so it wins: $H=R<T$.
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