AMC 10 · 2021 · #21
Grade 8 geometry-2dProblem
Let ABCDEF be an equiangular hexagon. The lines AB,CD, and EF determine a triangle with area 1923, and the lines BC,DE, and FA determine a triangle with area 3243. The perimeter of hexagon ABCDEF can be expressed as m+np, where m,n, and p are positive integers and p is not divisible by the square of any prime. What is m+n+p?
Pick an answer.
AMC 10 2021 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: An equiangular hexagon $ABCDEF$ has every interior angle equal to $120^\circ$. Extending alternate sides — first $AB, CD, EF$, then $BC, DE, FA$ — produces two triangles whose areas are $192\sqrt{3}$ and $324\sqrt{3}$. Find the perimeter of the hexagon in the form $m + n\sqrt{p}$ (with $p$ square-free) and compute $m + n + p$.
Givens: Hexagon $ABCDEF$ is equiangular, so each interior angle is $120^\circ$; Triangle from lines $AB, CD, EF$ has area $192\sqrt{3}$; Triangle from lines $BC, DE, FA$ has area $324\sqrt{3}$; Answer choices: (A) $47$, (B) $52$, (C) $55$, (D) $58$, (E) $63$
Unknowns: $m + n + p$ where the hexagon perimeter equals $m + n\sqrt{p}$
Understand
Restated: An equiangular hexagon $ABCDEF$ has every interior angle equal to $120^\circ$. Extending alternate sides — first $AB, CD, EF$, then $BC, DE, FA$ — produces two triangles whose areas are $192\sqrt{3}$ and $324\sqrt{3}$. Find the perimeter of the hexagon in the form $m + n\sqrt{p}$ (with $p$ square-free) and compute $m + n + p$.
Givens: Hexagon $ABCDEF$ is equiangular, so each interior angle is $120^\circ$; Triangle from lines $AB, CD, EF$ has area $192\sqrt{3}$; Triangle from lines $BC, DE, FA$ has area $324\sqrt{3}$; Answer choices: (A) $47$, (B) $52$, (C) $55$, (D) $58$, (E) $63$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #9 Solve an Easier Related Problem, #13 Convert to Algebra
Tool #1 (Diagram) — sketch the hexagon and extend the two triples of alternate sides until they cross; the picture immediately shows two triangles, each with three small "corner" triangles snipped at the hexagon's vertices. Tool #9 (Easier Problem) — a regular hexagon is the easy case ($192 = 324$), so we expect the answer for the general equiangular hexagon to still hinge on the side-length-to-area relation $s^2 \sqrt{3}/4$ for equilateral triangles. Tool #7 (Subproblems) — recover the two big triangle side lengths separately from the two areas; then read off the hexagon side lengths from the diagram. Tool #13 (Algebra) — the perimeter is just the sum of the two big triangle sides, since every hexagon side sits on one of the two big triangles.
Execute — Answer: C
8.G.A.5 Step 1 - Equiangular hexagon means every interior angle is $\tfrac{(6-2)\cdot 180^\circ}{6} = 120^\circ$.
- Extend the three alternate lines $AB, CD, EF$; at each pair of adjacent hexagon vertices the two exterior angles are $180^\circ - 120^\circ = 60^\circ$, so the little corner triangle snipped between two extended sides is equilateral, and the same is true for the three corners.
- So the BIG outer triangle has all $60^\circ$ angles — it is equilateral.
- The same argument applies to the triangle formed by $BC, DE, FA$.
💡 Sketch the hexagon and extend sides — every corner trims an equilateral $60^\circ$-$60^\circ$-$60^\circ$ chip, leaving a big equilateral triangle.
8.EE.A.2 Step 2 - Recover each big triangle's side length from its area.
- An equilateral triangle of side $s$ has area $\tfrac{s^2 \sqrt{3}}{4}$.
- For the first big triangle: $\tfrac{s_1^2 \sqrt{3}}{4} = 192\sqrt{3}$, so $s_1^2 = 768$, $s_1 = \sqrt{768} = 16\sqrt{3}$.
- For the second: $\tfrac{s_2^2 \sqrt{3}}{4} = 324\sqrt{3}$, so $s_2^2 = 1296$, $s_2 = 36$.
💡 Equilateral area formula run backward — area gives $s^2$, then square root gives $s$.
8.G.A.5 Step 3 - Read hexagon sides off the diagram.
- The big triangle with sides of length $s_2 = 36$ is the one whose sides contain $AB, CD, EF$.
- Along that triangle's three sides, the hexagon contributes segments $AB$, $CD$, $EF$, separated by the three equilateral corner chips.
- So $AB + CD + EF + (\text{three corner sides}) = 3 s_2 = 108$, but $AF + BC + DE$ ALSO sit as the three corner sides of this triangle (each corner chip shares its two non-hexagon sides with the BIG triangle and its third side with a hexagon edge), so $AB + CD + EF + (BC + DE + FA) = 3 s_2 = 108$...
- wait, recount: along the BIG triangle with sides on $AB, CD, EF$, each side is split into three: corner chip, hexagon edge, corner chip.
- The hexagon edges on it are $AB, CD, EF$ in turn; the corner chips have sides equal to the three OTHER hexagon edges $BC, DE, FA$.
- So the BIG triangle's side equals one hexagon edge plus two corner-chip sides.
- Summing over the three big-triangle sides: $3 s_2 = (AB + CD + EF) + 2(BC + DE + FA)$, i.e., $108 = P_1 + 2 P_2$ where $P_1 = AB+CD+EF$ and $P_2 = BC+DE+FA$.
- By the same logic on the other big triangle (sides on $BC, DE, FA$): $3 s_1 = 48\sqrt{3} = P_2 + 2 P_1$.
💡 Each big triangle's side splits into one hexagon edge plus two equilateral chip sides; sum around the triangle gives a linear relation.
4.OA.A.3 Step 4 - Sanity check with the easier related problem.
- A regular hexagon has $P_1 = P_2$ and both big triangles equal.
- Setting $P_1 = P_2 = P/2$ in our system gives $P_1 + 2 P_1 = 3 s_2 \Rightarrow s_2 = P_1$, which says each big triangle's side equals one hexagon edge — wait, that's only true if the corner chip sides are zero.
- Re-examine: in a REGULAR hexagon, the "big triangle" actually has side $= 2 \cdot \text{hexagon side}$ (the hexagon edge plus two corner chips of equal side length), so the relation $s = 1 \cdot \text{(hexagon edge)} + 2 \cdot \text{(chip side)}$ holds where chip side equals one hexagon edge.
- Both relations $P_1 + 2P_2 = 3 s_2$ and $2 P_1 + P_2 = 3 s_1$ are correct.
💡 Test the formula on a regular hexagon where everything collapses to one length — it works.
8.EE.C.8 Step 5 - Solve the linear system in $P_1, P_2$.
- From $P_1 + 2 P_2 = 108$ and $2 P_1 + P_2 = 48\sqrt{3}$: multiply the second equation by $2$ to get $4 P_1 + 2 P_2 = 96\sqrt{3}$; subtract the first to eliminate $P_2$: $3 P_1 = 96\sqrt{3} - 108$, so $P_1 = 32\sqrt{3} - 36$.
- Then $P_2 = (108 - P_1)/2 = (108 - 32\sqrt{3} + 36)/2 = (144 - 32\sqrt{3})/2 = 72 - 16\sqrt{3}$.
💡 Two equations, two unknowns — straight elimination.
5.NBT.B.5 Step 6 - The hexagon perimeter is the sum of all six sides: $P = P_1 + P_2 = (32\sqrt{3} - 36) + (72 - 16\sqrt{3}) = 36 + 16\sqrt{3}$.
- So $m = 36, n = 16, p = 3$, all positive integers with $p = 3$ square-free.
💡 Add the two triple-sums together to get the full perimeter; identify $m, n, p$ by inspection.
8.G.A.5 Equiangular hexagon means every interior angle is $\tfrac{(6-2)\cdot 180^\circ}{ 8.EE.A.2 Recover each big triangle's side length from its area. An equilateral triangle o 8.G.A.5 Read hexagon sides off the diagram. The big triangle with sides of length $s_2 = 4.OA.A.3 Sanity check with the easier related problem. A regular hexagon has $P_1 = P_2$ 8.EE.C.8 Solve the linear system in $P_1, P_2$. From $P_1 + 2 P_2 = 108$ and $2 P_1 + P_2 5.NBT.B.5 The hexagon perimeter is the sum of all six sides: $P = P_1 + P_2 = (32\sqrt{3} Review
Reasonableness: Each $P_i$ should be positive. $P_1 = 32\sqrt{3} - 36 \approx 55.4 - 36 = 19.4 > 0$ and $P_2 = 72 - 16\sqrt{3} \approx 72 - 27.7 = 44.3 > 0$, so the hexagon is geometrically realizable. The two big triangle side lengths $16\sqrt{3} \approx 27.7$ and $36$ are different, which matches the unequal areas $192\sqrt{3}$ vs $324\sqrt{3}$ (ratio $324/192 = 27/16$, square root $\sqrt{27}/4 = 3\sqrt{3}/4 \approx 1.30 = 36/27.7$ — consistent). Final answer $55$ matches choice (C) exactly.
Alternative: Tool #13 (Algebra) directly: let the hexagon sides be $a, b, c, d, e, f$ in order. Equiangular forces $a - d = e - b = c - f$ (a known relation from 120-degree turning), and the two big triangle sides are $a + b + c$ and $b + c + d$ etc., leading to the same linear system but with more bookkeeping. The diagram approach (Tool #1) keeps the equation count to two.
CCSS standards used (min grade 8)
4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Sanity-checking the formula on a regular hexagon by collapsing $P_1 = P_2$ and confirming the relation holds.)5.NBT.B.5Fluently multiply multi-digit whole numbers (Final arithmetic $36 + 16 + 3 = 55$ and intermediate products like $\sqrt{768} = 16\sqrt{3}$.)8.EE.A.2Use square root and cube root symbols to represent solutions (Solving $s_1^2 = 768$ and $s_2^2 = 1296$ to get $s_1 = 16\sqrt{3}, s_2 = 36$.)8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Solving the $2 \times 2$ system $P_1 + 2 P_2 = 108, 2 P_1 + P_2 = 48\sqrt{3}$ for $P_1, P_2$.)8.G.A.5Use informal arguments to establish facts about angle sum and exterior angles (Showing each interior angle is $120^\circ$, hence each corner chip and both big triangles are equilateral.)
⭐ This AMC 10 problem only needs Grade 8 angle reasoning and a $2 \times 2$ linear system you already know — extend the alternate sides, notice every corner chip and both big triangles are equilateral (since $180^\circ - 120^\circ = 60^\circ$), recover side lengths $s_1 = 16\sqrt{3}$ and $s_2 = 36$ from areas, then solve the two linear equations $P_1 + 2 P_2 = 108$ and $2 P_1 + P_2 = 48\sqrt{3}$ to get perimeter $= 36 + 16\sqrt{3}$, so $m + n + p = 55$.
⭐ This AMC 10 problem only needs Grade 8 angle reasoning and a $2 \times 2$ linear system you already know — extend the alternate sides, notice every corner chip and both big triangles are equilateral (since $180^\circ - 120^\circ = 60^\circ$), recover side lengths $s_1 = 16\sqrt{3}$ and $s_2 = 36$ from areas, then solve the two linear equations $P_1 + 2 P_2 = 108$ and $2 P_1 + P_2 = 48\sqrt{3}$ to get perimeter $= 36 + 16\sqrt{3}$, so $m + n + p = 55$.
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