AMC 10 · 2013 · #15
Grade 8 geometry-2dA wire is cut into two pieces, one of length a and the other of length b. The piece of length a is bent to form an equilateral triangle, and the piece of length b is bent to form a regular hexagon. The triangle and the hexagon have equal area. What is ba?
Pick an answer.
AMC 10 2013 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 wire is cut into two pieces of length $a$ and $b$. The piece of length $a$ is bent into an equilateral triangle, and the piece of length $b$ is bent into a regular hexagon. The two shapes have equal area. Find the ratio $\frac{a}{b}$.
Givens: The length-$a$ piece forms the whole perimeter of an equilateral triangle; The length-$b$ piece forms the whole perimeter of a regular hexagon; The triangle and the hexagon have equal area; Answer choices: (A) $1$, (B) $\frac{\sqrt{6}}{2}$, (C) $\sqrt{3}$, (D) $2$, (E) $\frac{3\sqrt{2}}{2}$
Unknowns: The ratio $\frac{a}{b}$ of the two wire pieces
Understand
Restated: A wire is cut into two pieces of length $a$ and $b$. The piece of length $a$ is bent into an equilateral triangle, and the piece of length $b$ is bent into a regular hexagon. The two shapes have equal area. Find the ratio $\frac{a}{b}$.
Givens: The length-$a$ piece forms the whole perimeter of an equilateral triangle; The length-$b$ piece forms the whole perimeter of a regular hexagon; The triangle and the hexagon have equal area; Answer choices: (A) $1$, (B) $\frac{\sqrt{6}}{2}$, (C) $\sqrt{3}$, (D) $2$, (E) $\frac{3\sqrt{2}}{2}$
Plan
Primary tool: #13 Convert to Algebra
Secondary: #7 Identify Subproblems, #4 Introduce a Variable
The words "equal area" are really an equation in disguise, so the main move (tool #13) is to write each area in terms of $a$ and $b$ and set them equal. Tool #7 (Identify Subproblems) splits the work into two clean pieces — find the triangle's area, then the hexagon's area — instead of juggling both shapes at once. Tool #4 (Introduce a Variable) turns each perimeter into a side length ($\frac{a}{3}$ and $\frac{b}{6}$) so the areas become expressions we can compare.
Execute — Answer: B
6.EE.B.6 Step 1 Turn wire lengths into sides
- The length-$a$ wire is bent into an equilateral triangle, so $a$ is its perimeter and each of the $3$ equal sides is $\frac{a}{3}$.
- The length-$b$ wire is bent into a regular hexagon, so $b$ is its perimeter and each of the $6$ equal sides is $\frac{b}{6}$.
- From now on we work with these side lengths, not the wire lengths.
💡 A regular shape splits its perimeter evenly, so dividing by the number of sides gives one side.
8.G.B.7 Step 2 Area of the triangle
- An equilateral triangle with side $s$ has area $\frac{\sqrt{3}}{4}s^2$: its altitude splits it into two $30$-$60$-$90$ right triangles, and the Pythagorean theorem gives height $\frac{\sqrt{3}}{2}s$, so the area is $\frac{1}{2}\cdot s\cdot\frac{\sqrt{3}}{2}s=\frac{\sqrt{3}}{4}s^2$.
- Put in $s=\frac{a}{3}$ to get the triangle's area.
💡 Dropping the altitude makes a right triangle, and the Pythagorean theorem pins down the height.
6.G.A.1 Step 3 Area of the hexagon
- A regular hexagon splits into $6$ small equilateral triangles that all share the center, each with side equal to the hexagon's side $\frac{b}{6}$.
- So the hexagon's area is $6$ times one small triangle's area, using the same $\frac{\sqrt{3}}{4}s^2$ formula.
- Multiply and simplify.
💡 A regular hexagon is just six identical equilateral triangles fanned around its center.
8.EE.C.7 Step 4 Set the areas equal
- Equal area means the two expressions are equal.
- Set them equal and cancel the common factor $\sqrt{3}$ from both sides, then compare the $a^2$ and $b^2$ pieces.
- Cross-multiplying the fractions gives a clean relationship between $a^2$ and $b^2$.
💡 "Equal area" is an equation, and the shared $\sqrt{3}$ cancels so only the squared lengths matter.
8.EE.A.2 Step 5 Take the square root
- We have $\frac{a^2}{b^2}=\frac{3}{2}$, so $\frac{a}{b}$ is the positive square root of $\frac{3}{2}$.
- Rationalize by multiplying inside the root by $\frac{2}{2}$: $\sqrt{\frac{3}{2}}=\sqrt{\frac{6}{4}}=\frac{\sqrt{6}}{2}$.
- So $\frac{a}{b}=\frac{\sqrt{6}}{2}$, which is choice (B).
💡 Squaring a ratio and then square-rooting are opposites, so the square root undoes the $a^2/b^2$.
6.EE.B.6 The length-$a$ wire is bent into an equilateral triangle, so $a$ is its perimete 8.G.B.7 An equilateral triangle with side $s$ has area $\frac{\sqrt{3}}{4}s^2$: its alti 6.G.A.1 A regular hexagon splits into $6$ small equilateral triangles that all share the 8.EE.C.7 Equal area means the two expressions are equal. Set them equal and cancel the co 8.EE.A.2 We have $\frac{a^2}{b^2}=\frac{3}{2}$, so $\frac{a}{b}$ is the positive square r Review
Reasonableness: The answer $\frac{\sqrt{6}}{2}\approx 1.22$ is a bit more than $1$, which makes sense: for equal areas the triangle (a less efficient shape) needs a little more wire than the hexagon, so $a>b$ and $\frac{a}{b}>1$. Checking size: $\left(\frac{a}{b}\right)^2=\frac{6}{4}=\frac{3}{2}$ exactly matches the area equation, and no other choice squares to $\frac{3}{2}$ — (A) gives $1$, (C) gives $3$, (D) gives $4$, (E) gives $\frac{9}{2}$ — so only (B) fits.
Alternative: Use the similarity shortcut. Both shapes are made of the same little equilateral triangle, so equal total area means the big triangle equals $6$ little ones — the big triangle has $6$ times a small triangle's area. Since area scales as the square of length, the big triangle's side is $\sqrt{6}$ times the small side, so its perimeter (and hence the ratio of wire lengths) also relates by lengths: the triangle side is $\frac{a}{3}$ and the hexagon side is $\frac{b}{6}$, giving $\frac{a/3}{b/6}=\sqrt{6}$, hence $\frac{2a}{b}=\sqrt{6}$ and $\frac{a}{b}=\frac{\sqrt{6}}{2}$.
CCSS standards used (min grade 8)
6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Writing each shape's side length as an expression in the wire length: $\frac{a}{3}$ for the triangle and $\frac{b}{6}$ for the hexagon.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Deriving the equilateral-triangle height $\frac{\sqrt{3}}{2}s$ to get the area formula $\frac{\sqrt{3}}{4}s^2$.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Decomposing the regular hexagon into $6$ equilateral triangles to compute its area.)8.EE.C.7Solve linear equations in one variable (Setting the two area expressions equal, cancelling $\sqrt{3}$, and solving to get $\frac{a^2}{b^2}=\frac{3}{2}$.)8.EE.A.2Use square root and cube root symbols to represent solutions (Taking the square root of $\frac{3}{2}$ and rationalizing to reach $\frac{a}{b}=\frac{\sqrt{6}}{2}$.)
⭐ Both shapes are built from the same little equilateral triangle, so "equal area" becomes one clean equation — set the areas equal, cancel, and square-root to get the ratio.
⭐ Both shapes are built from the same little equilateral triangle, so "equal area" becomes one clean equation — set the areas equal, cancel, and square-root to get the ratio.
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