AMC 10 · 2014 · #13
Grade 8 geometry-2dSix regular hexagons surround a regular hexagon of side length 1 as shown. What is the area of △ABC?
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 regular hexagon of side length $1$ is surrounded by six more regular hexagons, as drawn. Points $A$, $B$, $C$ are three vertices marked on the figure. Find the area of $\triangle ABC$.
Givens: The central hexagon and the six surrounding hexagons are all regular with side length $1$; The figure is drawn with exact coordinates: the central hexagon has a horizontal side from $(0,0)$ to $(10,0)$; The three vertices are $A=(0,0)$, $B=(30,\,17.3205\ldots)$, $C=(30,\,-17.3205\ldots)$ in drawing units; Answer choices: (A) $2\sqrt{3}$, (B) $3\sqrt{3}$, (C) $1+3\sqrt{2}$, (D) $2+2\sqrt{3}$, (E) $3+2\sqrt{3}$
Unknowns: The area of $\triangle ABC$
Understand
Restated: A regular hexagon of side length $1$ is surrounded by six more regular hexagons, as drawn. Points $A$, $B$, $C$ are three vertices marked on the figure. Find the area of $\triangle ABC$.
Givens: The central hexagon and the six surrounding hexagons are all regular with side length $1$; The figure is drawn with exact coordinates: the central hexagon has a horizontal side from $(0,0)$ to $(10,0)$; The three vertices are $A=(0,0)$, $B=(30,\,17.3205\ldots)$, $C=(30,\,-17.3205\ldots)$ in drawing units; Answer choices: (A) $2\sqrt{3}$, (B) $3\sqrt{3}$, (C) $1+3\sqrt{2}$, (D) $2+2\sqrt{3}$, (E) $3+2\sqrt{3}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #17 Visualize Spatial Relationships
The figure already hands us exact coordinates, so Tool #1 (Draw a Diagram) is really a coordinate-geometry engine: read the points off the drawing and let position do the work. First fix the scale so drawing units become real units. Then Tool #7 (Identify Subproblems) splits the area into two easy measurements — a base and a height — because $B$ and $C$ sit one directly above the other, making $BC$ a vertical segment and the height a plain horizontal distance. Tool #17 (Visualize Spatial Relationships) is the cross-check: seeing that the three vertices are the same distance apart tells us the triangle is equilateral, confirming the area a second way.
Execute — Answer: B
7.G.A.1 Step 1 Fix the scale of the drawing
- In the drawing the central hexagon has a side from $(0,0)$ to $(10,0)$, which is $10$ drawing units long.
- But that side is really length $1$.
- So the drawing is blown up by a factor of $10$: to get real lengths, divide every coordinate by $10$.
- This turns $A=(0,0)$, $B=(30,17.3205\ldots)$, $C=(30,-17.3205\ldots)$ into real-unit points.
💡 A scale drawing is honest about shape but lies about size, so you undo the blow-up before measuring.
8.NS.A.2 Step 2 Recognize the decimal as a radical
- The height number $1.73205\ldots$ is not random: it is the decimal for $\sqrt{3}$ (since $\sqrt{3}=1.7320508\ldots$).
- This makes sense — a regular hexagon of side $1$ has 'width' $\sqrt{3}$ across its flat sides.
- So in real units the three vertices are $A=(0,0)$, $B=(3,\sqrt{3})$, $C=(3,-\sqrt{3})$.
💡 A messy decimal is often a familiar square root wearing a disguise.
6.G.A.3 Step 3 Measure the base BC
- $B$ and $C$ have the same first coordinate, $x=3$, so segment $BC$ is vertical.
- The length of a vertical segment is just the gap between the two $y$-values: from $-\sqrt{3}$ up to $\sqrt{3}$.
- That gap is $2\sqrt{3}$.
- Use $BC$ as the base of the triangle.
💡 When two points sit on the same vertical line, their distance is simply the difference of heights.
6.G.A.3 Step 4 Measure the height from A
- The base $BC$ lies on the vertical line $x=3$.
- The height of the triangle is the perpendicular distance from the opposite vertex $A$ to that line.
- Since $A$ is at $x=0$ and the line is $x=3$, that horizontal distance is $3$.
💡 The distance from a point to a vertical line is just the difference in the sideways direction.
6.G.A.1 Step 5 Combine base and height into the area
- A triangle's area is half the base times the height.
- Here the base is $BC=2\sqrt{3}$ and the height is $3$.
- Multiplying gives $\tfrac12\cdot 2\sqrt{3}\cdot 3 = 3\sqrt{3}$.
- That matches choice (B).
💡 Half base times height turns two simple lengths into the whole area.
7.G.A.1 In the drawing the central hexagon has a side from $(0,0)$ to $(10,0)$, which is 8.NS.A.2 The height number $1.73205\ldots$ is not random: it is the decimal for $\sqrt{3} 6.G.A.3 $B$ and $C$ have the same first coordinate, $x=3$, so segment $BC$ is vertical. 6.G.A.3 The base $BC$ lies on the vertical line $x=3$. The height of the triangle is the 6.G.A.1 A triangle's area is half the base times the height. Here the base is $BC=2\sqrt Review
Reasonableness: Cross-check with side lengths. Using the Pythagorean distance, $AB=\sqrt{3^2+(\sqrt{3})^2}=\sqrt{9+3}=\sqrt{12}=2\sqrt{3}$, and by the up-down symmetry $AC=2\sqrt{3}$ too, while $BC=2\sqrt{3}$. All three sides equal $2\sqrt{3}$, so $\triangle ABC$ is equilateral. The area of an equilateral triangle of side $s$ is $\tfrac{\sqrt{3}}{4}s^2=\tfrac{\sqrt{3}}{4}\cdot(2\sqrt{3})^2=\tfrac{\sqrt{3}}{4}\cdot 12=3\sqrt{3}$, the same answer. Numerically $3\sqrt{3}\approx 5.20$, which is a believable size for a triangle spanning three hexagons, and it sits at choice (B); choice (A) $2\sqrt{3}\approx3.46$ would be too small.
Alternative: Solve it by dissection with no coordinates. The six thin triangular gaps left between the surrounding hexagons but inside $\triangle ABC$ can be slid together to build one more regular hexagon congruent to the others. So $\triangle ABC$ is made of exactly two regular hexagons' worth of area. A regular hexagon of side $1$ has area $\tfrac{3\sqrt{3}}{2}$, so two of them give $2\cdot\tfrac{3\sqrt{3}}{2}=3\sqrt{3}$, again answer (B).
CCSS standards used (min grade 8)
7.G.A.1Solve problems involving scale drawings of geometric figures (Reading the factor-of-$10$ scale from the central hexagon's side and converting drawing coordinates to real-unit coordinates.)8.NS.A.2Use rational approximations of irrational numbers to compare their size (Recognizing the coordinate $1.73205\ldots$ as the irrational number $\sqrt{3}$ so the vertices become exact.)6.G.A.3Draw polygons in the coordinate plane given coordinates for the vertices (Finding the vertical base $BC=2\sqrt{3}$ and the horizontal height $3$ directly from the vertex coordinates.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Combining base and height with the half-base-times-height rule to get the area $3\sqrt{3}$.)8.G.B.8Apply the Pythagorean theorem to find distance between two points in a coordinate system (Cross-checking the side lengths $AB=AC=BC=2\sqrt{3}$ to confirm the triangle is equilateral.)
⭐ Read the corner points from the picture, shrink them to real size, then multiply half the vertical base $2\sqrt{3}$ by the height $3$ to get the area $3\sqrt{3}$.
⭐ Read the corner points from the picture, shrink them to real size, then multiply half the vertical base $2\sqrt{3}$ by the height $3$ to get the area $3\sqrt{3}$.
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