2014 AMC 10B Problem 13

P11-15~4 minPrint

Triangles: Area & Pythagorean Quadrilaterals & Polygon Areas Add construction lines/points (drop altitudes, extend segments, connect centers)Exploit symmetry to reduce work or pair up objects

Six regular hexagons surround a regular hexagon of side length $1$ as shown. What is the area of $\triangle ABC$ ?

Show solution without answering

Key insight

Measure from the center of the middle hexagon: each corner of triangle ABC is 2 units away, and the three corners are evenly spaced, so the triangle is equilateral.

Solution

The whole picture is built around the center of the middle hexagon, so measure everything from there. Call that center $O$ .

Put in coordinates. Place $O$ at $(0, 0)$ with the middle hexagon's corners at $(1, 0)$ , $\left(\frac12, \frac{\sqrt3}{2}\right)$ , $\left(-\frac12, \frac{\sqrt3}{2}\right)$ , $(-1, 0)$ , $\left(-\frac12, -\frac{\sqrt3}{2}\right)$ , $\left(\frac12, -\frac{\sqrt3}{2}\right)$ . (A regular hexagon of side $1$ is six equilateral triangles of side $1$ meeting at the center, so each corner is $1$ from $O$ .)
Where are $A$ , $B$ , $C$ ? Each of them is a corner where two outer hexagons meet, sitting straight out from a corner of the middle hexagon. Two outer hexagons share an edge of length $1$ that starts at a corner of the middle hexagon and points straight away from $O$ . So each of $A$ , $B$ , $C$ is $1 + 1 = 2$ from $O$ , in the same direction as a corner of the middle hexagon.
Using every other corner, as in the figure: $A = (-2, 0)$ , $B = (1, \sqrt3)$ , $C = (1, -\sqrt3)$ . (Each is twice a corner of the middle hexagon.)
Area. $B$ and $C$ lie on the vertical line $x = 1$ , so take $BC$ as the base: its length is $\sqrt3 - (-\sqrt3) = 2\sqrt3$ .
The height is the horizontal distance from $A$ to that line: from $x = -2$ to $x = 1$ is $3$ .
Area $= \frac12 \times 2\sqrt3 \times 3 = 3\sqrt3$ .

The answer is $\boxed{\textbf{(B)}\ 3\sqrt{3}}$ .

Why this works

In a picture made of regular hexagons, the center of the middle one is the natural place to measure from, and every length is a whole number or a whole number times $\sqrt3$ . Once you know how far each corner of the triangle is from the center, the triangle is fixed. Choosing a base that lies on a vertical line makes the height easy to read off.

Alternative approach

Skip coordinates. The three corners $A$ , $B$ , $C$ are each $2$ from $O$ and spaced evenly around it, $120^\circ$ apart, so $ABC$ is equilateral. Drop a perpendicular from $O$ to side $AB$ : it makes a $30$ - $60$ - $90$ triangle with hypotenuse $OA = 2$ , so half of $AB$ is $\sqrt3$ and $AB = 2\sqrt3$ . The area of an equilateral triangle with side $s$ is $\frac{\sqrt3}{4}s^2 = \frac{\sqrt3}{4} \times 12 = 3\sqrt3$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Let $O$ be the center of the middle hexagon. Draw $OA$ , $OB$ , $OC$ .

Look at vertex $B$ . The ray from $O$ through a vertex $V$ of the central hexagon reaches $V$ after distance $1$ (a regular hexagon's circumradius equals its side). Beyond $V$ , two outer hexagons meet along a common edge that lies on the same ray by symmetry, and $B$ is the far end of that edge. So $OB = 1 + 1 = 2$ . Likewise $OA = OC = 2$ .

The three rays are spaced $120^\circ$ apart, so $\triangle ABC$ is equilateral with circumradius $2$ . Dropping the perpendicular from $O$ to $AB$ gives a $30$ - $60$ - $90$ triangle with hypotenuse $2$ , so half of $AB$ is $\sqrt3$ and $AB = 2\sqrt3$ .

The area is
$\frac{\sqrt3}{4}\,(2\sqrt3)^2 = \frac{\sqrt3}{4} \cdot 12 = 3\sqrt3.$

The answer is $\boxed{\textbf{(B)}\ 3\sqrt{3}}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Geometry, Chapter 8: Quadrilaterals
Introduction to Geometry, Chapter 19: Problem Solving Strategies in Geometry (§19.1)
Introduction to Algebra, Chapter 22: Special Manipulations (§22.3)