2011 AMC 10B Problem 20

P16-20~5 minPrint

Quadrilaterals & Polygon Areas Triangle Centers & Cevians Cut the figure into known shapes (triangles, rectangles, sectors)Exploit symmetry to reduce work or pair up objects

Rhombus $ABCD$ has side length $2$ and $\angle B = 120^\circ$ . Region $R$ consists of all points inside the rhombus that are closer to vertex $B$ than any of the other three vertices. What is the area of $R$ ?

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Key insight

The short diagonal BD splits the rhombus into two equilateral triangles of side 2. In each, the part closer to B than to the other corners is exactly one third.

Solution

First, the shape of the rhombus.

In a rhombus, neighboring angles add to $180^\circ$ . Since $\angle B = 120^\circ$ , the angles at $A$ and $C$ are $60^\circ$ .
Draw the diagonal $BD$ . Triangle $ABD$ has $AB = AD = 2$ and a $60^\circ$ angle at $A$ , so it is equilateral: all sides $2$ . The same is true of triangle $CBD$ .
So the rhombus is two equilateral triangles of side $2$ , glued along $BD$ .

Next, what "closer to $B$ " means.

The points exactly as far from $B$ as from $A$ form a line: the line through the midpoint of $AB$ at a right angle to $AB$ . Points on $B$ 's side of that line are closer to $B$ than to $A$ . The same goes for $B$ and $D$ , and for $B$ and $C$ .

Now look at one triangle, $ABD$ .

In an equilateral triangle, the three "halfway lines" for the three sides all meet at the center point $O$ of the triangle.
These lines cut the triangle into three matching pieces, one around each corner. The piece around $B$ holds the points closer to $B$ than to $A$ or $D$ . The three pieces are identical, so the piece around $B$ is one third of the triangle.
What about $C$ ? Every point of triangle $ABD$ is at least as close to $B$ as to $C$ (the halfway line between $B$ and $C$ passes through $D$ and stays outside triangle $ABD$ ), so $C$ does not cut anything off.
The area of an equilateral triangle with side $2$ is $\frac{\sqrt{3}}{4} \times 2^2 = \sqrt{3}$ . One third of it is $\frac{\sqrt{3}}{3}$ .

Put the halves together.

9. Triangle $CBD$ works the same way and gives another $\frac{\sqrt{3}}{3}$ . So the area of $R$ is
$\frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{3} = \frac{2\sqrt{3}}{3}.$

The answer is $\boxed{\textbf{(C)}\ \frac{2\sqrt{3}}{3}}$ .

Why this works

"Closer to one point than another" is always decided by the line halfway between them, at a right angle to the segment joining them. Splitting the rhombus into equilateral triangles lets symmetry do the work: each triangle divides evenly into three pieces, one per corner, so no coordinates are needed.

Alternative approach

Use coordinates: $B = (0, 0)$ , $D = (2, 0)$ , $A = (1, \sqrt{3})$ , $C = (1, -\sqrt{3})$ . The region $R$ is a five-sided shape with corners $B = (0, 0)$ , the midpoints of $BA$ and $BC$ at $(\tfrac{1}{2}, \pm\tfrac{\sqrt{3}}{2})$ , and the centers of the two triangles at $(1, \pm\tfrac{\sqrt{3}}{3})$ . Cut it along $BD$ into two matching four-sided pieces and add their areas; the total is $\frac{2\sqrt{3}}{3}$ .

Concise solution

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

Since $\angle B = 120^\circ$ , the angles at $A$ and $C$ are $60^\circ$ , so triangles $ABD$ and $CBD$ are both equilateral with side $2$ (in particular $BD = 2$ ). The rhombus is their union along $BD$ .

A point is closer to $B$ than to $A$ exactly when it lies on $B$ 's side of the perpendicular bisector of $AB$ , and similarly for $C$ and $D$ . Inside equilateral triangle $ABD$ , the perpendicular bisectors of $AB$ and $BD$ meet at the center $O$ of the triangle, and the points closer to $B$ than to $A$ or $D$ form the kite with vertices $B$ , the midpoint of $AB$ , $O$ , and the midpoint of $BD$ . By the threefold symmetry of the equilateral triangle, this kite is one third of the triangle. (The bisector of $BC$ does not cut into it: that bisector passes through $D$ and otherwise stays outside triangle $ABD$ .)

Triangle $ABD$ has area $\frac{\sqrt{3}}{4} \cdot 2^2 = \sqrt{3}$ , so the kite has area $\frac{\sqrt{3}}{3}$ . The same happens in triangle $CBD$ .

Total area of $R$ : $2 \cdot \frac{\sqrt{3}}{3} = \frac{2\sqrt{3}}{3}$ , so the answer is $\boxed{\textbf{(C)}\ \frac{2\sqrt{3}}{3}}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 8: Quadrilaterals
Introduction to Geometry, Chapter 7: Special Parts of a Triangle
Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2, §4.3)
Introduction to Algebra, Chapter 22: Special Manipulations (§22.3)