2016 AMC 10B Problem 23

P21-25~6 minPrint

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

In regular hexagon $ABCDEF$ , points $W$ , $X$ , $Y$ , and $Z$ are chosen on sides $\overline{BC}$ , $\overline{CD}$ , $\overline{EF}$ , and $\overline{FA}$ respectively, so lines $AB$ , $ZW$ , $YX$ , and $ED$ are parallel and equally spaced. What is the ratio of the area of hexagon $WCXYFZ$ to the area of hexagon $ABCDEF$ ?

Show solution without answering

Key insight

The three strips have equal height, but the hexagon is wider in the middle. The top and bottom strips are trapezoids worth 8/27 of the hexagon each, leaving 11/27.

Solution

The answer is a ratio, so any size works. Let each side be $1$ .

Step 1: picture the strips. Draw the hexagon with side $AB$ on top and side $ED$ on the bottom. Its height is $\sqrt3$ (two equilateral triangles of height $\frac{\sqrt3}{2}$ stacked). The four equally spaced lines cut it into three strips, each $\frac{\sqrt3}{3}$ tall. Hexagon $WCXYFZ$ is the middle strip; $C$ and $F$ are its left and right tips.

Step 2: how wide is line $ZW$ ?

1. The width of the hexagon grows steadily from $1$ at the top ( $AB$ ) to $2$ at the middle (the line through $F$ and $C$ ), which is $\frac{\sqrt3}{2}$ below the top.
2. Line $ZW$ is $\frac{\sqrt3}{3}$ below the top. Compare: $\frac{\sqrt3}{3} \div \frac{\sqrt3}{2} = \frac23$ . So $ZW$ is $\frac23$ of the way from the top to the middle.
3. The width grows by $1$ over that whole distance, so at $ZW$ it has grown by $\frac23$ :
$ZW = 1 + \tfrac23 = \tfrac53.$

Step 3: area of the top strip. It is a trapezoid with parallel sides $1$ and $\frac53$ and height $\frac{\sqrt3}{3}$ :
$\tfrac12\left(1 + \tfrac53\right) \cdot \tfrac{\sqrt3}{3} = \tfrac12 \cdot \tfrac83 \cdot \tfrac{\sqrt3}{3} = \tfrac{4\sqrt3}{9}.$
The bottom strip is the same shape, so the two outer strips together have area $\frac{8\sqrt3}{9}$ .

Step 4: subtract from the whole. The hexagon is six equilateral triangles of side $1$ , so its area is $6 \cdot \frac{\sqrt3}{4} = \frac{3\sqrt3}{2}$ . The middle strip is
$\tfrac{3\sqrt3}{2} - \tfrac{8\sqrt3}{9} = \tfrac{27\sqrt3}{18} - \tfrac{16\sqrt3}{18} = \tfrac{11\sqrt3}{18}.$

Step 5: the ratio.
$\frac{11\sqrt3/18}{27\sqrt3/18} = \frac{11}{27}.$

The answer is $\boxed{\textbf{(C)}\ \frac{11}{27}}$ .

Why this works

Equally spaced parallel lines make strips of the same height, so the only question is how wide each strip is. Along a straight slanted side, the width changes steadily, which makes the outer strips simple trapezoids. It is easier to find those two and subtract than to work with the six-sided middle piece.

Alternative approach

Count small triangles. Make the side $3$ and fill the hexagon with equilateral triangles of side $1$ : it holds $6 \cdot 9 = 54$ of them, in six rows. From top to bottom the rows hold $7, 9, 11, 11, 9, 7$ small triangles. Each strip is two rows. The top strip has $7 + 9 = 16$ , the bottom strip also $16$ , and the middle strip $11 + 11 = 22$ . The ratio is $\frac{22}{54} = \frac{11}{27}$ .

Concise solution

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

Let the side be $1$ and draw $AB$ at the top and $ED$ at the bottom, so the hexagon has height $\sqrt3$ . The four parallel lines cut it into three strips of height $\tfrac{\sqrt3}{3}$ ; hexagon $WCXYFZ$ is exactly the middle strip, with $C$ and $F$ its widest points.

The hexagon's width grows linearly from $1$ at $AB$ to $2$ at the line $FC$ through the center, which is $\tfrac{\sqrt3}{2}$ below $AB$ . Line $ZW$ is $\tfrac{\sqrt3}{3}$ below $AB$ , that is $\tfrac{2}{3}$ of the way down to $FC$ , so
$ZW = 1 + \tfrac23(2 - 1) = \tfrac53.$

The top strip $ABWZ$ is a trapezoid with parallel sides $1$ and $\tfrac53$ and height $\tfrac{\sqrt3}{3}$ :
$[ABWZ] = \tfrac12\left(1 + \tfrac53\right)\tfrac{\sqrt3}{3} = \tfrac{4\sqrt3}{9}.$
By symmetry the bottom strip has the same area, so together they cover $\tfrac{8\sqrt3}{9}$ .

The whole hexagon has area $\tfrac{3\sqrt3}{2}$ , so the middle strip has area
$\tfrac{3\sqrt3}{2} - \tfrac{8\sqrt3}{9} = \tfrac{27\sqrt3 - 16\sqrt3}{18} = \tfrac{11\sqrt3}{18},$
and the requested ratio is $\dfrac{11\sqrt3/18}{27\sqrt3/18} = \dfrac{11}{27}$ .

The answer is $\boxed{\textbf{(C)}\ \frac{11}{27}}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 8: Quadrilaterals
Introduction to Geometry, Chapter 3: Congruent Triangles
Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2, §4.3)
Introduction to Algebra, Chapter 22: Special Manipulations (§22.3)