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by **stupidityismygam** » Wed May 05, 2010 1:53 am

Let ABCDEF be a regular hexagon. Let G,H,I,J,K,L

be the midpoints of sides AB,BC,CD,DE,EF,AF

, respectively. The segments AH.BI,CJ,DK,EL,FG

bound a smaller regular hexagon. Let the ratio of the area of the smaller hexagon to the area of ABCDEF be expressed as a fraction $\frac{m}{n}$ where

and

are relatively prime positive integers. Find m+n

.

by **88bobcat** » Wed May 05, 2010 9:23 pm

stupidityismygam wrote:Let ABCDEF be a regular hexagon. Let be the midpoints of sides , respectively. The segments bound a smaller regular hexagon. Let the ratio of the area of the smaller hexagon to the area of ABCDEF be expressed as a fraction $\frac{m}{n}$ where and are relatively prime positive integers. Find .

The length of AH, from Law of Cosines, is $\displaystyle AH = \frac{\sqrt{7}}{2}$

The apothem $\eta$ of the inner, smaller triangle is drawn of the hexagons' center to AH.

From similar triangles, $\displaystyle \frac{\eta}{\displaystyle \frac{\sqrt{3}}{2}} = \frac{1}{\displaystyle \frac{\sqrt{7}}{2}}$ .

This means that the apothem length is $\displaystyle \eta = \sqrt{\frac{3}{7}}$

For an apothem of $\eta$ , the length of the side of the smaller hexagon is $\displaystyle \delta = \frac{2}{\sqrt{3}} \eta = \frac{2}{\sqrt{7}}$

Now the ratio of the area is the ratio of the squares of the sides of the hexagons:

$\displaystyle \frac{m}{n} = \frac{4/7}{1} = \frac{4}{7}$

$m + n = 4 + 7 = \boxed{11}$

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