https://data.artofproblemsolving.com...e10c907669.png

hi can someone please explain the solution to me please, thanks

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- Dec 1st 2013, 07:38 PMvegasgunnerhexagon
https://data.artofproblemsolving.com...e10c907669.png

hi can someone please explain the solution to me please, thanks - Dec 1st 2013, 07:45 PMLimpSpiderRe: hexagon
Do you want us to find

**and**explain the solution? Or just explain the solution without giving the solution? Where has your working taken you so far? - Dec 1st 2013, 07:54 PMvegasgunnerRe: hexagon
I just would like and explanation please

- Dec 1st 2013, 08:13 PMibduttRe: hexagon
Attachment 29839 with this much you should be able to complete

- Dec 2nd 2013, 04:13 PMvegasgunnerRe: hexagon
- Dec 2nd 2013, 04:58 PMHallsofIvyRe: hexagon
Good! I presume that you realized from idbutt's picture that the if we take the distance from the center to a vertex of the outer hexagon to be 1, then the distance from the center to the middle of a side of that hexagon is $\displaystyle \sqrt{3}/2$. And that is where the circle touches the hexagon so is a vertex of the inner hexagon. Since those are "equivalent" distances, the ratio of the areas is the square of the ratio of those distances: $\displaystyle \left(\frac{1}{\sqrt{3}/2}\right)^2= \frac{1}{3/4}= \frac{4}{3}$.