Cool walk through for geometric transforms

[Since its AMC 8 today - here’s a geometry walkthrough instead for the week]

[@sansu_original]

Find the area of the quadrilateral:  EFGN

Part 1 - Deal with triangle ENG

1. I’ve added the diameter of the dodecagon AG and then note that connecting the center to 2 vertices like E and G or  A and C form an equilateral triangle.  So AC = CE = EG = OG = OA = r.  Where r is the radius of the dodecagon.

2. From the inscribed angle  $ \angle{CAE} = 30 $ and from internal arcs $ \angle{CNE} =  \dfrac{\overparen{CE} + \overparen{AL}}{2} = 45 $

3. Dropping a perpendicular down at CM to build a 30-60 we get AC = r so $ CM = \dfrac{r}{2} $    and $AM = ME =  r \cdot \dfrac{\sqrt{3}}{2} $  Since CNM is an isosceles right triangle $ MN = CM =  \dfrac{r}{2} $

4.  We can now find the area of $ \triangle{ENG} $ in parts  ENG = NMG + MEG:

  $ \triangle{NMG} = \frac{1}{2}MN \cdot EG = \dfrac{r^2}{4} $

  $ \triangle{MEG} = \frac{1}{2}ME \cdot EG = \dfrac{\sqrt{3} \cdot r^2}{4} $

There are two things to note here.

• Since the area of the whole dodecagon is $ 3 \cdot r^2  $ one of the 12 wedges that make it up has area $ \dfrac{r^2}{4} $   which is the same as NMG.
• The area of any of the equilateral triangles with a side length of radius r is $ \dfrac{\sqrt{3} \cdot r^2}{4} $ which is the same as MEG.

Part 2: (Rearrange)

Its easy to show an equilateral triangle, wedge and edge triangle like EFG combine to a quarter 

of the whole dodecagon.

So $ \triangle{NMG} + \triangle{MEG} + \triangle{EFG} = \dfrac{3}{4} r^2 $

Note: Directly calculating EFG based on 15-75-90 side ratios we get $  \frac{1}{2} \cdot r \cdot  \dfrac{2 - \sqrt{3}}{2} \cdot r  =  \dfrac{r^2}{2} - \dfrac{\sqrt{3}}{4} \cdot r^2 $

And indeed: $ \dfrac{r^2}{4} + \dfrac{\sqrt{3}} {4} \cdot r^2 + \dfrac{r^2}{2} - \dfrac{\sqrt{3}}{4} \cdot r^2  = \dfrac{3}{4} r^2  $