Hello, xoluosox!
An equilateral triangle has area: .$\displaystyle A \:=\:\frac{\sqrt{3}}{4}s^2$
. . where $\displaystyle s$ is a side of the triangle.
Given a circle with radius $\displaystyle r$, and an inscribed regular hexagon.
Prove that the area of hexagon is a mean proportional between the areas
of the inscribed and circumscribed equilateral triangles. Code:
* o *
* o o *
* r o o
* o o *
o o r
* o o *
* * *
* *
* *
* *
* *
* * *
The inscribed hexagon is composed of six equilateral triangles with side $\displaystyle r.$
Its area is: .$\displaystyle A_{hex} \;=\;6 \times \frac{\sqrt{3}}{4}\,r^2 \;=\;\frac{3\sqrt{3}}{2}\,r^2$
Code:
* o *
* /|\ *
* / | \ *
* / |r \ *
/ | \
* / | \ *
* / r * r \ *
* / * * \ *
/ * * \
o- - - - - - - - -o
* *
* *
* * *
It can be shown that the side of this triangle is: .$\displaystyle s = \sqrt{3}\,r$
Then its area is: .$\displaystyle A_1 \;=\;\frac{\sqrt{3}}{4}\left(\sqrt{3}\,r\right)^2 \;=\;\frac{3\sqrt{3}}{4}r^2 $
Code:
o
/|\
/ | \
/ | \
/ | \
/ | \
/ * * * \
/* | *\
* | *
* | o *
/ | o \
/* | o r *\
/ * * * \
/ * * \
/ \
/ * * \
/ * * \
/ * * \
o - - - - - - - * * *- - - - - - - - o
It can be shown that the side of this triangle is: $\displaystyle 2\sqrt{3}\,r$
Its area is: .$\displaystyle A_2 \;=\;\frac{\sqrt{3}}{4}\,(2\sqrt{3}\,r)^2 \:=\:3\sqrt{3}\,r^2$
If $\displaystyle A_{hex}$ is the mean proportional between $\displaystyle A_1$ and $\displaystyle A_2$
. . then: .$\displaystyle \left(A_{hex}\right)^2 \;=\;\left(A_1\right)\cdot\left(A_2\right)$
We have: .$\displaystyle \left(\frac{3\sqrt{3}}{2}\,r^2\right)^2 \;=\;\left(\frac{3\sqrt{3}}{4}\,r^2\right)\cdot\le ft(3\sqrt{3}\,r^2\right)$
. . which simplifies to: .$\displaystyle \frac{27}{4}\,r^4 \;=\;\frac{27}{4}\,r^4$ . . . it's true!