1. ## Proof

Given a circle with radius t, 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. Urgent help please!

2. Hello, xoluosox!

An equilateral triangle has area: . $A \:=\:\frac{\sqrt{3}}{4}s^2$
. . where $s$ is a side of the triangle.

Given a circle with radius $r$, 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 $r.$

Its area is: . $A_{hex} \;=\;6 \times \frac{\sqrt{3}}{4}\,r^2 \;=\;\frac{3\sqrt{3}}{2}\,r^2$

Code:
* o *
*    /|\    *
*     / | \     *
*     /  |r \     *
/   |   \
*    /    |    \    *
*   /  r  *  r  \   *
*  /   *     *   \  *
/ *           * \
o- - - - - - - - -o
*               *
*           *
* * *
It can be shown that the side of this triangle is: . $s = \sqrt{3}\,r$

Then its area is: . $A_1 \;=\;\frac{\sqrt{3}}{4}\left(\sqrt{3}\,r\right)^2 \;=\;\frac{3\sqrt{3}}{4}r^2$

Code:
o
/|\
/ | \
/  |  \
/   |   \
/    |    \
/   * * *   \
/*     |     *\
*       |       *
*        |     o  *
/         |   o     \
/*         | o  r    *\
/ *         *         * \
/  *                   *  \
/                           \
/     *                 *     \
/       *               *       \
/          *           *          \
o - - - - - - - * * *- - - - - - - - o
It can be shown that the side of this triangle is: $2\sqrt{3}\,r$

Its area is: . $A_2 \;=\;\frac{\sqrt{3}}{4}\,(2\sqrt{3}\,r)^2 \:=\:3\sqrt{3}\,r^2$

If $A_{hex}$ is the mean proportional between $A_1$ and $A_2$

. . then: . $\left(A_{hex}\right)^2 \;=\;\left(A_1\right)\cdot\left(A_2\right)$

We have: . $\left(\frac{3\sqrt{3}}{2}\,r^2\right)^2 \;=\;\left(\frac{3\sqrt{3}}{4}\,r^2\right)\cdot\le ft(3\sqrt{3}\,r^2\right)$

. . which simplifies to: . $\frac{27}{4}\,r^4 \;=\;\frac{27}{4}\,r^4$ . . . it's true!