Hello, spred!
An equilateral triangle is inscribed in a circle,
and the circle is inscribed in another equilateral triangle.
Find the ratio of the area of the smaller triangle to that of the larger triangle. Code:
A
D ∆ - - - - - - - * ∆ * - - - - - - - ∆ E
\ * / \ * /
\ * / \ * /
\ * / \ * /
\ / \ /
\ * / \ * /
\ * / ∆ O \ * /
\* / | \ */
\ / | \ /
C ∆ - - - * - - - ∆ B
* D /
\* */
\ * * * /
\ /
\ /
\ /
\ /
\ /
∆
F
The center of the circle is $\displaystyle O.$
The inscribed triangle is $\displaystyle ABC.$
The circumscribed triangle is $\displaystyle D{E}F.$
$\displaystyle O$ is the centroid of $\displaystyle \Delta ABC$ and $\displaystyle \Delta D{E}F.$
Draw segment $\displaystyle OB.$
In right triangle $\displaystyle ODB,\,\text{ let }r = OD.$
. . Then the altitude $\displaystyle AD$ of $\displaystyle \Delta ABC$ is $\displaystyle 3r.$
Since $\displaystyle \angle ODB = 30^o\!:\;\;OB = 2r$
. . Then the altitude $\displaystyle DB$ of $\displaystyle \Delta D{E}F$ is $\displaystyle 6r$
The ratio of the altitudes is: .$\displaystyle 3r: 6r \:=\: 1: 2$
Therefore, the ratio of the area is: .$\displaystyle 1^2:2^2 \;=\;1:4$