## Inscribed in a Circle

An equilateral triangle and regular hexagon are inscribed in the same circle.   Find the ratio of the area of the triangle to that of the hexagon.
Source: mathcontest.olemiss.edu 12/7/2009

SOLUTION

First, rotate the equilateral triangle around center $C$ of the circle so that vertex $A$ coincides with a vertex of the regular hexagon. Draw segment $\overline{CD}$.

Show that triangle $ACB$ is congruent to triangle $ADB$
Triangle $ACB$ is congruent to triangle $ADB$ by ASA (angle side angle):

$\angle 1=\angle 2=30^{o}$    (see why below)
$AB=AB$    reflexive property
$\angle 3=\angle 4=30^{o}$    (see why below)

Since point $C$ is equidistant from the endpoints $A \textup{ and } B$ of segment $\overline{AB}$, $C$ is on the the perpendicular bisector of $\overline{AB}$. Furthermore, we can prove that $\overline{CM}$ is also the angle bisector of $\angle C$ by proving that right triangle $AMC$ is congruent to right triangle $BMC$.
Since $\angle C=120^{o}$ (center angle intercepting the same arc as the inscribed angle of an equilateral triangle), half of it is equal to $60^{o}$. Thus, $\angle 1=30^{o}$ and $\angle 3=30^{o}$.
Similarly, we can prove that $\angle 2=30^{o}$ and $\angle 4=30^{o}$.

Unit area
Let the area of triangle $ACB$ be the unit area. Then, we can visually determine that the equilateral triangle is made up of 3 units and the regular hexagon is made up of 6 units.

Thus, the ratio of the area of the equilateral triangle to the area of the regular hexagon is

$3:6 \textup{ or } 1:2$.

Answer: $1:2$.