AAPL wrote:**GMAT Prep**

What is the greatest possible area of a triangular region with one vertex at the center of a circle with radius one and the other two vertices on the circle?

$$A.\ \frac{ \sqrt{3}}{4} \,\,\,\,\,\,B.\ \frac{1}{2}\,\,\,\,\,\,C.\ \frac{\pi}{4}\,\,\,\,\,\,D.\ 1\,\,\,\,\,\,E.\ \sqrt{2}$$

$$?\,\,\, = \,\,{S_{\Delta ABC}}\,\,\max $$

Let C be the center of the circle with unitary radius.

Without loss of generality, we may (and will) assume point A is one unit at the right of point C (as shown in the figure on the left).

For the last vertex (B), **without loss of generality** (in terms of exploring possible areas) there are only two possibilities:

(1) B is in the arc AD (figure in the middle) or (2) B is in the arc DE (figure on the right)

In BOTH cases we have:
$${S_{\Delta ABC}} = {{AC \cdot h} \over 2} = {h \over 2}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,? = {1 \over 2}\,\,\,\,\,\,\left( {{\rm{when}}\,\,h = CD\,,\,\,{\rm{i}}{\rm{.e}}{\rm{.}},\,\,B = D} \right)$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,

Fabio.