slingfox wrote:

What is the greatest possible area of a triangular region with one vertex at the center of a circle of 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.

_________________

Fabio Skilnik :: GMATH method creator (Math for the GMAT)

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