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.
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Fabio Skilnik ::
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