Bunuel wrote:
Is the area of the triangle drawn in the circle (with center A) greater than \(\pi\) ?
(1) x = 36 degrees
(2) The radius of the circle is 1.
\({S_\Delta }\,\,\mathop > \limits^? \,\,\pi\)
(1) In the figure on the left presented below, all circles have center A and "same angle x" (shown in purple).
Please note that choosing convenient radii, we can create "circular sectors" with areas that are arbitrarily small, and arbitrarily large.
Consequence: the corresponding triangles (inside these circular sectors) can have areas that are also arbitrarily small, and arbitrarily large.
This is what we call a "GEOMETRIC BIFURCATION". This argument PROVES mathematically that (1) is insufficient!
(2) For any value of x (*) we have:
\(\left\{ \matrix{
\Delta \subset {\rm{circle}} \hfill \cr
\Delta \ne {\rm{circle}} \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{S_\Delta } < {S_{{\rm{circle}}}} = \pi {\left( 1 \right)^2}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle\)
(*) We may implicitly assume that x is positive and less than 180 degrees. (Think about that!)
The correct answer is (B), indeed.
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
_________________
Fabio Skilnik ::
GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here:
https://gmath.net