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 :: http://www.GMATH.net (Math for the GMAT)

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