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

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by fskilnik@GMATH » Wed Jan 09, 2019 8:42 am
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?
$$A.\ \frac{ \sqrt{3}}{4} \,\,\,\,\,\,B.\ \frac{1}{2}\,\,\,\,\,\,C.\ \frac{\pi}{4}\,\,\,\,\,\,D.\ 1\,\,\,\,\,\,E.\ \sqrt{2}$$
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$$?\,\,\, = \,\,{S_{\Delta ABC}}\,\,\max $$

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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by GMATGuruNY » Wed Jan 09, 2019 8:49 am
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?
$$A.\ \frac{ \sqrt{3}}{4}$$
$$B.\ \frac{1}{2}$$
$$C.\ \frac{\pi}{4}$$
$$D.\ 1$$
$$E.\ \sqrt{2}$$
OA B
Given two sides of a triangle, the greatest possible area will be achieved if the two sides form a RIGHT ANGLE and serve as the triangle's base and height.
For the problem above, I posted an illustrated solution here:
https://www.beatthegmat.com/largest-pos ... 74809.html
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by Scott@TargetTestPrep » Mon Jan 21, 2019 5:59 pm
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?
$$A.\ \frac{ \sqrt{3}}{4}$$
$$B.\ \frac{1}{2}$$
$$C.\ \frac{\pi}{4}$$
$$D.\ 1$$
$$E.\ \sqrt{2}$$
OA B
The triangle with the largest possible area, given the above constraints, is an isosceles right triangle. Thus, since the base and height of the triangle is 1, the maximum area is 1/2 x 1 x 1 = 1/2.

Answer: B

Scott Woodbury-Stewart
Founder and CEO
scott@targettestprep.com

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