What is the greatest possible area of a triangular region

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AAPL: Moderator; Posts: 2749; Joined: Sun Oct 29, 2017 2:08 pmFollowed by: 2 members

by AAPL » Tue Jan 08, 2019 4:14 am

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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

Source: Beat The GMAT — Problem Solving | Tue Jan 08, 2019 4:14 am

fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

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

$$?\,\,\, = \,\,{S_{\Delta ABC}}\,\,\max $$

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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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

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.
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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

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