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Three points are chosen independently an at random on the ci

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by BTGmoderatorDC » Thu Feb 14, 2019 3:45 am

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Difficulty—

Three points are chosen independently an at random on the circumference of a circle with radius r. What is the approximate probability that none of the three points lies more than a straight-line distance of r away from any other of the three points?

(A) 1/9
(B) 1/12
(C) 1/18
(D) 1/24
(E) 1/27

OA B

Source: Manhattan Prep
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Three points are chosen independently an at random on the ci

by fskilnik@GMATH » Thu Feb 14, 2019 4:37 pm
BTGmoderatorDC wrote:Three points are chosen independently an at random on the circumference of a circle with radius r. What is the approximate probability that none of the three points lies more than a straight-line distance of r away from any other of the three points?

(A) 1/9
(B) 1/12
(C) 1/18
(D) 1/24
(E) 1/27
Source: Manhattan Prep
Image

Very nice problem!

FOCUS : the probability of having all three points chosen (say blue, red and green, in that order) "sufficiently close".

1. The first point (blue) may be considered (without loss of generality) placed at the top of the circle (first figure).

2. The second point (red) must be placed in the arc of the circle shown in red, somewhere between the two nearer vertices of the top vertex of the regular hexagon (first figure).

Reason: a regular hexagon is composed of 6 equilateral triangles, in this case each side with length r (the radius of the circle that circumscribes the hexagon).

Conclusion: there is a 120/360 = 1/3 probability of choosing the second point favorably.

3. What about the third (green) point? Let´s start considering the two extremal scenarios:

> Extremal Case 1 (where the blue and red points coincide): in this case, we have again 1/3 of probability of choosing the third point in a place such that we have all three points "sufficiently close".

> Extremal Case 2 (where the blue and red points are most apart): in this case, we have 60/360 = 1/6 of probability of choosing the third point in a place such that we have all three points "sufficiently close".

> What about an "Intermediate Case"? If we move the red point away from the blue point, from the Extremal Case 1 until the Extremal Case 2, we have ALL (the infinite continuum possibilities) probabilities from 1/3 to 1/6 possible and, by symmetry, equiprobable!

Conclusion: we must agree that the probability of choosing the third point favorably must be the AVERAGE of 1/3 and 1/6, that is, (1/3 + 1/6)/2 = 1/4.


Finally, the answer we are looking for:

? = P(choosing the second point wisely) * P(choosing the third point wisely given that the second one was chosen wisely) = 1/3 * 1/4 = 1/12.


The correct answer is (B).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
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