Three points are randomly chosen on the circumference of a g

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

by fskilnik@GMATH » Mon Feb 18, 2019 12:50 pm

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GMATH practice exercise (Quant Class 19)

Three points are randomly chosen on the circumference of a given circle. What is the probability that the center of the circle lies inside the triangle whose vertices are at the three points?

(A) 1/3
(B) 1/4
(C) 1/5
(D) 2/5
(E) 2/7

Hint: the problem was created based on another given in the link hidden below:
[spoiler]https://www.beatthegmat.com/three-point ... tml#826897[/spoiler]

Answer: [spoiler]____(B)__[/spoiler]

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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Source: Beat The GMAT — Problem Solving | Mon Feb 18, 2019 12:50 pm

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

by fskilnik@GMATH » Tue Feb 19, 2019 11:21 am

fskilnik@GMATH wrote:GMATH practice exercise (Quant Class 19)

Three points are randomly chosen on the circumference of a given circle. What is the probability that the center of the circle lies inside the triangle whose vertices are at the three points?

(A) 1/3
(B) 1/4
(C) 1/5
(D) 2/5
(E) 2/7

Hint: the problem was created based on another ...

We will refer to the link given in the hint: https://www.beatthegmat.com/three-point ... tml#826897

FOCUS: P(center inside the triangle)

We may consider (without loss of generality) that the first point (blue) was chosen at the position shown in the first figure and that the second point (red) was chosen (without loss of generality) in the upper semicircle (red arc) also presented in the first figure.

In the second figure, we show one possibility of the first two points, to illustrate the fact that we will have a favorable scenario if and only if the third point (green) is chosen in the arc in green, obtained by two lines, each one defined by the center and one of the two first points.

Considering exactly the same reasoning presented in the exercise mentioned in the hint, letÂ´s consider the two extremal cases: when the first two points ("almost") coincide and when the first two points are ("almost") diametrically opposites.

The corresponding favorable probabilities are 0 and 1/2, respectively, hence (following the same rationale presented there) our FOCUS is their average: (0+1/2)/2 = 1/4.

The correct answer is (B).

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

Regards,
Fabio.

POST-MORTEM: it is possible to prove that the probabilities of having the first and second points coincidental or diametrically opposites are both ZERO (not approximately zero), therefore there is no trouble omitting the words "almost" presented in parentheses.
A subtle consequence: the answer obtained is not approximately right, it is exactly right. (Details are out-of-GMATÂ´s-scope.)

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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