Eight points are equally spaced on a circle. If 3 of the 8 points are to be selected at random,

BTGModeratorVI wrote: ↑

Wed Apr 22, 2020 11:10 am

Eight points are equally spaced on a circle. If 3 of the 8 points are to be selected at random, what is the probability that a triangle having the 3 points chosen as vertices will be a right triangle?

A) 1/14
B) 1/7
C) 3/14
D) 3/7
E) 6/7

Answer: D
Source: Princeton Review

No. of triangles possible, taking any 3 points out of 8 points = 8C3 = (8.7.6)/(1.2.3) = 56

Make 8 equally spaced points on a circle. Note that for specific three points to make a rightangled triangle, two points must make a diameter; any of the points as the third point form a right-angled triangle.

Let's take the first point 1. Its corresponding diagonal is Point 1 – Point 5. So, the no. of choices for the 3rd point is 6 (Points 2, 3, 4, 6, 7, and 8). So, from Point 1, a total of 6 right-angled triangles can be formed.

However, the count of 6 is double as when we consider that from Point 5, the same no. of 6 right-angled triangles would be formed. Thus, we must take this into account. Actual no. of right-angled triangles from one diameter = 6/2 = 3.

Since there are 8 points, the no. of right-angled triangles = 8*3 = 24

The required probability = 24/56 = 3/7

The correct answer: D

Hope this helps!

-Jay
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BTGModeratorVI wrote: ↑

Wed Apr 22, 2020 11:10 am

Eight points are equally spaced on a circle. If 3 of the 8 points are to be selected at random, what is the probability that a triangle having the 3 points chosen as vertices will be a right triangle?

A) 1/14
B) 1/7
C) 3/14
D) 3/7
E) 6/7

Answer: D
Source: Princeton Review

There are 8C3 = (8 x 7 x 6)/(3 x 2) = 8 x 7 = 56 ways to select 3 points from 8.

Now, let’s let the 8 points be A, B, C, D, E, F, G and H. Since these eight points are equally spaced on the circle. We see that each pair of consecutive points are separated by 360/8 = 45 degrees.

For any 3 chosen points to be the vertices of a right triangle, 2 of the vertices must be the endpoints of a diameter of the circle since the hypotenuse of a right triangle inscribed in a circle is a diameter of the circle. Therefore, if A is chosen, E has to be chosen, too. In other words, A and E need to be both chosen since AE is a diameter of the circle. If A and E are both chosen as the vertices of the triangle, then the last vertex can be any one of the remaining 6 points since the triangle will always be a right triangle once A and E are chosen. The same analogy can be made for B and F, C and G, and E and H since BF, CG and EH are diameters of the circle also.

Since there are 4 diameters (AE, BF, CG and EF) and each diameter has 6 ways to form a right triangle, there are 4 x 6 = 24 right triangles. Therefore, the probability that a triangle having the 3 points chosen as vertices will be a right triangle is 24/56 = 3/7.

Answer: D