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
