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

[BTGModeratorVI]

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

[Jay@ManhattanReview]

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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[swerve]

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

The total number of triangles will be \(8c3 = 8 \cdot 7\).

The total number of triangles with right angles would be where the diameter of the circle is one side of the triangle. So, 24 such triangles are possible.

Probability is \(\dfrac{3 \cdot 8}{8\cdot 7}=\dfrac{3}{7}\)

Regards!

[Scott@TargetTestPrep]

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