Problem Solving

BTGModeratorVI wrote: ↑

Fri Jul 03, 2020 7:21 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

Key property: An inscribed angle that contains (aka "holds") the DIAMETER will be a 90-degree angle.

For example, let's draw one of the diameters...
The inscribed angle that contains (aka "holds") the DIAMETER will be a 90-degree angle.
Likewise, this inscribed angle also contains (aka "holds") the DIAMETER, so it will also be a 90-degree angle. So, for the ONE PARTICULAR diameter (shown below)... ...we can see that, if we make any of the 6 points the 3rd vertex of the triangle, we will get a right triangle.

This means that, FOR EACH diameter in our circle, there are 6 points that will create a right triangle.

Since there are 4 diagonals altogether.... ....we know that the TOTAL number of right triangles possible = (4)(6) = 24

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Now we need to determine how many different triangles can be created by selecting 3 of the 8 points.
Since the order in which we select the 3 points does not matter, we can use COMBINATIONS.
We can select 3 points from 8 points in 8C3 ways

8C3 = (8)(7)(6)/(3)(2)(1) = 56

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So, P(we get a right triangle) = (total number of RIGHT triangles possible)/(total number of triangles possible)
= 24/56
= 3/7

Answer: D

Cheers,
Brent