BTGmoderatorDC wrote: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

OA D

Source: Princeton Review

One of the important properties of the circle: "A diameter subtends 90º angle at the circumference of the circle."

Draw a circle, number equally space 8 points on it, join those two points that are part of a diameter. They will be (1, 5); (2, 6); (3, 7); 4, 8). There will be 4 such diameters.

The diameter formed out of points 1 and 5 forms 6 right-angled triangles: ∆125; ∆135; ∆145; ∆155; ∆175; and ∆185. Thus, the total number of right-angles formed out of the 4 distinct diagonals = 4*6 = 24 right-angled triangles

Total number of all the possible triangles = 8C3 = 8.7.6 / 1.2.3 = 56

Thus, the probability that a triangle having the 3 points chosen as vertices will be a right triangle = 24/56 = 3/7

The correct answer: C

Hope this helps!

-Jay
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BTGmoderatorDC wrote: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

TOUGH question!!!

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

---------------------------------------
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

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

Answer: D