An obtuse triangle is a triangle that has an interior angle

Aren't all angles supposed to be acute for an acute triangle? With one point outside the semicircle, the triangle is no longer acute... Therefore to find probability of forming an obtuse triangle should be 1-prob. of forming an acute triangle.

Did I miss something??

What do we know about the angles of a triangle that is inscribed in a circle?

Can I get back to you on that.. ..

Sorry, you reminded me of my high-school teacher.. Couldn't resist...


Triangles in a circle...

1. Diameter subtends a right angle....
2. Angle at a point on the circle is 1/2 the internal arc of the circle it encompasses....

Can't some up with more.. but looks like I am missing something here.. let me think more.

May be i dont understand the question....but if i do the answer is 0 ? ...If we choose 3 points and draw a triangle one of the line segments becomes a chord and the other vertex lies on the circle. So the angle subtended by the line segment in consideration subtends twice the angle in the center of the circle. So the angle subtended by the chord on the circumference of the circle can never be greater than 90 ...so probability is 0.

Is that right ?

What if we draw a triangle between 12, 1 and 2?

Now I got it. This question was haunting me for last 24 hours. Atlast I nailed it.


First of all. If all three points are in the semicirle then the triangle will be obtuse.

Ways of selecting first point = 12
Ways of selecting 2nd point = 10 (You can select any point except the first and its opposite, for example if 12 selected then you can select any point other than 12 and 6)

Now comes the tricky part.
Average ways of selecting 3rd point is dependent on the second selection = 2 * (8+7+6+5+4) = 60/10 = 6
But there will be 5 more repeats of each combination of 3 points.
Total ways of selecting 3 points = 12C3 = 220

Prob = 12 * 10 * 6/220*6 = 6/11

We will take an example.
Let first selection is 12
1. Second selection is 1:
Third could be anything except these: opposite of 12 and 1 (i.e 6 and 7) and anything between them 6 and 7 (Nothing here) = 2,3,4,5,11,10,9,8 = 8 ways
2. Second selection is 2:
Third could be anything except these: opposite of 12 and 2 (i.e 6 and 8) and anything between 6 and 8 (i.e 7) = 1,3,4,5,11,10,9 = 7 ways
3. Second selection is 3:
Third could be anything except these: opposite of 12 and 3 (i.e 6 and 9) and anything between 6 and 9 (i.e 6,7,8,9) = 1,2,3,5,11,10 = 6 ways
4. Second selection is 4:
Third could be anything except these: opposite of 12 and 4 (i.e 6 and 10) and anything between 6 and 10 (i.e 6,7,8,9,10) = 1,2,3,5,11 = 5 ways
5. Second selection is 5:
Third could be anything except these: opposite of 12 and 11 (i.e 6 and 11) and anything between 6 and 11 (i.e 6,7,8,9,10,11) = 1,2,3,4 = 4 ways

Same way on the opposite side i.e when second selection is 11,10,9,8,7

Total average of selecting second and third = 2*(8+7+6+5+4) = 60
Total obtuse triangles = 12*60 = 720
But there are duplicates:
(12,1,2) there are 3! of duplicates of this.

So final prob = 12*60/(220*6) = 6/11

Kevin, your questions, especially this one, are too difficult to solve in 2 minutes.

I'm sorry. I hope that isn't the case in general. This one can be solved more easily, though. Imagine that the first number is chosen. Rotate clock so that number is at the 12 o'clock position for the sake of simplicity.

How can we form an obtuse triangle? The minimum distance between any two numbers must be at least 7. What is the probability that this occurs?

The two other points can be chosen in 11C2=55 ways.

Case I (other two points from 1 to 5) 5C2=10 choices

Case II (other two points from 7 to 11) 5C2=10 choices

Case III 11 and one of {1,2,3,4} or
.............. 10 and one of {1,2,3} or
..............9 and one of {1,2} or
..............8 and 1......................................10 choices

So probability that triangle is obtuse is 30/55=6/11