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An obtuse triangle is a triangle that has an interior angle

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An obtuse triangle is a triangle that has an interior angle [#permalink] New post 26 Jul 2006, 12:26
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An obtuse triangle is a triangle that has an interior angle that is greater than 90 degrees. Three integers from 1 to 12 are chosen at random and their corresponding hour marks on the attached clock are then joined to one another with straight lines(s). If a triangle is formed, what is the probability that this triangle is obtuse?

(A) 4/11 (B) 5/12 (C) 9/20 (D) 1/2 (E) 6/11
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Last edited by kevincan on 27 Jul 2006, 02:29, edited 1 time in total.
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 [#permalink] New post 26 Jul 2006, 15:14
Given a circle, for an angle to be obtuse atleast one of the points should lie outside the semicircle.

To look at the problem in the opposite way:

what is the probablity of picking points 3 such that all angles formed are acute?

P(Obtuse Angle) = 1 - P(Acute)

Prob of Acute = # of ways of picking 3 points within a semicircle/# of ways of picking 3 points out of 12

Note that we can form 4 semicircles from a circle :
{12, 1,2,3,4,5,6}, {3,4,5,6,7,8,9}, {6,7,8,9, 10,11,12} {9, 10,11,12,1,2}

# of ways of picking 3 points of 7 in each semicircle = 7x6x5/3x2 = 35

As there are 4 semicircles total = 35x4

# of ways of picking 3 points of 12 = 12x11x10/3x2 = 220

Prob(Acute) = 35x4/220 = 7/11

Therefore, P(Obtuse) = 1-7/11 = 4/11

Answer: A. 4/11
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 [#permalink] New post 26 Jul 2006, 15:22
Given a circle, for an angle to be obtuse atleast one of the points should lie outside the semicircle. Wouldn't this guarantee that the traingle is acute?
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 [#permalink] New post 26 Jul 2006, 15:25
kevincan wrote:
Given a circle, for an angle to be obtuse atleast one of the points should lie outside the semicircle. Wouldn't this guarantee that the traingle is acute?


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?? :?:
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 [#permalink] New post 26 Jul 2006, 15:39
What do we know about the angles of a triangle that is inscribed in a circle?
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 [#permalink] New post 26 Jul 2006, 15:46
Can I get back to you on that.. :P..

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.

kevincan wrote:
What do we know about the angles of a triangle that is inscribed in a circle?
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Re: PS: Obtuse Triangles Around the Clock [#permalink] New post 26 Jul 2006, 18:22
kevincan wrote:
An obtuse triangle is a triangle that has an interior angle that is greater than 90 degrees. Three integers from 1 to 12 are chosen at random and thir corresponding hour marks on the attached clock are then joined to one another with straight lines(s). If a triangle is formed, what is the probability that this triangle is obtuse?

(A) 4/11 (B) 5/12 (C) 9/20 (D) 1/2 (E) 6/11




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 ?
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 [#permalink] New post 26 Jul 2006, 22:10
What if we draw a triangle between 12, 1 and 2?
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 [#permalink] New post 27 Jul 2006, 08:26
Now I got it. This question was haunting me for last 24 hours. Atlast I nailed it.
:2gunfire: :2gunfire: :rocket :rocket

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.
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 [#permalink] New post 27 Jul 2006, 13:24
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

Last edited by kevincan on 27 Jul 2006, 13:31, edited 2 times in total.
  [#permalink] 27 Jul 2006, 13:24
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