Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

A Triangle in a Circle


Date: 05/26/2000 at 20:45:27
From: Simon Gamache
Subject: A triangle in a circle

Hi Dr. Math,

This problem was given to me by my math teacher, and I think I got the 
solution, but I'm not sure. Could you confirm my reasoning?

The problem: Suppose you have a circle, and you randomly place two 
points on the circumference (points A and B). What is the probability 
that a third point called C placed randomly on the circumference of 
the circle will form a triangle that will contain the center of the 
circle?

I found the following solution:

                     n
      Probability = ---
                    360

where n: the arc formed by points A and B (in degrees)

Indeed, suppose points A and B form a 60-degree arc. Then point C must 
be placed on the 60-degree arc opposite to the arc formed by points A 
and B in order to contain the center of the circle. (A picture speaks 
1000 words.) And we all know that there are 360 degrees in a circle. 
Is there anything wrong with my solution?


Date: 05/27/2000 at 03:28:37
From: Doctor Mike
Subject: Re: A triangle in a circle

Hi Simon,

Essentially you have it right, but you need to be a little more 
careful in stating your solution. When you say "the arc formed...", it 
is clear to me from what you have written afterward that you really 
mean "the smaller arc formed..." Always state a result precisely.

Also, the special case where points A and B are diametrically opposed 
(at opposite ends of a diameter line) is an interesting special case. 
The probability in this case depends on exactly what you mean by 
"contains the center" above. That is, does a triangle contain the 
center if the center is ON one of the legs of the triangle? Or does 
the center have to be IN the interior of the triangle? Maybe you 
should ask the teacher what was meant.

I agree with what you have written above the "point C must be..." but 
I do not agree that you have clearly presented the reasons why this is 
true. You leave a bit too much to the imagination and creativity of 
the person you are talking/writing to.

I agree that a picture is worth 1000 words in this case, but a clear 
and complete proof of your result needs to have a few more words from 
you. It's fine to use an example like 60 degrees to give the reader of 
your logical argument some help in understanding what you are saying. 
You could even say to imagine this 60-degree arc on a clock face from 
five o'clock to seven o'clock, but your words must be organized and 
clear enough to convince another person that your formula or theorem 
is true in ALL cases, not just for the one example given. The ability 
to present a logical argument is something that takes practice, and it 
will improve with practice.

Here is what I would use. Draw a circle and locate 2 arbitrary points 
A and B on the circle. Draw a diameter line from A through the center 
to the diametrically opposite point A'. Do the same from B to B'. Now, 
I believe it truly is clear from the picture that the third point C 
must be located on the smaller arc between A' and B' for triangle ABC 
to contain the circle center.
     
Now let's look at what parts of the circle's circumference we have:

   (1) The smaller arc between A and B which you say is n degrees
   (2) The smaller arc between B and A'
   (3) The smaller arc between A and B'
   (4) The smaller arc between A' and B', which is where the C point 
       must lie for the triangle to contain the circle center.
  
How do can we be *100 percent sure* that that arc (4) is n degrees? I 
would point out that the arcs (1) and (2) together form precisely half 
a circle since they are on one side of a diameter line. So since (1) 
is n degrees, then (2) must be 180-n degrees. Similarly, arc (3) is 
180-n degrees as well. Adding these up, the first 3 arcs together have 
(180-n)+(n)+(180-n) = 360-n degrees, and from this it is clear that 
arc (4) is n degrees. Do you see how what I have said really forces 
the reader to understand that the opposite arc also has n degrees? 
(The reader does not have the responsibility to think up a reason for 
that fact.) Now, the ending of the proof proceeds as you have done it.

You did well to discover the correct formula for the probability and 
to write down your reasons. I just want you to know what a 
mathematician must do to really nail down the truth of a theorem 
beyond a shadow of a doubt. I hope this helps. Thanks for writing.

- Doctor Mike, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Euclidean Geometry
College Probability
High School Euclidean/Plane Geometry
High School Probability

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/