# How many right angled triangles can a circle have?

Here's what I recall of the question from CNML Grade 11, 2010/2011 Contest #3, Question 7:

There are 2010 points on a circle, evenly spaced. Ford Prefect will* randomly choose three points on the circle. He will* connect these points to form a shape. What is the probability that the resulting shape will* form a right angled triangle?

I answered $\frac{1}{4} = 25\%$, but that's probably incorrect. (Right?)

When I got home, I thought it out in my head, and I got this:

$\frac{2010 * (2010/1005)}{2010 \choose 3}$

$\frac{2020050}{1351414120} = \frac{3015}{2017036} = 0.149476756984010201\%$

I'm probably wrong ...again. Can anyone tell how to get the right answer (if I'm not wrong :) )?

*in the past of the future of the perfect present present time double into ripple fluctuater byer doininger of the past future continuum...

EDIT: Realized my mistake in copying the question.

• it seems to me that probability is infinitely close to 0, as there are uncountably many points on the circle, though I am probably wrong, too. :) – InterestedGuest Dec 15 '10 at 6:16
• @TonyK: but does the fact that we can actually construct a right triangle in a circle contradict that? Or is it one of those thing where the probability is 0, but it's possible? – InterestedGuest Dec 15 '10 at 7:08
• @muntoo: I am not sure what you mean. The only way to inscribe a triangle in a circle is by having two vertices diametrically opposite. – Timothy Wagner Dec 15 '10 at 7:17
• @muntoo: See this en.wikipedia.org/wiki/Thales'_theorem – Timothy Wagner Dec 15 '10 at 7:30
• But muntoo...If you're using the past present pluperfect continuous tense, WTF is Ford Prefect doing in the year 2010? WTF is he doing here anyway? – TonyK Dec 15 '10 at 21:40

Let the points be $P_1$, $P_2$, and $P_3$. Letting $P_1$ be arbitrary, we win if $P_2$ is the point diametrically opposite to $P_1$, probability $1/2009$. Otherwise, with probability $2008/2009$, we still win if $P_3$ is diametrically opposite either $P_1$ or $P_2$, which it is with probability $2/2008$. So the probability of a win is $1/2009 + (2008/2009)(2/2008) = 3/2009$.

You can also get this as follows: The number of possible wins is the number of diameters times the number of remaining points, or 1005*2008. The number of possible triples is $2010\choose 3$. Dividing the first by the second gives $3/2009$.

Even easier: The probability that a given pair of points lie on a diameter is $1/2009$. With three points, you have three pairs, hence $3/2009$. Here it's OK to add probabilities because it's not possible to have overlap; if one of the pairs lies on a diameter, no other pair can lie on a diameter.

BTW, I do not agree with the comment thread that says we must consider the possibility that the points are not distinct. Sometimes there can be an ambiguity, but in common language "choose three points from 2010 points" means "without replacement". For example, I am not incorrect in saying that $n\choose r$ is the number of ways of choosing $r$ objects out of a set of $n$. Although I admit that if I were grading the test and this issue were pointed out to me, I might accept the "with replacement" answer as well.

The only other alternative answer I might accept is 1, i.e., that the points are certain to lie on a right triangle. After all, it is Ford Prefect, so I have no way of knowing whether the Infinite Improbability Drive has been activated.

• "I have no way of knowing whether the Infinite Improbability Drive has been activated." - I've already given this answer a +1 before you wrote that, and it's a shame I can't give another... – J. M. is a poor mathematician Dec 15 '10 at 12:25
• To be exact, Ford was trying to come up with the probability that Arthur's cup of not entirely unlike tea will entirely become tea when he would know whether the Improbability Drive would have been activated or not. – Mateen Ulhaq Dec 16 '10 at 0:28
• I'm pretty sure that the problem meant "choose different points", but it wasn't "clearly" said. – Mateen Ulhaq Dec 16 '10 at 0:31

Hint: You can choose $3$ points in ${2010}\choose {3}$ ways. To form a right angled triangle, we need to choose a pair of diametrically opposite points (how many ways to do this?) and then any third point will form a right angled triangle with the chosen pair.

• You have to be a bit careful: firstly, the problem didn't specify that the random points have to be distinct, so there are $2010^3$ ways of choosing those. Secondly, that doesn't quite give you the number of ways of choosing the points. – Alex B. Dec 15 '10 at 8:02
• @Alex: I agree with your first point, I assumed $3$ distinct points were chosen. I am not sure about your second: you can choose a pair of diametrically opposite points in 1050 ways and then choose any other point in 2008 ways. So you can form a right angled triangle with 3 distinct points in $2008\times 1050$ ways. I suppose I interpreted this question to mean how many right angles triangle can be formed using any three distinct of these $2010$ points without regards to the order in which they were chosen. – Timothy Wagner Dec 15 '10 at 8:09
• You calculate the number of triangles correctly, but not the number of ways of choosing a triangle. As you say, the order in which the points were chosen doesn't matter. There are 6 possible orders in which the points of a triangle can be chosen, so you have to multiply by those, if you want an honest probability. – Alex B. Dec 15 '10 at 8:14
• @Alex: Perhaps I am having a brain freeze here, but if you account for the order in which the points were chosen, wouldn't you multiply both the numerator and denominator by $6$. So the probability would be the same under either assumption. – Timothy Wagner Dec 15 '10 at 8:19
• Well, maybe it's me who has a brain freeze. We first count the number of all possible choices. That is $2010^3$, agreed? Now, how many of these choices produce a right-angled triangle? Suppose ABC is a right angled triangle. Then there are several routes to choosing it, by picking the points in different orders. The way we have counted all the choices, the choice (A,B,C) is different from (B,A,C), but they both produce a triangle of the desired shape. Ergo... As I say, it might well be me, who has a brain freeze. – Alex B. Dec 15 '10 at 8:44

Simplify it. There are 2010 points, but if you start with four points instead and pick 3 at random, you see the probability of getting a right triangle is 1 (draw it out, it helps). You can do it with six points too, and if you look at the probability for both figures you can create a formula to give the probability regardless of the number of points. 3/(n-1), where n is the number of points (I think it has to be even for this to work). With four points, 3/(4-1) = 1, six points 3/(6-1) = 3/5. If you have 2010 points then the probability would be 3/(2010-1) = 3/2009.