Triangle in a circle (vs triangle)

Geometry Level 3

Three points, says P , Q P, Q and R R are chosen uniformly at random from the circumference of a circle with centre O O . Let x x be the probability of triangle P Q R PQR contains the point O O . Find the value of 1000 x , \displaystyle\left\lfloor 1000x \right\rfloor, where \lfloor \cdot \rfloor denotes the floor function .


The answer is 250.

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3 solutions

Chan Lye Lee
Nov 26, 2018

Suppose P P and Q Q placed randomly. Join the points P P and O O and extend it to the point A A on the circumference. Do the similar to the segment Q O B QOB
If the triangle P Q R PQR contains the point O O , then the final point R R must be placed within the minor arc A B AB colored red (as shown). So the probability, x x is the length of minor arc A B AB divided by the circumference. Suppose the circumference is of length 1 unit. Then x x is the length of minor arc A B AB , which is the same as the length of minor arc P Q PQ .

Now fix the point P P and draw the segment P O A POA . By symmetry, the point Q Q is either above P A PA or below P A PA . Without loss of generality, let the point Q Q be above P A PA . As Q Q is chosen uniformly at random on the arc P A PA , averagely the point Q Q will fall on the center of the arc P A PA , which means that the length of minor arc P Q PQ is 1 4 \frac{1}{4} , and so the length of minor arc A B AB . This implies that x = 1 4 x=\frac{1}{4} and hence 1000 x = 250 \displaystyle\left\lfloor 1000x \right\rfloor =250 .

Jordan Cahn
Nov 27, 2018

A very similar problem was posts about one month ago . I'll copy the relevant portion of my solution below.


Given two points A , B A, B on a circle, the third point C C must lie in the red region below if A B C \triangle ABC is to contain the circle's center:

Now, we find the probability of this happening. Fix A A and B B on the circle, and let θ \theta be the angle between them ( A O B \angle AOB ). Then 0 < θ π 0<\theta\leq \pi . Note that the angle intercepting the red region in the diagram is precisely θ \theta . Thus the probability of a random point C C being placed in the red area is θ 2 π \frac{\theta}{2\pi} . We integrate to compute the average probability over all possible θ \theta : 1 π 0 0 π θ 2 π d θ = 1 π [ θ 2 4 π ] θ = 0 π = π 2 4 π 2 = 1 4 \frac{1}{\pi-0}\int_0^\pi \frac{\theta}{2\pi}\,\mathrm{d}\theta = \frac{1}{\pi}\left[\frac{\theta^2}{4\pi}\right]_{\theta=0}^\pi = \frac{\pi^2}{4\pi^2} = \frac{1}{4}

1000 × 1 4 = 250 = 250 \left\lfloor 1000\times\frac{1}{4} \right\rfloor = \left\lfloor 250 \right\rfloor = \boxed{250}

Great answer! This is much easier to understand than Parth Sankhe's answer.

Toby M - 2 years, 5 months ago
Parth Sankhe
Nov 26, 2018

Probability is simply the probability of an acute triangle from 3 random points on a circle.

= 1 π 0 0 π θ 2 π d θ = 0.25 =\frac {1}{π-0}\int ^{π}_{0}\frac {\theta }{2π}d\theta = 0.25

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