bonanova Posted March 16, 2014 Report Share Posted March 16, 2014 While our collective brain trust ponders the nature of triangles defined by three uniformly random points chosen inside a circle, specifically the mode of their areas, we ask another question. Recalling that the mean and median areal coverage of its circle's area are about 7.4% and 5.4%, what is the probability that a random triangle covers its circle's center? 1 1 Quote Link to comment Share on other sites More sharing options...

0 BMAD Posted March 30, 2014 Report Share Posted March 30, 2014 (edited) 25%? Edited March 30, 2014 by rookie1ja Quote Link to comment Share on other sites More sharing options...

0 Rob_G Posted March 24, 2014 Report Share Posted March 24, 2014 Given 2 random points on the circle (x_{1},y_{1}) and (x_{2},y_{2}) if the third point is on the arc between (-x_{1},-y_{1}) and (-x_{2},-y_{2}) then the triangle contains the center. So to simplify the problem I set the radius to 1 and set (x_{1},y_{1}) to (1,0). Looking at the top half of the circle (x_{2},y_{2}) will be of the form (x,sqrt(1-x^{2})). The angle between them is cos^{-1}(x)so the probablility that the triangle will contain the center is cos^{-1}(x)/360 (180?) or cos^{-1}(x)/2Pi (Pi?). This is where I've stopped and I'm not sure how to continue or if I'm on the right track. Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted March 24, 2014 Author Report Share Posted March 24, 2014 Given 2 random points on the circle (x_{1},y_{1}) and (x_{2},y_{2}) if the third point is on the arc between (-x_{1},-y_{1}) and (-x_{2},-y_{2}) then the triangle contains the center. So to simplify the problem I set the radius to 1 and set (x_{1},y_{1}) to (1,0). Looking at the top half of the circle (x_{2},y_{2}) will be of the form (x,sqrt(1-x^{2})). The angle between them is cos^{-1}(x)so the probablility that the triangle will contain the center is cos^{-1}(x)/360 (180?) or cos^{-1}(x)/2Pi (Pi?). This is where I've stopped and I'm not sure how to continue or if I'm on the right track. Sounds like an OK approach. You would need to do some (messy) integrals to account for the points being randomly chosen. I like to think of this as more of a logic puzzle than a math exercise. Is there a simpler way to think about the problem? Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted March 27, 2014 Author Report Share Posted March 27, 2014 What if the circle were divided into quadrants? Quote Link to comment Share on other sites More sharing options...

0 BMAD Posted March 28, 2014 Report Share Posted March 28, 2014 (edited) It would seem that if we fix one point randomly on the circle then as long as the angle (N) at that point towards the other two is 90>=N>0 then the triangle would cover the center. Assuming that an 180 degree angle is possible in a triangle, wouldn't that make it theoretically covered 50% of the time? I never get these right but ehh, why not try. Edited March 28, 2014 by BMAD Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted March 28, 2014 Author Report Share Posted March 28, 2014 Place the first point in one of the circle's quadrants. Investigate the sixteen cases that follow. Quote Link to comment Share on other sites More sharing options...

0 Rob_G Posted March 28, 2014 Report Share Posted March 28, 2014 (edited) About 35.3% or 20.55% Edited March 28, 2014 by Rob_G Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted March 30, 2014 Author Report Share Posted March 30, 2014 About 35.3% or 20.55% It's in that range. Can you nail it down? See my post #6. Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted March 31, 2014 Author Report Share Posted March 31, 2014 This answer is seen to apply for triangles covering the center of any containing shape that has four-fold rotational symmetry, including the circle, such as square and octagon. Does to hold for triangles contained by other shapes, say a pentagon? Quote Link to comment Share on other sites More sharing options...

0 BMAD Posted March 31, 2014 Report Share Posted March 31, 2014 are we assuming a regular pentagon? Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted April 1, 2014 Author Report Share Posted April 1, 2014 Yes. Basically I'm wondering whether the number of sides of the containing (regular) polygon matters. I'm working on it, don't have the answer yet. Quote Link to comment Share on other sites More sharing options...

0 BMAD Posted April 1, 2014 Report Share Posted April 1, 2014 it would seem that the answer is that it matters since a pentagon can't easily be divided into four identical pieces. Quote Link to comment Share on other sites More sharing options...

0 plainglazed Posted April 1, 2014 Report Share Posted April 1, 2014 i misunderstood this problem at first and thought the three points were on the perimeter of the circle. with this misunderstanding in mind: thought as a second point approached being directly opposite the first, the probability of a random triangle including the center neared 50%; and as the second point approached the first point, the probability of the center being included neared zero. Since the points were uniformly random, the average distance between them would be 1/4 of the circles circumference and for the center to be contained in the triangle generated by the third point, that point needed to be on the arc directly opposite that generated by the first two points - a 25% probability. When i reread the problem (correctly) it seemed the 25% might still hold as the points were uniformly random but my maths could not prove that. It made sense visually to me but cant really state why. The center would be included in a triangle where the third point lies within the sector directly opposite the sector defined by the first two points and the circles center. okay, i give up. So onto the polygons, seems the probability would be 25% if the opposite quadrants have 1/4 the perimeter, 1/4 the area, and are bilaterally symmetrical? EDIT: not sure on that opposite quadrants being bilaterally symmetrical bit or if all four quadrants do need to be equal Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted April 1, 2014 Author Report Share Posted April 1, 2014 i misunderstood this problem at first and thought the three points were on the perimeter of the circle. with this misunderstanding in mind: thought as a second point approached being directly opposite the first, the probability of a random triangle including the center neared 50%; and as the second point approached the first point, the probability of the center being included neared zero. Since the points were uniformly random, the average distance between them would be 1/4 of the circles circumference and for the center to be contained in the triangle generated by the third point, that point needed to be on the arc directly opposite that generated by the first two points - a 25% probability. When i reread the problem (correctly) it seemed the 25% might still hold as the points were uniformly random but my maths could not prove that. It made sense visually to me but cant really state why. The center would be included in a triangle where the third point lies within the sector directly opposite the sector defined by the first two points and the circles center. okay, i give up. So onto the polygons, seems the probability would be 25% if the opposite quadrants have 1/4 the perimeter, 1/4 the area, and are bilaterally symmetrical? EDIT: not sure on that opposite quadrants being bilaterally symmetrical bit or if all four quadrants do need to be equal I think for triangles with vertices either inside or on a circle the quadrant analysis applies and the result is the same. The only insight needed is to see that a line connecting points in opposite quadrants can be on either side of center with equal likelihood. I'm going to run simulations for other polygons (an equilateral-triangle containing shape seems interesting for starters) to see whether the result holds in every case. I started by dividing it into sixths and inspecting the pairs of opposing sixths. It certainly holds for a square container. Quote Link to comment Share on other sites More sharing options...

0 BMAD Posted April 1, 2014 Report Share Posted April 1, 2014 I think polygons are different because they are not equal distance to the center. Which is important in my approach and really matters in the forming of possible triangles across the center Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted April 1, 2014 Author Report Share Posted April 1, 2014 Possibly true so long as they do not have 4-fold rotational symmetry. I should have the triangle container simulated soon. Quote Link to comment Share on other sites More sharing options...

0 BMAD Posted April 1, 2014 Report Share Posted April 1, 2014 Probably only shapes that can be properly circumscribed Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted April 2, 2014 Author Report Share Posted April 2, 2014 The three altitude lines (light blue lines in the figure) divide an equilateral triangle into six triangles that are congruent if reflections are permitted. Label the triangles a-f. Construct a triangle that may cover the center (black dot) by placing a point in triangle a. From that point, lines can be drawn to points placed in each of the other triangles. Four of those lines pass with certainty to one side or the other of the center. (Black lines.) The lines connecting a with d, b with e and c with f, however, (red lines) pass by the center as shown with unknown probability p < 1. Consider the red line between b and e. It passes below the center, but if the points were moved up, a b-e line could pass above the center. It is these three red lines that are uncertain. In the case of a circle (or a square) quadrant symmetry dictates equal probabilities for a diagonal line passing the center point to one side or the other, so that p = 0.5. In the triangle case, it can easily be seen that p(b-e passing below center) is significantly greater than 0.5. Treating p as unknown for now, the 36 cases for placing points 2 and 3 (given the first point is in a) lead to the following probability table for creating a triangle that covers the center. Letting q = 1-p,point 3|V a b c d e f <= point 2a 0 0 0 pq 0 0b 0 0 0 p p 0c 0 0 0 p 1 pd pq p p pq q qe 0 p 1 q 0 0f 0 0 p q 0 0If we add the elements (they represent equally likely events) and divide by 36 we obtain.p(cover) = ( 3pq + 8p + 4q + 2 ) / 36This evaluates as follows. p p(cover) ----+------- 0.0 | 0.1667 0.1 | 0.1853 0.2 | 0.2022 0.3 | 0.2175 0.4 | 0.2311 0.5 | 0.2431 0.6 | 0.2533 0.7 | 0.2619 0.8 | 0.2689 0.9 | 0.2742 1.0 | 0.2778One million random pairs of points determined p = 0.69525... so that p(cover) = 0.2616...This is close to 0.25 as one would expect, since most points come from the central portion of the enclosing triangle. The difference between the triangle and a circle is the vertex areas, where increasingly fewer points lie, but points from which are more likely than central points to create covering triangles. Hence the covering probability is slightly greater than 0.25. From this reasoning a containing hexagon would admit a covering probability in excess of 0.25, but closer to it.So the shape of the enclosing figure determines the center-covering probability of a random triangle. Probably an enclosing triangle would have the greatest difference from 0.25 of any enclosing regular polygon. As the number of sides increases the polygon more closely approximates a circle. Quote Link to comment Share on other sites More sharing options...

0 plasmid Posted April 3, 2014 Report Share Posted April 3, 2014 (edited) A general case could in principle be solvable with an integral, but in practice it would be so messy that simulation would still be more feasible.The key is Rob_G's observation that if you're given any two points A and B, then point C will form a triangle ABC including the origin if and only if it falls anywhere in the cone formed by the origin with two rays going outward opposite of A and B. You can consider a polygon in polar coordinates -- place its center at (0, 0) and each of its vertices at a distance 1 from the center, so the vertices of an equilateral triangle would be at (1, 0), (1, 2pi/3), (1, 4pi/3). If you choose three points A, B, and C, they will enclose the origin if the angle of (C minus pi) lies between the angle of A and the angle of B (where "between" the angles of A and B refers to the shortest distance between them, ie if A is (x, pi/9) and B is (y, 12pi/13) then [C minus pi] would have to be greater than 12pi/13 or less than pi/9).You can thereby define the functionhasorigin(A, B, C)to be 1 if C minus pi is between A and B, and 0 otherwise.The probability that a point will fall at an angle T (for Theta) is proportional to the square of the distance from the center to the edge at angle T. For the triangle, we know that the distance to the edge equals 1 at the angles {0, 2pi/3, 4pi/3} (recall that's where I said the vertices are), equals 0.5 at the angles {pi/3, 3pi/3, and 5pi/3} (halfway between the vertices), and in general the distance to the edge is given byedgedist(T) = 0.5 / cos(angle away from nearest angle with length 0.5)And the probability that a randomly chosen point is at angle T isprob(T) = edgedist(T)^{2}Now you just have to take a triple integralInt[A=0 to 2pi] (prob(A) * Int (prob(B) * Int[C=0 to 2pi] (prob(C.) * hasorigin(A, B, C))))Which is easier said than done.Edit: It's actually more complicated than that hideous monstrosity... I left out that you would need to normalize the probabilities to sum to 1 instead of using raw squares of distances from the center. Edited April 3, 2014 by plasmid Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted April 4, 2014 Author Report Share Posted April 4, 2014 Edit: It's actually more complicated than that hideous monstrosity... I left out that you would need to normalize the probabilities to sum to 1 instead of using raw squares of distances from the center. You'd have the integrals you describe in the numerator of a fraction, divided by the same integrals without the HasOrigin part in the denominator. I think that would do the normalization. Ugh, agreed. 1 Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted April 6, 2014 Author Report Share Posted April 6, 2014 The three altitude lines (light blue lines in the figure) divide an equilateral triangle into six triangles that are congruent if reflections are permitted. Label the triangles a-f. Construct a triangle that may cover the center (black dot) by placing a point in triangle a. From that point, lines can be drawn to points placed in each of the other triangles. Four of those lines pass with certainty to one side or the other of the center. (Black lines.) The lines connecting a with d, b with e and c with f, however, (red lines) pass by the center as shown with unknown probability p < 1. Consider the red line between b and e. It passes below the center, but if the points were moved up, a b-e line could pass above the center. It is these three red lines that are uncertain. triangle in triangle.gifIn the case of a circle (or a square) quadrant symmetry dictates equal probabilities for a diagonal line passing the center point to one side or the other, so that p = 0.5. In the triangle case, it can easily be seen that p(b-e passing below center) is significantly greater than 0.5. Treating p as unknown for now, the 36 cases for placing points 2 and 3 (given the first point is in a) lead to the following probability table for creating a triangle that covers the center. Letting q = 1-p, point 3|V a b c d e f <= point 2a 0 0 0 pq 0 0b 0 0 0 p p 0c 0 0 0 p 1 pd pq p p pq q qe 0 p 1 q 0 0f 0 0 p q 0 0 If we add the elements (they represent equally likely events) and divide by 36 we obtain. p(cover) = ( 3pq + 8p + 4q + 2 ) / 36 This evaluates as follows. p p(cover) ----+------- 0.0 | 0.1667 0.1 | 0.1853 0.2 | 0.2022 0.3 | 0.2175 0.4 | 0.2311 0.5 | 0.2431 0.6 | 0.2533 0.7 | 0.2619 0.8 | 0.2689 0.9 | 0.2742 1.0 | 0.2778 One million random pairs of points determined p = 0.69525... so that p(cover) = 0.2616... This is close to 0.25 as one would expect, since most points come from the central portion of the enclosing triangle. The difference between the triangle and a circle is the vertex areas, where increasingly fewer points lie, but points from which are more likely than central points to create covering triangles. Hence the covering probability is slightly greater than 0.25. From this reasoning a containing hexagon would admit a covering probability in excess of 0.25, but closer to it. So the shape of the enclosing figure determines the center-covering probability of a random triangle. Probably an enclosing triangle would have the greatest difference from 0.25 of any enclosing regular polygon. As the number of sides increases the polygon more closely approximates a circle. Correcting an error in the simulation program, The probability that a random triangle inside an equilateral triangle covers its centroid is 0.24543 ... The average size of a random triangle inside any triangle is 1/12. Quote Link to comment Share on other sites More sharing options...

0 BMAD Posted April 7, 2014 Report Share Posted April 7, 2014 The three altitude lines (light blue lines in the figure) divide an equilateral triangle into six triangles that are congruent if reflections are permitted. Label the triangles a-f. Construct a triangle that may cover the center (black dot) by placing a point in triangle a. From that point, lines can be drawn to points placed in each of the other triangles. Four of those lines pass with certainty to one side or the other of the center. (Black lines.) The lines connecting a with d, b with e and c with f, however, (red lines) pass by the center as shown with unknown probability p < 1. Consider the red line between b and e. It passes below the center, but if the points were moved up, a b-e line could pass above the center. It is these three red lines that are uncertain. triangle in triangle.gifIn the case of a circle (or a square) quadrant symmetry dictates equal probabilities for a diagonal line passing the center point to one side or the other, so that p = 0.5. In the triangle case, it can easily be seen that p(b-e passing below center) is significantly greater than 0.5. Treating p as unknown for now, the 36 cases for placing points 2 and 3 (given the first point is in a) lead to the following probability table for creating a triangle that covers the center. Letting q = 1-p, point 3|V a b c d e f <= point 2a 0 0 0 pq 0 0b 0 0 0 p p 0c 0 0 0 p 1 pd pq p p pq q qe 0 p 1 q 0 0f 0 0 p q 0 0 If we add the elements (they represent equally likely events) and divide by 36 we obtain. p(cover) = ( 3pq + 8p + 4q + 2 ) / 36 This evaluates as follows. p p(cover) ----+------- 0.0 | 0.1667 0.1 | 0.1853 0.2 | 0.2022 0.3 | 0.2175 0.4 | 0.2311 0.5 | 0.2431 0.6 | 0.2533 0.7 | 0.2619 0.8 | 0.2689 0.9 | 0.2742 1.0 | 0.2778 One million random pairs of points determined p = 0.69525... so that p(cover) = 0.2616... This is close to 0.25 as one would expect, since most points come from the central portion of the enclosing triangle. The difference between the triangle and a circle is the vertex areas, where increasingly fewer points lie, but points from which are more likely than central points to create covering triangles. Hence the covering probability is slightly greater than 0.25. From this reasoning a containing hexagon would admit a covering probability in excess of 0.25, but closer to it. So the shape of the enclosing figure determines the center-covering probability of a random triangle. Probably an enclosing triangle would have the greatest difference from 0.25 of any enclosing regular polygon. As the number of sides increases the polygon more closely approximates a circle. Correcting an error in the simulation program, The probability that a random triangle inside an equilateral triangle covers its centroid is 0.24543 ... The average size of a random triangle inside any triangle is 1/12. I am assuming you mean the mean average. Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted April 7, 2014 Author Report Share Posted April 7, 2014 The average size of a random triangle inside any triangle is 1/12. I am assuming you mean the mean average. The average area of random triangles drawn inside any triangle T is 1/12 the area of of T. An affine transformation takes any given triangle into any other triangle while preserving relative (ratios of) areas. Quote Link to comment Share on other sites More sharing options...

0 BMAD Posted April 7, 2014 Report Share Posted April 7, 2014 That is a mouthful Quote Link to comment Share on other sites More sharing options...

## Question

## bonanova

While our collective brain trust ponders the nature of triangles defined

by three uniformly random points chosen inside a circle, specifically

the mode of their areas, we ask another question.

Recalling that the mean and median areal coverage of its circle's area are

about 7.4% and 5.4%, what is the probability that a random triangle covers

its circle's center?

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## 24 answers to this question

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