file:///C|/Misc/Yoav/Research/Math/ponder%2010-06V2%20final.txt

From: Yoav Raz [mailto:[email protected]] Sent: Wed, October 11, 2006 1:31 PM To: [email protected] Cc: Yoav Raz Subject: ANSWER: October 2006 Ponder This

------------------------------------------------------------Ponder This Challenge: Puzzle for October 2006. Revised, 10/3/06, 11:30AM ET: Consider 3 points chosen at random in the interior of a triangle. We ask what is the expected (average) area of the triangle determined by these 3 points as a fraction of the area of the triangle they are chosen in. It is fairly easy to see that this fraction is independent of the size or shape of the original triangle. So we may assume we are choosing points in the triangle, S, in the x-y plane with vertices (0,0), (1,0) and (0,1). Let R be the triangle determined by the three random points. The value we are asking for can be computed by answering the following questions. 1. Consider the minimum triangle (with edges parallel to the edges of S), T, containing the 3 random points. What is the ratio of the expected area of T to the area of S. 2. There are 2 configurations of T and the 3 random points which occur with positive probability. These are: Case A - One random point is a vertex of T, another lies along the opposite side and the third is in the interior of T. Case B - Each edge of T contains one of the 3 random points. What is the probability of occurrence of each of these cases? 3. What is the ratio of the expected area of R to the area of T for each of the cases A and B above? 4. Combining the answers to questions 1-3, what is the ratio of the expected area of a triangle, R, formed by choosing 3 points at random in another triangle, S, to the area of S.

------------------------------------------------------------Revised and corrected. file:///C|/Misc/Yoav/Research/Math/ponder%2010-06V2%20final.txt (1 of 8) [2/21/2008 11:23:19 PM]

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TXT file attached if format not kept in mail.

Summary ------1. 5/14 2. P(A)=3/5 P(B)=2/5 3. For A 2/9; For B 1/4 4. (5/14) * ( (3/5)*(2/9) + (2/5)*(1/4) ) = 1/12

revised solution ----------------

IBM Ponder Oct 2006

We use 1. Infinitesimal dx as the measure or weight for every degree of freedom value of x (parameter, variable) below. All objects below will typically have measure dxdxdy (3D). Averages will be calculated by integration over them and dividing by total object weight (in most cases same integral but with integrand 1). 2. (INT x = a to b)...dx for integrals; (INT)...dx when limits are clear from context; (INT)^3...dxdydz etc. for multiple. 3. S, T are the triangles in the problem with angles A B C clockwise where A is the right angle. 4. A(R) = Area of triangle R (non-negative; it will be indicated when computed negative, and abs() will be taken. ) A(S) = 1/2 file:///C|/Misc/Yoav/Research/Math/ponder%2010-06V2%20final.txt (2 of 8) [2/21/2008 11:23:19 PM]

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Part 1 E = Expected A(T)/A(S) --------------------------------Let T be A(r,t), B(r,t+s-r), C(s,t) s=0 to 1; r=0 to s; t=0 to 1-s+r Each value of r, s, t defines a distinct T inside S. A(T) = (s-r)^2/2 For three random points to define T, the weight of T is proportional to side^3 or (s-r)^3 Weighted area = = (INT s=0 to 1)(INT r=0 to s)(INT t=0 to 1-s+r) (s-r)^5/2 dtdrds = 1/336 Total weight = = (INT s=0 to 1)(INT r=0 to s)(INT t=0 to 1-s+r) (s-r)^3 dtdrds] = 1/60 E = (60/336)/(1/2) = 5/14 []

Part 2 - Probabilities of A and B ---------------------------------Let the random triangle R be defined by points 1(x1,y1), 2, 3 where triangle T is now (0,0) (0,1) (1,0) The cases A and B are covered by the following table (The Table)

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B A1 A2 A3 --------------------------------------------------------------------------x1= 0 0 0 1 y1= y 0=
Thus the probabilities of A and B are 3/5 and 2/5 respectively. (A1, A2, A3 have equal probabilities of 1/5) []

Part 3 - Average area of R for each case ---------------------------------------As in part 2 Let the random triangle R be defined by points 1(x1,y1), 2, 3 where triangle T is (0,0) (0,1) (1,0)

We use the area formula file:///C|/Misc/Yoav/Research/Math/ponder%2010-06V2%20final.txt (4 of 8) [2/21/2008 11:23:19 PM]

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|x1 y1 1| | | (*) 2*A(R) = |x2 y2 1| = x1(y2-y3)+x2(y3-y1)+x3(y1-y2) | | |x3 y3 1| A(R) is possitive if 1,2,3 are counter-clockwise. We'll adjust sign if otherwise. Let A1av, A2av, A3av, Bav be the average areas of R for the four cases A1,...,B respectively. Aav is the total average area for case A.

Case B -----Substituting in (*) above from the Table -2*A(R) = -t*y+x*y-x*(1-t)

(clockwise, negative expression)

-(weighted area for B) = 0.5 * (INT from 0 to 1)^3 (-t*y+x*y-x*(1-t) ) dxdydt = -0.5 * (1/4 - 1/4 + 1/2 -1/4) = -1/8 Total weight for case B is 1 (see part 2), so Bav = 1/8 Expected ratio EB = Bav/A(T) = 1/4 []

Case A1 - (0,0) is fixed in R ----------------------------In order to be consistent with the area sign We integrate only for a point 3(x,y) that is below the line y = x*(1-t)/t The case for point 3 above that line is symmetric, and we multiply the result by 2 for file:///C|/Misc/Yoav/Research/Math/ponder%2010-06V2%20final.txt (5 of 8) [2/21/2008 11:23:19 PM]

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(total) weighted area. -2*A(R) = -t*y+t*x-x

by (*) and Table (clock-wise and negative)

Integration regions below the line: 0 =< t =< 1 0 =< x =< t and 0 =< y =< x*(1-t)/t t =< x =< 1 and 0 =< y =< 1-x

-(weighted area of A1) = = 2*0.5*[ (INT t=0 to 1) [(INT x=0 to t)(INT y=0 to x*(1-t)/t) (-t*y+t*x-x) dydx + (INT x=t to 1)(INT y=0 to 1-x) (-t*y+t*x-x)] dydx ] dt = -1/180 -1/20 = -1/18 The total weight of A1 is 1/2 (part 2 above), and A1av = (1/18)/(1/2) = 1/9 []

Case A2 - (0,1) is fixed in R -----------------------------

First, (i) point 3(x,y) is below the line y = 1 - x/t -2*A(R) = t*y-t+x

by (*) and Table (clockwise and negative)

-(weighted area of A2(i) ) = = 0.5*(INT t=0 to 1)(INT x=0 to t)(INT y=0 to 1-x/t) (t*y-t+x) dydxdt = -1/36

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Then (ii) when point 3(x,y) is above the line y = 1 - x/t 2*A(R) = t*y-t+x

(counter-clockwise and positive)

We have two integration regions:

0 =< t =< 1 0 =< x =< t and

0 =< y =< 1-x/t

t =< x =< 1 and 1-x/t =< y =< 1-x (weighted area of A2(ii) ) = = 0.5*(INT t=0 to 1)(INT x=0 to t)(INT y=1-x/t to 1) (t*y-t+x) dydxdt +0.5*(INT t=0 to 1)(INT x=t to 1)(INT y=0 to 1-x) (t*y-t+x) dydxdt = 1/360 + 1/40 = 1/36 (not a coincidence...)

(total weighted area of A2) = 1/36 + 1/36 = 1/18 The total weight of A2 is 1/2 (part 2 above), and A2av = (1/18)/(1/2) = 1/9

(Conjecture: True for all triangles)

[]

Case A3 - (1,0) is fixed in R ----------------------------Same as case A2 (by symmetry; not using any conjecture) A3av = (1/18)/(1/2) = 1/9 []

Case A = A1,2,3 file:///C|/Misc/Yoav/Research/Math/ponder%2010-06V2%20final.txt (7 of 8) [2/21/2008 11:23:19 PM]

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--------------Aav = 1/9 (same weight to 1,2,3) Expected ratio EA = Aav/A(T) = 2/9 []

Part 4 - Summary ---------------Average of A(R)/A(S) = E * ( EA*3 + EB*2) /5 = (5/14) * (3*2/9+2*1/4)/5 = 1/12 []

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