Modeling with Maple some Geometric Probabilities Eugen J. Ionascu Columbus State University February 25th, 2011 http://math.colstate.edu/ejionascu/
joint work with Gariel Prajitura The College at Brockport, State University of New York, US
[email protected]
Abstract: In a recent paper (see arXiv:1009.0890), we studied some new versions of an old geometric probability question: The Broken Stick Problem (1854)[20]: A spaghetti stick, dropped on the floor, breaks at random into three pieces. What is the probability that the three parts obtained are the sides of a triangle? One may easily arrive to the conclusion that, with a reasonable interpretation of the probability distribution, the answer to this problem is 1/4. For some similar questions, we found pretty interesting exact answers, but in some cases we only obtained experimental data. In this talk, we present these investigations and show the model and the Maple program we used that gave us various experimental probability frequencies. These kinds of experiments are very easy to construct and can be used in helping our students check their work or develop an intuition about a geometric probability question.
√ (b) α, β, γ< 3/2
(a)
Figure 1: Distances α = OP, β = OM, and γ = ON , add up to
√
3.
The idea of considering an equilateral triangle. √ α = y, β =
3(1 + x) − y and γ = 2
♦ notice that α+β+γ =
√
3.
√
3(1 − x) − y . 2
(1)
Maple Program for simulation >with(plots):with(plottools):Digits:=30: > trianglecondition:=proc(a,b,c) local x,u,v,w; x:=false; u:=a+b-c;v:=a+c-b;w:=b+c-a; if u>0 and v>0 and w>0 then x:=true;fi; x; end: > rt:=sqrt(3.):O1:=[-1,0]:O2:=[1,0]:O3:=[0,rt]: > pltr:=display(polygon([O1, O2, O3]), color=yellow, linestyle=dash, thickness=2): >triangleproblem:=proc(n,m) local i,alpha,beta,gamma,r,x1,x2,x3, u,v,s,A,nf; nf:=0; A:={}; for i from 1 to n do r:=rand(1..m);x1:=r()/m;x2:=r()/m; if x1+x2 2max(u, v, w). If the triangle exists, it is unique. Moreover, the triangle is acute, if and only if u2 + v 2 + w2 < 6 min(u2, v 2, w2).
Theorem 3 Given the hypothesis of the stick problem, the probability that there exists a triangle whose medians are α, β and γ is 14 . Moreover, the probability that this triangle is acute equals 8 1 5 − ln ≈ 0.0722202059. 3 9 5
‘number of trials 20000 Probability‘ || (.72250000000000000000e–1) 1.6 1.4 1.2 1 y 0.8 0.6 0.4 0.2
–1
–0.5
0
0.5
1
x
√ 1 ( 9x2 +10−1), x∈(−1/4,1/4) y> √ 3 3
1 5 8 − ln ≈ 0.0722202059. 3 9 5
Altitudes of a Triangle, General and Acute situation Theorem 4 A stick is broken into three pieces at random (as described earlier). (i) The probability that the three segments are the heights of a triangle is equal to ! √ √ 3+ 5 4 −5 . 3 5 ln 25 2 (ii) The probability that α , β and γ are the heights of an acute triangle is equal to
√ Z 1−2 3 0
√ √ 2 6− 3 7
2
√
2
√
15t − 6 3t + 9 − 12t(2t − 2 3t + 3)
1 2
12
dt ≈ 0.07744388 .
Triangles with α, β and γ as heights ! √ √ Area(R1 ) 4 3+ 5 √ = 3 5 ln −5 25 2 3 √ √ 15x2 + 12 3 3 − y= 5 5
‘number of trials 30000 Probability‘ || (.23730000000000000000) 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2
–1
–0.5
0
0.5
1
Acute triangles with α, β and γ as heights
‘number of trials 30000 Probability‘ || (.77166666666666666667e–1) 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2
–1
–0.5
0
Area(R1 ) Area(R2 )
0.5
≈3
1
(a) Excircles
(b) Acute triangles with α, β and γ as radii of excircles, Region bounded by ellipses
Figure 2: Circles tangent to the sides
Radii of excircles.
Theorem 5 If u, v, and w are greater than 0 then there is a unique triangle such that ra = u, rb = v, and rc = w. Moreover, this triangle is acute if and only if uv + vw + wu > max{u2, v 2, w2}. Corollary Under the hypothesis of the stick problem, the probability that the three segments are the radii of the excircles of a triangle is equal to 1. Moreover, the probability that this triangle be an acute triangle is: √ √ 24 7 14 2 P = arcsin( ) − ≈ 0.3449830931 49 8 7
(a) Distances to sides from the circumcenter
(b) The incircle
Figure 3:
SPECIAL CASES Theorem 6 Consider a triangle ABC and let O be its circumcenter. Denote the distances of O to the sides BC, AC, and AB, by u, v and w respectively. (i) The radius R, of the circle circumscribed to the triangle ABC, satisfies the equation R3 − (u2 + v 2 + w2)R − 2uvw = 0, if 4ABC is acute;
(2)
and (obviously) 1
R = (u2 + v 2 + w2) 2 , if 4ABC is a right triangle.
(4)
(ii) Given three positive real numbers u, v, and w, there exists one and only one acute triangle with the distances of the circumcenter to the sides equal to u, v and w. The previous statement is true if one changes the adjective acute to obtuse. (iii) The equation (2) has infinitely many integer solutions (u, v, w, R) ∈ N4 such that u, v, and w are all different. Theorem 7 If u, v, and w are positive quantities then there is a unique triangle such that the distances from the vertices to the center of the incircle are equal to u, v, and w respectively. Theorem 8 Given u, v and w three positive real numbers, the triangle insured by Theorem 7 is acute if and only if
√ 2u2vw + u2(v 2 + w2) − v 2w2 > 0, √ 2uv 2w + v 2(u2 + w2) − u2w2 > 0, and √2uvw2 + w2(u2 + v 2) − u2v 2 > 0.
(5)
In the context of the broken stick problem if u = α, v = β and w = γ, the probability that the triangle given by Theorem 7 is acute is approximately 0.1962.
(a) Orthocenter, {α, β, γ}={ HA, HB, HC}
(b) {α, β, γ}={ ha , `a , ma }
Figure 4: Triangle ABC, ha = u, `a = v, and ma = w.
Theorem 9 If 0 < u < v < w then there is a triangle such that the altitude, the angle bisector, and the median from one of the vertices of the triangle equal u, v, and w respectively. Moreover, this triangle is acute if and only if p u v 4 − 3u2(v 2 − u2) uv 2
