Download Problem A6 - Art of Problem Solving

Problem A6
Let n be given, n ≥ 4 and suppose that P1 , P2 , …, Pn are n randomly, independently and
uniformly chosen points on a circle. Consider the convex n-gon whose vertices are the
Pi . What is the probability that at least one of the vertex angles of this polygon is acute?
Solution:
We regard the points as stones that have yet to be placed on the circle. We will let ∠ Pi
denote the angle of the polygon which has point Pi as its vertex, and xi be the measure
of ∠ Pi in degrees. If A and B are two randomly chosen points on the circle, then we’ll

let AB be the arc that starts at A and ends at B in the counterclockwise direction. One
natural question that should occur is this. Suppose we want ∠ Pi to be acute. How
should we place the other stones on the circle to make ∠ Pi acute? If ∠ Pi is acute, then
by definition 0 < ∠ Pi < 90 . We must now recall a theorem from high school
geometry. In the picture below, suppose that the angle whose vertex is on the circle has

degree measure x. Then the degree measure of AB is 2x.
B
2x
A
x
Hence if ∠ Pi is acute, then 0 < xi < 90 and 0 < 2 xi < 180 . The latter inequality
implies that if we want ∠ Pi to be acute, then we must place all other stones on an arc
that does not contain Pi and whose degree measure is strictly less than 180 .
Another natural question that should occur is this. How many acute angles can this
polygon have? Without loss of generality, suppose that both ∠ P1 and ∠ Pi are acute for
some i ∈ {3, …, n – 1}. The arc subtended by ∠ P1 starts at P2 , contains the points P3 ,
…, Pn 1 , and ends at Pn . The arc subtended by ∠ Pi starts at Pi 1 , contains the points
Pi  2 , …, Pn , P1 , …, Pi  2 , and ends at Pi 1 . Since any point that is on the circle is
contained in at least one of the two arcs, the sum of the lengths of the two arcs is at least
the circumference of the circle. On the other hand, both angles are acute which means
that 0 < x1 < 180 and 0 < xi < 180 . These two inequalities imply that the length of
each arc is strictly less than half the circumference of the circle. So the sum of the
lengths of the two arcs must be strictly less than the circumference of the circle. We have
just generated a contradiction with our assumption. This shows that if the polygon has at
least one acute angle, then it either has exactly one acute angle or two acute angles
occurring consecutively.
If the polygon has any acute angles, then there is a unique pair of vertices Pi and Pi 1 in
counterclockwise order for which ∠ Pi is not acute and ∠ Pi 1 is acute. Without loss of
generality, let us call the angle that is not acute ∠ P1 and the angle that is acute ∠ P2 .

Let L be the random variable that is the length of P1 P2 . Let Qi be the point on the circle
that is diametrically opposite of Pi . Suppose the circle has radius R, for some R > 0. The
 

length of Q1Q2 is the same as the length of P1 P2 . In order to make ∠ P1 not acute, we
 
must place at least one stone on Q2 P1 . In order to make ∠ P2 acute, it is necessary that
 
we place the other n − 2 stones on Q1 P1 . However, we cannot place any stone on Q1
itself since that would make ∠ P2 a right angle. We’ll first find the probability that this
polygon has exactly one acute angle. Then we’ll find the probability that it has two acute
angles occurring consecutively. These are two mutually exclusive cases. Refer to the
picture below.
P2
P1
Q1
Q2
In order for this polygon to have exactly one acute angle, it is necessary that at least two
 
of the stone get placed on Q1Q2 . We must keep in mind that we also need to place at
 
least one stone on Q2 P1 to ensure that ∠ P1 is not acute. For a given choice of which two
stones act as P1 and P2 , the probability that this convex polygon has exactly one acute
angle is
1
n 3
2
n4
 n  2  R  l   l 
 n  2  R  l   l 




 + 
 
 +…+
 1  2R   2R 
 2  2R   2R 
n 5
3
n4
2
n2
 n  2  R  l   l 
 n  2  R  l   l 
l 
 R  l




−

 
 + 
 
 =
2R 
 2R
 n  5  2R   2R 
 n  4  2R   2R 
0
n2
n 3
1
n2
0
 n  2  R  l   l 
 n  2  R  l   l 
 n  2  R  l   l 




 
 − 
 
 − 
 
 =
 0  2R   2R 
 n  3  2R   2R 
 n  2  2R   2R 
n2
n2
1
(n  2)l (R  l ) n 3
 R  l 
 l 
−
−

 −
2 n2
(2R) n  2
 2R 
 2R 
given L = l. L takes values on the interval (0, πR). By the laws of probability we can see
that the probability that the polygon has exactly one acute angle is
n2
n2
 1
(n  2)l (R  l ) n 3  R  l  
 l 

  n2   2R   (2R) n2   2R   P( L  l )
l( 0,R )  2

for a given choice of which two stones act as P1 and P2 . Unfortunately this expression is
nonsensical for two reasons. Firstly, L is a continuous random variable which means that
P(L = l ) = 0. Secondly it does not make sense to add over all l ∈ (0, πR). For now,
think of P(L = l ) as some number that is infinitesimally small instead of 0. Our desired
expression should be some sort of an integral since L is a continuous random variable.
So let’s approximate the area under the curve
 x 
−

 2R 
1
2 n2
n2
(n  2) x(R  x) n 3
 R  x 
−
−

n2
(2R)
 2R 
n2
with m rectangles for some positive integer m. Each rectangle will have width πR/m.
Our approximation is
n2
n2
 1
(n  2)( jR / m)(R  jR / m) n 3  R  jR / m  
 jR / m 







 .

m j 1  2 n2  2R 
2R
(2R) n 2

 
R
m
We can think of the width of the jth rectangle as 2πR×P(( j − 1)πR/m < L < jπR/m). If we
let m approach infinity, then the intervals become infinitesimally small so then it makes
sense to speak of something like P(L = l ). So our desired probability is
n2
n2
 1
(n  2)( jR / m)(R  jR / m) n3  R  jR / m   1
 jR / m 

lim



 

 n 2  2R 
m  m
2R
(2R) n 2

  2R
j 1  2
n2
n2
R
1
1 R  1
(n  2) x(R  x) n 3  R  x  
 x 
=



dx = n 1  dx −


n

2
n

2

2 R 0
2R 0  2
(2R)
 2R 
 2R  
R n  2
R
1
(n  2) R
1
x dx −
x(R  x) n 3 dx −
(R  x) n  2 dx .
n 1 0
n 1 0
n 1 0
(2R)
(2R)
(2R)
R
m
The first integral is
1
2
n 1
R 
R
0
dx =
1
2
n 1
1
R
R
x0 =
2
n 1
R
R  0 =
R
1
= n 1 .
2 R
2
n 1
The second integral is
1
(2R) n 1

R
0
x n  2 dx =
1
(2R)
n 1
(n  1)
x n 1
R
0
=
1
(2R)
n 1
(R)
(n  1)
n 1

 0 n 1 =
1
(R) n 1
= n 1
.
n 1
2 (n  1)
(2R) (n  1)
We can use integration by parts to evaluate
(n  2)
(2R) n 1

R
0
x(R  x) n 3 dx .
If c is some constant, then we can let u = x and dv = (c  x) n3 dx to see that
n 3
 x(c  x) dx =  udv = uv −  vdu =
(c  x) n1
−
.
(n  1)( n  2)
Then the third integral is
 x (c  x ) n  2
−
n2
 (c  x ) n  2
 x (c  x ) n  2
dx
=
 n2
n2
(n  2)
(2R) n 1

R
0
R
x(R  x)
n 3
(n  2)   x(R  x) n 2
(R  x) n 1 
dx =


 =
n2
(n  1)( n  2)  0
(2R) n 1 
(n  2)   R(R  R) n  2 (R  R) n 1   0(R  0) n  2
(R  0) n 1  


 =



n2
(n  1)( n  2) 
n2
(n  1)( n  2)  
(2R) n 1 

1
(n  2) 
(R) n 1  
(n  2)  (R) n1 





 = n 1
0

0

0

=
.
n 1 
n 1 



(n  1)( n  2)   (2R)  (n  1)( n  2)  2 (n  1)
(2R) 

And the fourth integral is
1
(2R) n 1

R
0
R
(R  x)
n2
 1  (R  R) n1
 1  (R  x) n 1 

dx =
−

 =
n 1
(2R) n 1  n  1  0
(2R) n1 
1
(R) n 1
1 
(R) n1 
(R  0) n 1 
 =


=
=
.
0

n

1
n

1
n 1 
2 (n  1)
n  1  (2R) (n  1)
n  1  (2R) 
So the probability of the convex polygon having exactly one acute angle is
1
2
n 1
−
2
n 1
1
1
1
3
n 1 3
1
− n 1
− n 1
= n 1 − n 1
= n 1
=
(n  1)
2 (n  1)
2 (n  1)
2 (n  1)
2 (n  1)
2
n4
2 (n  1)
n 1
for a given choice of which two stones act as P1 and P2 . Finally, there are n(n − 1) ways
to choose which stones act as P1 and P2 so the probability that the polygon has exactly
one acute angle is
n(n  1)
n4
n 2  4n
=
.
2 n 1 (n  1)
2 n 1
If we want the polygon to have two acute angles occurring consecutively, then we can
 
place no more than one stone on Q1Q2 . That is, we must either place all remaining stones
 
 
on Q2 P1 or place only one stone on Q1Q2 . The probability of having only one of the
 
remaining stones placed on Q1Q2 is
 n  2  l   R  l 




 1  2R   2R 
1
n 3
=
(n  2)l (R  l ) n 3
(2R) n  2
 
given L = l. The probability of having all remaining stones placed on Q2 P1 is
 R  l 


 2R 
n2
given L = l. So the probability of the polygon having two consecutive acute angles is
 (n  2)l (R  l ) n 3  R  l  n  2 

  P( L  l )
  (2R) n2
 2R  
l( 0 ,R ) 
for a given choice of which two stones act as P1 and P2 . If we use the same type of
reasoning we did previously, then the nonsensical expression is really just
1 R  (n  2) x(R  x) n 3  R  x 


2R 0 
(2R) n 2
 2R 
1
(2R) n 1

R
dx = (n  2)
x(R  x) n 3 dx +
n

1

0

(2R)

R
1
1
2
1
n2
0 (R  x) dx = 2 n1 (n  1) + 2 n1 (n  1) = 2 n1 (n  1) = 2 n2 (n  1) .
n2
There are n(n − 1) ways to choose which stones act as P1 and P2 so the polygon has two
consecutive acute angles with probability
n(n  1)
2
n2
n
1
= n2 .
2
(n  1)
Finally, we can say that the probability this polygon has at least one acute angle is
n
2n
n 2  4n
n 2  4n
n 2  4n  2n n 2  2n
+
=
+
=
=
.
2 n2
2 n 1
2 n 1
2 n 1
2 n 1
2 n 1