Prove that if, in a semi circle of radius 1, five points A , B , C , D and E are taken in consecutive order?

Question

and E are taken in consecutive order, then:
(AB)^2 +( BC)^2 + (CD)^2 + (DE)^2 + (AB)(BC)(CD) +
+ (BC)(CD)(DE) < 4

Answer 1

Ironduke: for your second example the correct answer is 3.9996 < 4. I think you were calculating the upper bound that you found, rather than the value we are actually looking at - which shows that your upper bound is not sufficient.

Working from the statement that the chord length must be less than the arc length won't be good enough, because as you have in your second case where one angle approaches π, the chord length approaches 2 but the arc length approaches π. We need to show that 2 is the upper bound in this case.

Note that (AB)^2 = 2 (1 - cos a), and similarly for the others. From cos 2x = 1 - 2 sin^2 x we can derive
AB = 2 sin (a/2), BC = 2 sin (b/2), etc.

We can assume without loss of generality that A and E are at the ends of the semicircle; if they're not, moving them to the ends will increase AB and DE, so if we can prove the statement in this case it will be true for the general case as well. Don't know if this helps, but it simplifies things a little.

Suppose B and C are fixed (and A and E, obviously), and we'll vary the position of D in order to maximise the LHS of the inequality. So a and b are fixed, and d = π - a - b - c.

LHS = f(a, b, c) = 4(sin^2 (a/2) + sin^2 (b/2) + sin^2 (c/2) + sin^2 ((π-a-b-c)/2)) + 8 sin (b/2) sin (c/2) (sin (a/2) + sin ((π-a-b-c)/2))
∂f/∂c = 4 sin (c/2) cos (c/2) - 4 sin ((π-a-b-c)/2) cos ((π-a-b-c)/2)
= 2 sin c - 2 sin (π-a-b-c)
= 0 <=> sin c = sin (π-a-b-c)
<=> c = π-a-b-c or c = a+b+c
<=> c = (π-a-b)/2 or a+b = 0

If a+b = 0 then we must have a = b = 0 (since both are nonnegative) and we get
f(c) = 4(sin^2 (c/2) + sin^2 ((π-c)/2)) = 4(sin^2 (c/2) + cos^2 (c/2)) = 4.
This is to be expected since A, B, and C are at one end of the semicircle, E is at the other, so D is a point projected on the arc by the diameter CE and hence angle CDE is a right angle, and CD^2 + DE^2 = CE^2 = 2^2 = 4.

In the non-degenerate case (implied since we want strict inequality), we have c = (π-a-b)/2 and
∂2f/∂c^2 = 2 cos c + 2 cos (π-a-b-c)
= 4 cos ((π-a-b)/2)
Since 0 < a+b < π, 0 < (π-a-b)/2 < π/2, so cos ((π-a-b)/2) > 0, so this value of c minimises f.
So the maximum value of f must be at the extremes of c: c = 0 or c = π-a-b.
If c = 0 we have f(a, b, c) = 4(sin^2 (a/2) + sin^2 (b/2) + sin^2 ((π-a-b)/2))
If c = π-a-b we have f(a, b, c) = 4(sin^2 (a/2) + sin^2 (b/2) + sin^2 ((π-a-b)/2)) + 8 sin (a/2) sin (b/2) sin ((π-a-b)/2)
Since 8 sin (a/2) sin (b/2) sin ((π-a-b)/2) will always be non-negative, the maximum value for f will occur when c = π-a-b, i.e. when D sits on top of E.

By symmetry it follows that (with A and E fixed at the ends as before) for any positioning of C and D, the maximum value of f will occur when B sits on top of A.

So we only need to consider the case where a = d = 0, b is unknown and c = π-b. But this is essentially the same as the case we considered earlier with a = b = 0, and results in f being a constant 4.

So the maximum possible value of f is 4, in the degenerate case where some points are at one end of the semicircle and some are at the other. In the non-degenerate case f will be strictly less than 4.

I'm sure there's a much faster way to get this result using a geometrical interpretation, but I haven't been able to spot it, so you get the clunky calculus method instead. ;-)

Answer 2

Let O be the center of the semicircle
Draw OA,OB,OC,OD,OE
Let angle AOB = a, angle BOC = b, angle COD = c and angle DOE = d. All angles are in radians.
Let r = radius = 1
Now arc AB = ra =a
Thus AB < a since a straight line is shortest distance from A to B. Similarly BC
Thus AB*BC*CD < abc and BC*CD*DE < bcd
Further AB^2
Hence AB^2+BC^2+CD^2+DE^2+AB*BC*CD+
BC*CD*DE
a+b+c+d= So pick the case where a=b=c=d= pi/4.
We get 3.523688714 which is <4

Now take the case where a = pi- .03 and b=c=d = .01
We get 9.807798868 which is >4

So it appears we have found a counter example and the hypothesis is incorrect.