I have checked this out with Sketchpad, but I'm stumped on how to prove it.

For an acute triangle, s > 2R.

Any help will be greatly appreciated.

This can be shown in three steps.

Show that every closed borken line of perimeter 1 can be enclosed by a circle of radius 1/4. (Chose two points on the line that divide the perimeter into two halves. Then consider a circle with the two points defining a diameter.)

Show that if a triangle is entirely inside a circle C of radius S then the circumradius R < S. (The arc of the circumcircle that may be outside C is less than 180°, because the triangle is acute. Therefore, there is a circumdiameter which is entirely inside C.)

By 1, a triangle of perimeter p can be encircled by a circle of radius p/4. By 2, its circumradius R satisfies R < p/4. If 2s = p, you get 2R < s.

Thanks Alex.

Dear Gerald Brown and Alex,

I have found one another proof using some well known interesting facts:
Suppose in acute triangle ABC:
AD = diameter of circumcircle of ABC
M = midpoint of BC

1. Prove that ABDC is convex by the fact: a triangle is acute if and only if circumcenter is inside the triangle.
From that we can show: angle(BDC) > 90 (1)
(This result requires that triangle ABC must be acute)

2. Prove that: in a triangle the median from obtuse angle is less than half of its side.
Combine with (1), we can show: MD < BC/2 (2)

3. Prove that: in any triangle any median is less than average of two adjacent sides.
Therefore: AM < (AB + AC)/2 (3)

From (2) and (3) we have:
2*R = AD < AM + MD < (AB + AC + BC)/2 = s what we must prove.

Best regards,
Bui Quang Tuan