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 Subject: "Semiperimeter and Circumradius Inequality" Previous Topic | Next Topic
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Bractals
Member since Jun-9-03
Oct-14-07, 01:49 PM (EST)    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.

alexb
Charter Member
2100 posts
Oct-14-07, 09:08 PM (EST)    1. "RE: Semiperimeter and Circumradius Inequality"
In response to message #0

 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. Bractals
Member since Jun-9-03
Oct-15-07, 07:05 AM (EST)    2. "RE: Semiperimeter and Circumradius Inequality"
In response to message #1

 Thanks Alex.

Bui Quang Tuan
Member since Jun-23-07
Oct-15-07, 08:31 AM (EST)    3. "RE: Semiperimeter and Circumradius Inequality"
In response to message #0

 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 ABCM = midpoint of BC1. 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

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