For the Mathematicians of the TFP - Tilted Forum Project Discussion Community

JadziaDax · 07-08-2003, 02:43 PM

Given three concentric circles in the plane, prove that (up to rotation and reflection) there exists a unique triangle of maximum area having exactly one vertex on each circle, respectively.

spasblitz · 07-08-2003, 04:53 PM

holy shit, that's all i have to say about that !

Peetster · 07-08-2003, 05:27 PM

The three vertices will be refered to as points A, B and C. A will exist on the innermost circle, B on the middle circle and C on the outer circle.

Since rotation and reflection are excluded, placement of A is arbitrary.

In order to maximize volume, B must be placed exactly opposite A on the middle circle. Earlier proofs conclude that volume to circumference is maximized as the shape approaches a circle, or as the number of vertices approaches infinity. (Can I use other proofs, or do you want this from scratch?)

Connect A, the center of concentricity, and B with a straight line. Draw a line at a right angle at the center. Mark the intersection with the outer circle C. This triangle is unique, and maximizes volume.

Any movements of A or B around their respective circles will narrow the triangle and reduce volume. Any movement of C will skew the triangle and reduce volume.

QED? Or should I call C^2 = A^2 + B^2 - 2AB*cos(c) as my next witness?

MacGnG · 07-08-2003, 08:44 PM

sorry cant do these kindsa problems without a graph and a textbook

cheerios · 07-09-2003, 01:07 AM

why is it unique, Peetster?

Jadz, algebra, please? I SUCK at geometry! :P

JadziaDax · 07-09-2003, 01:13 AM

Quote:

Originally posted by Peetster
(Can I use other proofs, or do you want this from scratch?

From scratch, please...

and almost... there is some information missing from this proof to completely convince me...

Peetster · 07-09-2003, 06:44 AM

Sub-proof 1: An isoseles triangle maximizes the volume to perimeter ratio.

Let x, y and z be the sides of the triangle, with x equal y for an isoseles.

Perimeter P1= x+y+z
since x=y
P1=2x+z

Area A1=1/4zx + 1/4 zy
since x=y
A1=1/2zx

Therefore (I don't know the ASCII code for the three dots)

A1/P1 = (1/2zx) / (2x+z)
= zx / 2(2x+z)
= zx / (4x+2z)

OK, I'm thinking this might not be the right direction to go since I must now calculate the differences in the perimiter based on movement around the circle.

Peetster · 07-09-2003, 03:43 PM

It must be an isoseles triangle, which is composed of two right triangles. Each hypotenuse is = sqrt(1/2x^2 + y^2).

If the point is moved the hypotenuse becomes = sqrt(1/2x^2 + y^2 - 2xy*Cos(a)). Since any movement away from an isoseles triangle adds a term which reduces the length of the hypotenuse, it must be the largest triangle.

Close?

07-08-2003, 02:43 PM	#1 (permalink)
JadziaDax Loser Location: who the fuck cares?	For the Mathematicians of the TFP Given three concentric circles in the plane, prove that (up to rotation and reflection) there exists a unique triangle of maximum area having exactly one vertex on each circle, respectively. Last edited by JadziaDax; 07-08-2003 at 08:53 PM..

07-08-2003, 04:53 PM	#2 (permalink)
spasblitz Upright Location: Illinois	holy shit, that's all i have to say about that !

07-08-2003, 05:27 PM	#3 (permalink)
Peetster Right Now Location: Home	The three vertices will be refered to as points A, B and C. A will exist on the innermost circle, B on the middle circle and C on the outer circle. Since rotation and reflection are excluded, placement of A is arbitrary. In order to maximize volume, B must be placed exactly opposite A on the middle circle. Earlier proofs conclude that volume to circumference is maximized as the shape approaches a circle, or as the number of vertices approaches infinity. (Can I use other proofs, or do you want this from scratch?) Connect A, the center of concentricity, and B with a straight line. Draw a line at a right angle at the center. Mark the intersection with the outer circle C. This triangle is unique, and maximizes volume. Any movements of A or B around their respective circles will narrow the triangle and reduce volume. Any movement of C will skew the triangle and reduce volume. QED? Or should I call C^2 = A^2 + B^2 - 2AB*cos(c) as my next witness?

07-08-2003, 08:44 PM	#4 (permalink)
MacGnG Know Where!	sorry cant do these kindsa problems without a graph and a textbook

07-09-2003, 01:07 AM	#5 (permalink)
cheerios Banned Location: 'bout 2 feet from my iMac	why is it unique, Peetster? Jadz, algebra, please? I SUCK at geometry! :P

07-09-2003, 06:44 AM	#7 (permalink)
Peetster Right Now Location: Home	Sub-proof 1: An isoseles triangle maximizes the volume to perimeter ratio. Let x, y and z be the sides of the triangle, with x equal y for an isoseles. Perimeter P1= x+y+z since x=y P1=2x+z Area A1=1/4zx + 1/4 zy since x=y A1=1/2zx Therefore (I don't know the ASCII code for the three dots) A1/P1 = (1/2zx) / (2x+z) = zx / 2(2x+z) = zx / (4x+2z) OK, I'm thinking this might not be the right direction to go since I must now calculate the differences in the perimiter based on movement around the circle.

07-09-2003, 03:43 PM	#8 (permalink)
Peetster Right Now Location: Home	It must be an isoseles triangle, which is composed of two right triangles. Each hypotenuse is = sqrt(1/2x^2 + y^2). If the point is moved the hypotenuse becomes = sqrt(1/2x^2 + y^2 - 2xy*Cos(a)). Since any movement away from an isoseles triangle adds a term which reduces the length of the hypotenuse, it must be the largest triangle. Close?