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   Minimal triangles in convex figures
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   Author  Topic: Minimal triangles in convex figures  (Read 1566 times)
Deedlit
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Minimal triangles in convex figures  
« on: Jun 5th, 2005, 3:47pm »
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Show that, for any convex figure S in the plane, it is possible to choose four points in S such that any three of the four points form a triangle with area at least 1/4 that of S.
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SWF
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Re: Minimal triangles in convex figures  
« Reply #1 on: Jul 1st, 2005, 7:25pm »
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It is not difficult to see that it is possible to sandwich the shape between two parallel lines touching the perimeter of shape at points A and B, such that line AB divides the shape into two equal areas.  
 
Let W be the area of the shape. Construct the perpendicular bisector, CD, of AB with points C and D lying on the boundary of the shape. CD goes through point M, the midpoint of AB. Triangles AMC and BCM must have the same area, call it Y=LAB*LCM/4.
 
Since the shape is convex the maximum possible area of the portion of the shape on the C side of AB is LAB*LCM, and therefore  
 
LAB*LCM >= W/2
 
Divide by 4 and note left side equals Y:
 
LAB*LCM/4 = Y >= W/8
 
Triangles AMC and BCM both have an area greater than W/8.  By similar reasoning triangles ADM and DBM also have areas greater than W/8.  
 
All four ways of forming a triangle from three of the four points, A, B, C, and D results a triangle made up of two triangles with area greater than or equal to W/8, for a combined area greater than or equal to W/4.
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Deedlit
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Re: Minimal triangles in convex figures  
« Reply #2 on: Aug 2nd, 2005, 8:46pm »
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Sigh - another post that I really needed to reply to much earlier.
 
on Jul 1st, 2005, 7:25pm, SWF wrote:
It is not difficult to see that it is possible to sandwich the shape between two parallel lines touching the perimeter of shape at points A and B, such that line AB divides the shape into two equal areas.  

 
It's not difficult, but I'm inclined to say that this needs proof.
 
Quote:

Since the shape is convex the maximum possible area of the portion of the shape on the C side of AB is LAB*LCM, and therefore  
 
LAB*LCM >= W/2

 
Unfortunately, this isn't true.  Imagine the convex figure to be very wide, and AB to be on a slant.  CD is likely to be much shorter than the width of the figure.
« Last Edit: Aug 2nd, 2005, 8:46pm by Deedlit » IP Logged
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