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Quarilateral vertex reflection (Posted on 2021-08-04) Difficulty: 3 of 5
Each vertex of the convex quadrilateral of area S is reflected symmetrically with respect to the diagonal not containing this vertex. Let's denote the area of the resulting quadrilateral by S'. Prove that S'/S <3.

No Solution Yet Submitted by Danish Ahmed Khan    
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Some Thoughts Proof for rectangles Comment 1 of 1
Let the rectangle diagonal = 2 and center it at the origin with A=(1,0) and B in the first quadrant.  Call angle AOB = a thus 

A=(1,0), B=(cos a, sin a), C=(-1,0), D=(-cos a, -sin a)
S=[ABCD]=2 sin a

and the reflection images
A'=(cos 2a, sin 2a), B'=(cos a, -sin a), C'=(-cos 2a, -sin 2a), D'=(-cos a, sin a)
S'=[A'B'C'D']=2 sin 3a.

S'/S = sin(3a)/sin(a) = 2cos(2x)+1

This clearly never exceeds 3.  It approaches 3 as a approaches 0.





  Posted by Jer on 2021-08-06 12:57:12
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