| No Solution Yet | Submitted by Danish Ahmed Khan |
| No Rating |
Proof for rectangles
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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
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 |
