Re: Two Olympiad Problems
Good answer Tom .
Here is another solution :
Fix two points A and B and a variable point C on the line parallel to the line AB . Clearly the area of ABC is fixed but we wish to minimize the distance AC + CB . Let A' be the reflection of A on the parallel line and so A'C + CB is minimized when we have a straight line which happens when C is on the perpendicular bisector of AB . Now if we fix B and C and let A be the variable point , then it's immediately clear that the shortest perimeter occurs when we have an equilateral triangle .
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