Two pictures have equal area. Their frames, which have a thickness of \(1\) inch, also have equal area.

Prove that the two frames are identical.

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## Comments

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TopNewestFirst, cut out all the square \(1\) inch corners. Then if \(a,b\) and \(c,d\) are the dimensions of the pictures, we then see that the perimeters and areas have to be the same. The rest follows.

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Prove that if the perimeters and areas are equal, then the rectangles must be congruent.

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If a+b=c+d and ab=cd, then a-b=c-d, therefore a=c and b=d.

I'll try to think of a more clever proof, if nobody else improves on this.

Okay, maybe I'll post a graphic a bit later to illustrate this. Given 2 rectangles of equal areas. Overlap them so that they share a common corner. Then, excluding the union of the 2 rectangles, we have 2 smaller rectangles, the areas of which have to be the same. If the perimeters of both rectangles are the same, then both of the smaller rectangles have to have a side of the same length. This means that the other sides they have must also be of the same length. This means that the 2 rectangles are congruent.

Picture Frames

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I recommend downloading Geogebra, it's free the quality is very good.

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