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# Equal Areas of Pictures and Frames

Imgur

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.

Note by Daniel Liu
3 years, 6 months ago

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First, 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.

- 3 years, 6 months ago

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

- 3 years, 6 months ago

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

- 3 years, 6 months ago

Technically $$a-b=\pm(c-d)$$ which gives $$a=c,b=d$$ or $$a=d,b=c$$.

- 3 years, 6 months ago

yeah something like that

- 3 years, 6 months ago

What geometry diagram-making tool do you use? It looks a little tacky.

- 3 years, 6 months ago

I used colored pencils and then I put the picture into my Underwood typewriter and clacked on some letters and then I scanned it and put it here. Just now downloaded and tried out Geogebra. Maybe I'll get the hang of it one day.

- 3 years, 6 months ago