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Math
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2 \times 3
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2^{34}
234
a_{i-1}
ai−1
\frac{2}{3}
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\sqrt{2}
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\sum_{i=1}^3
∑i=13
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Comments
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.
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.
@Daniel Liu
–
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.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
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.
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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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a−b=±(c−d) which gives a=c,b=d or a=d,b=c.
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I recommend downloading Geogebra, it's free the quality is very good.
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