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What type of triangle has the maximum perimeter for a certain area?

Suppose I start with a rectangle of that area (axb=Z). I can stretch one dimension of the rectangle until infinity, reducing the other side, and decompose it into two halves that create an isosceles triangle of the same area but larger perimeter (since the diagonal will always be greater than b, therefore larger than 2b and the base will be equal (a/2)x2). Is this the way to the maximum perimeter possible or is there another construction that I'm missing?

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    $\begingroup$ Your construction shows that there can be no maximum perimeter. $\endgroup$ – Michael Biro Dec 13 '14 at 15:39
  • $\begingroup$ Yes, the perimeter tends towards infinity, that is easily showable. But how do I prove that an isosceles triangle is the optimal solution instead of a mystical 4-sided triangle or something? $\endgroup$ – Aniceto Dec 13 '14 at 18:23
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    $\begingroup$ What optimal solution? For any isosceles triangle you choose, I can find a scalene triangle with the same area and larger perimeter. Then you can find an isosceles triangle better than mine, and I can find a scalene one better than yours... It's turtles all the way down. $\endgroup$ – Michael Biro Dec 13 '14 at 18:25
  • $\begingroup$ Just realized what was wrong about my reasoning. If the perimeter tends towards infinity, it's irrelevant what the shape of the triangle is. It just can't be equilateral. Thanks. $\endgroup$ – Aniceto Dec 13 '14 at 19:09
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For any triangle with known base and height, you can move the vertex on a line parallel to the base without changing the height, hence the area. Then send the vertex arbitrarily far :)

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There is no maximum perimeter as you can keep on increasing the perimeter without increasing the area.

The area of a triangle is given by $\frac12bh$. This means that if you halve the base and double the height, the area remains constant (though the perimeter has increased). You could divide the base by 10 and multiply the height by 10, and still have the same area, but significantly greater perimeter.

In the same way, you can progress till infinity.

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  • $\begingroup$ Yes, that shows that the perimeter of that triangle will tends towards infinity. But how to do I prove that it is the optimal solution instead of a right triangle or a scalene one? $\endgroup$ – Aniceto Dec 13 '14 at 18:24
  • $\begingroup$ @Aniceto What do you really mean? $\endgroup$ – Sawarnik Dec 13 '14 at 18:29
  • $\begingroup$ Just realized what was wrong about my reasoning. If the perimeter tends towards infinity, it's irrelevant what the shape of the triangle is. It just can't be equilateral. $\endgroup$ – Aniceto Dec 13 '14 at 19:04

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