# Construct any regular polygon that has the same area as the sum of $n$ given triangles

Original question: Construct any regular (or similar-scaled to a given) geometric shape that has the same area as given triangle?

My idea is application of generalized Pythagora's theorem. Euclid Elements Book VI. $$31$$,

I know it's possible to construct such square and equilateral triangle, but are there some other creative ideas?

I'm looking for a marked ruler and compass construction, or a construction using two (marked) triangles and compass because it really doesn't matter in sense of problem solvability (is that the word?)

EDIT: Thinking thinking, Euclid Elements Book VI. $$18$$, $$\Rightarrow$$ Every (regular) polygon can be visualized as multiple triangles

And, after that apply Euclid Elements Book VI. $$25$$, although I'm not sure if there exist a simpler way to construct a triangle that has the same area and is similar to another given triangle ? (every center triangle of a polygon is equal and has $$\frac{1}{n}$$ Area of an original triangle)

the answer is obviously long, so I'm not looking for complete answer, just for $$1$$ tranfsormation of triangle into regular $$n$$-gon. $$n>4$$

• If you mean with straightedge and compass, the half circle is impossible because $\pi$ is not constructible. – Gae. S. May 27 at 21:34
• Thanks, I'll edit the question. Silly me! – bcurious May 27 at 21:36

You can construct any regular polygon with 3 or more sides if you know the area, A, of the triangle you are trying to construct a similar scaled geometric shape from. Using $$A$$, create a regular polygon with n sides of side length:

$$s = \sqrt{\frac{4A\tan(\frac{180}{n})}{n}}$$

and with interior angle being:

$$a = 180 - \frac{360}{n}$$

• interesting formula, but i think it is not constructible with straightedge and compass. I edited the question according to your answer. – bcurious May 28 at 5:34
• Using a pencil, straight edge, and a compass you technically can because you know side lengths and angles. But if you want to use the original triangle as a basis then yeah this is different – justaguy May 28 at 7:27
• area A is not given implicitly – bcurious May 28 at 8:19
• I don't agree with this. For instance, the angle $\frac{360^\circ}7$ is provably impossible to contruct with compass and straightedge. – Gae. S. May 28 at 14:54
• This is fair, if you really want me to I can limit this answer to polygons with nonrepeating rational values for angles and sidelengths – justaguy May 28 at 15:47