I want to find the area of the perfect triangle, i.e. a triangle with no particularity whatsoever :
- no side shall be equal to another,
- no right angle,
- no obtuse angle.
So I gave myself a segment $[A,B]$ of length $x$ and I drew all the lines and circles which would provide the perfect triangle from happening.
Now I want to put the point $C$ to complete the triangle in one of the two spots $C1$ or $C2$, so he is as far as possible of the forbidden lines and circles.
Remark : I don't want to put it too far north, to avoid the triangle looking particular.
I am trying to calculate the areas of the triangles $ABC1$ and $ABC2$.
Any ideas would be much appreciated.
$\endgroup$ 41 Answer
$\begingroup$For circle 1 (see picture below), if $r$ is the radius of the small circle and $AB=x$, from Pythagoras' theorem applied to triangles $AC_1H_1$ and $BC_1H_1$ we have: $$ \overline{C_1H_1}^2=(x-r)^2-r^2=(x+r)^2-(x-r)^2. $$ Solving we get $r=x/6$ and $\overline{C_1H_1}=\sqrt{2/3}x$, so the area of triangle $ABC_1$ is $\sqrt{1/6}\,x^2$.
A similar reasoning can be made for circle 2, leading to a radius $r=x/8$ and an area of $\sqrt{6}\,x^2/8$.
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