The hardest geometry question with "a triangle" and "a circle" - Circle intersecting triangle equally in 5 parts I received this question long time ago from one of my old friends who is mathematician/physicist. He called it the hardest geometry question with "a triangle" and "a circle". I am not sure if he know the answer or not. Here it is.




Note: These picture are just the example of possible answer. The actual answer might look like one of these or not like any of these.
The circle intersects the triangle and divides the area into 5 parts. Each 5 parts of the red-fill areas have equal area. The radius of circle is 1.
That is all the information the question gives. Then the question asks these two things.

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*Find the distance from the center of the circle to the centroid of the triangle. If there are more than 1 possible solutions, find longest and shortest possible the distance from the center of the circle to the centroid of the triangle.

*Find length of outer perimeter of the new intersecting shape. If there are more than 1 possible solutions, find longest and shortest possible length of outer perimeter of new intersecting shape.

Note : If you think that there is only 1 possible solutions that the triangle is isosceles, you also have to proof that there is really only 1 possible solutions.
I have to say that I don't even know how to start to solve this question. The question doesn't even specify the type of the triangle. I think the triangle has to be isosceles triangle, but I remember that the question doesn't specify it.
Edit: I realize where I get this question from. And for now, I doubt if the triangle has to be isosceles or not.
Edit(2) : Due to some technical reason, the account of the real original poster is merged to my account. That is why there is a delay of the update on this question. I do contact and take to the real original poster. Here is why he edit this question.
After sometimes, I decide to change the original question a bit by adding more picture. The original question is actually tricky that give only the first image and trick anyone who try to solve it that that is how the answer look like. R. J. Mathar and John Bentin answer is good but that isn't what the original question ask for.
Edit(3) : The question might not have analytical solution. So, if there is no analytical solution, numerical solution is acceptable.
 A: There are enough degrees of freedom in this problem to choose the triangle to be isosceles and symmetrically placed with respect to the circle, as you have done, for simplicity. Choose units so that the radius of the circle is $1$. Cut the diagram symmetrically in half with a line through the centre of the circle and the apex of the triangle, which perpendicularly bisects the base of the triangle.
Let us measure angles around the circle anticlockwise from the downward ray of the bisector. Say the base cuts the circle at a point with circular angle $\alpha$, while the right-hand apical side meets the circle where the corresponding angle is $\theta$. From the given equality of areas, we get
$$\alpha-\cos\alpha\sin\alpha=\tfrac13\pi.\qquad\qquad(1)$$After some comparison of areas, from the given conditions, we can obtain
$$(\tfrac23\pi-\theta+\cos\alpha\sin\theta)^2=\pi(\tfrac23\pi-\theta+\cos\theta\sin\theta).\quad(2)$$
Now, mixed algebraic–trigonometric equations like these cannot be solved by Euclidean geometry. The best we can do is use a numerical approach, such as Newton–Raphson, first by finding $\alpha$ to the required degree of accuracy from eqn 1 and then solving eqn 2 using this value of $\alpha$.
Uniqueness of solution: Equation 1 may be written
$$\sin2\alpha=2\alpha-\tfrac23\pi.$$
For $0<\alpha<\frac13\pi$, the LHS is positive while the RHS is negative. In the interval from $\frac13\pi$ to $\frac12\pi$, the LHS continually decreases, with negative gradient, from $\frac12\surd3\approx0.866$ to $0$, while the RHS correspondingly increases, with positive gradient, from $0$ to $\frac13\pi\approx1.047. $ Hence there is a unique solution for $\alpha$ in this interval. Between $\frac12\pi$ and $\pi$, there is again a mismatch of sign between the sides. Therefore the solution for $\alpha$ is unique: Newton–Raphson gives $\alpha\approx1.302663$, with $\cos\alpha\approx0.264932$.
By the orientation of the triangle, $\theta>\alpha$. In the range $\alpha<\theta<\frac12\pi$, the LHS of eqn 2 is no more than $(\frac23\pi-\alpha+\cos\alpha)^2\approx1.12$, while the RHS decreases to a minimum of $\frac16\pi^2\approx1.645$. So the solution lies above $\theta=\frac12\pi$, where the LHS decreases (tangentially) to its minimum value $0$ at $\theta=\beta\approx2.293159$. At that point, the RHS is negative, so the solution is in the range $\frac12\pi$ to $\beta$. In this range, the LHS is increasing while the RHS is decreasing. It follows that the solution for $\theta$ is unique.
The general case: Five parameters need to be fixed to specify a triangle in a plane, given a reference orientation, but only four constraints in the conditions are given for the areas. Hence there is a one-dimensional space of solutions if we drop the requirement that the triangle is isosceles. A manageable case is the one where two sides trisect the circle while the other lies outside the circle. While I don't think that this would be significantly harder to address than the symmetric case, it naturally forms a separate question, which should be posted as such. The asymmetric case where a vertex lies inside the circle looks ugly to me, and not much fun to answer. The borderline case where a vertex lies on the circle is as easy to deal with as the symmetric case, with a specific solution. Again, that should be posted as a separate question.
A: We prove here the non-unicity of the required triangles. It is clear that if two circular segments on a circle have same area, also have same length of chord. In the unitary circle we are concerned with two disjoint circular segments having area $\dfrac{\pi}{3}$ because the circle has area equal to $\pi$. The involved chord have length, noted $a$, given by the system
$$\dfrac{\theta-\sin(\theta)}{2}=\dfrac{\pi}{3}\\a=2\sin\left(\frac{\theta}{2}\right)$$ from which we have the data for the problem to solve (see at attached figure)
$$\begin{cases}\theta\approx 2.60533\\ a\approx 1.92853529\end{cases}$$

To determine the outer region having area equal to $\dfrac{\pi}{3}$ we have five unknowns, , $\alpha,\beta, \gamma,x$ and $y$ for which we have the following system of five equations.
$$\begin{cases}2\theta+\beta+\gamma=2\pi\\\\\alpha=\dfrac{\gamma-\beta}{2}\\\\(a+y)y=(a+x)x\\\\\dfrac{xy\sin(\alpha)-(\beta-\sin(\beta))}{2}=\dfrac{\pi}{3}\\\\(2\sin(\beta))^2=x^2+y^2-2xy\cos(\alpha)\end{cases}$$
From which $\gamma=\pi-\theta+\alpha=0.536262653+\alpha$ and $\beta=\pi-\theta-\alpha=0.536262653-\alpha$.
Put for confort $A=0.536262653$ and $B=\dfrac{2\pi}{3}=2.0943951$ so we have  the system  of three unknowns.
$$\begin{cases}(a+y)y=(a+x)x\\\\{xy\sin(\alpha)-A+\alpha+\sin(A-\alpha)}=B\\\\\left(2\sin(A-\alpha)\right)^2=x^2+y^2-2xy\cos(\alpha)\end{cases}$$
Note that $y=x$ because the chords $\overline{AB}=\overline{CD}$ (or because $(a+y)y=(a+x)x\iff(y-x)(y+x-a)=0)$ so we have the resultant in $\alpha$
$$x^2=\dfrac{B+A-\alpha-\sin(A-\alpha)}{\sin(\alpha)}=\dfrac{2\sin^2(A-\alpha)}{1-\cos(\alpha)}$$ i.e.
$$\dfrac{2.6306577-\alpha-\sin(0.5362626-\alpha)}{\sin(\alpha)}=\dfrac{2\sin^2(0.5362626-\alpha)}{1-\cos(\alpha)}$$ from where we get  $\alpha\approx 0.203$ radiands.
This value of $\alpha$ determines the position of point $P$ in the figure because the chord $\overline{AB}$ should have constant length $a\approx 1.92853529$ (optionnally one can calculate $x$ having $\alpha$). This position of vertex $P$ is unique because if the point $A$ change of position then the outer region $PBD$ becomes larger or smaller than $\dfrac{\pi}{3}$.
Regarding the other outer part of the triangle, that one with two vertices, the mean value theorem ensures the existence of a lot of possible shapes according with the angle we consider between two sides of the triangle. The effective calculation I suppose can be made in analogous way to that for region $BDP$.
