Inside a circle: Four triangles with equal area, 5 unknown angles

Question

In this circle, I have four triangles equal area: $$A_1=A_2=A_3=A_4$$ and 5 unknown angles.Is it possible to find the value these angles?

Given $$\alpha=?$$ $$\beta=?$$ $$\gamma=?$$ $$\delta=?$$ $$\varepsilon=?$$

Answer 1

Following Matthew Daly's comment there is not a unique answer to the problem. $A_1$ and $A_2$ have the same area. If we make $\alpha$ small, the area of $A_1$ and $A_2$ are also small. We can then slide the point where $\delta$ is and the point of $A_4$ on the chord from angle $\beta$ to make the areas of $A_3$ and $A_4$ match $A_1$ and $A_2$. There will be a solution for all $\alpha$ over a range starting at $0$. It stops when the sides of $A_3$ and $A_4$ pass through each other. One could compute the angles where that happens, where the sides of $A_3$ and $A_4$ are coincident.

Answer 2

As mentioned in @MatthewDaly's comment and @RossMillikan's answer, the configuration is not unique, so the values of the angles are not fixed.

Here's an animation taking $\alpha$ from $0^\circ$ to $45^\circ$.

There's no need to show $\triangle A_2$, whose area is automatically equal to that of $\triangle A_1$; also, $\angle \beta = \tfrac12\angle \alpha$, so no need to show that, either. I've also substituted an out-of-the-way angle for $\delta$ in $\triangle A_3$. (Retrieving $\delta$ is easy enough, I suspect.) And I've included the counterpart angle for $\triangle A_4$.

Answer 3

The angles can be express in terms of $\alpha$: $$\beta=\color{red}{0.5\alpha} \quad \text{(inscribed and central angles)}\\ \varepsilon=\color{red}{\frac12(90^\circ-\alpha)} \quad \text{(inscribed and central angles (which is complementary))}\\ \gamma=\color{red}{45^\circ} \quad \text{(inscribed angle and central angle of $90^\circ$)}\\ $$ Since $\beta+\varepsilon=45^\circ$, the longest side of $A_4$ can be found from the isosceles right triangle to be $R\sqrt{2}$.

The base of $A_4$ (call it $x$, which is equal to the base of $A_3$) can be expressed through area: $$S=\frac12R\sqrt{2}x\sin \varepsilon \Rightarrow x=\frac{\sqrt2S}{R\sin(45^\circ-0.5\alpha)}$$ The upper right side of $A_3$ (call it $y$) can be found from Cosine theorem: $$y^2=2R^2-2R^2\sin (90^\circ-\alpha)$$ And finally, $\delta$ can be found from the Sine theorem: $$\frac{y}{\sin \delta}=\frac{x}{\sin(\delta +45^\circ)}.$$ However, the solution looks cumbersome!