# Can an angle between the subdividing segments and the edges of a triangle be determined only by interior angles and the intersection of the segments?

I'm working on a robotics project, where a body $$(D)$$ is to be linked to three anchor points $$(A, B, C)$$ with segments of variable length. The movement envelope is the area of the $$\triangle ABC$$, which itself is of an arbitrary, but fixed shape and size.

Now, I want to retain the orientation of $$D$$ during movement, and I pondered whether this was possible without using electronics and extra actuators for compensating for the rotation around an anchor point that would arise from referencing one of the connecting segments as an origin axis.
To do this, I would need to find an angle $$\kappa$$ that compensates for this rotation, say, around $$A$$. Simply put, I want to find an angle between a parallel of $$AC$$, and $$DA$$, based only on the angles between the segments $$\overline{AD},\overline{BD},\overline{CD}$$, labeled $$\alpha$$, $$\beta$$, and $$\gamma$$ and the known internal angles of $$\triangle ABC$$.

As is readily apparent on the image, this angle $$\kappa$$ is a so-called "Z-angle" of $$\eta$$, so the problem boils down to finding a single angle between one of the three connecting segments $$\overline{AD},\overline{BD},\overline{CD}$$ and one of the edges of the triangle.

At first glance, it seemed to me that the six angles would determine the result uniquely, but I've been thinking around in circles and starting to doubt myself. There are obviously some redundancies in these six parameters, like how $$\alpha,\beta,\gamma$$ are really only two angles subdividing a full rotation, as are $$\delta, \varepsilon, \zeta$$. I'm still confident this should be possible, it seems the triangle can't have any other shape or orientation, given the six (four) angles.
(Edit: Mucking around in Fusion360 has revealed that the size of the triangle is not constrained, however that still leaves the sought angles constant, which leads me to believe this problem is fully constrained!)
However, solving a linear system of equations on the internal angles of the three smaller triangles yields results where at least one of the segment-edge angles are a parameter. For example
\begin{align}\gamma + \delta - a_1 + b_1 - 180 &= 0\\\beta + \varepsilon - b_1 + c_1 - 180& = 0\\\alpha + \zeta + a_1 - c_1 - 180 &= 0\end{align} (where $$a_1, b_1, c_1$$ are the clockwise facing angles between the segments and edges, a_1 corresponding to $$\eta$$ on the second image) yields results like $$b_1 = a_1 - \gamma - \delta + 180$$ and $$c_1 = a_1 - \beta - \gamma - \delta - \varepsilon + 360$$
In all of these, another unknown segment-edge angle pops up, suggesting the solution is not unique. Can somebody help me understand the possibility/impossibility of this problem, and where my thinking went wrong? Is there some clever identity I'm missing?

• Just calculate the circumcenter X of ACD and the circumcenter Y of BCD. Then D is symmetric to C with respect to the line XY. Commented May 12, 2020 at 23:34
• These circumcenters can be easily found. For instance, X is (the unique) point on perpendicular bisector of AC such that the angle AXC is $360^\circ - 2\alpha$ Commented May 12, 2020 at 23:38
• @timon92 I did that, and what you say is true, but sadly I fail to see how these two points, the bisectors and AXC get me closer to expressing $\eta$ (or ?) in terms of $\alpha, \beta, \gamma, \delta, \epsilon, \zeta$. Commented May 14, 2020 at 14:17
• Also, to construct the circumcenters, one would have to have access to either the length of the segments DA, DB, DC, or an angle at a vertex opposing these segments (in ADC, BDA, CDB respectively), none of which I have access to in the problem. Commented May 14, 2020 at 14:54
• The construction shows that point D and the angle eta are uniquely determined. I think there is no "nice" formula for eta. However, using trigonometry, you can find an equation connecting eta with the other angles. Using sine rule we get: sin eta/sin alpha = DC/AC, sin beta/sin CBD = BC/DC, sin epsilon/sin delta = AC/BC. Multiplying these and noting that CBD = 360-alpha-beta-zeta-eta, we get the following: sin eta sin beta sin epsilon = sin alpha sin delta sin(360-alpha-beta-zeta-eta). So all you have to do is to solve this equation for eta. Let me know if you need more assistance Commented May 14, 2020 at 15:24

Using only angles in a triangle is not enough to tell anything.
If you only know the inner star by angles $$\alpha,\beta$$ then a simple rotation and translation can be made on point $$D$$ to get point $$D'$$ and the star still goes through vertices $$A,B,C$$ with different segment lengths.

A length or other no-angle condition is needed.

Consider this sketch:

Using the Law of Sins: $$\frac {\lVert AC \rVert}{\sin\ \alpha} \ = \ \frac {\lVert DC \rVert}{\sin\ \vartheta _1}\ \ \implies \ \ \lVert DC \rVert = \frac {\lVert AC \rVert}{\sin\ \alpha} \ {\sin\ \vartheta _1}$$

$$\frac {\lVert BC \rVert}{\sin\ \beta} \ = \ \frac {\lVert DC \rVert}{\sin\; \theta _4}\ \ \implies \ \ \lVert DC \rVert = \frac {\lVert BC \rVert}{\sin\ \beta} \ {\sin\ \vartheta _4}$$

So from both expressions of $$\lVert DC \rVert$$

$$\sin\ \theta _1 \ = \ \frac {\lVert BC \rVert}{\lVert AC \rVert} \frac {\sin\ \alpha}{\sin\ \beta} \ \sin \vartheta _4$$

Because the ABC shape is constant we know angle $$\angle ACB$$. Let's play with angles:

$$\vartheta _4 \ =\ \frac {\pi}{2} - \beta - \theta _3$$ $$\vartheta _3 = \angle ACB - \theta _2$$ $$\vartheta _2 = \frac {\pi}{2} - \alpha - \vartheta _1$$ So $$\theta _4 \ =\ \frac {\pi}{2} - \beta - \angle ACB + \frac {\pi}{2} - \alpha - \theta _1 \ \ \ =\ \pi - \beta - \angle ACB - \alpha - \vartheta _1 \ =\ k-\vartheta _1$$ Where $$k= \pi - \beta - \angle ACB - \alpha$$

Apply 'sin' and expressing the sin of a difference:

$$\sin\ \vartheta _4 \ =\ \sin(k-\vartheta _1) \ =\ \sin\ k\ \cos\ \vartheta _1 \ -\ \cos\ k\ \sin\ \vartheta _1$$

We could transform this to get other expression, just with $$sin \theta _1$$ unknown: $$\ =\ \sin\ k\ \sqrt{1-\sin^2\vartheta_1} \ -\ \cos\ k\ \sin\ \vartheta _1$$ but we really don't need it.

Now we have $$\sin \vartheta _4$$ defined, let's subsitute it in the former expression for $$\sin \vartheta _1$$

$$\sin \ \vartheta _1 \ =\ \frac {\lVert BC \rVert}{\lVert AC \rVert} \frac {\sin\ \alpha}{\sin\ \beta} (\sin\ k\ \cos\ \vartheta _1 \ -\ \cos\ k\ \sin\ \vartheta _1)$$

The last step is solving this function for $$\vartheta _1$$

$$\sin\ \vartheta_1 \ =\ F \cos\ \vartheta _1 - J \sin \ \vartheta _1$$

Where $$F$$ and $$J$$ are calculated directly from the given data: $$F\ =\ \frac {\lVert BC \rVert}{\lVert AC \rVert} \frac {\sin\ \alpha}{\sin\ \beta} sin\ k$$ $$J\ =\ \frac {\lVert BC \rVert}{\lVert AC \rVert} \frac {\sin\ \alpha}{\sin\ \beta} \cos\ k$$

So: $$F \cos\ \vartheta _1 = (1+J)\sin \ \vartheta _1$$ $$\tan\ \vartheta _1 \ =\ \frac {F}{(1+J)}$$

Be aware of the case $$J= \ -1$$ wich means that $$\vartheta _1 = \ \pi/2$$

• Nice complete answer, could you please elaborate a bit on the motivation behind the introduction of F, and J? k I could follow as a sum identity, but not the last couple of steps. Commented May 19, 2020 at 22:30
• @MártonPósa Done. I've edited my answer. Commented May 19, 2020 at 23:12
• @Cheesecake Is there any particular reason for switching over to the $\vartheta$ symbol towards the end of the calculation? Commented May 22, 2020 at 17:37
• @MártonPósa, this font is more common, at least in most of the literature I've seen so far, so it's more about estetics. My aim was to add var in every place, but I missed a few. I corrected it. Commented May 22, 2020 at 18:20