How many "outward" triangles are needed on a circle so that when one is removed, each triangle consistently overlaps with two triangles on the circle?

Question

Consider a circle, and a set of iscosceles triangles where the angles are specified. These triangles are placed on the circle with one point such that the bisecting line is perpendicular to a line parallel to the circle at the point. How many triangles would you need at the minimum for a specified angle such that when 1 triangle is taken away and the triangles are made arbitrary large that each triangle always overlap with 2 other triangles.

When the triangles are not really triangles and the angle between the two outward lines of the triangle are 180 degrees to each other than the answer to this question I think is 5, 3 if none are taken away. I think this because with four such "triangles" all lines are perpendicular and only cross within the circle not outside of it. I however don't really know how to approach this problem when the angles become smaller.

Answer 1

Let's call the angle at the "top" of the isosceles triangle $\delta$. If you put two of them on a circle with midpoint $M$ at points $A$ and $B$ in the way you described it, you get the following situation:

Let $\alpha$ be the angle at $M$ between $A$ and $B$. The prolonged radii from $M$ to $A/B$ split the respective angle $\delta$ in half. Let $A_1$ be a point on the side of the triangle at $A$ that leans toward the other triangle, and the same for $B_1$.

Then we have $\angle MAA_1 = 180° -\delta/2$ and $\angle MBB_1=180-\delta/2$. That intersection point $X$ exist (and thus the triangles will be overlapping) if and only if

$$\alpha + (180° -\delta/2) + (180° -\delta/2) < 360°.$$

Because if $X$ exists, the $\angle AXB$ must be positive and if added to the left hand side of the above inequality give $360°$ (sum if inner angles in quadrilateral $MAXB$). If the left hand side is exactly $360°$, then lines $AA_1$ and $BB_1$ are parallel. If the left hand sider is bigger than $360°$, those lines meet, but "on the other side" of $A/B$, which doesn't cause the triangles to overlap.

Simplifying the above leads to

$$ \alpha < \delta$$

as the necessary and sufficient condition.

For your problem, assume you have $n$ of your triangles, placed at points $A_1, A_2, \ldots, A_n$ in clockwise order around the circle. Let's call the center angles $\angle A_iMA_{i+1}=: \alpha_i$, for $i=1,2,\ldots,n$, where $A_{n+1}=A_1$. Since the $A_i$ go once around the circle, we have

$$\alpha_1 + \alpha_2 + \ldots + \alpha_n = 360°.$$

If we remove one of the triangles, say the one at $A_i$, then suddenly in clockwise order the triangle at $A_{i-1}$ needs to overlap on it's clockwise side with the triangle at $A_{i+1}$. But we know what the necessary and sufficient condition is for that to happen:

$$\angle A_{i-1}MA_{i+1} = \alpha_{i-1} + \alpha_i < \delta.$$

That must be true for all $i=1,2,\ldots,n$, with $\alpha_0 = \alpha_n$.

If we add all these $n$ inequalities up, each term $\alpha_k$ will appear exactly twice, so we get

$$2(\alpha_1 + \alpha_2 + \ldots + \alpha_n) < n\delta,$$

but we know what the left hand side is, it is $2\times360°=720°$. That leads us to the necessray condition that

$$n > \frac{720°}\delta.$$

But that condition is also sufficient! We can simply choose

$$\alpha_1=\alpha_2=\ldots=\alpha_n = \frac{360°}n,$$

and then for all $i=1,2,\ldots,n$ we have

$$\alpha_{i-1} + \alpha_i = \frac{360°}n + \frac{360°}n = \frac{720°}n < \frac{720°}{\frac{720°}\delta} = \delta,$$

which means that even after removing the triangle at $A_i$, the triangles at $A_{i-1}$ and $A_{i+1}$ still intersect.

So the answer to your problem is, that you need at least $\lfloor \frac{720°}{\delta}\rfloor +1$ triangles.

Answer 2

Here is a figure showing how to transform this issue in order for it to be easily mastered, using inversion :

Fig. 1 : case with $n=10$ triangles where $\alpha = \frac{\pi}{2}$.

Indeed, standard inversion (with the unit circle as the inversion circle and power $1$) defined with complex numbers in the following way :

$$I(z)=\frac{1}{\overline{z}}$$

(see here) transforms each triangle (considered as unbounded) with angle $\alpha$ (such as the red one) into a lens or a petal, the intersection of two circles (such as the blue one) with the same angle $\alpha$ that can be found at three places ; indeed inversion is a conformal mapping, i.e., it preserves angles.

These petals constitute a "flower". The condition you want to be fulfilled is that the petals of this flower are dense enough (as is the case here) in the sense that petal $\# k$ has a non void intersection with petal $\# (k+2)$ for any $k$.

Without entering into a full fledge proof, let us consider the case at hand (which is simple because there is an even number of triangles) : the petals of even rank (or odd rank) must share a common area ; such is the case iff

$$5 \alpha > 2 \pi \ \iff \ 10 \alpha > 4 \pi \ \iff \ n \alpha > 4 \pi $$

(recall : $n$ is the number of triangles).

A result which is equivalent to the condition given in the answer of @Ingix :

$$n \ge \lfloor \frac{4 \pi}{\alpha}\rfloor+1$$