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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.

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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.

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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:

enter image description here

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.

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  • $\begingroup$ It's my turn to say : please have a look at my answer... $\endgroup$
    – Jean Marie
    Commented Jan 4 at 18:39
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Here is a figure showing how to transform this issue in order for it to be easily mastered, using inversion :

enter image description here

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 :

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

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