Perimeter change when moving one point of a triangle along a circle

Question

I am interested in the perimeter $P(\Delta')$ of a modified version of a triangle $\Delta = (A,B,C)$. I chose point $B$ to be the point I want to shift towards the inside (or on the boundary) of $\Delta$. I chose a distance $r<h$ where $h$ is the height of the triangle. $B$ can be moved to an arbitrary point $E$ which is of distance $r$ to $B$ and lays inside or on the boundary of $\Delta$. This means that $E$ must also lay on the circle of radius $r$ with the center point $B$. I denote $\Delta'=(A,E,C)$. What is the perimeter $P(\Delta')$ of the new triangle given the original side lengths, $r$ (and probably some representation of the arc length to E)? In what range does $P(\Delta)-P(\Delta')$ lie in?

I attached a small figure which visualizes both triangles.

Edit

Based on the comments, I added some notation:

Assuming $B=(0,0)$, then we have $E=(x,y)$ with $x=\lambda_A x_A + \lambda_C x_C$ and $y=\lambda_A y_A + \lambda_C y_C$ ($\lambda_A+\lambda_C = 1$).

This gives us the following set of equations:

$$r=\sqrt{(0-x)^2 + (0-y)^2} = \sqrt{(\lambda_A x_A + \lambda_C x_C)^2 + (\lambda_A y_A + \lambda_C y_C)^2}$$ $$|AE| = \sqrt{(x-x_A)^2+(y-y_A)^2}$$ $$|CE| = \sqrt{(x-x_C)^2+(y-y_C)^2}$$

Moreover, we have: $$|AB| = \sqrt{x_A^2 + y_A^2}$$ $$|CB| = \sqrt{x_C^2 + y_C^2}$$

The perimeter change is in fact the difference $|AB|+ |CB| - |AE| - |CE|$. I would like to show that $r \leq |AB|+ |CB| - |AE| - |CE|$. I struggle with the proof because I still have six variables $x_A,x_C,y_A,y_C,\lambda_A$ and $\lambda_C$.

Background

This question arises from a computer science problem called lawn mowing/milling, where you are given a cutter (in my case a circular shape) and a Polygon to cover. I noticed that given a tour of the circular cutter one can change the cutter from a circular shape with a radius $s$ to a square with side length $2s$. Afterward, the tour can be shortened a bit. I have already proven a specific length $r$ that I can move the points of the tour "inwards" to shorten the tour. My goal is to find out how much shorter the final tour gets when I modify it. In my example, $A-B-C$ are part of the tour and $B$ is a point which can be moved inward by $r$ to a point $E$. My tour will later be $A-E-C$. I want to find out how much the tour is shortened after my transformation and I hope that it will be shortened by at least $r$.

Answer 1

We can find perimeter change as a function of parameter $ \theta$

Let the circle be centered at (0,0) and radius r.

Let two points on a vertical line be $ ( h,-p),(h,q) $

slant line lengths sum for $\theta_1,$ $ L1=$

$$ = \sqrt { r\cos \theta_1 -h)^2} + \sqrt{( r \sin \theta_1 +p)^2}$$ $$ + \sqrt { r\cos \theta_1 -h)^2} + \sqrt{( r \sin \theta_1-q)^2}$$ For $\theta_2$, $L_2=$ $$ = \sqrt { (r\cos \theta_2 -h)^2} + \sqrt{( r \sin \theta_2 +p)^2}$$ $$ + \sqrt { (r\cos \theta_2 -h)^2} + \sqrt{( r \sin \theta_2-q)^2}$$

The perimeter change is difference $ L_2- L_1$ where $L_2$ pair is marked in red. It depends on two $\theta_ s$ , the constants are $( h,p,q,r).$

Answer 2

(Expanding a comment.)

I would like to show that $$r \leq |AB|+|CB|−|AE|−|CE| \tag1$$

Without additional conditions, $(1)$ may not be true. Writing the inequality as $$|AE|+|CE|\leq |AB|+|CB|-r \tag{1'}$$ leads us to consider the locus of points $P$ satisfying the corresponding equality: $$|AP|+|CP|=|AB|+|CB|-r \tag2$$ This is an ellipse with foci $A$ and $C$, with major radius $\frac12(|AB|+|CB|-r)$. This ellipse separates the points with a larger sum from those with a smaller sum. The described points $E$, then, must lie on the arc where $\bigcirc B$, with radius $r$, overlaps this ellipse; depending upon $r$, this arc may-or-may-not include all (or even any!) points of the arc overlapping the triangle itself.

For instance, this first figure shows a situation with points $E_-$, $E_0$, $E_+$ on $\bigcirc B$ and inside $\triangle ABC$, but respectively inside, on, and outside the ellipse:

That is, we have $$\begin{align} |AE_-|+|CE_-| < |AB|+|BC|-r \quad\to\quad r < |AB|+|BC|-|AE_-|-|CE_-|\\ |AE_0\;|+|CE_0\,| = |AB|+|BC|-r \quad\to\quad r = |AB|+|BC|-|AE_0\,|-|CE_0\;|\\ |AE_+|+|CE_+| > |AB|+|BC|-r \quad\to\quad r > |AB|+|BC|-|AE_+|-|CE_+| \end{align} \tag3$$ so that only $E_-$ and $E_0$ satisfy $(1)$.

With appropriate conditions, we can have a circumstance where the circle-triangle overlap is contained within the ellipse (which seems to be the case OP envisions):

On the other hand, we can have a case where the circle doesn't meet the ellipse at all, so all points on the circle are in the $E_+$ family that don't satisfy $(1)$:

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For a given triangle, the threshold $r$-values that keep the circular arc within the ellipse can be determined by intersecting the circle with the ellipse. Unfortunately, conic-conic intersections like this lead to quartic equations, which are exceedingly messy to solve in general.

Unless/until OP provides additional constraints or context, this is about as far as this investigation will go.

Answer 3

This answer deals with the minimal perimeter issue.

Using slightly different notation, we have a known $\triangle ABC$ and given radius $r_a$ of the circle $\mathcal{A}(A,r_a)$. The circle intersects with $AB$, $AC$ at $D,\ E$, respectively. For the point $F$ along the interior arc $DE$, and the point $G=AF\cap BC$, the minimal perimeter of $\triangle FBC$ is achieved when $F$ is the tangential point, where the ellipse focused at $B,C$ touches the circle $\mathcal{A}$, which also means that $\angle GFB=\angle CFG=\theta$.

Let $|BF|=u,\ |CF|=v$, $|BG|=at,\ |CG|=a(1-t)$ for some $t\in(0,1)$.

Then using the cosine rule for $\triangle ABG$, Stewart’s theorem for $\triangle FBC$ and the properties of the bisector of $\angle CFB$, we can express $t$ in terms of the side lengths $a,b,c$ of $\triangle ABC$ and the radius $r_a$ as the root of the quartic

\begin{align} q_4\,t^4+q_3\,t^3+q_2\,t^2+q_1\,t+q_0&=0 \tag{1}\label{1} , \end{align}

where

\begin{align} q_4 &= (b^2-c^2)^2-4\,r_a^2\,a^2, \\ q_3 &= 4\,c^2\,(b^2-c^2)-4\,r_a^2\,(b^2-2\,a^2-c^2), \\ q_2 &= r_a^2\,(4\,b^2-8\,c^2-5\,a^2)+2\,c^2\,(3\,c^2-b^2), \\ q_1 &= r_a^2\,(a^2-b^2+5\,c^2)-4\,c^4, \\ q_0 &= c^2\,(c^2-r_a^2) \tag{2}\label{2} , \end{align} if the vertices $A,B,C$ in counter-clockwise orientation.

The real roots of \eqref{1} need to be checked for which one gives the actual minimum.

For the attached image, I've got all four real roots, two of them were greater than $1$, one gives the minimal perimeter, and, unfortunately, another one was also in a valid range, but the corresponding perimeter was not minimal.

Also, the curved red, green and blue lines, shown in the image,
emanating from the vertices $A,B$ and $C$, respectively, illustrate the corresponding locus of the points of minimal perimeter for possible values of $r_a,\ r_b$ and $r_c$ for the circles $\mathcal{A}(A,r_a)$, $\mathcal{B}(B,r_b)$ and $\mathcal{C}(C,r_c)$.

Naturally, if the angles of $\triangle ABC$ are less than $120^\circ$, all this curves intersect at the Fermat–Torricelli point $T$, which provides the minimum of $|TA|+|TB|+|TC|$.

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Edit

On the other hand, it is much simpler to find the corresponding radius $r_a$ given some $t\in(0,1)$:

\begin{align} r_a &= \left| \frac{t^2 b^2-(1-t)^2\,c^2} {(1-2t)\sqrt{(c^2-a^2\,t)(1-t)+b^2\,t}} \right| . \end{align}

Note that $t=\tfrac12$ is a special (and simpler) case, which must agree only with $b=c$, and such condition is better considered separately.

For $b=c$, the equation \eqref{1}

factors out as

\begin{align} (1-2t)^2(b^4-r_a^2(b^2-a^2\,t\,(1-t))) &=0 , \end{align}

and the proper root is $t=\tfrac12$, so, for any valid value of $r_a$, $G$ must be the midpoint of $BC$, to get the minimal perimeter of $\triangle FBC$, as expected for the isosceles $b=c$ case.

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Edit 2

Actually, the valid range of $t$ would be a subset of either $(0,\tfrac12)$ or $(\tfrac12,1)$, depending of the side lengths $b$ and $c$.