# Geometry problem using circular arcs

The construction for the problem is as follows:

• Given some circular arc $$A$$ centred at $$C$$ with an angle $$\theta \geq 180^{\circ}$$ and endpoints $$a,b$$, take some arbitrary point $$t$$ inside the region bounded by the arc and the segment $$ab$$.

• Construct two arcs, $$A_1, A_2$$ within $$A$$ by taking the intersection of the perpendicular bisector of $$at$$ / $$tb$$ and the segment $$aC$$ / $$Cb$$. The endpoints of the arcs are at $$a, t$$ and $$t, b$$ respectively, and the centre at the aforementioned intersection point.

• The problem is to show that $$\mathbf{A_1 + A_2 \leq A}$$ for any choice of $$\mathbf{t}$$.

Construction:

Some initial points are that clearly if $$t$$ is on $$A$$, then $$A_1, A_2$$ just make up the entire arc, and then if $$t$$ is on the segment $$ab$$ then you can show using similar triangles that $$A_1 + A_2 = A$$ in this case. While this problem seems intuitively true, I haven't been able to come up with any way of proving it definitively.

• When you write $A_1+A_2 \le A$ that refers to the lengths of the arcs? Commented Oct 13, 2021 at 12:55
• yes, the two smaller arcs that are contained inside the larger arc. We want to show that the sum of the two smaller arcs is leq the larger arc. Commented Oct 14, 2021 at 0:35
• @burbank. Nice question but I kind of confused. There are maybe some typos. At the second point why would need the perpendicular bisector of $aC$ and $Cb$? Cause I don't see the relevance of that in the picture. And why are names of the end points are $a,b$ while your picture says that they are $A,B$? Commented Oct 15, 2021 at 3:01
• @burbank. And what are the points $F$ and $G$ in the picture, are they also arbitrary points on the bisector? And also, are $\beta$ and $\alpha$ required to be greater than $180^\circ$? Commented Oct 15, 2021 at 3:07
• @burbank. I think I should've pointed it out in my previous comment. I know that $G$ and $F$ are the center of the circular arcs, I am confused that why are they not on $at$ and $tb$? Commented Oct 15, 2021 at 3:18

$$\hspace{1.5cm}$$

The picture above illustrates how $$A_1 + A_2$$ depends on the position of $$T$$. For every $$T$$ there is a point on the vertical axis with the same sum of the arcs. Therefore we can consider a unit circle with center $$C = (0,c)$$ that intersects the horizontal axis at $$(0,\pm x)$$ where $$x=\sqrt{1-c^2}$$ and place $$T$$ at $$(0,y)$$

$$\hspace{3.5cm}$$

The length of the arcs is then given by $$A = 2\arccos{(x)} + \pi$$ and $$A_1 = A_2 = r \cdot \alpha$$ where

$$\begin{matrix} \displaystyle r = \sqrt{a^2 + \left(y - c \left(\frac{a}{x} + 1 \right) \right)^2} \\ \displaystyle \alpha = \arccos{\left( \frac{a + x}{1 - c^2} \, \frac{a + cx(c-y)}{r^2} \right)} \end{matrix} \qquad a = \frac{x}{2} \, \frac{(c-y)^2 - 1}{1 + c(y-c)}$$

So your problem becomes just algebra $$\arccos{(x)} + \frac{\pi}{2} > r \cdot \begin{cases} \alpha \\ (2 \pi - \alpha)\end{cases} \begin{matrix} \text{if} \; 0 < y - c < 1 \\ \text{if} \; 0 < y < c \end{matrix}$$

As mentioned in the OP, the equality $$A_1+A_2=A$$ is obtained when $$t\in ab$$. Thales theorem guarantees in this case that $$AG+FB=CB=AC$$ and that $$\alpha=\beta=\theta$$.

If $$t\notin ab$$, then clearly both $$\alpha$$ and $$\beta$$ are lesser than $$\theta$$. If $$\alpha$$ were equal to $$\theta$$, then $$tG$$ would be parallel to $$CB$$; and as $$tg=tA$$, that would imply that $$t \in ab$$, which is false by assumption. The same argument can be used with $$\beta$$: if equal to $$\theta$$, then $$tF$$ would be parallel to $$AC$$; and as $$tF=tB$$, it would imply that $$t\in ab$$, which is false by assumption.

The above means that $$AG+FB$$ must be greater than $$AC=CB$$ to achieve $$A_1+A_2=A$$. As $$AG=Gt$$ and $$FB=Ft$$, it is the same as stating that $$Gt+Ft$$ must be greater than $$AC=CB$$. I would bet that this is not possible, but I lack a formal proof.

If you choose $$t$$ on the arc $$A$$, then $$At$$ and $$Bt$$ are chords of the circle, and you get $$G=C$$ and $$|AG|=|AC|$$, respectively $$F=C$$ and $$|BC|=|BF|$$. Then assuming you take $$A_2$$ (resp. $$A_1$$) as the smaller arc if $$t$$ is above the line $$BC$$ (resp. $$AC$$), the arcs $$A_1$$ and $$A_2$$ are disjoint except their endpoints, and their union is exactly $$A$$.

I'm not sure what happens as you start with $$t$$ on the arc $$A$$, closer to $$B$$ than to the point $$A$$, then pull $$t$$ towards the point $$A$$. The angle $$\beta$$ remains constant as you move $$t$$ towards point $$A$$, so the length of the arc $$A_1$$ decreases linearly with $$|At|$$, but I'm not sure how to show $$|A_2|$$ doesn't increase faster than $$|A_1|$$ decreases.

(Using upper-case letters for all point names ...) Define $$\alpha := \angle TAB$$, $$\beta := \angle TBA$$, $$\gamma :=\frac12\angle ACB$$, and $$\gamma':=\pi-\gamma$$. (Note that, for $$T$$ strictly inside the big sector of $$\bigcirc C$$, we have $$\alpha+\gamma<\pi$$ and $$\beta+\gamma<\pi$$, so that $$\alpha<\gamma'$$ and $$\beta<\gamma'$$. Also, $$\alpha+\beta<\pi$$.)
A little angle chasing shows that our desired arcs have measures (not lengths) $$\stackrel{\frown}{AB}=2(\pi-\gamma)=2\gamma' \quad \stackrel{\frown}{AT}=2(\pi-\alpha-\gamma)=2(\gamma'-\alpha) \quad \stackrel{\frown}{BT}=2(\gamma'-\beta) \tag1$$ Also, we find, for $$r$$ the radius of $$\bigcirc C$$, $$|\overline{AG}|=\frac{r\sin\beta\sin\gamma}{\sin(\alpha+\beta)\sin(\alpha+\gamma)}\qquad |\overline{BF}|=\frac{r\sin\alpha\sin\gamma}{\sin(\alpha+\beta)\sin(\beta+\gamma)} \tag2$$ (Note that the denominators are positive since each angle-sum is less than $$\pi$$.)
OP's conjectured inequality of arc lengths (for $$T$$ strictly inside the sector) can therefore be manipulated thusly: \begin{align} 0 &\stackrel{?}{<} |\stackrel{\frown}{AB}| - |\stackrel{\frown}{AT}|-|\stackrel{\frown}{BT}| \tag3\\[1em] \iff \quad 0 &\stackrel{?}{<} 2r\gamma' - \frac{2r(\gamma'-\alpha)\sin\beta\sin\gamma}{\sin(\alpha+\beta)\sin(\alpha+\gamma)} -\frac{2r(\gamma'-\beta)\sin\alpha\sin\gamma}{\sin(\alpha+\beta)\sin(\beta+\gamma)} \tag4\\[1em] \iff \quad 0 &\stackrel{?}{<} \gamma'\sin\alpha\sin\beta\sin(\alpha+\beta+2\gamma) \\ &\qquad+ \alpha\sin\beta\sin\gamma\sin(\beta+\gamma) \tag5 \\ &\qquad+\beta\sin\alpha\sin\gamma\sin(\alpha+\gamma) \\[1em] \iff \quad 0 &\stackrel{?}{<} \alpha\,\frac{\sin(\beta+\gamma)}{\sin\alpha} +\beta\,\frac{\sin(\alpha+\gamma)}{\sin\beta} +\gamma'\,\frac{\sin(\alpha+\beta+2\gamma)}{\sin\gamma} \tag6 \end{align}
The only part of $$(6)$$ that becomes negative (and thus poses a threat to the inequality) is the factor $$\sin(\alpha+\beta+2\gamma)$$, when $$2\gamma > \pi-\alpha-\beta$$; that is, when $$\angle ACB > \angle ATB$$, which occurs when $$T$$ lies outside $$\bigcirc ABC$$. Whether that last term can ever overwhelm the first two and make the sum negative is unclear.
(By the way, here's a sanity check: When $$T$$ is on the arc, we know $$\pi-\alpha-\beta=\angle ATB=\frac12\angle ACB=\gamma$$. So, the right-hand side of $$(6)$$ reduces to $$\alpha+\beta-\gamma'$$, and thus to $$0$$, consistent with the fact that the inequality $$(3)$$ becomes an equality in this case.)