Geometry problem using circular arcs

Question

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.

Answer 1

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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 $(0,y)$

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

Answer 2

Partial answer

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.

Answer 3

Another partial answer

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.

Answer 4

Yet another partial answer.

(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.)

That's about as far as my analysis has gone.