Mathematical Cavalcade revisited

I found this problem in Brian Bolt’s well-known book, entitled “Mathematical Cavalcade”. The book is a collection of interesting and very amusing recreational mathematical problems. This is the only puzzle in the book where I spotted a flaw. Hints and solutions are given in a commentary at the back of the book. There, the answer stated is $42$ . As we shall see, a better solution exists. figure 1

Let $O$ be the center of the circular beaker’s base (figure 1). The radius of the circle is $R= \dfrac{8}{\pi }$ . Brilli the ant can reach the base of the beaker, say at point $D$ , between $B$ and $E$ ( $AE$ is tangent to the circle).
Denote the length of $\overset\frown{BD}$ by $s$ and let $\angle AOD=\theta$ . Then, $\theta =\dfrac{s}{R}$ and by cosine rule $A{{D}^{2}}=A{{O}^{2}}+{{R}^{2}}-2AO\cdot R\cdot \cos \theta$ $\Rightarrow AD=\sqrt{{{\left( 25+\dfrac{8}{\pi } \right)}^{2}}+{{\left( \dfrac{8}{\pi } \right)}^{2}}-2\left( 25+\dfrac{8}{\pi } \right)\cdot \dfrac{8}{\pi }\cdot \cos \left( \dfrac{\pi s}{8} \right)} \ \ \ \ s\in I:=\left[ 0,\ \dfrac{8}{\pi }\cdot {{\cos }^{-1}}\dfrac{8}{25\pi +8} \right] \approx \left[ 0,\ 3.76 \right] \ \ \ \ \ (1)$

If we flatten out half of the curved surface of the beaker we come with a rectangle $BFGH$ , with point $C$ on $FG$ (figure 2). Reflect point $C$ on $GH$ to point ${C}'$ .

figure 2 Suppose Brilli the ant takes the route $D\to P\to C$ .
Then, $DP+PC=DP+P{C}'\ge D{C}'=\sqrt{{{\left( 8-s \right)}^{2}}+{{15}^{2}}} \ \ \ \ \ (2)$

Combining $(1)$ and $(2)$ , we have the minimum distance from $A$ to $C$ as a function of $s$ :

$f\left( s \right)=\sqrt{{{\left( 25+\dfrac{8}{\pi } \right)}^{2}}\text{ }+\text{ }{{\left( \dfrac{8}{\pi } \right)}^{2}}\text{ }-\text{ }2\cdot \left( 25+\dfrac{8}{\pi } \right)\cdot \dfrac{8}{\pi }\cdot \cos \left( \dfrac{s\cdot \pi }{8} \right)}+\sqrt{\left( 8\text{ }-\text{ }s \right){}^\text{2}\text{ }+\text{ }225},\ \ \ \ s\in I$ The derivative of $f$ is

figure 3 ${f}'\left( s \right)=\dfrac{\left( 8+25\pi \right)\cdot \sin \left( \dfrac{\pi s}{8} \right)}{\sqrt{-16\left( 8+25\pi \right)\cdot \cos \left( \dfrac{\pi s}{8} \right)+625{{\pi }^{2}}+400\pi +128}}-\dfrac{8-s}{\sqrt{{{s}^{2}}-16x+289}}$ ${f}'\left( 0 \right)=-\frac{8}{17}<0$ and this ensures that there exists an interval ${{I}_{1}}:= \left[ 0,\ {{s}_{0}} \right) \subseteq I$ , where the continuous function ${f}'$ outputs negative values, therefore $f$ is decreasing in $I_1$ .

In other words, for $s\in I_1$ , $s\ne 0$ ,
$f\left( s \right)<f\left( 0 \right)=42$ In fact, at ${{s}_{1}}\approx 1.01$ $f$ gives its minimum, which is $f\left( {{s}_{1}} \right)\approx 41.77$ (figure 3).

Hence, the correct answer is $\boxed{x<42}$ .

I used the theory that the shortest distance between two points is a line: $d = \sqrt{40^2 + 8^2 }=40.8<41.77$ . What am I missing here?

Cool problem!

The shortest path length must be less than $25+\dfrac {16}{2}+(9-6)$ or $36$ cm. Hence it is obviously less than $42$ cm.

Ok My answer it's pretty different from all of this, first I flattened out the beaker then I noticed that the point c was in the middle so I use the hypotenuse=√9²+8²=12 then to that I added the 25cm that is between point A to point B and the result was 12+25=37 so the answer will be x<42 But please help because I think I'm wrong :'(