Escape From Trigonometric Maze

The path is the nested function

$Tan(ArcCot(Csch(ArcSinh(Cot(ArcSin(Tanh(ArcCsch(x))))))))=x$

so the following rooms are passed through

$1, 2, 7, 12, 13, 14, 19, 24, 25$

which totals $117$

Let $gd(x)$ be the Gudermannian Function . Then we have the following identities

1) $Sin(gd(x))=Tanh(x)$
2) $Cot(gd(x))=Csch(x)$
3) $Tan(gd(x))=Sinh(x)$

From 1) and 2) we have

$gd(x)=ArcSin(Tanh(x))$
$Cot(gd(ArcCsch(x))=x$
$Cot(ArcSin(Tanh(ArcCsch(x)))=x$

From 3) and 2) we have

$Tan(gd(ArcSinh(x))=x$
$gd(x)=ArcCot(Csch(x))$
$Tan(ArcCot(Csch(ArcSinh(x))))=x$

Combining the two, we have

$Tan(ArcCot(Csch(ArcSinh(Cot(ArcSin(Tanh(ArcCsch(x)))))))=x$

so if one starts out, for example, as $x=0.5$ , one would go through the following values

$0.5, 1.44364, 0.894427, 1.10715, 0.5, 0.481212, 2., 0.463648, 0.5$

and escaping as $0.5$ . The only values $x$ cannot start with is $i, -i$ , or $0$ as that would leave you indeterminate.

For a heuristic in seeking the correct answer, I looked for a path that had equal numbers of "trig and inverse trig" as well as "hyperbolic trig and inverse hyperbolic trig" functions in the appropriate orders such that I could handle "cancelling pairs" from the inside-out as needed (Recognizing that I know identities that allow me to deal with tan(arccos(x)) and arcsinh(cosh(x)) and the like). For example, if we code "trig" as $t$ and "inverse trig" as $t^{-1}$ , hyperbolic trig as $h$ and $h^{-1}$ likewise, I sought paths which would simplify by cancelling only adjacent terms with opposite powers.

Following a quick color-coding it was quick to see that no such path could travel through blocks 3 or 8 (due to the number of inverse hyperbolic segments "blocking" the rest of the route) or through blocks 16 or 17 for a similar reason with trigonometric functions. This left

It turns out that there were only two such paths to worry about:

1 2 7 12 13 14 19 24 25

and

1 6 11 12 13 14 19 24 25

Checking both of them through identities, as many have demonstrated already, shows that the first is correct. If I end up with the time I might demonstrate exactly which identities in which order yield the "least annoying" computations, but this should be apparent from the "cancellations" above.

My primary reason in writing this solution, which is not a complete solution, is to provide one quick heuristic for looking for candidate paths, since this narrowed 70 possible paths down to 2.

Note that $\tan(\cot^{-1}x) = x^{-1} \hspace{2cm} \mathrm{cosech}\,(\sinh^{-1}x) = x^{-1}$ for nonzero real $x$ . Thus $\tan\left(\cot^{-1}\left(\mathrm{cosech}\left(\sinh^{-1}(x)\right)\right)\right) \; = \; x$ for nonzero real $x$ . We only now need to note that $\tanh\big(\mathrm{cosech}^{-1}(x)\big) \; = \; \frac{\mathrm{sgn}(x)}{\sqrt{x^2+1}} \hspace{2cm} \cot\left(\sin^{-1} \left(\tfrac{\mathrm{sgn}(x)}{\sqrt{x^2+1}}\right)\right) \; = \; x$ for real nonzero $x$ , so we see that the correct sequence is $1\;\;2\;\;7\;\;12\;\;13\;\;14\;\;19\;\;24\;\;25$ making the answer $\boxed{117}$ . The inverse trigonometric and inverse hyperbolic functions require careful argument choices to extend to the complex plane, but analytic continuation will be enough to show that $\tan \circ \cot^{-1} \circ\,\mathrm{cosech}\,\circ \sinh^{-1} \hspace{1cm} \mathrm{and} \hspace{1cm} \cot \circ \sin^{-1} \circ \tanh \circ\,\mathrm{cosech}^{-1}$ both act as the identity on $\mathbb{C} \backslash \{0,i,-i\}$ (and not just on the nonzero reals), which is what we want.

I just chose a path which nests inverse function and the function or vice versa together.

The nesting of two is within same type i.e. trigonometric and hyperbolic. Pairing sequence is (1) arccsch, tanh, (2) arcsin, cot , (3) arcsinh, csch and (4) arccot, tan.

Testing with Wolfram Alpha gave positive result. Summing all the digits answer is 1+2+7+12+13+14+19+24+25=117

Here is a Matlab / Octave code for solving this problem:

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dx=0.01;
x=-10+dx/2:dx:10+dx/2;

fun=cell(5,5);
fun{1,1} = x;
fun{1,2} = @(x) acsch(x);
fun{1,3} = @(x) sec(x);
fun{1,4} = @(x) acosh(x);
fun{1,5} = @(x) cot(x);
fun{2,1} = @(x) cos(x);
fun{2,2} = @(x) tanh(x);
fun{2,3} = @(x) atanh(x);
fun{2,4} = @(x) csc(x);
fun{2,5} = @(x) acsc(x);
fun{3,1} = @(x) acos(x);
fun{3,2} = @(x) asin(x);
fun{3,3} = @(x) cot(x);
fun{3,4} = @(x) asinh(x);
fun{3,5} = @(x) asech(x);
fun{4,1} = @(x) cosh(x);
fun{4,2} = @(x) atan(x);
fun{4,3} = @(x) asec(x);
fun{4,4} = @(x) csch(x);
fun{4,5} = @(x) sin(x);
fun{5,1} = @(x) acoth(x);
fun{5,2} = @(x) sech(x);
fun{5,3} = @(x) sinh(x);
fun{5,4} = @(x) acot(x);
fun{5,5} = @(x) tan(x);

comb=perms([0 0 0 0 1 1 1 1]);
comb=unique(comb,'rows');

for i_comb=1:size(comb,1)
    row=1;column=1;
    f=fun{1,1};
    cum_sum=1;
    for i_move=1:8
        if comb(i_comb,i_move)
            row=row+1;
        else
            column=column+1;
        end
        f=fun{row,column}(f);
        cum_sum = cum_sum+5*(row-1)+column;
    end
    if max(abs(f-x))<1e-10
        label={'right','down'};
        disp(label(comb(i_comb,:)+1))
        disp(['Total sum is: ' num2str(cum_sum)])
    end
end

Because each two functions like $ArcCsch$ & $Csch$ is just same as $x \times 1$ which is $x$ .

The path way number is ${1,2,7,12,13,14,19,24,25}$

And the answer is $\boxed{\large{117}}$

How to INCORRECTLY solve this problem: 1) find the highest possible value of your track:1+6+11+16+21+22+23+24+25=149 2) find the lowest possible value of your track: 1+2+3+4+5+10+15+20+25=85 3) find the average of these values: (149+85)/2=117

Probability says this value should be pretty close... and what do you know!