Squares are <3

Since we are only interested in ratios of distance, we can arbitrarily set $BN=1$ , so that $NC=2$ and the side of the square is $3$ .

Defining $x=MB$ will give us $MN=\sqrt{1+x^2}$

Drawing a line through $N$ parallel to $MD$ will produce a trapezoid $DMNP$ with $\angle PDM=\angle DMN$ and $\angle MNP=\angle NPD$ .

This means $DP=MN=\sqrt{1+x^2}$ and $PC=3-\sqrt{1+x^2}$

Triangles $ADM$ and $CNP$ are similar, since all their respective sides are parallel, so

$\tan(\angle ADM)=\dfrac{3-x}{3}=\tan(\angle CNP)=\dfrac {3-\sqrt{1+x^2}}{2}$

This is an equation in $x$ , namely $\dfrac{3-x}{3}= \dfrac{3-\sqrt{1+x^2}}{2}$

There are two solutions: $x=\dfrac{12}{5}$ and $x=0$ .

Inly the first solution provides a non-degenerate triangle MBN, so $x=\dfrac{12}{5}$

Therefore $\tan(\angle ADM)=\dfrac{3-x}{3}=\dfrac15=\boxed{0.2}$

Without loss of generality let $|AB| = 3$ and $|AM| = x$ . Then $|DN| = 1$ and $|DM| = 3 - x$ . Now since $\angle NMB = \angle MBC = \angle AMB = \tan^{-1}\left(\dfrac{3}{x}\right)$ , we need to find $x \lt 3$ such that

$\angle AMB + \angle NMB + \angle DMN = \pi \Longrightarrow 2\tan^{-1}\left(\dfrac{3}{x}\right) + \tan^{-1}\left(\dfrac{1}{3 - x}\right) = \pi \Longrightarrow$

$2\tan^{-1}\left(\dfrac{3}{x}\right) = \pi - \tan^{-1}\left(\dfrac{1}{3 - x}\right)$ .

Taking the tan of both sides gives us that

$\dfrac{2 \times \dfrac{3}{x}}{1 - \dfrac{9}{x^{2}}} = - \dfrac{1}{3 - x} \Longrightarrow \dfrac{6x}{x^{2} - 9} = \dfrac{1}{x - 3} \Longrightarrow 6x = x + 3 \Longrightarrow x = \dfrac{3}{5}$ .

Then $\tan(\angle ABM) = \dfrac{x}{3} = \dfrac{1}{5} = \boxed{0.2}$ .

Let's call $\overline{DN}=a \Rightarrow \overline{NC}=2a$

$\overline{BC}=\overline{AB}=\overline{AD}=3a$

$MD=x \Rightarrow AM=3a-x$

Triangles MDN and CGN are similar and triangle BMG is isosceles, therefore $\overline{MG}=\overline{BG} \Rightarrow \sqrt{x^2+a^2}+2\sqrt{x^2+a^2}=3a+2x$

Solving $x$ results in $x_1=0$ and $x_2=\frac {15}{2}a$

$x_1=0$ is not compatible with the initial assumptions and the only solution is $x_2=\frac {15}{2}a$

Therefore $\overline{AM}=\frac{3}{5}a \Rightarrow \tan{\angle{ABM}}=\frac{\overline{AM}}{\overline{AB}}=\frac{1}{5}$

Construction: Produce $MN$ and $BC$ towards right such that they intersect at a point $G$ .

Since triangles $MDN$ and $NCG$ are right angled and angle $DNM\ =$ angle $CNG$ , they are similar.

Therefore, we have $\frac{CN}{ND}=\frac{CG}{DM}=\frac{NG}{NM}$

Subsequently, $\frac{2k}{k}=\frac{CG}{DM}=\frac{NG}{NM}$

$\implies CG = 2DM$ & $NG = 2NM$ .

Let the side of the square be $a$ , then immediately we have $k = \frac{a}{3}$ .

Since angle $NMB =$ angle $MBC$ , we have $MG = BG$

i.e. $MN + NG = BC + CG$ .

$\implies MN + 2MN = a + 2DM \implies 3MN = a + 2(a-AM)$

$\implies 3\sqrt{DN^2+DM^2}=\ 3a-2AM$

$\implies 3\sqrt{\left(\frac{a}{3}\right)^2+\left(a-AM\right)^2}=\ AM\left(3\left(\frac{a}{AM}\right)-2\right)$

$\implies 3AM\sqrt{\left(\frac{1}{3}\left(\frac{a}{AM}\right)\right)^2+\left(\frac{a}{AM}-1\right)^2}=\ AM\left(3\left(\frac{a}{AM}\right)-2\right)$

$\implies 3\sqrt{\left(\frac{1}{3}\left(\frac{a}{AM}\right)\right)^2+\left(\frac{a}{AM}-1\right)^2}=\ 3\frac{a}{AM}-2$

From triangle $ABM$ , $\tan \theta = \frac{a}{AM}$ , therefore:

$\implies 3\sqrt{\left(\frac{1}{3}\tan \theta \right)^2+\left(\tan \theta -1\right)^2}=\ 3\tan \theta -2$

$\implies 9\left(\frac{1}{9}\tan ^2\theta +\tan ^2\theta -2\tan \theta +1\right)=\left(\ 3\tan \theta -2\right)^2$

$\implies \tan ^2\theta -6\tan \theta +5=0 \implies \tan \theta = 1$ or $\tan \theta = 5$ .

Note that if $\tan \theta = 1 \implies \theta = \frac{\pi}{4}$ , then triangle $MDN$ will degenerate. But it wont't degenerate for $\tan \theta = 5$ . So the former is neglected and the latter is taken.

Clearly, $\tan ABM = \tan(\frac{\pi}{2} - \theta) = \cot \theta = \frac{1}{\tan \theta} = \boxed{\frac{1}{5}}$

---

This question is an altered version of the $2^{nd}$ question from 2017's South-East Asian Mathematical Olympiad sample paper for Seniors (Code F).

Suppose the side length of the square is 3, and that $B$ is at the origin. Extend $BC$ and $MN$ to their intersection point, which I will call $Q$ , and label $M=(x,3)$ . Then triangle $BQM$ is isosceles, and $Q=(9-2x,0)$ . $BQ \cong MQ$ so

$9-2x = \sqrt{9+(9-3x)^2}$ ,

which becomes

$5x^2 - 18x + 9 = 0$ ,

which has the two solutions $x=3$ and $x = \frac{3}{5}$ . The non-degeneracy hypothesis causes us to discard $x=3$ . The tangent of the desired angle is therefore

$\frac{\frac{3}{5}}{3} = \frac{1}{5}$ .

Let $\alpha = \angle ABM$ . Then $\angle AMB = \angle CBM = \angle BMN = 90^\circ - \alpha;\ \ \ \angle DMN = 180^\circ - 2(90^\circ - \alpha) = 2\alpha.$ Let the side of the square be $z$ , then $AM = z\tan\alpha;\ \ \ MD = z(1 - \tan\alpha);\ \ \ ND = \frac z 3.$ We have $\tan \angle DMN = \frac{ND}{MD}\ \ \ \ \therefore\ \ \ \ \tan 2\alpha = \frac{z}{3z(1- \tan\alpha)}$ $\ \ \ \ \therefore\ \ \ \ \frac{2\tan\alpha}{1-\tan^2 \alpha} = \frac{1}{3(1-\tan \alpha)} \ \ \ \ \therefore\ \ \ \ \frac{2\tan\alpha}{(1-\tan \alpha)(1 + \tan\alpha)} = \frac{1}{3(1-\tan \alpha)}$ $\ \ \ \ \therefore\ \ \ \ 2\cdot 3\tan\alpha = 1 + \tan\alpha \ \ \ \ \therefore\ \ \ \ 5\tan \alpha = 1 \ \ \ \ \therefore\ \ \ \ \tan\alpha = \boxed{\frac 1 5}.$

Got it correct on a complete whim. I don't know advanced math but I took the ratio of DN : AM as 1:0.6 AB = 3DN That gave me the ratio tan of angle ABM as 0.6 ÷ 3 = 0.2 Sorry if this answer isn't good enough.

I couldn't resist finding a solution which exploited the isosceles triangle - which isn't there but can soon be constructed!

Extend MN and BC to meet at a new point E.

Notice that $\triangle MEB$ is isosceles and that the right-angled triangles $\triangle MDN$ and $\triangle ECN$ are similar with a ratio of 1 to 2.

Suppose the square has side length three, and let $AM=x$ so that $MD=3-x$ and by Pythagoras $MN=\sqrt{(x^2-6x+10)}$

The similarity of the two right angles lets us to scale these lengths up to get

$CE=6-2x$ and $NE=2\sqrt{(x^2-6x+10)}$

Now, remembering that $BC=3$ we can equate the two sides of the isosceles triangle to get

$9-2x=3\sqrt{(x^2-6x+10)}$

Straight forward manipulation converts this to the quadratic

$5x^2-18x+9=0$

which has two solutions $x=3$ and $x=\frac{3}{5}$

The first of these is the degenerate solution which we are to ignore.

Referring to the diagram, and using $AB=3$ we see that the required tangent is $\frac{x}{AB}=\boxed{\frac{1}{5}}$

Let $DN = x$ and $MD = y$ and $\angle MBC = \theta$

$\implies \dfrac{3x}{3x - y} = \tan\theta$ and $\dfrac{x}{y} = \tan(180 - 2\theta) = -\tan(2\theta) = \dfrac{2 \tan\theta}{\tan^2\theta - 1} \implies$

$\dfrac{x}{y} = \dfrac{6x (3x - y)}{y (6x - y)} \implies 6x - y = 18x - 6y \implies y = \dfrac{12x}{5}$

$\implies \tan\theta = 5 \implies \tan \angle ABM = \tan(90 - \theta) = \cot\theta = \dfrac{1}{5} = \boxed{0.2}$