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$O$ is intersection of diagonals of the square $ABCD$. If $M$ and $N$ are midpoints of the segments $OB$ and $CD$ respectively, find the value of $\angle ANM$.

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Here is my approach:

Assuming the length of the square is $a$. We have $\tan(\angle AND)=2$ and I draw a perpendicular segment from $M$ to $NC$ and calling the intersection point $H$ then $\tan (\angle MNH)=\dfrac{\frac34a}{\frac a4}=3$ ($MH$ can be found by Thales Theorem in $\triangle BDC$) Hence

$$\angle ANM=180^{\circ}-(\tan^{-1}2+\tan^{-1}3)=45^{\circ}$$


I'm looking for other approaches to solve this problem if it is possible.

Intuitively, If I drag the point $N$ to $D$ and $M$ to $O$ (the angle is clearly $45^{\circ}$ here) then by moving $N$ from $D$ to $C$ and $M$ from $O$ to $B$ with constant speed, I think the angle remain $45^{\circ}$. But I don't know how to prove it.

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    $\begingroup$ You can use the law of cosines to get $\cos x=\frac{\sqrt 2}{2}$ $\endgroup$ Oct 18, 2021 at 1:17

6 Answers 6

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Notice that $\triangle OAM \sim \triangle DAN$, $\angle OAM = \angle DAN$.

So, $\angle NAM = 45^\circ$.

Drop a perp from $N$ to diagonal $BD$.

Notice that $\triangle TMN \cong \triangle OAM$

So, $AM = MN$ leading to $\angle ANM = 45^\circ$.

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    $\begingroup$ Alternative ending: from similarity we obtain $\angle ADM=\angle ANM$ and hence, $AMND$ is a cyclic quadrilateral. Thus, $\angle ANM=\angle ADM=45^\circ$. $\endgroup$
    – richrow
    Oct 19, 2021 at 5:57
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enter image description hereLet a side of the the square be 1 unit.

Let the circle ADN cut BD at Z and AC at Z'.

Note that DN = NC; $AC = BD = \sqrt 2$ and by symmetry, BZ = CZ' = x, say.

By power of a point, $x \times \sqrt 2 = 1 \times 0.5$. This means $ x = \dfrac {\sqrt 2}{4}$.

That is Z is your M and hence, $\angle ANM = \angle ADB = 45^0$

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Not sure how rigorous it is, but here's an alternate "dragging the point" explanation using complex numbers:

Let $\mathbf{a}=(1,0)$ and $\mathbf{b}=(0,-1)$ be unit vectors in the complex plane. As shown in Figure 1, letting $\mathbf{c}=1/2(\mathbf{a}+\mathbf{b})$, then, by the paralellogram law for addition, $\angle \mathbf{0}\mathbf{b}\mathbf{c}=\pi/4$.

For $0\leq t\leq 1$, consider the lines $$\mathscr{L}(t)=Re^{\theta i}\mathbf{b}=Re^{\left(\theta+\frac{3\pi}{2}\right)i}$$ and $$\mathscr{M}(t)=Re^{\theta i}\mathbf{a}=i\mathscr{L}(t),$$ where $R=\sqrt{1+t^2}$ and $\theta=\arcsin\left(\frac{t}{\sqrt{1+t^2}}\right).$ The lines $\mathscr{L}$ and $\mathscr{M}$ are perpendicular, $|\mathscr{L}(t)|=|\mathscr{M}(t)|$, and $$\mathscr{N}(t)=\frac{1}{2}\big(\mathscr{L}(t)+\mathscr{M}(t)\big)=\frac{1}{2}\big((1+i)\mathscr{L}(t)\big)=\frac{\sqrt{2}}{2}e^{\frac{\pi i}{4}}\mathscr{L}(t)$$ is also a line. We need only show that $\mathscr{N}$ is the diagonal line between $\mathbf{c}$ and $\mathbf{a}$ and we're done. This follows from the fact that $\mathscr{L}(1)=1-i$ and $\mathscr{M}(1)=1+i$, so that $\mathscr{N}(1)=\mathbf{a}$. Hence, for all $t$, $$\angle\mathbf{0}\mathscr{\mathscr{L}(t)}\mathscr{\mathscr{N}(t)}=\frac{\pi}{4}. \qquad \square$$

enter image description here $\tag{Fig. 1}$

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Hint: $\;\triangle ANJ\,$ is an isosceles right triangle.

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[ EDIT ] The following is about the second part of OP's question.

Intuitively, If I drag the point $N$ to $D$ and $M$ to $O$  [...]  then by moving $N$ from $D$ to $C$ and $M$ from $O$ to $B$ with constant speed, I think the angle remain $45^{\circ}$.

Let $\,AB=1\,$, $\,DN=z \in \left[0, \frac{1}{2}\right]\,$ and $\,\frac{OM}{OB}=\frac{DN}{DC}\,$ so that $\,z=\frac{1}{2}\,$ corresponds to the original diagram and $\,z=0\,$ corresponds to $\,M \mapsto O, N \mapsto D\,$. Then $\,OM = \frac{OB}{DC} \,DN = \frac{z}{\sqrt{2}}$, and:

$$ \tan\left(\angle AND\right) = \frac{1}{z} \\ \tan\left(\angle MNC\right)=\frac{1+z}{1-z} = -\,\frac{1+\frac{1}{z}}{1 - \frac{1}{z}} = -\,\frac{\tan\left(\frac{\pi}{4}\right)+\tan\left(\angle AND\right)}{1-\tan\left(\frac{\pi}{4}\right)\,\tan\left(\angle AND\right)} \\ = \tan\left(\pi -\left(\frac{\pi}{4}+\angle AND\right)\right) $$

It follows that $\,\angle MNC + \frac{\pi}{4} + \angle AND = \pi\,$, and therefore $\,x = \frac{\pi}{4}\,$ regardless of $\,z\,$.

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    $\begingroup$ Thanks! I really like your answer too, but unfortunately I can't check mark more than one answer. $\endgroup$
    – Etemon
    Oct 19, 2021 at 9:43
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Image

Note that $\small \triangle ADN$ is a right triangle with perpendicular sides in the ratio $1:2$.

Draw $\small NT$ parallel to $\small CA$.

From $\small \triangle DOC$, $\small DT=TO$ (midpoint theorem).

Thus, can you see $\small TM=2\ TN$?

Therefore $\small \triangle MTN$ is also a right triangle with perpendicular sides in the ratio $1:2$.

$\small \therefore\triangle ADN\sim\triangle MTN\implies\angle DAN=\angle TMN\implies ADNM\text{ is cyclic}$

$\small \implies x= \angle ANM=\angle ADM=45^\circ$

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If you rotate the square $ABCD$ clock-wise $90^{\circ}$ around the center $O$ of the square $ABCD$, then vertex $B$ is rotated to vertex $D$, which means that the segment $OB$ is rotated to the segment $OD$, which means that the midpoint $M$ of $OB$ is mapped to the midpoint $K$ of $OD$.

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Consequently, since under the rotation, the vertex $A$ is rotated to the vertex $B$, the segment $AM$ is rotated to the segment $BK$, which means that $AM = BK$ and $AM \, \perp\, BK$.

Furthermore, $KN$ is a midsegment in the triangle $COD$, so $KN = \frac{1}{2}\,OC$ and $KN \, || \, OC$. However, $$BM = \frac{1}{2}\,OB = \frac{1}{2}\,OC = KN \,\, \text{ and } \,\, KN \, || \, BM$$ which means that $BMNK$ is a parallelogram and thus $$MN = BK \,\, \text{ and } \,\, MN \, || \, BK$$ Therefore, $$MN = BK = AM \,\, \text{ and }\,\, MN \, \perp \, AM$$ Consequently, $$\angle\, AMN = 90^{\circ} \,\, \text{ and } \,\, AM = MN$$ which means that the triangle $AMN$ is equilateral right-angled triangle, so $$\angle\, ANM = \angle \, NAM = 45^{\circ}$$

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