# The 31st Nordic Mathematical Contest 

## Monday, 3 April 2017

## Solutions

Problem 1 Let $n$ be a positive integer. Show that there exist positive integers $a$ and $b$ such that:

$$
\frac{a^{2}+a+1}{b^{2}+b+1}=n^{2}+n+1
$$

Solution 1 Let $P(x)=x^{2}+x+1$. We have $P(n) P(n+1)=\left(n^{2}+n+1\right)\left(n^{2}+3 n+3\right)=$ $n^{4}+4 n^{3}+7 n^{2}+6 n+3$. Also, $P\left((n+1)^{2}\right)=n^{4}+4 n^{3}+7 n^{2}+6 n+3$. By choosing $a=(n+1)^{2}$ and $b=n+1$ we get $P(a) / P(b)=P(n)$ as desired.

Problem 2 Let $a, b, \alpha, \beta$ be real numbers such that $0 \leq a, b \leq 1$, and $0 \leq \alpha, \beta \leq \frac{\pi}{2}$. Show that if

$$
a b \cos (\alpha-\beta) \leq \sqrt{\left(1-a^{2}\right)\left(1-b^{2}\right)}
$$

then

$$
a \cos \alpha+b \sin \beta \leq 1+a b \sin (\beta-\alpha)
$$

Solution 2 The condition can be rewritten as

$$
a b \cos (\alpha-\beta)=a b \cos \alpha \cos \beta+a b \sin \alpha \sin \beta \leq \sqrt{\left(1-a^{2}\right)\left(1-b^{2}\right)}
$$

Set $x=a \cos \alpha, y=b \sin \beta, z=b \cos \beta, t=a \sin \alpha$. We can now rewrite the condition as

$$
x z+y t \leq \sqrt{\left(1-x^{2}-t^{2}\right)\left(1-y^{2}-z^{2}\right)}
$$

whereas the inequality we need to prove now looks like

$$
x+y \leq 1+x y-z t
$$

Since $x, y, z, t \geq 0$, and $1+x y-z t=1+a b \sin (\beta-\alpha) \geq 0$, we can square both sides of both inequalities, and get equivalent ones. After a couple of cancelations the condition yields

$$
2 x y z t \leq 1-x^{2}-y^{2}-z^{2}-t^{2}+x^{2} y^{2}+z^{2} t^{2}
$$

so that

$$
x^{2}+y^{2}+z^{2}+t^{2} \leq(x y-z t)^{2}+1
$$

which is equivalent to

$$
x^{2}+y^{2}+z^{2}+t^{2}+2 x y-2 z t \leq(1+x y-z t)^{2}
$$

or

$$
(x+y)^{2}+(z-t)^{2} \leq(1+x y-z t)^{2}
$$

Since $(x+y)^{2} \leq(x+y)^{2}+(z-t)^{2}$, the desired inequality follows.

Problem 3 Let $M$ and $N$ be the midpoints of the sides $A C$ and $A B$, respectively, of an acute triangle $A B C, A B \neq A C$. Let $\omega_{B}$ be the circle centered at $M$ passing through $B$, and let $\omega_{C}$ be the circle centered at $N$ passing through $C$. Let the point $D$ be such that $A B C D$ is an isosceles trapezoid with $A D$ parallel to $B C$. Assume that $\omega_{B}$ and $\omega_{C}$ intersect in two distinct points $P$ and $Q$. Show that $D$ lies on the line $P Q$.

Solution 3 Let $E$ be such that $A B E C$ is a parallelogram with $A B \| C E$ and $A C \| B E$, and let $\omega$ be the circumscribed circle of $\triangle A B C$ with centre $O$.

It is known that the radical axis of two circles is perpendicular to the line connecting the two centres. Since $B E \perp M O$ and $C E \perp N O$, this means that $B E$ and $C E$ are the radical axes of $\omega$ and $\omega_{B}$, and of $\omega$ and $\omega_{C}$, respectively, so $E$ is the radical centre of $\omega$, $\omega_{B}$, and $\omega_{C}$.

![](https://cdn.mathpix.com/cropped/2024_06_06_5b88ce828e68ab241c51g-2.jpg?height=867&width=664&top_left_y=1304&top_left_x=709)

Now as $B E=A C=B D$ and $C E=A B=C D$ we find that $B C$ is the perpendicular bisector of $D E$. Most importantly we have $D E \perp B C$. Denote by $t$ the radical axis of $\omega_{B}$ and $\omega_{C}$, i.e. $t=P Q$. Then since $t \perp M N$ we find that $t$ and $D E$ are parallel. Therefore since $E$ lies on $t$ we get that $D$ also lies on $t$.

Alternative solution Reflect $B$ across $M$ to a point $B^{\prime}$ forming a parallelogram $A B C B^{\prime}$. Then $B^{\prime}$ lies on $\omega_{B}$ diagonally opposite $B$, and since $A B^{\prime} \| B C$ it lies on $A D$. Similarly reflect $C$ across $N$ to a point $C^{\prime}$, which satisfies analogous properties. Note that $C B^{\prime}=A B=C D$, so we find that triangle $C D B^{\prime}$ and similarly triangle $B D C^{\prime}$ are isosceles.

Let $B^{\prime \prime}$ and $C^{\prime \prime}$ be the orthogonal projections of $B$ and $C$ onto $A D$. Since $B B^{\prime}$ is a diameter of $\omega_{B}$ we get that $B^{\prime \prime}$ lies on $\omega_{B}$, and similarly $C^{\prime \prime}$ lies on $\omega_{C}$. Moreover $B B^{\prime \prime}$ is an altitude of the isosceles triangle $B D C^{\prime}$ with $B D=B C^{\prime}$, hence it coincides with the median from $B$, so $B^{\prime \prime}$ is in fact the midpoint of $D C^{\prime}$. Similarly $C^{\prime \prime}$ is the midpoint of $D B^{\prime}$. From this we get

$$
2=\frac{D C^{\prime}}{D B^{\prime \prime}}=\frac{D B^{\prime}}{D C^{\prime \prime}}
$$

which rearranges as $D C^{\prime} \cdot D C^{\prime \prime}=D B^{\prime} \cdot D B^{\prime \prime}$. This means that $D$ has same the power with respect to $\omega_{B}$ and $\omega_{C}$, hence it lies on their radical axis $P Q$.

Problem 4 Find all integers $n$ and $m, n>m>2$, and such that a regular $n$-sided polygon can be inscribed in a regular $m$-sided polygon so that all the vertices of the $n$-gon lie on the sides of the $m$-gon.

Solution 4 It works only for $n=2 m$, and for $m=3$ and $n=4$.

To begin with let's see why it works for $n=2 m$. For a $2 m$-gon we can choose two points on each side, symmetrically, so that the distance between the two of them is equal to the distance between two close points on adjacent sides.

For $n=4$ and $m=3$ we need to inscribe a square in an equilateral triangle, by choosing two vertices of the square symmetrically on one of the sides of the triangle. It is easy to calculate the side length of the square so that its remaining two vertices lie on the remaining two sides of the triangle.

We need to show that it cannot be done in any other way. One side of the $m$-gon can contain at most two of the vertices of the $n$-gon, so that $n \leq 2 m$. For $n \geq m+2$ at least two of the sides of the $m$-gon must contain two of the vertices of the $n$-gon each. By symmetry the midpoints, and thus the perpendicular bisectors, of such a side of the $m$-gon and of the side of the $n$-gon it contains must coincide. If two such sides of the $m$-gon are not opposite to each other the corresponding perpendicular bisectors will intersect, and we can deduce that the centres of the circumscribed circles of the $m$-gon and of the $n$-gon must coincide. If two such sides are opposite, then the centres of the circumscribed circles will coincide with the midpoint of the segment between the midpoints of the sides, and thus the two circumscribed circles will once again have the same centre.

Denote the radii of the two circumscribed circles by $R$ and $r$, where $R>r$. The smaller circle intersects the sides of the $m$-gon at $2 m$ points, among which are the possible vertices for the $n$-gon. Denote these points by $P_{1}, P_{2}, \ldots, P_{2 m}$ clockwise, where $P_{1}$ and $P_{2}$
are vertices of the $n$-gon. If the side length of the $n$-gon is $s$, we now have $\left|P_{1} P_{2}\right|=s$, and thus $\left|P_{3} P_{4}\right|=s$. If only one of the points $P_{3}$ and $P_{4}$ is a vertex in the $n$-gon, it would have to be $P_{4}$, but the distance between $P_{2}$ and $P_{4}$ is greater than $s$, which means $P_{2} P_{4}$ cannot be a side. We can now deduce that both $P_{3}$ and $P_{4}$ are vertices of the $n$-gon, and by symmetry all the $2 m$ points of intersection will be vertices of the $n$-gon, so that $n=2 m$.

We now need to handle the case $n=m+1>4$. Denote the vertices of the $m$-gon by $Q_{1}, Q_{2}, \ldots, Q_{m}$ clockwise. Now only one of the sides of the $m$-gon contains two of the vertices of the $n$-gon, let this side be $Q_{m} Q_{1}$. Denote the vertices of the $(m+1)$-gon by $P_{1}, P_{2}, \ldots, P_{m+1}$ clockwise, where $P_{1}$ and $P_{m+1}$ lie on the side $Q_{m} Q_{1}$ of the $m$-gon. Let $\alpha=\pi-2 \pi /(m+1)$, and $\beta=\pi-2 \pi / m$ be the angles of the $n$-gon and the $m$ gon, respectively. We now get a number of triangles $P_{1} Q_{1} P_{2}, P_{2} Q_{2} P_{3}, \ldots P_{m} Q_{m} P_{m+1}$. We begin by establishing a connection between the sizes of the angles. To begin with we have $\angle Q_{1} P_{1} P_{2}=\pi-\alpha$, and $\angle P_{1} Q_{1} P_{2}=\beta$, so that $\angle P_{1} P_{2} Q_{1}=\alpha-\beta$. We now proceed to get $\angle Q_{2} P_{2} P_{3}=\pi-\alpha-(\alpha-\beta)$, and $\angle P_{2} Q_{2} P_{3}=\beta$, which gives $\angle P_{2} P_{3} Q_{2}=2(\alpha-\beta)$, and so on. If now $\gamma=\alpha-\beta=2 \pi / m(m+1)$, we have $\angle P_{k} P_{k+1} Q_{k}=k \gamma$, and $\angle Q_{k} P_{k} P_{k+1}=$ $(m+1-k) \gamma$ for $k=1,2, \ldots, m$.

Since $s$ is the side length of the $n$-gon, i.e. $s=\left|P_{1} P_{2}\right|=\left|P_{1} P_{2}\right|=\cdots=\left|P_{m+1} P_{1}\right|$, according to the law of sines we get

$$
\frac{s}{\sin \beta}=\frac{\left|P_{k} Q_{k}\right|}{\sin \left(P_{k} P_{k+1} Q_{k}\right)}=\frac{\left|Q_{k} P_{k+1}\right|}{\sin \left(Q_{k} P_{k} P_{k+1}\right)}
$$

i.e.

$$
\frac{s}{\sin \beta}=\frac{\left|P_{k} Q_{k}\right|}{\sin (k \gamma)}=\frac{\left|Q_{k} P_{k+1}\right|}{\sin ((m+1-k) \gamma)}
$$

Since $\left|Q_{k} Q_{k+1}\right|=\left|Q_{k} P_{k+1}\right|+\left|P_{k+1} Q_{k+1}\right|$, we get

$$
\sin ((k+1) \gamma)+\sin ((m+1-k) \gamma)=\frac{\sigma \sin \beta}{s}
$$

where $\sigma=\left|Q_{1} Q_{2}\right|=\left|Q_{2} Q_{3}\right|=\cdots=\left|Q_{m} Q_{1}\right|$ is the side length of the $m$-gon.

Using the above for $k=1$ and $k=2$, we get

$$
\sin 2 \gamma+\sin m \gamma=\sin 3 \gamma+\sin (m-1) \gamma
$$

For $m \geq 4$ the angles $3 \gamma$ and $(m-1) \gamma$ are both in the interval between $2 \gamma$ and $m \gamma$, which means we can't have equality as the sine function is concave (convex from above) in the interval $[0, \pi / 2]$. We therefore deduce that it is impossible to inscribe an $(m+1)$-gon in an $m$-gon for $m \geq 4$.