INMO 2016

If $\alpha$ and $\beta$ are both integers, then

This proves one implication.

Observe that $\alpha+\beta=m$ and $\alpha \beta=n$. We use the property of integer function: $x-1<[x] \leq x$ for any real number $x$. Thus

Since $m$ and $n$ are positive integers, both $\alpha$ and $\beta$ must be positive. If $m \geq 2$, we observe that there is no square between $m^{2}-2$ and $m^{2}$. Hence, either $m=1$ or $[m \alpha]+[m \beta]=m^{2}$. If $m=1$, then $\alpha+\beta=1$ implies that both $\alpha$ and $\beta$ are positive reals smaller than 1. Hence $n=\alpha \beta$ cannot be a positive integer. We conclude that $[m \alpha]+[m \beta]=m^{2}$.

Putting $m=\alpha+\beta$ in this relation, we get

Using $[x+k]=[x]+k$ for any real number $x$ and integer $k$, this reduces to

This shows that $\alpha^{2}$ and $\beta^{2}$ are both integers. On the other hand,

Thus

is a rational number. Since $\alpha+\beta=m$ is a rational number, it follows that both $\alpha$ and $\beta$ are rational numbers. However, both $\alpha^{2}$ and $\beta^{2}$ are integers. Hence each of $\alpha$ and $\beta$ is an integer.

In a permutation of $(1,2,3, \ldots, n)$, two inversions can occur in only one of the following two ways: (A) Two disjoint consecutive pairs are interchanged:

(B) Each block of three consecutive integers can be permuted in any of the following 2 ways:

Consider case (A). For $j=1$, there are $n-3$ possible values of $k$; for $j=2$, there are $n-4$ possibilities for $k$ and so on. Thus the number of permutations with two inversions of this type is

In case (B), we see that there are $n-2$ permutations of each type, since $k$ can take values from $1$ to $n-2$. Hence we get $2(n-2)$ permutations of this type. Finally, the number of permutations with two inversions is

Let $AD = h$, $PD = y$ and $BD = DC = a$. We observe that $BP^2 = a^2 + y^2$. Moreover, $PE = PA \sin \angle DAC = (h - y) \frac{DC}{AC} = \frac{a(h - y)}{b}$, where $b = AC = AB$. Using $AP / PD = (h - y) / y$, we obtain $y = h / (1 + \lambda)$. Thus

But $(h - y) = \lambda y = \lambda h / (1 + \lambda)$ and $b^2 = a^2 + h^2$. Thus we obtain

Using $m = a / h$ and $z = m^2(1 + \lambda)$, this simplifies to

Dividing by $z$, this gives

However $z + (1 / z) \geq 2$ for any positive real number $z$. Thus $\lambda^3 - \lambda^2 - 4 \geq 0$. This may be written in the form $(\lambda - 2)(\lambda^2 + \lambda + 2) \geq 0$. But $\lambda^2 + \lambda + 2 > 0$. (For example, one may check that its discriminant is negative.) Hence $\lambda \geq 2$. If $\lambda = 2$, then $z + (1 / z) = 2$ and hence $z = 1$. This gives $m^2 = 1 / 3$ or $\tan (A / 2) = m = 1 / \sqrt{3}$. Thus $A = 60^\circ$ and hence $ABC$ is equilateral.

Conversely, if triangle $ABC$ is equilateral, then $m = \tan (A / 2) = 1 / \sqrt{3}$ and hence $z = (1 + \lambda) / 3$. Substituting this in the equation satisfied by $z$, we obtain

This may be written in the form $(\lambda - 2)(3\lambda^3 + 6\lambda^2 + 8\lambda + 8) = 0$. Here the second factor is positive because $\lambda > 0$. We conclude that $\lambda = 2$.

Solution 1.

We begin with the observation that

and similar bounds for $y^{2}+y z+z^{2}, z^{2}+z x+x^{2}$. Thus

Thus it is sufficient to prove that

Equivalently, we need to prove that

However, we note that

Thus the required inequality takes the form

This follows from AM-GM inequalities;

Solution 2.

Let us introduce $x+y=c, y+z=a$ and $z+x=b$. Then $a, b, c$ are the sides of a triangle. If $s=(a+b+c) / 2$, then it is easy to calculate $x=s-a, y=s-b, z=s-c$ and $x+y+z=s$. We also observe that $x^{2}+x y+y^{2}=(x+y)^{2}-x y=c^{2}-\frac{1}{4}(c+a-b)(c+b-a)=\frac{3}{4} c^{2}+\frac{1}{4}(a-b)^{2} \geq \frac{3}{4} c^{2}$. Moreover, $x y+y z+z x=(s-a)(s-b)+(s-b)(s-c)+(s-c)(s-a)$. Thus it is sufficient to prove that

But, $\sum(s-a)(s-b)=r(4 R+r)$, where $r, R$ are respectively the in-radius, the circum-radius of the triangle whose sides are $a, b, c$, and $a b c=4 R r s$. Thus the inequality reduces to

This is simply $2 r \leq R$. This follows from $I O^{2}=R(R-2 r)$, where $I$ is the incentre and $O$ the circumcentre.

Solution 3.

If we set $x=\lambda a, y=\lambda b, z=\lambda c$, then the inequality changes to

This shows that we may assume $x+y+z=1$. Let $\alpha=x y+y z+z x$. We see that

Thus

Thus we need to prove that $\alpha^{2} \leq 3\left(\alpha^{2}-\alpha^{3}-x y z\right)$. This reduces to

However

so that $2-3 \alpha \geq 1$. Thus it suffices to prove that $3 x y z \leq \alpha^{2}$. But

Note that $IA_{1} = IB_{1} = IC_{1} = 2r$, where $r$ is the in-radius of the triangle $ABC$. Hence $I$ is the circum-centre of the triangle $A_{1}B_{1}C_{1}$.

Let $K$ be the point of intersection of $IB_{1}$ and $AC$. Then $IK = r$, $IA = 2r$ and $\angle IKA = 90^{\circ}$. It follows that $\angle IAK = 30^{\circ}$ and hence $\angle IAB_{1} = 60^{\circ}$. Thus $AIB_{1}$ is an equilateral triangle. Similarly triangle $AIC_{1}$ is also equilateral. We hence obtain $AB_{1} = AC_{1} = AI = IB_{1} = IC_{1} = 2r$.

We also observe that $\angle B_{1}IC_{1} = 120^{\circ}$ and $IB_{1}AC_{1}$ is a rhombus. Thus $\angle B_{1}AC_{1} = 120^{\circ}$ and by concyclicity $\angle A_{1} = 60^{\circ}$. Since $AB_{1} = AC_{1}$, $A$ is the midpoint of the arc $B_{1}AC_{1}$. It follows that $A_{1}A$ bisects $\angle A_{1}$ and $I_{1}$ lies on the line $A_{1}A$. This implies that

Since $\angle B_{1}IC_{1} = 120^{\circ}$, we conclude that $B_{1}, I, I_{1}, C_{1}$ are concyclic. (Further $A$ is the centre.)

We begin with the standard factorisation

Thus we have $y^{2}-2y+2 = p^{m}$ and $y^{2}+2y+2 = p^{n}$ for some positive integers $m$ and $n$ such that $m+n = x$. Since $y^{2}-2y+2 < y^{2}+2y+2$, we have $m < n$ so that $p^{m}$ divides $p^{n}$. Thus $y^{2}-2y+2$ divides $y^{2}+2y+2$. Writing $y^{2}+2y+2 = y^{2}-2y+2 + 4y$, we infer that $y^{2}-2y+2$ divides $4y$ and hence $y^{2}-2y+2$ divides $4y^{2}$. But

Thus $y^{2}-2y+2$ divides $8(y-1)$. Since $y^{2}-2y+2$ divides both $4y$ and $8(y-1)$, we conclude that it also divides $8$. This gives $y^{2}-2y+2 = 1, 2, 4$ or $8$. If $y^{2}-2y+2 = 1$, then $y = 1$ and $y^{4}+4 = 5$, giving $p = 5$ and $x = 1$. If $y^{2}-2y+2 = 2$, then $y^{2}-2y = 0$ giving $y = 2$. But then $y^{4}+4 = 20$ is not the power of a prime. The equations $y^{2}-2y+2 = 4$ and $y^{2}-2y+2 = 8$ have no integer solutions. We conclude that $(p, x, y) = (5, 1, 1)$ is the only solution.

Alternatively, using $y^{2}-2y+2 = p^{m}$ and $y^{2}+2y+2 = p^{n}$, we may get

If $m > 0$, then $p$ divides $4$ or $y$. If $p$ divides $4$, then $p = 2$. If $p$ divides $y$, then $y^{2}-2y+2 = p^{m}$ shows that $p$ divides $2$ and hence $p = 2$. But then $2^{x} = y^{4}+4$, which shows that $y$ is even. Taking $y = 2z$, we get $2^{x-2} = 4z^{4}+1$. This implies that $z = 0$ and hence $y = 0$, which is a contradiction. Thus $m = 0$ and $y^{2}-2y+2 = 1$. This gives $y = 1$ and hence $p = 5, x = 1$.