INMO 2020

Suppose that $m$ is an integer root of $x^{4}-a x^{3}-b x^{2}-c x-d=0$. As $d \neq 0$, we have $m \neq 0$.

Suppose now that $m>0$. Then $m^{4}-a m^{3}=b m^{2}+c m+d>0$ and hence $m>a \geq d$. On the other hand $d=m\left(m^{3}-a m^{2}-b m-c\right)$ and hence $m$ divides $d$, so $m \leq d$, a contradiction.

If $m<0$, then writing $n=-m>0$ we have $n^{4}+a n^{3}-b n^{2}+c n-d=n^{4}+n^{2}(a n-b)+(c n-d)>0$, a contradiction.

This proves that the given polynomial has no integer roots.

Solution 1.

We show that $T_{n}-n$ is even. Note that the subsets $\{1\},\{2\}, \cdots,\{n\}$ are good. Among the other good subsets, let $A$ be the collection of subsets with an integer average which belongs to the subset, and let $B$ be the collection of subsets with an integer average which is not a member of the subset. Then there is a bijection between $A$ and $B$, because removing the average takes a member of $A$ to a member of $B$; and including the average in a member of $B$ takes it to its inverse. So $T_{n}-n=|A|+|B|$ is even.

Solution 2.

Let $S=\{1,2, \ldots, n\}$. For a subset $A$ of $S$, let $\bar{A}=\{n+1-a \mid a \in A\}$. We call a subset $A$ symmetric if $\bar{A}=A$. Note that the arithmetic mean of a symmetric subset is $(n+1)/2$. Therefore, if $n$ is even, then there are no symmetric good subsets, while if $n$ is odd then every symmetric subset is good.

If $A$ is a proper good subset of $S$, then so is $\bar{A}$. Therefore, all the good subsets that are not symmetric can be paired. If $n$ is even then this proves that $t_{n}$ is even. If $n$ is odd, we have to show that there are odd number of symmetric subsets. For this, we note that a symmetric subset contains the element $(n+1)/2$ if and only if it has odd number of elements. Therefore, for any natural number $k$, the number of symmetric subsets of size $2k$ equals the number of symmetric subsets of size $2k+1$. The result now follows since there is exactly one symmetric subset with only one element.

Let $R$ be the circumradius of $\triangle ABC$ and $\Delta$ its area. We have $OD = R \cos A$ and $DC = \frac{a}{2}$, so

Again $HE = 2R \cos C \cos A$ and $EA = c \cos A$. Hence

Further

Equating (1) and (2) we get $\tan A = 4 \sin C \cos C$. And equating (1) and (3), and using this relation we get

Since $\cos A \neq 0$ we get $3 = 2t(-4t + 5)$ where $t = \sin^2 C$. This implies $(4t - 3)(2t - 1) = 0$ and therefore, since $\sin C > 0$, we get $\sin C = \sqrt{3}/2$ or $\sin C = 1/\sqrt{2}$. Because $\triangle ABC$ is acute, it follows that $\widehat{C} = \pi/3$ or $\pi/4$.

We observe that the given conditions are satisfied in an equilateral triangle, so $\widehat{C} = \pi/3$ is a possibility. Also, the conditions are satisfied in a triangle where $\widehat{C} = \pi/4$, $\widehat{A} = \tan^{-1} 2$ and $\widehat{B} = \tan^{-1} 3$. Therefore $\widehat{C} = \pi/4$ is also a possibility.

Thus the two possible values of $\widehat{C}$ are $\pi/3$ and $\pi/4$.

Let

Then $f(t) = k t$ for some constant $k$. Note that $k a = f(a) = (a - x)(a - y)(a - z) \leq 0$ and hence $k \leq 0$. Similarly, $k c = f(c) = (c - x)(c - y)(c - z) \geq 0$ and hence $k \geq 0$. Combining the two, it follows that $k = 0$ and that $f(a) = f(c) = 0$. These equalities imply that $a = x$ and $c = z$, and then it also follows that $b = y$.

Solution 1.

We begin with the following lemma:

Lemma: Let $XYZ$ be a triangle with $\angle XYZ = 90 + \alpha$. Construct an isosceles triangle $XEZ$, externally on the side $XZ$, with base angle $\alpha$. Then $E$ is the circumcentre of $\triangle XYZ$.

Proof of the Lemma: Draw $ED \perp XZ$. Then $DE$ is the perpendicular bisector of $XZ$. We also observe that $\angle XED = \angle ZED = 90 - \alpha$. Observe that $E$ is on the perpendicular bisector of $XZ$. Construct the circumcircle of $XYZ$. Draw the perpendicular bisector of $XY$ and let it meet $DE$ in $F$. Then $F$ is the circumcentre of $\triangle XYZ$. Join $XF$. Then $\angle XFD = 90 - \alpha$. But we know that $\angle XED = 90 - \alpha$. Hence $E = F$.

Let $r_1$, $r_2$ and $r$ be the inradii of the triangles $ABD$, $CBD$ and $ABC$ respectively. Join $PD$ and $DQ$. Observe that $\angle PDQ = 90^{\circ}$. Hence

Let $s_1 = (AB + BD + DA)/2$. Observe that $BD = ca/b$ and $AD = \sqrt{AB^2 - BD^2} = \sqrt{c^2 - \left(\frac{ca}{b}\right)^2} = c^2/b$. This gives $s_1 = cs/b$. But $r_1 = s_1 - c = (c/b)(s-b) = cr/b$. Similarly, $r_2 = ar/b$. Hence

Consider $\triangle PIQ$. Observe that $\angle PIQ = 90 + (B/2) = 135^{\circ}$. Hence $PQ$ subtends $90^{\circ}$ on the circumference of the circumcircle of $\triangle PIQ$. But we have seen that $\angle PDQ = 90^{\circ}$. Now construct a circle with $PQ$ as diameter. Let it cut $AC$ again in $K$. It follows that $\angle PKQ = 90^{\circ}$ and the points $P, D, K, Q$ are concyclic. We also notice $\angle KPQ = \angle KDQ = 45^{\circ}$ and $\angle PQK = \angle PDA = 45^{\circ}$.

Thus $PKQ$ is an isosceles right-angled triangle with $KP = KQ$. Therefore $KP^2 + KQ^2 = PQ^2 = 2r^2$ and hence $KP = KQ = r$.

Now $\angle PIQ = 90 + 45$ and $\angle PKQ = 2 \times 45^{\circ} = 90^{\circ}$ with $KP = KQ = r$.

Hence $K$ is the circumcentre of $\triangle PIQ$.

Solution 2.

Here we use computation to prove that the point of contact $K$ of the incircle with $AC$ is the circumcentre of $\triangle PIQ$. We show that $KP = KQ = r$.

Let $r_1$ and $r_2$ be the inradii of triangles $ABD$ and $CBD$ respectively. Draw $PL \perp AC$ and $QM \perp AC$. If $s_1$ is the semiperimeter of $\triangle ABD$, then $AL = s_1 - BD$.

But

Hence $s_1 = cs/b$. This gives $r_1 = s_1 - c = cr/b$, $AL = s_1 - BD = c(s-a)/b$. Hence $KL = AK - AL = (s-a) - \frac{c(s-a)}{b} = \frac{(b-c)(s-a)}{b}$.

We observe that $2r^2 = \frac{(c+a-b)^2}{2} = \frac{c^2 + a^2 + b^2 - 2bc - 2ab + 2ca}{2} = (b^2 - ba - bc + ac) = (b-c)(b-a)$.

Thus $KL = ra/b$. Finally,

Thus $KP = r$. Similarly, $KQ = r$. This gives $KP = KI = KQ = r$ and therefore $K$ is the circumcentre of $\triangle PIQ$.

For any prime $p$, let $\nu_{p}(n)$ be the maximum power of $p$ dividing $n$; i.e., $p^{\nu_{p}(n)}$ divides $n$ but not a higher power. Let $r$ be the length of the non-periodic part of the infinite decimal expansion of $1 / n$.

Write

We show that $r = \max \left(\nu_{2}(n), \nu_{5}(n)\right)$.

Let $a$ and $b$ be the numbers $a_{1} a_{2} \cdots a_{r}$ and $b = b_{1} b_{2} \cdots b_{s}$ respectively. (Here $a_{1}$ and $b_{1}$ can both be $0$.) Then

Thus we get $10^{r}\left(10^{s} - 1\right) = n\left(\left(10^{s} - 1\right)a + b\right)$. It shows that $r \geq \max \left(\nu_{2}(n), \nu_{5}(n)\right)$. Suppose $r > \max \left(\nu_{2}(n), \nu_{5}(n)\right)$. Then $10$ divides $b - a$. Hence the last digits of $a$ and $b$ are equal: $a_{r} = b_{s}$. This means

This contradicts the definition of $r$. Therefore $r = \max \left(\nu_{2}(n), \nu_{5}(n)\right)$.