| # The $4^{\text {th }}$ Romanian Master of Mathematics Competition - Solutions Day 1: Friday, February 25, 2011, Bucharest |
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| Problem 1. Prove that there exist two functions |
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| $$ |
| f, g: \mathbb{R} \rightarrow \mathbb{R} |
| $$ |
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| such that $f \circ g$ is strictly decreasing, while $g \circ f$ is strictly increasing. |
| (Poland) Andrzej KomisArsKi \& Marcin Kuczma |
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| ## Solution. Let |
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| $$ |
| \begin{aligned} |
| & \cdot A=\bigcup_{k \in \mathbb{Z}}\left(\left[-2^{2 k+1},-2^{2 k}\right) \bigcup\left(2^{2 k}, 2^{2 k+1}\right]\right) \\ |
| & \cdot B=\bigcup_{k \in \mathbb{Z}}\left(\left[-2^{2 k},-2^{2 k-1}\right) \bigcup\left(2^{2 k-1}, 2^{2 k}\right]\right) |
| \end{aligned} |
| $$ |
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| Thus $A=2 B, B=2 A, A=-A, B=-B, A \cap B=\varnothing$, and finally $A \cup B \cup\{0\}=\mathbb{R}$. Let us take |
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| $$ |
| f(x)=\left\{\begin{array}{lll} |
| x & \text { for } & x \in A \\ |
| -x & \text { for } & x \in B \\ |
| 0 & \text { for } & x=0 |
| \end{array}\right. |
| $$ |
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| Take $g(x)=2 f(x)$. Thus $f(g(x))=f(2 f(x))=-2 x$ and $g(f(x))=2 f(f(x))=2 x$. |
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| Problem 2. Determine all positive integers $n$ for which there exists a polynomial $f(x)$ with real coefficients, with the following properties: |
| (1) for each integer $k$, the number $f(k)$ is an integer if and only if $k$ is not divisible by $n$; |
| (2) the degree of $f$ is less than $n$. |
| (Hungary) GÉza Kós |
| Solution. We will show that such polynomial exists if and only if $n=1$ or $n$ is a power of a prime. |
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| We will use two known facts stated in Lemmata 1 and 2. |
| Lemma 1. If $p^{a}$ is a power of a prime and $k$ is an integer, then $\frac{(k-1)(k-2) \ldots\left(k-p^{a}+1\right)}{\left(p^{a}-1\right)!}$ is divisible by $p$ if and only if $k$ is not divisible by $p^{a}$. |
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| Proof. First suppose that $p^{a} \mid k$ and consider |
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| $$ |
| \frac{(k-1)(k-2) \cdots\left(k-p^{a}+1\right)}{\left(p^{a}-1\right)!}=\frac{k-1}{p^{a}-1} \cdot \frac{k-2}{p^{a}-2} \cdots \frac{k-p^{a}+1}{1} |
| $$ |
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| In every fraction on the right-hand side, $p$ has the same maximal exponent in the numerator as in the denominator. |
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| Therefore, the product (which is an integer) is not divisible by $p$. |
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| Now suppose that $p^{a} \nmid k$. We have |
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| $$ |
| \frac{(k-1)(k-2) \cdots\left(k-p^{a}+1\right)}{\left(p^{a}-1\right)!}=\frac{p^{a}}{k} \cdot \frac{k(k-1) \cdots\left(k-p^{a}+1\right)}{\left(p^{a}\right)!} . |
| $$ |
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| The last fraction is an integer. In the fraction $\frac{p^{a}}{k}$, the denominator $k$ is not divisible by $p^{a}$. |
| Lemma 2. If $g(x)$ is a polynomial with degree less than $n$ then |
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| $$ |
| \sum_{\ell=0}^{n}(-1)^{\ell}\binom{n}{\ell} g(x+n-\ell)=0 |
| $$ |
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| Proof. Apply induction on $n$. For $n=1$ then $g(x)$ is a constant and |
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| $$ |
| \binom{1}{0} g(x+1)-\binom{1}{1} g(x)=g(x+1)-g(x)=0 |
| $$ |
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| Now assume that $n>1$ and the Lemma holds for $n-1$. Let $h(x)=g(x+1)-g(x)$; the degree of $h$ is less than the degree of $g$, so the induction hypothesis applies for $g$ and $n-1$ : |
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| $$ |
| \begin{gathered} |
| \sum_{\ell=0}^{n-1}(-1)^{\ell}\binom{n-1}{\ell} h(x+n-1-\ell)=0 \\ |
| \sum_{\ell=0}^{n-1}(-1)^{\ell}\binom{n-1}{\ell}(g(x+n-\ell)-g(x+n-1-\ell))=0 \\ |
| \binom{n-1}{0} g(x+n)+\sum_{\ell=1}^{n-1}(-1)^{\ell}\left(\binom{n-1}{\ell-1}+\right. \\ |
| \left.\binom{n-1}{\ell}\right) g(x+n-\ell)-(-1)^{n-1}\binom{n-1}{n-1} g(x)=0 \\ |
| \sum_{\ell=0}^{n}(-1)^{\ell}\binom{n}{\ell} g(x+n-\ell)=0 |
| \end{gathered} |
| $$ |
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| Lemma 3. If $n$ has at least two distinct prime divisors then the greatest common divisor of $\binom{n}{1},\binom{n}{2}, \ldots,\binom{n}{n-1}$ is 1 . |
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| Proof. Suppose to the contrary that $p$ is a common prime divisor of $\binom{n}{1}, \ldots,\binom{n}{n-1}$. In particular, $p \left\lvert\,\binom{ n}{1}=n\right.$. Let $a$ be the exponent of $p$ in the prime factorization of $n$. Since $n$ has at least two prime divisors, we have $1<p^{a}<n$. Hence, $\binom{n}{p^{a}-1}$ and $\binom{n}{p^{a}}$ are listed among $\binom{n}{1}, \ldots,\binom{n}{n-1}$ and thus $p \left\lvert\,\binom{ n}{p^{a}}\right.$ and $p \left\lvert\,\binom{ n}{p^{a}-1}\right.$. But then $p$ divides $\binom{n}{p^{a}}-\binom{n}{p^{a}-1}=\binom{n-1}{p^{a}-1}$, which contradicts Lemma 1. |
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| Next we construct the polynomial $f(x)$ when $n=1$ or $n$ is a power of a prime. |
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| For $n=1, f(x)=\frac{1}{2}$ is such a polynomial. |
| If $n=p^{a}$ where $p$ is a prime and $a$ is a positive integer then let |
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| $$ |
| f(x)=\frac{1}{p}\binom{x-1}{p^{a}-1}=\frac{1}{p} \cdot \frac{(x-1)(x-2) \cdots\left(x-p^{a}+1\right)}{\left(p^{a}-1\right)!} . |
| $$ |
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| The degree of this polynomial is $p^{a}-1=n-1$. |
| The number $\frac{(k-1)(k-2) \cdots\left(k-p^{a}+1\right)}{\left(p^{a}-1\right)!}$ is an integer for any integer $k$, and, by Lemma 1 , it is divisible by $p$ if and only if $k$ is not divisible by $p^{a}=n$. |
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| Finally we prove that if $n$ has at least two prime divisors then no polynomial $f(x)$ satisfies $(1,2)$. Suppose that some polynomial $f(x)$ satisfies (1,2), and apply Lemma 2 for $g=f$ and $x=-k$ where $1 \leq k \leq n-1$. We get that |
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| $$ |
| \binom{n}{k} f(0)=\sum_{0 \leq \ell \leq n, \ell \neq k}(-1)^{k-\ell}\binom{n}{\ell} f(-k+\ell) |
| $$ |
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| Since $f(-k), \ldots, f(-1)$ and $f(1), \ldots, f(n-k)$ are all integers, we conclude that $\binom{n}{k} f(0)$ is an integer for every $1 \leq k \leq n-1$. |
| By dint of Lemma 3, the greatest common divisor of $\binom{n}{1},\binom{n}{2}, \ldots,\binom{n}{n-1}$ is 1 . Hence, there will exist some integers $u_{1}, u_{2}, \ldots, u_{n-1}$ for which $u_{1}\binom{n}{1}+\cdots+u_{n-1}\binom{n}{n-1}=1$. Then |
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| $$ |
| f(0)=\left(\sum_{k=1}^{n-1} u_{k}\binom{n}{k}\right) f(0)=\sum_{k=1}^{n-1} u_{k}\binom{n}{k} f(0) |
| $$ |
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| is a sum of integers. This contradicts the fact that $f(0)$ is not an integer. So such polynomial $f(x)$ does not exist. |
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| Alternative Solution. (I. Bogdanov) We claim the answer is $n=p^{\alpha}$ for some prime $p$ and nonnegative $\alpha$. |
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| Lemma. For every integers $a_{1}, \ldots, a_{n}$ there exists an integervalued polynomial $P(x)$ of degree $<n$ such that $P(k)=a_{k}$ for all $1 \leq k \leq n$. |
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| Proof. Induction on $n$. For the base case $n=1$ one may set $P(x)=a_{1}$. For the induction step, suppose that the polynomial $P_{1}(x)$ satisfies the desired property for all $1 \leq k \leq n-1$. Then set $P(x)=P_{1}(x)+\left(a_{n}-P_{1}(n)\right)\binom{x-1}{n-1}$; since $\binom{k-1}{n-1}=0$ for $1 \leq k \leq n-1$ and $\binom{n-1}{n-1}=1$, the polynomial $P(x)$ is a sought one. |
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| Now, if for some $n$ there exists some polynomial $f(x)$ satisfying the problem conditions, one may choose some integer-valued polynomial $P(x)$ (of degree $<n-1$ ) coinciding with $f(x)$ at points $1, \ldots, n-1$. The difference $f_{1}(x)=$ $f(x)-P(x)$ also satisfies the problem conditions, therefore we may restrict ourselves to the polynomials vanishing at points $1, \ldots, n-1-$ that are, the polynomials of the form $f(x)=c \prod_{i=1}^{n-1}(x-i)$ for some (surely rational) constant $c$. |
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| Let $c=p / q$ be its irreducible form, and $q=\prod_{j=1}^{d} p_{j}^{\alpha_{j}}$ be the prime decomposition of the denominator. |
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| 1. Assume that a desired polynomial $f(x)$ exists. Since $f(0)$ is not an integer, we have $9 \nmid(-1)^{n-1}(n-1)$ ! and hence $p_{j}^{\alpha_{j}} \nmid(-1)^{n-1}(n-1)$ ! for some $j$. Hence |
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| $$ |
| \prod_{i=1}^{n-1}\left(p_{j}^{\alpha_{j}}-i\right) \equiv(-1)^{n-1}(n-1)!\not \equiv 0 \quad\left(\bmod p_{j}^{\alpha_{j}}\right) |
| $$ |
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| therefore $f\left(p_{i}^{\alpha_{i}}\right)$ is not integer, too. By the condition (i), this means that $n \mid p_{i}^{\alpha_{i}}$, and hence $n$ should be a power of a prime. |
| 2. Now let us construct a desired polynomial $f(x)$ for any power of a prime $n=p^{\alpha}$. We claim that the polynomial |
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| $$ |
| f(x)=\frac{1}{p}\binom{x-1}{n-1}=\frac{n}{p x}\binom{x}{n} |
| $$ |
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| fits. Actually, consider some integer $x$. From the first representation, the denominator of the irreducible form of $f(x)$ may be 1 or $p$ only. If $p^{\alpha} \nmid x$, then the prime decomposition of the fraction $n /(p x)$ contains $p$ with a nonnegative exponent; hence $f(x)$ is integer. On the other hand, if $n=p^{\alpha} \mid x$, then the numbers $x-1, x-2, \ldots, x-(n-1)$ contain the same exponents of primes as the numbers $n-1, n-2, \ldots, 1$ respectively; hence the number |
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| $$ |
| \binom{x-1}{n-1}=\frac{\prod_{i=1}^{n-1}(x-i)}{\prod_{i=1}^{n-1}(n-i)} |
| $$ |
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| is not divisible by $p$. Thus $f(x)$ is not an integer. |
| Problem 3. A triangle $A B C$ is inscribed in a circle $\omega$. A variable line $\ell$ chosen parallel to $B C$ meets segments $A B$, $A C$ at points $D, E$ respectively, and meets $\omega$ at points $K, L$ (where $D$ lies between $K$ and $E$ ). Circle $\gamma_{1}$ is tangent to the segments $K D$ and $B D$ and also tangent to $\omega$, while circle $\gamma_{2}$ is tangent to the segments $L E$ and $C E$ and also tangent to $\omega$. Determine the locus, as $\ell$ varies, of the meeting point of the common inner tangents to $\gamma_{1}$ and $\gamma_{2}$. |
| (Russia) VASily Mokin \& Fedor IvLev |
| Solution. Let $P$ be the meeting point of the common inner tangents to $\gamma_{1}$ and $\gamma_{2}$. Also, let $b$ be the angle bisector of $\angle B A C$. Since $K L \| B C, b$ is also the angle bisector of $\angle K A L$. |
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| Let $\mathfrak{H}$ be the composition of the symmetry $\mathfrak{S}$ with respect to $b$ and the inversion $\mathfrak{I}$ of centre $A$ and ratio $\sqrt{A K \cdot A L}$ (it is readily seen that $\mathfrak{S}$ and $\mathfrak{I}$ commute, so since $\mathfrak{S}^{2}=\mathfrak{I}^{2}=$ id, then also $\mathfrak{H}^{2}=\mathrm{id}$, the identical transformation). The elements of the configuration interchanged by $\mathfrak{H}$ are summarized in Table I. |
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| Let $O_{1}$ and $O_{2}$ be the centres of circles $\gamma_{1}$ and $\gamma_{2}$. Since the circles $\gamma_{1}$ and $\gamma_{2}$ are determined by their construction (in a unique way), they are interchanged by $\mathfrak{H}$, therefore the rays $A O_{1}$ and $A O_{2}$ are symmetrical with respect |
| to $b$. Denote by $\rho_{1}$ and $\rho_{2}$ the radii of $\gamma_{1}$ and $\gamma_{2}$. Since $\angle O_{1} A B=\angle O_{2} A C$, we have $\rho_{1} / \rho_{2}=A O_{1} / A O_{2}$. On the other hand, from the definition of $P$ we have $O_{1} P / O_{2} P=$ $\rho_{1} / \rho_{2}=A O_{1} / A O_{2}$; this means that $A P$ is the angle bisector of $\angle O_{1} A O_{2}$ and therefore of $\angle B A C$. |
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| The limiting, degenerated, cases are when the parallel line passes through $A$ - when $P$ coincides with $A$; respectively when the parallel line is $B C$ - when $P$ coincides with the foot $A^{\prime} \in B C$ of the angle bisector of $\angle B A C$ (or any other point on $B C$ ). By continuity, any point $P$ on the open segment $A A^{\prime}$ is obtained for some position of the parallel, therefore the locus is the open segment $A A^{\prime}$ of the angle bisector $b$ of $\angle B A C$. |
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| | point $K$ | $\longleftrightarrow$ | point $L$ | |
| | :---: | :---: | :---: | |
| | line $K L$ | $\longleftrightarrow$ | circle $\omega$ | |
| | ray $A B$ | $\longleftrightarrow$ | ray $A C$ | |
| | point $B$ | $\longleftrightarrow$ | point $E$ | |
| | point $C$ | $\longleftrightarrow$ | point $D$ | |
| | segment $B D$ | $\longleftrightarrow$ | segment $E C$ | |
| | $\operatorname{arc} B K$ | $\longleftrightarrow$ | segment $E L$ | |
| | $\operatorname{arc} C L$ | $\longleftrightarrow$ | segment $D K$ | |
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| TABLE I: Elements interchanged by $\mathfrak{H}$. |
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