exam: USAMO

problem: Let $a, b, c$ be positive real numbers such that $a+b+c=4 \sqrt[3]{a b c}$. Prove that  $$ 2(a b+b c+c a)+4 \min \left(a^{2}, b^{2}, c^{2}\right) \geq a^{2}+b^{2}+c^{2} . $$

solution: WLOG let $c=\min (a, b, c)=1$ by scaling. The given inequality becomes equivalent to  $$ 4 a b+2 a+2 b+3 \geq(a+b)^{2} \quad \forall a+b=4(a b)^{1 / 3}-1 $$  Now, let $t=(a b)^{1 / 3}$ and eliminate $a+b$ using the condition, to get  $$ 4 t^{3}+2(4 t-1)+3 \geq(4 t-1)^{2} \Longleftrightarrow 0 \leq 4 t^{3}-16 t^{2}+16 t=4 t(t-2)^{2} $$  which solves the problem. Equality occurs only if $t=2$, meaning $a b=8$ and $a+b=7$, which gives  $$ \{a, b\}=\left\{\frac{7 \pm \sqrt{17}}{2}\right\} $$  with the assumption $c=1$. Scaling gives the curve of equality cases.


problem: Find all functions $f:(0, \infty) \rightarrow(0, \infty)$ such that  $$ f\left(x+\frac{1}{y}\right)+f\left(y+\frac{1}{z}\right)+f\left(z+\frac{1}{x}\right)=1 $$  for all $x, y, z>0$ with $x y z=1$.

solution: The main part of the problem is to show all solutions are linear. As always, let $x=b / c$, $y=c / a, z=a / b$ (classical inequality trick). Then the problem becomes  $$ \sum_{\mathrm{cyc}} f\left(\frac{b+c}{a}\right)=1 $$  Let $f(t)=g\left(\frac{1}{t+1}\right)$, equivalently $g(s)=f(1 / s-1)$. Thus $g:(0,1) \rightarrow(0,1)$ which satisfies $\sum_{\mathrm{cyc}} g\left(\frac{a}{a+b+c}\right)=1$, or equivalently  $$ g(a)+g(b)+g(c)=1 \quad \forall a+b+c=1 . $$   Claim - The function $g$ is linear.  - First, whenever $a+b \leq 1$ we have  $$ 1-g(1-(a+b))=g(a)+g(b)=2 g\left(\frac{a+b}{2}\right) $$  Hence $g$ obeys Jensen's functional equation over $(0,1 / 2)$.  - Define $h:[0,1] \rightarrow \mathbb{R}$ by $h(t)=g\left(\frac{2 t+1}{8}\right)-(1-t) \cdot g(1 / 8)-t \cdot g(3 / 8)$, then $h$ satisfies Jensen's functional equation too over $[0,1]$. We have also arranged that $h(0)=h(1)=0$, hence $h(1 / 2)=0$ as well. - Since  $$ h(t)=h(t)+h(1 / 2)=2 h(t / 2+1 / 4)=h(t+1 / 2)+h(0)=h(t+1 / 2) $$  for any $t<1 / 2$, we find $h$ is periodic modulo $1 / 2$. It follows one can extend $\widetilde{h}$ by  $$ \widetilde{h}: \mathbb{R} \rightarrow \mathbb{R} \quad \text { by } \quad \widetilde{h}(t)=h(t-\lfloor t\rfloor) $$  and still satisfy Jensen's functional equation. Because $\widetilde{h}(0)=0$, it's well-known this implies $\widetilde{h}$ is additive (because $\widetilde{h}(x+y)=2 \widetilde{h}((x+y) / 2)=\widetilde{h}(x)+\widetilde{h}(y)$ for any real numbers $x$ to $y$ ).  But $\widetilde{h}$ is bounded below on $[0,1]$ since $g \geq 0$, and since $\widetilde{h}$ is also additive, it follows (well-known) that $\widetilde{h}$ is linear. Thus $h$ is the zero function. So, the function $g$ is linear over $[1 / 8,3 / 8]$; thus we may write $g(x)=k x+\ell$, valid for $1 / 8 \leq x \leq 3 / 8$.  Since $3 g(1 / 3)=1$, it follows $k+3 \ell=1$. For $0<x<1 / 8$ we have $g(x)=2 g(0.15)-g(0.3-x)=2(0.15 k+\ell)-(k(0.3-x)+\ell)=$ $k x+\ell$, so $g$ is linear over $(0,3 / 8)$ as well. Finally, for $3 / 8<x<1$, we use the given equation  $$ 1=g\left(\frac{1-x}{2}\right)+g\left(\frac{1-x}{2}\right)+g(x) \Longrightarrow g(x)=1-2\left(k \cdot \frac{1-x}{2}+\ell\right)=k x+\ell $$  since $\frac{1-x}{2}<\frac{5}{16}<\frac{3}{8}$. Thus $g$ is linear over all. Putting this back in, we deduce that $g(x)=k x+\frac{1-k}{3}$ for some $k \in[-1 / 2,1]$, and so  $$ f(x)=\frac{k}{x+1}+\frac{1-k}{3} $$  for some $k \in[-1 / 2,1]$. All such functions work.


problem: Let $n \geq 2$ be an integer, and let $\left\{a_{1}, \ldots, a_{m}\right\}$ denote the $m=\varphi(n)$ integers less than $n$ and relatively prime to $n$. Assume that every prime divisor of $m$ also divides $n$. Prove that $m$ divides $a_{1}^{k}+\cdots+a_{m}^{k}$ for every positive integer $k$.

solution: For brevity, given any $n$, we let $A(n)=\{1 \leq x \leq n, \operatorname{gcd}(x, n)=1\}$ (thus $|A(n)|=\varphi(n)$ ). Also, let $S(n, k)=\sum_{a \in A(n)} a^{k}$.  We will prove the stronger statement (which eliminates the hypothesis on $n$ ). Claim - Let $n \geq 2$ be arbitrary (and $k \geq 0$ ). If $p \mid n$, then  $$ \nu_{p}(\varphi(n)) \leq \nu_{p}(S(n, k)) $$  We start with the special case where $n$ is a prime power.  ## Lemma  Let $p$ be prime, $e \geq 1, k \geq 0$. We always have  $$ S\left(p^{e}, k\right)=\sum_{x \in A\left(p^{e}\right)} x^{k} \equiv 0 \quad\left(\bmod p^{e-1}\right) $$   $$ S\left(p^{e}, k\right) \equiv 1+g^{k}+g^{2 k}+\cdots+g^{\left(\varphi\left(p^{e}\right)-1\right) k} \equiv \frac{g^{\varphi\left(p^{e}\right) k}-1}{g^{k}-1} $$  If $p-1 \nmid k$, then the denominator is not divisible by $p$ and hence the entire expression is $0\left(\bmod p^{e}\right)$. In the other case where $p-1 \mid k$, since $\nu_{p}\left(\varphi\left(p^{e}\right)\right)=e-1$, the exponent lifting lemma implies  $$ \nu_{p}\left(\left(g^{k}\right)^{\varphi\left(p^{e}\right)}-1\right)=\nu_{p}\left(g^{k}-1\right)+(e-1) $$  and so the conclusion is true here too.  ## Corollary  We have $\nu_{p}\left(1^{k}+\cdots+t^{k}\right) \geq \nu_{p}(t)-1$ for any $k, t, p$.   Now the idea is to add primes $q$ one at a time to $n$, starting from the base case $n=p^{e}$. So, formally we proceed by induction on the number of prime divisors of $n$. We'll also assume $k \geq 1$ in what follows since the base case $k=0$ is easy.  - First, suppose we want to go from $n$ to $n q$ where $q \nmid n$. In that case $\varphi(n q)$ gained $\nu_{p}(q-1)$ factors of $p$ and then we need to show $\nu_{p}(S(n q, k)) \geq \nu_{p}(\varphi(n))+\nu_{p}(q-1)$. The trick is to write  $$ A(n q)=\{a+n h \mid a \in A(n) \text { and } h=0, \ldots, q-1\} \backslash q A(n) $$  and then expand using binomial theorem:  $$ \begin{aligned} S(n q, k) & =\sum_{a \in A(n)} \sum_{h=0}^{q-1}(a+n h)^{k}-\sum_{a \in A(n)}(q a)^{k} \\ & =-q^{k} S(n, k)+\sum_{a \in A(n)} \sum_{h=0}^{q-1} \sum_{j=0}^{k}\left[\binom{k}{j} a^{k-j} n^{j} h^{j}\right] \\ & =-q^{k} S(n, k)+\sum_{j=0}^{k}\left[\binom{k}{j} n^{j}\left(\sum_{a \in A(n)} a^{k-j}\right)\left(\sum_{h=0}^{q-1} h^{j}\right)\right] \\ & =-q^{k} S(n, k)+\sum_{j=0}^{k}\left[\binom{k}{j} n^{j} S(n, k-j)\left(\sum_{h=1}^{q-1} h^{j}\right)\right] \\ & =\left(q-q^{k}\right) S(n, k)+\sum_{j=1}^{k}\left[\binom{k}{j} n^{j} S(n, k-j)\left(\sum_{h=1}^{q-1} h^{j}\right)\right] \end{aligned} $$  We claim every term here has enough powers of $p$. For the first term, $S(n, k)$ has at least $\nu_{p}(\varphi(n))$ factors of $p$; and we have the $q-q^{k}$ multiplier out there. For the other terms, we apply induction to $S(n, k-j)$; moreover $\sum_{h=1}^{q-1} h^{j}$ has at least $\nu_{p}(q-1)-1$ factors of $p$ by corollary, and we get one more factor of $p$ (at least) from $n^{j}$.  - On the other hand, if $q$ already divides $n$, then this time  $$ A(n q)=\{a+n h \mid a \in A(n) \text { and } h=0, \ldots, q-1\} $$  and we have no additional burden of $p$ to deal with; the same calculation gives  $$ S(n q, k)=q S(n, k)+\sum_{j=1}^{k}\left[\binom{k}{j} n^{j} S(n, k-j)\left(\sum_{h=1}^{q-1} h^{j}\right)\right] $$  which certainly has enough factors of $p$ already.  Remark. A curious bit about the problem is that $\nu_{p}(\varphi(n))$ can exceed $\nu_{p}(n)$, and so it is not true that the residues of $A(n)$ are well-behaved modulo $\varphi(n)$.  As an example, let $n=2 \cdot 3 \cdot 7 \cdot 13=546$, so $m=\varphi(n)=1 \cdot 2 \cdot 6 \cdot 12=144$. Then $A(n)$ contains 26 elements which are $1 \bmod 9$ and 23 elements which are $4 \bmod 9$.  Remark. The converse of the problem is true too (but asking both parts would make this too long for exam).


problem: Let $p$ be a prime, and let $a_{1}, \ldots, a_{p}$ be integers. Show that there exists an integer $k$ such that the numbers  $$ a_{1}+k, a_{2}+2 k, \ldots, a_{p}+p k $$  produce at least $\frac{1}{2} p$ distinct remainders upon division by $p$.

solution: For each $k=0, \ldots, p-1$ let $G_{k}$ be the graph on $\{1, \ldots, p\}$ where we join $\{i, j\}$ if and only if  $$ a_{i}+i k \equiv a_{j}+j k \quad(\bmod p) \Longleftrightarrow k \equiv-\frac{a_{i}-a_{j}}{i-j} \quad(\bmod p) . $$  So we want a graph $G_{k}$ with at least $\frac{1}{2} p$ connected components. However, each $\{i, j\}$ appears in exactly one graph $G_{k}$, so some graph has at most $\frac{1}{p}\binom{p}{2}=\frac{1}{2}(p-1)$ edges (by "pigeonhole"). This graph has at least $\frac{1}{2}(p+1)$ connected components, as desired.  Remark. Here is an example for $p=5$ showing equality can occur:  $$ \left[\begin{array}{lllll} 0 & 0 & 3 & 4 & 3 \\ 0 & 1 & 0 & 2 & 2 \\ 0 & 2 & 2 & 0 & 1 \\ 0 & 3 & 4 & 3 & 0 \\ 0 & 4 & 1 & 1 & 4 \end{array}\right] . $$  Ankan Bhattacharya points out more generally that $a_{i}=i^{2}$ is sharp in general.


problem: Let $A B C D$ be a convex cyclic quadrilateral with $E=\overline{A C} \cap \overline{B D}, F=\overline{A B} \cap \overline{C D}$, $G=\overline{D A} \cap \overline{B C}$. The circumcircle of $\triangle A B E$ intersects line $C B$ at $B$ and $P$, and the circumcircle of $\triangle A D E$ intersects line $C D$ at $D$ and $Q$. Assume $C, B, P, G$ and $C$, $Q, D, F$ are collinear in that order. Let $M=\overline{F P} \cap \overline{G Q}$. Prove that $\angle M A C=90^{\circ}$.

solution: We present three general routes. (The second route, using the fact that $\overline{A C}$ is an angle bisector, has many possible variations.) \l First solution (Miquel points). This is indeed a Miquel point problem, but the main idea is to focus on the self-intersecting cyclic quadrilateral $P B Q D$ as the key player, rather than on the given $A B C D$.  Indeed, we will prove that $A$ is its Miquel point; this follows from the following two claims.  Claim - The self-intersecting quadrilateral $P Q D B$ is cyclic.   Claim - Point $E$ lies on line $P Q$.  ![](https://cdn.mathpix.com/cropped/2024_11_19_3f8866632dd9408ace0ag-10.jpg?height=804&width=724&top_left_y=1617&top_left_x=649)  To finish, let $H=\overline{P D} \cap \overline{B Q}$. By properties of the Miquel point, we have $A$ is the foot from $H$ to $\overline{C E}$. But also, points $M, A, H$ are collinear by Pappus theorem on $\overline{B P G}$ and $\overline{D Q F}$, as desired.


solution: We present three general routes. (The second route, using the fact that $\overline{A C}$ is an angle bisector, has many possible variations.) 【 Second solution (projective). We start with a synthetic observation. Claim - The line $\overline{A C}$ bisects $\angle P A D$ and $\angle B A Q$.  There are three ways to finish from here:  - (Michael Kural) Suppose the external bisector of $\angle P A D$ and $\angle B A Q$ meet lines $B C$ and $D C$ at $X$ and $Y$. Then  $$ -1=(G P ; X C)=(F D ; Y C) $$  which is enough to imply that $\overline{X Y}, \overline{G Q}, \overline{P F}$ are concurrent (by so-called prism lemma).  - (Daniel Liu) Alternatively, apply the dual Desargues involution theorem to complete quadrilateral GQFPCM, through the point $A$. This gives that an involutive pairing of  $$ (A C, A M)(A P, A Q)(A G, A F) $$  This is easier to see if we project it onto the line $\ell$ through $C$ perpendicular to $\overline{A C}$; if we let $P^{\prime}, Q^{\prime}, G^{\prime}, F^{\prime}$ be the images of the last four lines, we find the involution coincides with negative inversion through $C$ with power $\sqrt{C P^{\prime} \cdot C Q^{\prime}}$ which implies that $\overline{A M} \cap \ell$ is an infinity point, as desired.


solution: We present three general routes. (The second route, using the fact that $\overline{A C}$ is an angle bisector, has many possible variations.) 【 Third solution (inversion, Andrew Wu). Noting that $C E \cdot C A=C P \cdot C B=C Q \cdot C D$, we perform an inversion at $C$ swapping these pairs of points. The point $G$ is mapped to a point $G^{*}$ ray $C B$ for which $Q E G^{*} C$ is cyclic, but then  $$ \measuredangle C G^{*} E=\measuredangle C Q E=\measuredangle C Q P=\measuredangle D B C=\measuredangle C B E $$  and so we conclude $E B=E G^{*}$. Similarly, $E D=E F^{*}$. Finally, $M^{*}=\left(C G^{*} D\right) \cap\left(C F^{*} B\right) \neq C$, and we wish to show that $\angle E M^{*} C=90^{\circ}$. ![](https://cdn.mathpix.com/cropped/2024_11_19_3f8866632dd9408ace0ag-11.jpg?height=800&width=575&top_left_y=1830&top_left_x=746)  Note that $M^{*}$ is the center of the spiral similarity sending $\overline{B G^{*}}$ to $\overline{F^{*} D}$. Hence it also maps the midpoint $K$ of $B G^{*}$ to the midpoint $L$ of $\overline{F^{*} E}$. Consequently, $M^{*}$ lies on the circumcircle $K L C$ as well. In other words, $E L C K M^{*}$ is a cyclic pentagon with circumdiameter $\overline{C E}$, as desired.


problem: Let $a_{n}$ be the number of permutations $\left(x_{1}, \ldots, x_{n}\right)$ of $(1, \ldots, n)$ such that the ratios $x_{k} / k$ are all distinct. Prove that $a_{n}$ is odd for all $n \geq 1$.

solution: The first idea:  ## Lemma  If a permutation $x$ works, so does the inverse permutation. Thus it suffices to consider permutations $x$ in which all cycles have length at most 2 . Of course, there can be at most one fixed point (since that gives the ratio 1 ), and hence exactly one if $n$ is odd, none if $n$ is even.  We consider the graph $K_{n}$ such that the edge $\{i, j\}$ is labeled with $i / j$ (for $i<j$ ). The permutations we're considering are then equivalent to maximal matchings of this $K_{n}$. We call such a matching fantastic if it has an all of distinct edge labels.  Now the second insight is that if edges $a b$ and $c d$ have the same label for $a<b$ and $c<d$, then so do edges $a c$ and $b d$. Thus:  Definition. Given a matching $\mathcal{M}$ as above we say the neighbors of $\mathcal{M}$ are those other matchings obtained as follows: for each label $\ell$, we take some disjoint pairs of edges (possibly none) with label $\ell$ and apply the above switching operation (in which we replace $a b$ and $c d$ with $a c$ and $b d)$.  This neighborship relation is reflexive, and most importantly it is symmetric (because one can simply reverse the moves). But it is not transitive.  The second observation is that: Claim - The matching $\mathcal{M}$ has an odd number of neighbors (including itself) if and only if it is fantastic.  If we pick $k$ disjoint pairs and swap them, the number of ways to do this is $\binom{n_{e}}{2 k}(2 k-1)!!$, and so the total number of ways to perform operations on the edges labeled $\ell$ is  $$ \sum_{k}\binom{n_{\ell}}{2 k}(2 k-1)!!\equiv \sum_{k}\binom{n_{\ell}}{2 k}=2^{n_{\ell}-1} \quad(\bmod 2) $$  This is even if and only if $n_{\ell}>1$. Finally, note that the number of neighbors of $\mathcal{M}$ is the product across all $\ell$ of the above. So it is odd if and only if each factor is odd, if and only if $n_{\ell}=1$ for every $\ell$.  To finish, consider a huge simple graph $\Gamma$ on all the maximal matchings, with edge relations given by neighbor relation (we don't consider vertices to be connected to themselves). Observe that:  - Fantastic matchings correspond to isolated vertices (of degree zero, with no other neighbors) of $\Gamma$. - The rest of the vertices of $\Gamma$ have odd degrees (one less than the neighbor count) - The graph $\Gamma$ has an even number of vertices of odd degree (this is true for any simple graph, see "handshake lemma"). - The number of vertices of $\Gamma$ is odd, namely $(2\lceil n / 2\rceil-1)!!$