For any two different real numbers $x$ and $y$, we define $D(x, y)$ to be the unique integer $d$ satisfying $2^{d} \leqslant|x-y|<2^{d+1}$. Given a set of reals $\mathcal{F}$, and an element $x \in \mathcal{F}$, we say that the scales of $x$ in $\mathcal{F}$ are the values of $D(x, y)$ for $y \in \mathcal{F}$ with $x \neq y$.  Let $k$ be a given positive integer. Suppose that each member $x$ of $\mathcal{F}$ has at most $k$ different scales in $\mathcal{F}$ (note that these scales may depend on $x$ ). What is the maximum possible size of $\mathcal{F}$ ? (Italy)

Solution: We first construct a set $\mathcal{F}$ with $2^{k}$ members, each member having at most $k$ different scales in $\mathcal{F}$. Take $\mathcal{F}=\left\{0,1,2, \ldots, 2^{k}-1\right\}$. The scale between any two members of $\mathcal{F}$ is in the set $\{0,1, \ldots, k-1\}$.  We now show that $2^{k}$ is an upper bound on the size of $\mathcal{F}$. For every finite set $\mathcal{S}$ of real numbers, and every real $x$, let $r_{\mathcal{S}}(x)$ denote the number of different scales of $x$ in $\mathcal{S}$. That is, $r_{\mathcal{S}}(x)=|\{D(x, y): x \neq y \in \mathcal{S}\}|$. Thus, for every element $x$ of the set $\mathcal{F}$ in the problem statement, we have $r_{\mathcal{F}}(x) \leqslant k$. The condition $|\mathcal{F}| \leqslant 2^{k}$ is an immediate consequence of the following lemma. Lemma. Let $\mathcal{S}$ be a finite set of real numbers, and define  $$ w(\mathcal{S})=\sum_{x \in \mathcal{S}} 2^{-r_{\mathcal{S}}(x)} $$  Then $w(\mathcal{S}) \leqslant 1$. Proof. Induction on $n=|\mathcal{S}|$. If $\mathcal{S}=\{x\}$, then $r_{\mathcal{S}}(x)=0$, so $w(\mathcal{S})=1$. Assume now $n \geqslant 2$, and let $x_{1}<\cdots<x_{n}$ list the members of $\mathcal{S}$. Let $d$ be the minimal scale between two distinct elements of $\mathcal{S}$; then there exist neighbours $x_{t}$ and $x_{t+1}$ with $D\left(x_{t}, x_{t+1}\right)=d$. Notice that for any two indices $i$ and $j$ with $j-i>1$ we have $D\left(x_{i}, x_{j}\right)>d$, since  $$ \left|x_{i}-x_{j}\right|=\left|x_{i+1}-x_{i}\right|+\left|x_{j}-x_{i+1}\right| \geqslant 2^{d}+2^{d}=2^{d+1} . $$  Now choose the minimal $i \leqslant t$ and the maximal $j \geqslant t+1$ such that $D\left(x_{i}, x_{i+1}\right)=$ $D\left(x_{i+1}, x_{i+2}\right)=\cdots=D\left(x_{j-1}, x_{j}\right)=d$.  Let $E$ be the set of all the $x_{s}$ with even indices $i \leqslant s \leqslant j, O$ be the set of those with odd indices $i \leqslant s \leqslant j$, and $R$ be the rest of the elements (so that $\mathcal{S}$ is the disjoint union of $E, O$ and $R$ ). Set $\mathcal{S}_{O}=R \cup O$ and $\mathcal{S}_{E}=R \cup E$; we have $\left|\mathcal{S}_{O}\right|<|\mathcal{S}|$ and $\left|\mathcal{S}_{E}\right|<|\mathcal{S}|$, so $w\left(\mathcal{S}_{O}\right), w\left(\mathcal{S}_{E}\right) \leqslant 1$ by the inductive hypothesis.  Clearly, $r_{\mathcal{S}_{O}}(x) \leqslant r_{\mathcal{S}}(x)$ and $r_{\mathcal{S}_{E}}(x) \leqslant r_{\mathcal{S}}(x)$ for any $x \in R$, and thus  $$ \begin{aligned} \sum_{x \in R} 2^{-r_{\mathcal{S}}(x)} & =\frac{1}{2} \sum_{x \in R}\left(2^{-r_{\mathcal{S}}(x)}+2^{-r_{\mathcal{S}}(x)}\right) \\ & \leqslant \frac{1}{2} \sum_{x \in R}\left(2^{-r_{\mathcal{S}_{O}}(x)}+2^{-r_{\mathcal{S}_{E}}(x)}\right) \end{aligned} $$  On the other hand, for every $x \in O$, there is no $y \in \mathcal{S}_{O}$ such that $D_{\mathcal{S}_{O}}(x, y)=d$ (as all candidates from $\mathcal{S}$ were in $E$ ). Hence, we have $r_{\mathcal{S}_{O}}(x) \leqslant r_{\mathcal{S}}(x)-1$, and thus  $$ \sum_{x \in O} 2^{-r \mathcal{S}_{\mathcal{S}}(x)} \leqslant \frac{1}{2} \sum_{x \in O} 2^{-r s_{O}(x)} $$  Similarly, for every $x \in E$, we have  $$ \sum_{x \in E} 2^{-r_{\mathcal{S}}(x)} \leqslant \frac{1}{2} \sum_{x \in E} 2^{-r_{\mathcal{S}_{E}}(x)} $$  We can then combine these to give  $$ \begin{aligned} w(S) & =\sum_{x \in R} 2^{-r_{\mathcal{S}}(x)}+\sum_{x \in O} 2^{-r_{\mathcal{S}}(x)}+\sum_{x \in E} 2^{-r_{\mathcal{S}}(x)} \\ & \leqslant \frac{1}{2} \sum_{x \in R}\left(2^{-r_{\mathcal{S}_{O}}(x)}+2^{-r_{\mathcal{S}_{E}}(x)}\right)+\frac{1}{2} \sum_{x \in O} 2^{-r_{\mathcal{S}_{O}}(x)}+\frac{1}{2} \sum_{x \in E} 2^{-r_{\mathcal{S}_{E}}(x)} \\ & \left.=\frac{1}{2}\left(\sum_{x \in \mathcal{S}_{O}} 2^{-r_{S_{O}}(x)}+\sum_{x \in \mathcal{S}_{E}} 2^{-r s_{S_{E}}(x)}\right) \quad \text { (since } \mathcal{S}_{O}=O \cup R \text { and } \mathcal{S}_{E}=E \cup R\right) \\ & \left.\left.=\frac{1}{2}\left(w\left(\mathcal{S}_{O}\right)+w\left(\mathcal{S}_{E}\right)\right)\right) \quad \text { (by definition of } w(\cdot)\right) \\ & \leqslant 1 \quad \text { (by the inductive hypothesis) } \end{aligned} $$  which completes the induction. Comment 1. The sets $O$ and $E$ above are not the only ones we could have chosen. Indeed, we could instead have used the following definitions:  Let $d$ be the maximal scale between two distinct elements of $\mathcal{S}$; that is, $d=D\left(x_{1}, x_{n}\right)$. Let $O=\left\{x \in \mathcal{S}: D\left(x, x_{n}\right)=d\right\}$ (a 'left' part of the set) and let $E=\left\{x \in \mathcal{S}: D\left(x_{1}, x\right)=d\right\}$ (a 'right' part of the set). Note that these two sets are disjoint, and nonempty (since they contain $x_{1}$ and $x_{n}$ respectively). The rest of the proof is then the same as in Solution 1.  Comment 2. Another possible set $\mathcal{F}$ containing $2^{k}$ members could arise from considering a binary tree of height $k$, allocating a real number to each leaf, and trying to make the scale between the values of two leaves dependent only on the (graph) distance between them. The following construction makes this more precise.  We build up sets $\mathcal{F}_{k}$ recursively. Let $\mathcal{F}_{0}=\{0\}$, and then let $\mathcal{F}_{k+1}=\mathcal{F}_{k} \cup\left\{x+3 \cdot 4^{k}: x \in \mathcal{F}_{k}\right\}$ (i.e. each half of $\mathcal{F}_{k+1}$ is a copy of $F_{k}$ ). We have that $\mathcal{F}_{k}$ is contained in the interval $\left[0,4^{k+1}\right.$ ), and so it follows by induction on $k$ that every member of $F_{k+1}$ has $k$ different scales in its own half of $F_{k+1}$ (by the inductive hypothesis), and only the single scale $2 k+1$ in the other half of $F_{k+1}$.  Both of the constructions presented here have the property that every member of $\mathcal{F}$ has exactly $k$ different scales in $\mathcal{F}$. Indeed, it can be seen that this must hold (up to a slight perturbation) for any such maximal set. Suppose there were some element $x$ with only $k-1$ different scales in $\mathcal{F}$ (and every other element had at most $k$ different scales). Then we take some positive real $\epsilon$, and construct a new set $\mathcal{F}^{\prime}=\{y: y \in \mathcal{F}, y \leqslant x\} \cup\{y+\epsilon: y \in \mathcal{F}, y \geqslant x\}$. We have $\left|\mathcal{F}^{\prime}\right|=|\mathcal{F}|+1$, and if $\epsilon$ is sufficiently small then $\mathcal{F}^{\prime}$ will also satisfy the property that no member has more than $k$ different scales in $\mathcal{F}^{\prime}$.  This observation might be used to motivate the idea of weighting members of an arbitrary set $\mathcal{S}$ of reals according to how many different scales they have in $\mathcal{S}$.


Let $A B C$ be a triangle. Circle $\Gamma$ passes through $A$, meets segments $A B$ and $A C$  again at points $D$ and $E$ respectively, and intersects segment $B C$ at $F$ and $G$ such that $F$ lies between $B$ and $G$. The tangent to circle $B D F$ at $F$ and the tangent to circle $C E G$ at $G$ meet at point $T$. Suppose that points $A$ and $T$ are distinct. Prove that line $A T$ is parallel to $B C$. (Nigeria)

Solution: Notice that $\angle T F B=\angle F D A$ because $F T$ is tangent to circle $B D F$, and moreover $\angle F D A=\angle C G A$ because quadrilateral $A D F G$ is cyclic. Similarly, $\angle T G B=\angle G E C$ because $G T$ is tangent to circle $C E G$, and $\angle G E C=\angle C F A$. Hence,  $$ \angle T F B=\angle C G A \text { and } \quad \angle T G B=\angle C F A $$  ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-058.jpg?height=624&width=1326&top_left_y=930&top_left_x=365)  Triangles $F G A$ and $G F T$ have a common side $F G$, and by (1) their angles at $F, G$ are the same. So, these triangles are congruent. So, their altitudes starting from $A$ and $T$, respectively, are equal and hence $A T$ is parallel to line $B F G C$.  Comment. Alternatively, we can prove first that $T$ lies on $\Gamma$. For example, this can be done by showing that $\angle A F T=\angle A G T$ using (1). Then the statement follows as $\angle T A F=\angle T G F=\angle G F A$.


Let $A B C$ be an acute-angled triangle and let $D, E$, and $F$ be the feet of altitudes from $A, B$, and $C$ to sides $B C, C A$, and $A B$, respectively. Denote by $\omega_{B}$ and $\omega_{C}$ the incircles of triangles $B D F$ and $C D E$, and let these circles be tangent to segments $D F$ and $D E$ at $M$ and $N$, respectively. Let line $M N$ meet circles $\omega_{B}$ and $\omega_{C}$ again at $P \neq M$ and $Q \neq N$, respectively. Prove that $M P=N Q$. (Vietnam)

Solution: Denote the centres of $\omega_{B}$ and $\omega_{C}$ by $O_{B}$ and $O_{C}$, let their radii be $r_{B}$ and $r_{C}$, and let $B C$ be tangent to the two circles at $T$ and $U$, respectively. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-059.jpg?height=801&width=1014&top_left_y=659&top_left_x=521)  From the cyclic quadrilaterals $A F D C$ and $A B D E$ we have  $$ \angle M D O_{B}=\frac{1}{2} \angle F D B=\frac{1}{2} \angle B A C=\frac{1}{2} \angle C D E=\angle O_{C} D N, $$  so the right-angled triangles $D M O_{B}$ and $D N O_{C}$ are similar. The ratio of similarity between the two triangles is  $$ \frac{D N}{D M}=\frac{O_{C} N}{O_{B} M}=\frac{r_{C}}{r_{B}} . $$  Let $\varphi=\angle D M N$ and $\psi=\angle M N D$. The lines $F M$ and $E N$ are tangent to $\omega_{B}$ and $\omega_{C}$, respectively, so  $$ \angle M T P=\angle F M P=\angle D M N=\varphi \quad \text { and } \quad \angle Q U N=\angle Q N E=\angle M N D=\psi $$  (It is possible that $P$ or $Q$ coincides with $T$ or $U$, or lie inside triangles $D M T$ or $D U N$, respectively. To reduce case-sensitivity, we may use directed angles or simply ignore angles $M T P$ and $Q U N$.)  In the circles $\omega_{B}$ and $\omega_{C}$ the lengths of chords $M P$ and $N Q$ are  $$ M P=2 r_{B} \cdot \sin \angle M T P=2 r_{B} \cdot \sin \varphi \quad \text { and } \quad N Q=2 r_{C} \cdot \sin \angle Q U N=2 r_{C} \cdot \sin \psi $$  By applying the sine rule to triangle $D N M$ we get  $$ \frac{D N}{D M}=\frac{\sin \angle D M N}{\sin \angle M N D}=\frac{\sin \varphi}{\sin \psi} $$  Finally, putting the above observations together, we get  $$ \frac{M P}{N Q}=\frac{2 r_{B} \sin \varphi}{2 r_{C} \sin \psi}=\frac{r_{B}}{r_{C}} \cdot \frac{\sin \varphi}{\sin \psi}=\frac{D M}{D N} \cdot \frac{\sin \varphi}{\sin \psi}=\frac{\sin \psi}{\sin \varphi} \cdot \frac{\sin \varphi}{\sin \psi}=1, $$  so $M P=N Q$ as required.


In triangle $A B C$, let $A_{1}$ and $B_{1}$ be two points on sides $B C$ and $A C$, and let $P$ and $Q$ be two points on segments $A A_{1}$ and $B B_{1}$, respectively, so that line $P Q$ is parallel to $A B$. On ray $P B_{1}$, beyond $B_{1}$, let $P_{1}$ be a point so that $\angle P P_{1} C=\angle B A C$. Similarly, on ray $Q A_{1}$, beyond $A_{1}$, let $Q_{1}$ be a point so that $\angle C Q_{1} Q=\angle C B A$. Show that points $P, Q, P_{1}$, and $Q_{1}$ are concyclic. (Ukraine)

Solution: Throughout the solution we use oriented angles. Let rays $A A_{1}$ and $B B_{1}$ intersect the circumcircle of $\triangle A C B$ at $A_{2}$ and $B_{2}$, respectively. By  $$ \angle Q P A_{2}=\angle B A A_{2}=\angle B B_{2} A_{2}=\angle Q B_{2} A_{2} $$  points $P, Q, A_{2}, B_{2}$ are concyclic; denote the circle passing through these points by $\omega$. We shall prove that $P_{1}$ and $Q_{1}$ also lie on $\omega$. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-060.jpg?height=615&width=858&top_left_y=909&top_left_x=602)  By  $$ \angle C A_{2} A_{1}=\angle C A_{2} A=\angle C B A=\angle C Q_{1} Q=\angle C Q_{1} A_{1}, $$  points $C, Q_{1}, A_{2}, A_{1}$ are also concyclic. From that we get  $$ \angle Q Q_{1} A_{2}=\angle A_{1} Q_{1} A_{2}=\angle A_{1} C A_{2}=\angle B C A_{2}=\angle B A A_{2}=\angle Q P A_{2} $$  so $Q_{1}$ lies on $\omega$. It follows similarly that $P_{1}$ lies on $\omega$.


Solution: First consider the case when lines $P P_{1}$ and $Q Q_{1}$ intersect each other at some point $R$.  Let line $P Q$ meet the sides $A C$ and $B C$ at $E$ and $F$, respectively. Then  $$ \angle P P_{1} C=\angle B A C=\angle P E C $$  so points $C, E, P, P_{1}$ lie on a circle; denote that circle by $\omega_{P}$. It follows analogously that points $C, F, Q, Q_{1}$ lie on another circle; denote it by $\omega_{Q}$.  Let $A Q$ and $B P$ intersect at $T$. Applying Pappus' theorem to the lines $A A_{1} P$ and $B B_{1} Q$ provides that points $C=A B_{1} \cap B A_{1}, R=A_{1} Q \cap B_{1} P$ and $T=A Q \cap B P$ are collinear.  Let line $R C T$ meet $P Q$ and $A B$ at $S$ and $U$, respectively. From $A B \| P Q$ we obtain  $$ \frac{S P}{S Q}=\frac{U B}{U A}=\frac{S F}{S E} $$  so  $$ S P \cdot S E=S Q \cdot S F $$  ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-061.jpg?height=792&width=1072&top_left_y=204&top_left_x=498)  So, point $S$ has equal powers with respect to $\omega_{P}$ and $\omega_{Q}$, hence line $R C S$ is their radical axis; then $R$ also has equal powers to the circles, so $R P \cdot R P_{1}=R Q \cdot R Q_{1}$, proving that points $P, P_{1}, Q, Q_{1}$ are indeed concyclic.  Now consider the case when $P P_{1}$ and $Q Q_{1}$ are parallel. Like in the previous case, let $A Q$ and $B P$ intersect at $T$. Applying Pappus' theorem again to the lines $A A_{1} P$ and $B B_{1} Q$, in this limit case it shows that line $C T$ is parallel to $P P_{1}$ and $Q Q_{1}$.  Let line $C T$ meet $P Q$ and $A B$ at $S$ and $U$, as before. The same calculation as in the previous case shows that $S P \cdot S E=S Q \cdot S F$, so $S$ lies on the radical axis between $\omega_{P}$ and $\omega_{Q}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-061.jpg?height=800&width=1043&top_left_y=1413&top_left_x=512)  Line $C S T$, that is the radical axis between $\omega_{P}$ and $\omega_{Q}$, is perpendicular to the line $\ell$ of centres of $\omega_{P}$ and $\omega_{Q}$. Hence, the chords $P P_{1}$ and $Q Q_{1}$ are perpendicular to $\ell$. So the quadrilateral $P P_{1} Q_{1} Q$ is an isosceles trapezium with symmetry axis $\ell$, and hence is cyclic.  Comment. There are several ways of solving the problem involving Pappus' theorem. For example, one may consider the points $K=P B_{1} \cap B C$ and $L=Q A_{1} \cap A C$. Applying Pappus' theorem to the lines $A A_{1} P$ and $Q B_{1} B$ we get that $K, L$, and $P Q \cap A B$ are collinear, i.e. that $K L \| A B$. Therefore, cyclicity of $P, Q, P_{1}$, and $Q_{1}$ is equivalent to that of $K, L, P_{1}$, and $Q_{1}$. The latter is easy after noticing that $C$ also lies on that circle. Indeed, e.g. $\angle(L K, L C)=\angle(A B, A C)=\angle\left(P_{1} K, P_{1} C\right)$ shows that $K$ lies on circle $K L C$.  This approach also has some possible degeneracy, as the points $K$ and $L$ may happen to be ideal.


Let $P$ be a point inside triangle $A B C$. Let $A P$ meet $B C$ at $A_{1}$, let $B P$ meet $C A$ at $B_{1}$, and let $C P$ meet $A B$ at $C_{1}$. Let $A_{2}$ be the point such that $A_{1}$ is the midpoint of $P A_{2}$, let $B_{2}$ be the point such that $B_{1}$ is the midpoint of $P B_{2}$, and let $C_{2}$ be the point such that $C_{1}$ is the midpoint of $P C_{2}$. Prove that points $A_{2}, B_{2}$, and $C_{2}$ cannot all lie strictly inside the circumcircle of triangle $A B C$. (Australia)

Solution: Since  $$ \angle A P B+\angle B P C+\angle C P A=2 \pi=(\pi-\angle A C B)+(\pi-\angle B A C)+(\pi-\angle C B A), $$  at least one of the following inequalities holds:  $$ \angle A P B \geqslant \pi-\angle A C B, \quad \angle B P C \geqslant \pi-\angle B A C, \quad \angle C P A \geqslant \pi-\angle C B A . $$  Without loss of generality, we assume that $\angle B P C \geqslant \pi-\angle B A C$. We have $\angle B P C>\angle B A C$ because $P$ is inside $\triangle A B C$. So $\angle B P C \geqslant \max (\angle B A C, \pi-\angle B A C)$ and hence  $$ \sin \angle B P C \leqslant \sin \angle B A C . $$  Let the rays $A P, B P$, and $C P$ cross the circumcircle $\Omega$ again at $A_{3}, B_{3}$, and $C_{3}$, respectively. We will prove that at least one of the ratios $\frac{P B_{1}}{B_{1} B_{3}}$ and $\frac{P C_{1}}{C_{1} C_{3}}$ is at least 1 , which yields that one of the points $B_{2}$ and $C_{2}$ does not lie strictly inside $\Omega$.  Because $A, B, C, B_{3}$ lie on a circle, the triangles $C B_{1} B_{3}$ and $B B_{1} A$ are similar, so  $$ \frac{C B_{1}}{B_{1} B_{3}}=\frac{B B_{1}}{B_{1} A} $$  Applying the sine rule we obtain  $$ \frac{P B_{1}}{B_{1} B_{3}}=\frac{P B_{1}}{C B_{1}} \cdot \frac{C B_{1}}{B_{1} B_{3}}=\frac{P B_{1}}{C B_{1}} \cdot \frac{B B_{1}}{B_{1} A}=\frac{\sin \angle A C P}{\sin \angle B P C} \cdot \frac{\sin \angle B A C}{\sin \angle P B A} . $$  Similarly,  $$ \frac{P C_{1}}{C_{1} C_{3}}=\frac{\sin \angle P B A}{\sin \angle B P C} \cdot \frac{\sin \angle B A C}{\sin \angle A C P} $$  Multiplying these two equations we get  $$ \frac{P B_{1}}{B_{1} B_{3}} \cdot \frac{P C_{1}}{C_{1} C_{3}}=\frac{\sin ^{2} \angle B A C}{\sin ^{2} \angle B P C} \geqslant 1 $$  using (*), which yields the desired conclusion.  Comment. It also cannot happen that all three points $A_{2}, B_{2}$, and $C_{2}$ lie strictly outside $\Omega$. The same proof works almost literally, starting by assuming without loss of generality that $\angle B P C \leqslant \pi-\angle B A C$ and using $\angle B P C>\angle B A C$ to deduce that $\sin \angle B P C \geqslant \sin \angle B A C$. It is possible for $A_{2}, B_{2}$, and $C_{2}$ all to lie on the circumcircle; from the above solution we may derive that this happens if and only if $P$ is the orthocentre of the triangle $A B C$, (which lies strictly inside $A B C$ if and only if $A B C$ is acute).


Solution: Define points $A_{3}, B_{3}$, and $C_{3}$ as in   Observe that $\triangle P B C_{3} \sim \triangle P C B_{3}$. Let $X$ be the point on side $P B_{3}$ that corresponds to point $C_{1}$ on side $P C_{3}$ under this similarity. In other words, $X$ lies on segment $P B_{3}$ and satisfies $P X: X B_{3}=P C_{1}: C_{1} C_{3}$. It follows that  $$ \angle X C P=\angle P B C_{1}=\angle B_{3} B A=\angle B_{3} C B_{1} . $$  Hence lines $C X$ and $C B_{1}$ are isogonal conjugates in $\triangle P C B_{3}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-063.jpg?height=812&width=807&top_left_y=1207&top_left_x=533)  Let $Y$ be the foot of the bisector of $\angle B_{3} C P$ in $\triangle P C B_{3}$. Since $P C_{1}<C_{1} C_{3}$, we have $P X<X B_{3}$. Also, we have $P Y<Y B_{3}$ because $P B_{1}<B_{1} B_{3}$ and $Y$ lies between $X$ and $B_{1}$. By the angle bisector theorem in $\triangle P C B_{3}$, we have $P Y: Y B_{3}=P C: C B_{3}$. So $P C<C B_{3}$ and it follows that $\angle P B_{3} C<\angle C P B_{3}$. Now since $\angle P B_{3} C=\angle B B_{3} C=\angle B A C$, we have  $$ \angle B A C<\angle C P B_{3} . $$  Similarly, we have  $$ \angle C B A<\angle A P C_{3} \text { and } \angle A C B<\angle B P A_{3}=\angle B_{3} P A . $$  Adding these three inequalities yields $\pi<\pi$, and this contradiction concludes the proof.


Solution: Choose coordinates such that the circumcentre of $\triangle A B C$ is at the origin and the circumradius is 1 . Then we may think of $A, B$, and $C$ as vectors in $\mathbb{R}^{2}$ such that  $$ |A|^{2}=|B|^{2}=|C|^{2}=1 $$  $P$ may be represented as a convex combination $\alpha A+\beta B+\gamma C$ where $\alpha, \beta, \gamma>0$ and $\alpha+\beta+\gamma=1$. Then  $$ A_{1}=\frac{\beta B+\gamma C}{\beta+\gamma}=\frac{1}{1-\alpha} P-\frac{\alpha}{1-\alpha} A, $$  so  $$ A_{2}=2 A_{1}-P=\frac{1+\alpha}{1-\alpha} P-\frac{2 \alpha}{1-\alpha} A $$  Hence  $$ \left|A_{2}\right|^{2}=\left(\frac{1+\alpha}{1-\alpha}\right)^{2}|P|^{2}+\left(\frac{2 \alpha}{1-\alpha}\right)^{2}|A|^{2}-\frac{4 \alpha(1+\alpha)}{(1-\alpha)^{2}} A \cdot P $$  Using $|A|^{2}=1$ we obtain  $$ \frac{(1-\alpha)^{2}}{2(1+\alpha)}\left|A_{2}\right|^{2}=\frac{1+\alpha}{2}|P|^{2}+\frac{2 \alpha^{2}}{1+\alpha}-2 \alpha A \cdot P $$  Likewise  $$ \frac{(1-\beta)^{2}}{2(1+\beta)}\left|B_{2}\right|^{2}=\frac{1+\beta}{2}|P|^{2}+\frac{2 \beta^{2}}{1+\beta}-2 \beta B \cdot P $$  and  $$ \frac{(1-\gamma)^{2}}{2(1+\gamma)}\left|C_{2}\right|^{2}=\frac{1+\gamma}{2}|P|^{2}+\frac{2 \gamma^{2}}{1+\gamma}-2 \gamma C \cdot P $$  Summing (1), (2) and (3) we obtain on the LHS the positive linear combination  $$ \text { LHS }=\frac{(1-\alpha)^{2}}{2(1+\alpha)}\left|A_{2}\right|^{2}+\frac{(1-\beta)^{2}}{2(1+\beta)}\left|B_{2}\right|^{2}+\frac{(1-\gamma)^{2}}{2(1+\gamma)}\left|C_{2}\right|^{2} $$  and on the RHS the quantity  $$ \left(\frac{1+\alpha}{2}+\frac{1+\beta}{2}+\frac{1+\gamma}{2}\right)|P|^{2}+\left(\frac{2 \alpha^{2}}{1+\alpha}+\frac{2 \beta^{2}}{1+\beta}+\frac{2 \gamma^{2}}{1+\gamma}\right)-2(\alpha A \cdot P+\beta B \cdot P+\gamma C \cdot P) . $$  The first term is $2|P|^{2}$ and the last term is $-2 P \cdot P$, so  $$ \begin{aligned} \text { RHS } & =\left(\frac{2 \alpha^{2}}{1+\alpha}+\frac{2 \beta^{2}}{1+\beta}+\frac{2 \gamma^{2}}{1+\gamma}\right) \\ & =\frac{3 \alpha-1}{2}+\frac{(1-\alpha)^{2}}{2(1+\alpha)}+\frac{3 \beta-1}{2}+\frac{(1-\beta)^{2}}{2(1+\beta)}+\frac{3 \gamma-1}{2}+\frac{(1-\gamma)^{2}}{2(1+\gamma)} \\ & =\frac{(1-\alpha)^{2}}{2(1+\alpha)}+\frac{(1-\beta)^{2}}{2(1+\beta)}+\frac{(1-\gamma)^{2}}{2(1+\gamma)} \end{aligned} $$  Here we used the fact that  $$ \frac{3 \alpha-1}{2}+\frac{3 \beta-1}{2}+\frac{3 \gamma-1}{2}=0 . $$  We have shown that a linear combination of $\left|A_{1}\right|^{2},\left|B_{1}\right|^{2}$, and $\left|C_{1}\right|^{2}$ with positive coefficients is equal to the sum of the coefficients. Therefore at least one of $\left|A_{1}\right|^{2},\left|B_{1}\right|^{2}$, and $\left|C_{1}\right|^{2}$ must be at least 1 , as required.  Comment. This proof also works when $P$ is any point for which $\alpha, \beta, \gamma>-1, \alpha+\beta+\gamma=1$, and $\alpha, \beta, \gamma \neq 1$. (In any cases where $\alpha=1$ or $\beta=1$ or $\gamma=1$, some points in the construction are not defined.)  This page is intentionally left blank


Let $A B C D E$ be a convex pentagon with $C D=D E$ and $\angle E D C \neq 2 \cdot \angle A D B$. Suppose that a point $P$ is located in the interior of the pentagon such that $A P=A E$ and $B P=B C$. Prove that $P$ lies on the diagonal $C E$ if and only if $\operatorname{area}(B C D)+\operatorname{area}(A D E)=$ $\operatorname{area}(A B D)+\operatorname{area}(A B P)$. (Hungary)

Solution: Let $P^{\prime}$ be the reflection of $P$ across line $A B$, and let $M$ and $N$ be the midpoints of $P^{\prime} E$ and $P^{\prime} C$ respectively. Convexity ensures that $P^{\prime}$ is distinct from both $E$ and $C$, and hence from both $M$ and $N$. We claim that both the area condition and the collinearity condition in the problem are equivalent to the condition that the (possibly degenerate) right-angled triangles $A P^{\prime} M$ and $B P^{\prime} N$ are directly similar (equivalently, $A P^{\prime} E$ and $B P^{\prime} C$ are directly similar). ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-066.jpg?height=535&width=549&top_left_y=806&top_left_x=756)  For the equivalence with the collinearity condition, let $F$ denote the foot of the perpendicular from $P^{\prime}$ to $A B$, so that $F$ is the midpoint of $P P^{\prime}$. We have that $P$ lies on $C E$ if and only if $F$ lies on $M N$, which occurs if and only if we have the equality $\angle A F M=\angle B F N$ of signed angles modulo $\pi$. By concyclicity of $A P^{\prime} F M$ and $B F P^{\prime} N$, this is equivalent to $\angle A P^{\prime} M=\angle B P^{\prime} N$, which occurs if and only if $A P^{\prime} M$ and $B P^{\prime} N$ are directly similar. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-066.jpg?height=306&width=441&top_left_y=1663&top_left_x=813)  For the other equivalence with the area condition, we have the equality of signed areas $\operatorname{area}(A B D)+\operatorname{area}(A B P)=\operatorname{area}\left(A P^{\prime} B D\right)=\operatorname{area}\left(A P^{\prime} D\right)+\operatorname{area}\left(B D P^{\prime}\right)$. Using the identity $\operatorname{area}(A D E)-\operatorname{area}\left(A P^{\prime} D\right)=\operatorname{area}(A D E)+\operatorname{area}\left(A D P^{\prime}\right)=2$ area $(A D M)$, and similarly for $B$, we find that the area condition is equivalent to the equality  $$ \operatorname{area}(D A M)=\operatorname{area}(D B N) . $$  Now note that $A$ and $B$ lie on the perpendicular bisectors of $P^{\prime} E$ and $P^{\prime} C$, respectively. If we write $G$ and $H$ for the feet of the perpendiculars from $D$ to these perpendicular bisectors respectively, then this area condition can be rewritten as  $$ M A \cdot G D=N B \cdot H D $$  (In this condition, we interpret all lengths as signed lengths according to suitable conventions: for instance, we orient $P^{\prime} E$ from $P^{\prime}$ to $E$, orient the parallel line $D H$ in the same direction, and orient the perpendicular bisector of $P^{\prime} E$ at an angle $\pi / 2$ clockwise from the oriented segment $P^{\prime} E$ - we adopt the analogous conventions at $B$.) ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-067.jpg?height=529&width=575&top_left_y=204&top_left_x=758)  To relate the signed lengths $G D$ and $H D$ to the triangles $A P^{\prime} M$ and $B P^{\prime} N$, we use the following calculation. Claim. Let $\Gamma$ denote the circle centred on $D$ with both $E$ and $C$ on the circumference, and $h$ the power of $P^{\prime}$ with respect to $\Gamma$. Then we have the equality  $$ G D \cdot P^{\prime} M=H D \cdot P^{\prime} N=\frac{1}{4} h \neq 0 . $$  Proof. Firstly, we have $h \neq 0$, since otherwise $P^{\prime}$ would lie on $\Gamma$, and hence the internal angle bisectors of $\angle E D P^{\prime}$ and $\angle P^{\prime} D C$ would pass through $A$ and $B$ respectively. This would violate the angle inequality $\angle E D C \neq 2 \cdot \angle A D B$ given in the question.  Next, let $E^{\prime}$ denote the second point of intersection of $P^{\prime} E$ with $\Gamma$, and let $E^{\prime \prime}$ denote the point on $\Gamma$ diametrically opposite $E^{\prime}$, so that $E^{\prime \prime} E$ is perpendicular to $P^{\prime} E$. The point $G$ lies on the perpendicular bisectors of the sides $P^{\prime} E$ and $E E^{\prime \prime}$ of the right-angled triangle $P^{\prime} E E^{\prime \prime}$; it follows that $G$ is the midpoint of $P^{\prime} E^{\prime \prime}$. Since $D$ is the midpoint of $E^{\prime} E^{\prime \prime}$, we have that $G D=\frac{1}{2} P^{\prime} E^{\prime}$. Since $P^{\prime} M=\frac{1}{2} P^{\prime} E$, we have $G D \cdot P^{\prime} M=\frac{1}{4} P^{\prime} E^{\prime} \cdot P^{\prime} E=\frac{1}{4} h$. The other equality $H D \cdot P^{\prime} N$ follows by exactly the same argument. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-067.jpg?height=727&width=672&top_left_y=1664&top_left_x=726)  From this claim, we see that the area condition is equivalent to the equality  $$ \left(M A: P^{\prime} M\right)=\left(N B: P^{\prime} N\right) $$  of ratios of signed lengths, which is equivalent to direct similarity of $A P^{\prime} M$ and $B P^{\prime} N$, as desired.


Solution: Along the perpendicular bisector of $C E$, define the linear function  $$ f(X)=\operatorname{area}(B C X)+\operatorname{area}(A X E)-\operatorname{area}(A B X)-\operatorname{area}(A B P), $$  where, from now on, we always use signed areas. Thus, we want to show that $C, P, E$ are collinear if and only if $f(D)=0$. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-068.jpg?height=433&width=512&top_left_y=503&top_left_x=772)  Let $P^{\prime}$ be the reflection of $P$ across line $A B$. The point $P^{\prime}$ does not lie on the line $C E$. To see this, we let $A^{\prime \prime}$ and $B^{\prime \prime}$ be the points obtained from $A$ and $B$ by dilating with scale factor 2 about $P^{\prime}$, so that $P$ is the orthogonal projection of $P^{\prime}$ onto $A^{\prime \prime} B^{\prime \prime}$. Since $A$ lies on the perpendicular bisector of $P^{\prime} E$, the triangle $A^{\prime \prime} E P^{\prime}$ is right-angled at $E$ (and $B^{\prime \prime} C P^{\prime}$ similarly). If $P^{\prime}$ were to lie on $C E$, then the lines $A^{\prime \prime} E$ and $B^{\prime \prime} C$ would be perpendicular to $C E$ and $A^{\prime \prime}$ and $B^{\prime \prime}$ would lie on the opposite side of $C E$ to $D$. It follows that the line $A^{\prime \prime} B^{\prime \prime}$ does not meet triangle $C D E$, and hence point $P$ does not lie inside $C D E$. But then $P$ must lie inside $A B C E$, and it is clear that such a point cannot reflect to a point $P^{\prime}$ on $C E$.  We thus let $O$ be the centre of the circle $C E P^{\prime}$. The lines $A O$ and $B O$ are the perpendicular bisectors of $E P^{\prime}$ and $C P^{\prime}$, respectively, so  $$ \begin{aligned} \operatorname{area}(B C O)+\operatorname{area}(A O E) & =\operatorname{area}\left(O P^{\prime} B\right)+\operatorname{area}\left(P^{\prime} O A\right)=\operatorname{area}\left(P^{\prime} B O A\right) \\ & =\operatorname{area}(A B O)+\operatorname{area}\left(B A P^{\prime}\right)=\operatorname{area}(A B O)+\operatorname{area}(A B P), \end{aligned} $$  and hence $f(O)=0$. Notice that if point $O$ coincides with $D$ then points $A, B$ lie in angle domain $C D E$ and $\angle E O C=2 \cdot \angle A O B$, which is not allowed. So, $O$ and $D$ must be distinct. Since $f$ is linear and vanishes at $O$, it follows that $f(D)=0$ if and only if $f$ is constant zero - we want to show this occurs if and only if $C, P, E$ are collinear. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-068.jpg?height=556&width=1254&top_left_y=1906&top_left_x=408)  In the one direction, suppose firstly that $C, P, E$ are not collinear, and let $T$ be the centre of the circle $C E P$. The same calculation as above provides  $$ \operatorname{area}(B C T)+\operatorname{area}(A T E)=\operatorname{area}(P B T A)=\operatorname{area}(A B T)-\operatorname{area}(A B P) $$  $$ f(T)=-2 \operatorname{area}(A B P) \neq 0 $$  Hence, the linear function $f$ is nonconstant with its zero is at $O$, so that $f(D) \neq 0$. In the other direction, suppose that the points $C, P, E$ are collinear. We will show that $f$ is constant zero by finding a second point (other than $O$ ) at which it vanishes. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-069.jpg?height=410&width=1098&top_left_y=366&top_left_x=479)  Let $Q$ be the reflection of $P$ across the midpoint of $A B$, so $P A Q B$ is a parallelogram. It is easy to see that $Q$ is on the perpendicular bisector of $C E$; for instance if $A^{\prime}$ and $B^{\prime}$ are the points produced from $A$ and $B$ by dilating about $P$ with scale factor 2, then the projection of $Q$ to $C E$ is the midpoint of the projections of $A^{\prime}$ and $B^{\prime}$, which are $E$ and $C$ respectively. The triangles $B C Q$ and $A Q E$ are indirectly congruent, so  $$ f(Q)=(\operatorname{area}(B C Q)+\operatorname{area}(A Q E))-(\operatorname{area}(A B Q)-\operatorname{area}(B A P))=0-0=0 $$  The points $O$ and $Q$ are distinct. To see this, consider the circle $\omega$ centred on $Q$ with $P^{\prime}$ on the circumference; since triangle $P P^{\prime} Q$ is right-angled at $P^{\prime}$, it follows that $P$ lies outside $\omega$. On the other hand, $P$ lies between $C$ and $E$ on the line $C P E$. It follows that $C$ and $E$ cannot both lie on $\omega$, so that $\omega$ is not the circle $C E P^{\prime}$ and $Q \neq O$.  Since $O$ and $Q$ are distinct zeroes of the linear function $f$, we have $f(D)=0$ as desired. Comment 1. The condition $\angle E D C \neq 2 \cdot \angle A D B$ cannot be omitted. If $D$ is the centre of circle $C E P^{\prime}$, then the condition on triangle areas is satisfied automatically, without having $P$ on line $C E$.  Comment 2. The "only if" part of this problem is easier than the "if" part. For example, in the second part of Solution 2, the triangles $E A Q$ and $Q B C$ are indirectly congruent, so the sum of their areas is 0 , and $D C Q E$ is a kite. Now one can easily see that $\angle(A Q, D E)=\angle(C D, C B)$ and $\angle(B Q, D C)=\angle(E D, E A)$, whence area $(B C D)=\operatorname{area}(A Q D)+\operatorname{area}(E Q A)$ and $\operatorname{area}(A D E)=$ $\operatorname{area}(B D Q)+\operatorname{area}(B Q C)$, which yields the result.  Comment 3. The origin of the problem is the following observation. Let $A B D H$ be a tetrahedron and consider the sphere $\mathcal{S}$ that is tangent to the four face planes, internally to planes $A D H$ and $B D H$ and externally to $A B D$ and $A B H$ (or vice versa). It is known that the sphere $\mathcal{S}$ exists if and only if area $(A D H)+\operatorname{area}(B D H) \neq \operatorname{area}(A B H)+\operatorname{area}(A B D)$; this relation comes from the usual formula for the volume of the tetrahedron.  Let $T, T_{a}, T_{b}, T_{d}$ be the points of tangency between the sphere and the four planes, as shown in the picture. Rotate the triangle $A B H$ inward, the triangles $B D H$ and $A D H$ outward, into the triangles $A B P, B D C$ and $A D E$, respectively, in the plane $A B D$. Notice that the points $T_{d}, T_{a}, T_{b}$ are rotated to $T$, so we have $H T_{a}=H T_{b}=H T_{d}=P T=C T=E T$. Therefore, the point $T$ is the centre of the circle $C E P$. Hence, if the sphere exists then $C, E, P$ cannot be collinear.  If the condition $\angle E D C \neq 2 \cdot \angle A D B$ is replaced by the constraint that the angles $\angle E D A, \angle A D B$ and $\angle B D C$ satisfy the triangle inequality, it enables reconstructing the argument with the tetrahedron and the tangent sphere. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-070.jpg?height=1069&width=1306&top_left_y=825&top_left_x=381)


Let $I$ be the incentre of acute-angled triangle $A B C$. Let the incircle meet $B C, C A$, and $A B$ at $D, E$, and $F$, respectively. Let line $E F$ intersect the circumcircle of the triangle at $P$ and $Q$, such that $F$ lies between $E$ and $P$. Prove that $\angle D P A+\angle A Q D=\angle Q I P$. (Slovakia)

Solution: Let $N$ and $M$ be the midpoints of the arcs $\widehat{B C}$ of the circumcircle, containing and opposite vertex $A$, respectively. By $\angle F A E=\angle B A C=\angle B N C$, the right-angled kites $A F I E$ and $N B M C$ are similar. Consider the spiral similarity $\varphi$ (dilation in case of $A B=A C$ ) that moves $A F I E$ to $N B M C$. The directed angle in which $\varphi$ changes directions is $\angle(A F, N B)$, same as $\angle(A P, N P)$ and $\angle(A Q, N Q)$; so lines $A P$ and $A Q$ are mapped to lines $N P$ and $N Q$, respectively. Line $E F$ is mapped to $B C$; we can see that the intersection points $P=E F \cap A P$ and $Q=E F \cap A Q$ are mapped to points $B C \cap N P$ and $B C \cap N Q$, respectively. Denote these points by $P^{\prime}$ and $Q^{\prime}$, respectively. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-071.jpg?height=689&width=1512&top_left_y=883&top_left_x=272)  Let $L$ be the midpoint of $B C$. We claim that points $P, Q, D, L$ are concyclic (if $D=L$ then line $B C$ is tangent to circle $P Q D$ ). Let $P Q$ and $B C$ meet at $Z$. By applying Menelaus' theorem to triangle $A B C$ and line $E F Z$, we have  $$ \frac{B D}{D C}=\frac{B F}{F A} \cdot \frac{A E}{E C}=-\frac{B Z}{Z C}, $$  so the pairs $B, C$ and $D, Z$ are harmonic. It is well-known that this implies $Z B \cdot Z C=Z D \cdot Z L$. (The inversion with pole $Z$ that swaps $B$ and $C$ sends $Z$ to infinity and $D$ to the midpoint of $B C$, because the cross-ratio is preserved.) Hence, $Z D \cdot Z L=Z B \cdot Z C=Z P \cdot Z Q$ by the power of $Z$ with respect to the circumcircle; this proves our claim.  By $\angle M P P^{\prime}=\angle M Q Q^{\prime}=\angle M L P^{\prime}=\angle M L Q^{\prime}=90^{\circ}$, the quadrilaterals $M L P P^{\prime}$ and $M L Q Q^{\prime}$ are cyclic. Then the problem statement follows by  $$ \begin{aligned} \angle D P A+\angle A Q D & =360^{\circ}-\angle P A Q-\angle Q D P=360^{\circ}-\angle P N Q-\angle Q L P \\ & =\angle L P N+\angle N Q L=\angle P^{\prime} M L+\angle L M Q^{\prime}=\angle P^{\prime} M Q^{\prime}=\angle P I Q . \end{aligned} $$


Solution: Define the point $M$ and the same spiral similarity $\varphi$ as in the previous solution. (The point $N$ is not necessary.) It is well-known that the centre of the spiral similarity that maps $F, E$ to $B, C$ is the Miquel point of the lines $F E, B C, B F$ and $C E$; that is, the second intersection of circles $A B C$ and $A E F$. Denote that point by $S$.  By $\varphi(F)=B$ and $\varphi(E)=C$ the triangles $S B F$ and $S C E$ are similar, so we have  $$ \frac{S B}{S C}=\frac{B F}{C E}=\frac{B D}{C D} $$  By the converse of the angle bisector theorem, that indicates that line $S D$ bisects $\angle B S C$ and hence passes through $M$.  Let $K$ be the intersection point of lines $E F$ and $S I$. Notice that $\varphi$ sends points $S, F, E, I$ to $S, B, C, M$, so $\varphi(K)=\varphi(F E \cap S I)=B C \cap S M=D$. By $\varphi(I)=M$, we have $K D \| I M$. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-072.jpg?height=669&width=706&top_left_y=839&top_left_x=675)  We claim that triangles $S P I$ and $S D Q$ are similar, and so are triangles $S P D$ and $S I Q$. Let ray $S I$ meet the circumcircle again at $L$. Note that the segment $E F$ is perpendicular to the angle bisector $A M$. Then by $\angle A M L=\angle A S L=\angle A S I=90^{\circ}$, we have $M L \| P Q$. Hence, $\widetilde{P L}=\widetilde{M Q}$ and therefore $\angle P S L=\angle M S Q=\angle D S Q$. By $\angle Q P S=\angle Q M S$, the triangles $S P K$ and $S M Q$ are similar. Finally,  $$ \frac{S P}{S I}=\frac{S P}{S K} \cdot \frac{S K}{S I}=\frac{S M}{S Q} \cdot \frac{S D}{S M}=\frac{S D}{S Q} $$  shows that triangles $S P I$ and $S D Q$ are similar. The second part of the claim can be proved analogously.  Now the problem statement can be proved by  $$ \angle D P A+\angle A Q D=\angle D P S+\angle S Q D=\angle Q I S+\angle S I P=\angle Q I P . $$


Solution: Denote the circumcircle of triangle $A B C$ by $\Gamma$, and let rays $P D$ and $Q D$ meet $\Gamma$ again at $V$ and $U$, respectively. We will show that $A U \perp I P$ and $A V \perp I Q$. Then the problem statement will follow as  $$ \angle D P A+\angle A Q D=\angle V U A+\angle A V U=180^{\circ}-\angle U A V=\angle Q I P . $$  Let $M$ be the midpoint of arc $\widehat{B U V C}$ and let $N$ be the midpoint of arc $\widehat{C A B}$; the lines $A I M$ and $A N$ being the internal and external bisectors of angle $B A C$, respectively, are perpendicular. Let the tangents drawn to $\Gamma$ at $B$ and $C$ meet at $R$; let line $P Q$ meet $A U, A I, A V$ and $B C$ at $X, T, Y$ and $Z$, respectively.  As in Solution 1, we observe that the pairs $B, C$ and $D, Z$ are harmonic. Projecting these points from $Q$ onto the circumcircle, we can see that $B, C$ and $U, P$ are also harmonic. Analogously, the pair $V, Q$ is harmonic with $B, C$. Consider the inversion about the circle with centre $R$, passing through $B$ and $C$. Points $B$ and $C$ are fixed points, so this inversion exchanges every point of $\Gamma$ by its harmonic pair with respect to $B, C$. In particular, the inversion maps points $B, C, N, U, V$ to points $B, C, M, P, Q$, respectively.  Combine the inversion with projecting $\Gamma$ from $A$ to line $P Q$; the points $B, C, M, P, Q$ are projected to $F, E, T, P, Q$, respectively. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-073.jpg?height=1006&width=1384&top_left_y=1082&top_left_x=339)  The combination of these two transformations is projective map from the lines $A B, A C$, $A N, A U, A V$ to $I F, I E, I T, I P, I Q$, respectively. On the other hand, we have $A B \perp I F$, $A C \perp I E$ and $A N \perp A T$, so the corresponding lines in these two pencils are perpendicular. This proves $A U \perp I P$ and $A V \perp I Q$, and hence completes the solution.


The incircle $\omega$ of acute-angled scalene triangle $A B C$ has centre $I$ and meets sides $B C$, $C A$, and $A B$ at $D, E$, and $F$, respectively. The line through $D$ perpendicular to $E F$ meets $\omega$ again at $R$. Line $A R$ meets $\omega$ again at $P$. The circumcircles of triangles $P C E$ and $P B F$ meet again at $Q \neq P$. Prove that lines $D I$ and $P Q$ meet on the external bisector of angle $B A C$. (India)

Solution: Step 1. The external bisector of $\angle B A C$ is the line through $A$ perpendicular to $I A$. Let $D I$ meet this line at $L$ and let $D I$ meet $\omega$ at $K$. Let $N$ be the midpoint of $E F$, which lies on $I A$ and is the pole of line $A L$ with respect to $\omega$. Since $A N \cdot A I=A E^{2}=A R \cdot A P$, the points $R$, $N, I$, and $P$ are concyclic. As $I R=I P$, the line $N I$ is the external bisector of $\angle P N R$, so $P N$ meets $\omega$ again at the point symmetric to $R$ with respect to $A N-i . e$ at $K$.  Let $D N$ cross $\omega$ again at $S$. Opposite sides of any quadrilateral inscribed in the circle $\omega$ meet on the polar line of the intersection of the diagonals with respect to $\omega$. Since $L$ lies on the polar line $A L$ of $N$ with respect to $\omega$, the line $P S$ must pass through $L$. Thus it suffices to prove that the points $S, Q$, and $P$ are collinear. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-074.jpg?height=875&width=1350&top_left_y=1176&top_left_x=361)  Step 2. Let $\Gamma$ be the circumcircle of $\triangle B I C$. Notice that  $$ \begin{aligned} & \angle(B Q, Q C)=\angle(B Q, Q P)+\angle(P Q, Q C)=\angle(B F, F P)+\angle(P E, E C) \\ &=\angle(E F, E P)+\angle(F P, F E)=\angle(F P, E P)=\angle(D F, D E)=\angle(B I, I C) \end{aligned} $$  so $Q$ lies on $\Gamma$. Let $Q P$ meet $\Gamma$ again at $T$. It will now suffice to prove that $S, P$, and $T$ are collinear. Notice that $\angle(B I, I T)=\angle(B Q, Q T)=\angle(B F, F P)=\angle(F K, K P)$. Note $F D \perp F K$ and $F D \perp B I$ so $F K \| B I$ and hence $I T$ is parallel to the line $K N P$. Since $D I=I K$, the line $I T$ crosses $D N$ at its midpoint $M$. Step 3. Let $F^{\prime}$ and $E^{\prime}$ be the midpoints of $D E$ and $D F$, respectively. Since $D E^{\prime} \cdot E^{\prime} F=D E^{\prime 2}=$ $B E^{\prime} \cdot E^{\prime} I$, the point $E^{\prime}$ lies on the radical axis of $\omega$ and $\Gamma$; the same holds for $F^{\prime}$. Therefore, this radical axis is $E^{\prime} F^{\prime}$, and it passes through $M$. Thus $I M \cdot M T=D M \cdot M S$, so $S, I, D$, and $T$ are concyclic. This shows $\angle(D S, S T)=\angle(D I, I T)=\angle(D K, K P)=\angle(D S, S P)$, whence the points $S, P$, and $T$ are collinear, as desired. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-075.jpg?height=1089&width=992&top_left_y=198&top_left_x=532)  Comment. Here is a longer alternative proof in step 1 that $P, S$, and $L$ are collinear, using a circular inversion instead of the fact that opposite sides of a quadrilateral inscribed in a circle $\omega$ meet on the polar line with respect to $\omega$ of the intersection of the diagonals. Let $G$ be the foot of the altitude from $N$ to the line $D I K L$. Observe that $N, G, K, S$ are concyclic (opposite right angles) so  $$ \angle D I P=2 \angle D K P=\angle G K N+\angle D S P=\angle G S N+\angle N S P=\angle G S P, $$  hence $I, G, S, P$ are concyclic. We have $I G \cdot I L=I N \cdot I A=r^{2}$ since $\triangle I G N \sim \triangle I A L$. Inverting the circle $I G S P$ in circle $\omega$, points $P$ and $S$ are fixed and $G$ is taken to $L$ so we find that $P, S$, and $L$ are collinear.


Solution: We start as in   Step 1. Let $A R$ meet the circumcircle $\Omega$ of $A B C$ again at $X$. The lines $A R$ and $A K$ are isogonal in the angle $B A C$; it is well known that in this case $X$ is the tangency point of $\Omega$ with the $A$-mixtilinear circle. It is also well known that for this point $X$, the line $X I$ crosses $\Omega$ again at the midpoint $M^{\prime}$ of arc $B A C$.  Step 2. Denote the circles $B F P$ and $C E P$ by $\Omega_{B}$ and $\Omega_{C}$, respectively. Let $\Omega_{B}$ cross $A R$ and $E F$ again at $U$ and $Y$, respectively. We have  $$ \angle(U B, B F)=\angle(U P, P F)=\angle(R P, P F)=\angle(R F, F A) $$  so $U B \| R F$. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-076.jpg?height=1009&width=1474&top_left_y=883&top_left_x=248)  Next, we show that the points $B, I, U$, and $X$ are concyclic. Since  $$ \angle(U B, U X)=\angle(R F, R X)=\angle(A F, A R)+\angle(F R, F A)=\angle\left(M^{\prime} B, M^{\prime} X\right)+\angle(D R, D F) $$  it suffices to prove $\angle(I B, I X)=\angle\left(M^{\prime} B, M^{\prime} X\right)+\angle(D R, D F)$, or $\angle\left(I B, M^{\prime} B\right)=\angle(D R, D F)$. But both angles equal $\angle(C I, C B)$, as desired. (This is where we used the fact that $M^{\prime}$ is the midpoint of $\operatorname{arc} B A C$ of $\Omega$.)  It follows now from circles $B U I X$ and $B P U F Y$ that  $$ \begin{aligned} \angle(I U, U B)=\angle(I X, B X)=\angle\left(M^{\prime} X, B X\right)= & \frac{\pi-\angle A}{2} \\ & =\angle(E F, A F)=\angle(Y F, B F)=\angle(Y U, B U) \end{aligned} $$  so the points $Y, U$, and $I$ are collinear. Let $E F$ meet $B C$ at $W$. We have  $$ \angle(I Y, Y W)=\angle(U Y, F Y)=\angle(U B, F B)=\angle(R F, A F)=\angle(C I, C W) $$  so the points $W, Y, I$, and $C$ are concyclic.  Similarly, if $V$ and $Z$ are the second meeting points of $\Omega_{C}$ with $A R$ and $E F$, we get that the 4-tuples $(C, V, I, X)$ and $(B, I, Z, W)$ are both concyclic.  Step 3. Let $Q^{\prime}=C Y \cap B Z$. We will show that $Q^{\prime}=Q$. First of all, we have  $$ \begin{aligned} & \angle\left(Q^{\prime} Y, Q^{\prime} B\right)=\angle(C Y, Z B)=\angle(C Y, Z Y)+\angle(Z Y, B Z) \\ & =\angle(C I, I W)+\angle(I W, I B)=\angle(C I, I B)=\frac{\pi-\angle A}{2}=\angle(F Y, F B), \end{aligned} $$  so $Q^{\prime} \in \Omega_{B}$. Similarly, $Q^{\prime} \in \Omega_{C}$. Thus $Q^{\prime} \in \Omega_{B} \cap \Omega_{C}=\{P, Q\}$ and it remains to prove that $Q^{\prime} \neq P$. If we had $Q^{\prime}=P$, we would have $\angle(P Y, P Z)=\angle\left(Q^{\prime} Y, Q^{\prime} Z\right)=\angle(I C, I B)$. This would imply  $$ \angle(P Y, Y F)+\angle(E Z, Z P)=\angle(P Y, P Z)=\angle(I C, I B)=\angle(P E, P F), $$  so circles $\Omega_{B}$ and $\Omega_{C}$ would be tangent at $P$. That is excluded in the problem conditions, so $Q^{\prime}=Q$. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-077.jpg?height=966&width=1468&top_left_y=1070&top_left_x=251)  Step 4. Now we are ready to show that $P, Q$, and $S$ are collinear. Notice that $A$ and $D$ are the poles of $E W$ and $D W$ with respect to $\omega$, so $W$ is the pole of $A D$. Hence, $W I \perp A D$. Since $C I \perp D E$, this yields $\angle(I C, W I)=\angle(D E, D A)$. On the other hand, $D A$ is a symmedian in $\triangle D E F$, so $\angle(D E, D A)=\angle(D N, D F)=\angle(D S, D F)$. Therefore,  $$ \begin{aligned} \angle(P S, P F)=\angle(D S, D F)=\angle(D E, D A)= & \angle(I C, I W) \\ & =\angle(Y C, Y W)=\angle(Y Q, Y F)=\angle(P Q, P F), \end{aligned} $$  which yields the desired collinearity.


Let $\mathcal{L}$ be the set of all lines in the plane and let $f$ be a function that assigns to each line $\ell \in \mathcal{L}$ a point $f(\ell)$ on $\ell$. Suppose that for any point $X$, and for any three lines $\ell_{1}, \ell_{2}, \ell_{3}$ passing through $X$, the points $f\left(\ell_{1}\right), f\left(\ell_{2}\right), f\left(\ell_{3}\right)$ and $X$ lie on a circle.  Prove that there is a unique point $P$ such that $f(\ell)=P$ for any line $\ell$ passing through $P$. (Australia)

Solution: We provide a complete characterisation of the functions satisfying the given condition.  Write $\angle\left(\ell_{1}, \ell_{2}\right)$ for the directed angle modulo $180^{\circ}$ between the lines $\ell_{1}$ and $\ell_{2}$. Given a point $P$ and an angle $\alpha \in\left(0,180^{\circ}\right)$, for each line $\ell$, let $\ell^{\prime}$ be the line through $P$ satisfying $\angle\left(\ell^{\prime}, \ell\right)=\alpha$, and let $h_{P, \alpha}(\ell)$ be the intersection point of $\ell$ and $\ell^{\prime}$. We will prove that there is some pair $(P, \alpha)$ such that $f$ and $h_{P, \alpha}$ are the same function. Then $P$ is the unique point in the problem statement.  Given an angle $\alpha$ and a point $P$, let a line $\ell$ be called $(P, \alpha)$-good if $f(\ell)=h_{P, \alpha}(\ell)$. Let a point $X \neq P$ be called $(P, \alpha)$-good if the circle $g(X)$ passes through $P$ and some point $Y \neq P, X$ on $g(X)$ satisfies $\angle(P Y, Y X)=\alpha$. It follows from this definition that if $X$ is $(P, \alpha)$ good then every point $Y \neq P, X$ of $g(X)$ satisfies this angle condition, so $h_{P, \alpha}(X Y)=Y$ for every $Y \in g(X)$. Equivalently, $f(\ell) \in\left\{X, h_{P, \alpha}(\ell)\right\}$ for each line $\ell$ passing through $X$. This shows the following lemma. Lemma 1. If $X$ is $(P, \alpha)$-good and $\ell$ is a line passing through $X$ then either $f(\ell)=X$ or $\ell$ is $(P, \alpha)$-good. Lemma 2. If $X$ and $Y$ are different $(P, \alpha)$-good points, then line $X Y$ is $(P, \alpha)$-good. Proof. If $X Y$ is not $(P, \alpha)$-good then by the previous Lemma, $f(X Y)=X$ and similarly $f(X Y)=Y$, but clearly this is impossible as $X \neq Y$.  Lemma 3. If $\ell_{1}$ and $\ell_{2}$ are different $(P, \alpha)$-good lines which intersect at $X \neq P$, then either $f\left(\ell_{1}\right)=X$ or $f\left(\ell_{2}\right)=X$ or $X$ is $(P, \alpha)$-good. Proof. If $f\left(\ell_{1}\right), f\left(\ell_{2}\right) \neq X$, then $g(X)$ is the circumcircle of $X, f\left(\ell_{1}\right)$ and $f\left(\ell_{2}\right)$. Since $\ell_{1}$ and $\ell_{2}$ are $(P, \alpha)$-good lines, the angles  $$ \angle\left(P f\left(\ell_{1}\right), f\left(\ell_{1}\right) X\right)=\angle\left(P f\left(\ell_{2}\right), f\left(\ell_{2}\right) X\right)=\alpha $$  so $P$ lies on $g(X)$. Hence, $X$ is $(P, \alpha)$-good. Lemma 4. If $\ell_{1}, \ell_{2}$ and $\ell_{3}$ are different $(P, \alpha)$ good lines which intersect at $X \neq P$, then $X$ is $(P, \alpha)$-good. Proof. This follows from the previous Lemma since at most one of the three lines $\ell_{i}$ can satisfy $f\left(\ell_{i}\right)=X$ as the three lines are all $(P, \alpha)$-good. Lemma 5. If $A B C$ is a triangle such that $A, B, C, f(A B), f(A C)$ and $f(B C)$ are all different points, then there is some point $P$ and some angle $\alpha$ such that $A, B$ and $C$ are $(P, \alpha)$-good points and $A B, B C$ and $C A$ are $(P, \alpha)-\operatorname{good}$ lines. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-079.jpg?height=415&width=612&top_left_y=207&top_left_x=728)  Proof. Let $D, E, F$ denote the points $f(B C), f(A C), f(A B)$, respectively. Then $g(A)$, $g(B)$ and $g(C)$ are the circumcircles of $A E F, B D F$ and $C D E$, respectively. Let $P \neq F$ be the second intersection of circles $g(A)$ and $g(B)$ (or, if these circles are tangent at $F$, then $P=F$ ). By Miquel's theorem (or an easy angle chase), $g(C)$ also passes through $P$. Then by the cyclic quadrilaterals, the directed angles  $$ \angle(P D, D C)=\angle(P F, F B)=\angle(P E, E A)=\alpha $$  for some angle $\alpha$. Hence, lines $A B, B C$ and $C A$ are all $(P, \alpha)$-good, so by Lemma $3, A, B$ and $C$ are $(P, \alpha)$-good. (In the case where $P=D$, the line $P D$ in the equation above denotes the line which is tangent to $g(B)$ at $P=D$. Similar definitions are used for $P E$ and $P F$ in the cases where $P=E$ or $P=F$.)  Consider the set $\Omega$ of all points $(x, y)$ with integer coordinates $1 \leqslant x, y \leqslant 1000$, and consider the set $L_{\Omega}$ of all horizontal, vertical and diagonal lines passing through at least one point in $\Omega$. A simple counting argument shows that there are 5998 lines in $L_{\Omega}$. For each line $\ell$ in $L_{\Omega}$ we colour the point $f(\ell)$ red. Then there are at most 5998 red points. Now we partition the points in $\Omega$ into 10000 ten by ten squares. Since there are at most 5998 red points, at least one of these squares $\Omega_{10}$ contains no red points. Let $(m, n)$ be the bottom left point in $\Omega_{10}$. Then the triangle with vertices $(m, n),(m+1, n)$ and $(m, n+1)$ satisfies the condition of Lemma 5 , so these three vertices are all $(P, \alpha)$-good for some point $P$ and angle $\alpha$, as are the lines joining them. From this point on, we will simply call a point or line good if it is $(P, \alpha)$-good for this particular pair $(P, \alpha)$. Now by Lemma 1, the line $x=m+1$ is good, as is the line $y=n+1$. Then Lemma 3 implies that $(m+1, n+1)$ is good. By applying these two lemmas repeatedly, we can prove that the line $x+y=m+n+2$ is good, then the points $(m, n+2)$ and $(m+2, n)$ then the lines $x=m+2$ and $y=n+2$, then the points $(m+2, n+1),(m+1, n+2)$ and $(m+2, n+2)$ and so on until we have prove that all points in $\Omega_{10}$ are good.  Now we will use this to prove that every point $S \neq P$ is good. Since $g(S)$ is a circle, it passes through at most two points of $\Omega_{10}$ on any vertical line, so at most 20 points in total. Moreover, any line $\ell$ through $S$ intersects at most 10 points in $\Omega_{10}$. Hence, there are at least eight lines $\ell$ through $S$ which contain a point $Q$ in $\Omega_{10}$ which is not on $g(S)$. Since $Q$ is not on $g(S)$, the point $f(\ell) \neq Q$. Hence, by Lemma 1 , the line $\ell$ is good. Hence, at least eight good lines pass through $S$, so by Lemma 4, the point $S$ is good. Hence, every point $S \neq P$ is good, so by Lemma 2 , every line is good. In particular, every line $\ell$ passing through $P$ is good, and therefore satisfies $f(\ell)=P$, as required.


Solution: Note that for any distinct points $X, Y$, the circles $g(X)$ and $g(Y)$ meet on $X Y$ at the point $f(X Y) \in g(X) \cap g(Y) \cap(X Y)$. We write $s(X, Y)$ for the second intersection point of circles $g(X)$ and $g(Y)$. Lemma 1. Suppose that $X, Y$ and $Z$ are not collinear, and that $f(X Y) \notin\{X, Y\}$ and similarly for $Y Z$ and $Z X$. Then $s(X, Y)=s(Y, Z)=s(Z, X)$. Proof. The circles $g(X), g(Y)$ and $g(Z)$ through the vertices of triangle $X Y Z$ meet pairwise on the corresponding edges (produced). By Miquel's theorem, the second points of intersection of any two of the circles coincide. (See the diagram for Lemma 5 of Solution 1.)  Now pick any line $\ell$ and any six different points $Y_{1}, \ldots, Y_{6}$ on $\ell \backslash\{f(\ell)\}$. Pick a point $X$ not on $\ell$ or any of the circles $g\left(Y_{i}\right)$. Reordering the indices if necessary, we may suppose that $Y_{1}, \ldots, Y_{4}$ do not lie on $g(X)$, so that $f\left(X Y_{i}\right) \notin\left\{X, Y_{i}\right\}$ for $1 \leqslant i \leqslant 4$. By applying the above lemma to triangles $X Y_{i} Y_{j}$ for $1 \leqslant i<j \leqslant 4$, we find that the points $s\left(Y_{i}, Y_{j}\right)$ and $s\left(X, Y_{i}\right)$ are all equal, to point $O$ say. Note that either $O$ does not lie on $\ell$, or $O=f(\ell)$, since $O \in g\left(Y_{i}\right)$.  Now consider an arbitrary point $X^{\prime}$ not on $\ell$ or any of the circles $g\left(Y_{i}\right)$ for $1 \leqslant i \leqslant 4$. As above, we see that there are two indices $1 \leqslant i<j \leqslant 4$ such that $Y_{i}$ and $Y_{j}$ do not lie on $g\left(X^{\prime}\right)$. By applying the above lemma to triangle $X^{\prime} Y_{i} Y_{j}$ we see that $s\left(X^{\prime}, Y_{i}\right)=O$, and in particular $g\left(X^{\prime}\right)$ passes through $O$.  We will now show that $f\left(\ell^{\prime}\right)=O$ for all lines $\ell^{\prime}$ through $O$. By the above note, we may assume that $\ell^{\prime} \neq \ell$. Consider a variable point $X^{\prime} \in \ell^{\prime} \backslash\{O\}$ not on $\ell$ or any of the circles $g\left(Y_{i}\right)$ for $1 \leqslant i \leqslant 4$. We know that $f\left(\ell^{\prime}\right) \in g\left(X^{\prime}\right) \cap \ell^{\prime}=\left\{X^{\prime}, O\right\}$. Since $X^{\prime}$ was suitably arbitrary, we have $f\left(\ell^{\prime}\right)=O$ as desired.


Solution: Notice that, for any two different points $X$ and $Y$, the point $f(X Y)$ lies on both $g(X)$ and $g(Y)$, so any two such circles meet in at least one point. We refer to two circles as cutting only in the case where they cross, and so meet at exactly two points, thus excluding the cases where they are tangent or are the same circle. Lemma 1. Suppose there is a point $P$ such that all circles $g(X)$ pass through $P$. Then $P$ has the given property. Proof. Consider some line $\ell$ passing through $P$, and suppose that $f(\ell) \neq P$. Consider some $X \in \ell$ with $X \neq P$ and $X \neq f(\ell)$. Then $g(X)$ passes through all of $P, f(\ell)$ and $X$, but those three points are collinear, a contradiction.  Lemma 2. Suppose that, for all $\epsilon>0$, there is a point $P_{\epsilon}$ with $g\left(P_{\epsilon}\right)$ of radius at most $\epsilon$. Then there is a point $P$ with the given property. Proof. Consider a sequence $\epsilon_{i}=2^{-i}$ and corresponding points $P_{\epsilon_{i}}$. Because the two circles $g\left(P_{\epsilon_{i}}\right)$ and $g\left(P_{\epsilon_{j}}\right)$ meet, the distance between $P_{\epsilon_{i}}$ and $P_{\epsilon_{j}}$ is at most $2^{1-i}+2^{1-j}$. As $\sum_{i} \epsilon_{i}$ converges, these points converge to some point $P$. For all $\epsilon>0$, the point $P$ has distance at most $2 \epsilon$ from $P_{\epsilon}$, and all circles $g(X)$ pass through a point with distance at most $2 \epsilon$ from $P_{\epsilon}$, so distance at most $4 \epsilon$ from $P$. A circle that passes distance at most $4 \epsilon$ from $P$ for all $\epsilon>0$ must pass through $P$, so by Lemma 1 the point $P$ has the given property.  Lemma 3. Suppose no two of the circles $g(X)$ cut. Then there is a point $P$ with the given property. Proof. Consider a circle $g(X)$ with centre $Y$. The circle $g(Y)$ must meet $g(X)$ without cutting it, so has half the radius of $g(X)$. Repeating this argument, there are circles with arbitrarily small radius and the result follows by Lemma 2.  Lemma 4. Suppose there are six different points $A, B_{1}, B_{2}, B_{3}, B_{4}, B_{5}$ such that no three are collinear, no four are concyclic, and all the circles $g\left(B_{i}\right)$ cut pairwise at $A$. Then there is a point $P$ with the given property. Proof. Consider some line $\ell$ through $A$ that does not pass through any of the $B_{i}$ and is not tangent to any of the $g\left(B_{i}\right)$. Fix some direction along that line, and let $X_{\epsilon}$ be the point on $\ell$ that has distance $\epsilon$ from $A$ in that direction. In what follows we consider only those $\epsilon$ for which $X_{\epsilon}$ does not lie on any $g\left(B_{i}\right)$ (this restriction excludes only finitely many possible values of $\epsilon$ ).  Consider the circle $g\left(X_{\epsilon}\right)$. Because no four of the $B_{i}$ are concyclic, at most three of them lie on this circle, so at least two of them do not. There must be some sequence of $\epsilon \rightarrow 0$ such that it is the same two of the $B_{i}$ for all $\epsilon$ in that sequence, so now restrict attention to that sequence, and suppose without loss of generality that $B_{1}$ and $B_{2}$ do not lie on $g\left(X_{\epsilon}\right)$ for any $\epsilon$ in that sequence.  Then $f\left(X_{\epsilon} B_{1}\right)$ is not $B_{1}$, so must be the other point of intersection of $X_{\epsilon} B_{1}$ with $g\left(B_{1}\right)$, and the same applies with $B_{2}$. Now consider the three points $X_{\epsilon}, f\left(X_{\epsilon} B_{1}\right)$ and $f\left(X_{\epsilon} B_{2}\right)$. As $\epsilon \rightarrow 0$, the angle at $X_{\epsilon}$ tends to $\angle B_{1} A B_{2}$ or $180^{\circ}-\angle B_{1} A B_{2}$, which is not 0 or $180^{\circ}$ because no three of the points were collinear. All three distances between those points are bounded above by constant multiples of $\epsilon$ (in fact, if the triangle is scaled by a factor of $1 / \epsilon$, it tends to a fixed triangle). Thus the circumradius of those three points, which is the radius of $g\left(X_{\epsilon}\right)$, is also bounded above by a constant multiple of $\epsilon$, and so the result follows by Lemma 2 .  Lemma 5. Suppose there are two points $A$ and $B$ such that $g(A)$ and $g(B)$ cut. Then there is a point $P$ with the given property. Proof. Suppose that $g(A)$ and $g(B)$ cut at $C$ and $D$. One of those points, without loss of generality $C$, must be $f(A B)$, and so lie on the line $A B$. We now consider two cases, according to whether $D$ also lies on that line.


Solution: For any point $X$, denote by $t(X)$ the line tangent to $g(X)$ at $X$; notice that $f(t(X))=X$, so $f$ is surjective.  Step 1: We find a point $P$ for which there are at least two different lines $p_{1}$ and $p_{2}$ such that $f\left(p_{i}\right)=P$.  Choose any point $X$. If $X$ does not have this property, take any $Y \in g(X) \backslash\{X\}$; then $f(X Y)=Y$. If $Y$ does not have the property, $t(Y)=X Y$, and the circles $g(X)$ and $g(Y)$ meet again at some point $Z$. Then $f(X Z)=Z=f(Y Z)$, so $Z$ has the required property.  We will show that $P$ is the desired point. From now on, we fix two different lines $p_{1}$ and $p_{2}$ with $f\left(p_{1}\right)=f\left(p_{2}\right)=P$. Assume for contradiction that $f(\ell)=Q \neq P$ for some line $\ell$ through $P$. We fix $\ell$, and note that $Q \in g(P)$.  Step 2: We prove that $P \in g(Q)$. Take an arbitrary point $X \in \ell \backslash\{P, Q\}$. Two cases are possible for the position of $t(X)$ in relation to the $p_{i}$; we will show that each case (and subcase) occurs for only finitely many positions of $X$, yielding a contradiction.  Case 2.1: $t(X)$ is parallel to one of the $p_{i}$; say, to $p_{1}$. Let $t(X) \operatorname{cross} p_{2}$ at $R$. Then $g(R)$ is the circle $(P R X)$, as $f(R P)=P$ and $f(R X)=X$. Let $R Q$ cross $g(R)$ again at $S$. Then $f(R Q) \in\{R, S\} \cap g(Q)$, so $g(Q)$ contains one of the points $R$ and $S$.  If $R \in g(Q)$, then $R$ is one of finitely many points in the intersection $g(Q) \cap p_{2}$, and each of them corresponds to a unique position of $X$, since $R X$ is parallel to $p_{1}$.  If $S \in g(Q)$, then $\angle(Q S, S P)=\angle(R S, S P)=\angle(R X, X P)=\angle\left(p_{1}, \ell\right)$, so $\angle(Q S, S P)$ is constant for all such points $X$, and all points $S$ obtained in such a way lie on one circle $\gamma$ passing through $P$ and $Q$. Since $g(Q)$ does not contain $P$, it is different from $\gamma$, so there are only finitely many points $S$. Each of them uniquely determines $R$ and thus $X$. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-083.jpg?height=843&width=909&top_left_y=207&top_left_x=568)  So, Case 2.1 can occur for only finitely many points $X$. Case 2.2: $t(X)$ crosses $p_{1}$ and $p_{2}$ at $R_{1}$ and $R_{2}$, respectively. Clearly, $R_{1} \neq R_{2}$, as $t(X)$ is the tangent to $g(X)$ at $X$, and $g(X)$ meets $\ell$ only at $X$ and $Q$. Notice that $g\left(R_{i}\right)$ is the circle $\left(P X R_{i}\right)$. Let $R_{i} Q$ meet $g\left(R_{i}\right)$ again at $S_{i}$; then $S_{i} \neq Q$, as $g\left(R_{i}\right)$ meets $\ell$ only at $P$ and $X$. Then $f\left(R_{i} Q\right) \in\left\{R_{i}, S_{i}\right\}$, and we distinguish several subcases. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-083.jpg?height=824&width=866&top_left_y=1501&top_left_x=595)  Subcase 2.2.1: $f\left(R_{1} Q\right)=S_{1}, f\left(R_{2} Q\right)=S_{2}$; so $S_{1}, S_{2} \in g(Q)$. In this case we have $0=\angle\left(R_{1} X, X P\right)+\angle\left(X P, R_{2} X\right)=\angle\left(R_{1} S_{1}, S_{1} P\right)+\angle\left(S_{2} P, S_{2} R_{2}\right)=$ $\angle\left(Q S_{1}, S_{1} P\right)+\angle\left(S_{2} P, S_{2} Q\right)$, which shows $P \in g(Q)$.  Subcase 2.2.2: $f\left(R_{1} Q\right)=R_{1}, f\left(R_{2} Q\right)=R_{2}$; so $R_{1}, R_{2} \in g(Q)$. This can happen for at most four positions of $X$ - namely, at the intersections of $\ell$ with a line of the form $K_{1} K_{2}$, where $K_{i} \in g(Q) \cap p_{i}$.  Subcase 2.2.3: $f\left(R_{1} Q\right)=S_{1}, f\left(R_{2} Q\right)=R_{2}$ (the case $f\left(R_{1} Q\right)=R_{1}, f\left(R_{2} Q\right)=S_{2}$ is similar). In this case, there are at most two possible positions for $R_{2}$ - namely, the meeting points of $g(Q)$ with $p_{2}$. Consider one of them. Let $X$ vary on $\ell$. Then $R_{1}$ is the projection of $X$ to $p_{1}$ via $R_{2}, S_{1}$ is the projection of $R_{1}$ to $g(Q)$ via $Q$. Finally, $\angle\left(Q S_{1}, S_{1} X\right)=\angle\left(R_{1} S_{1}, S_{1} X\right)=$ $\angle\left(R_{1} P, P X\right)=\angle\left(p_{1}, \ell\right) \neq 0$, so $X$ is obtained by a fixed projective transform $g(Q) \rightarrow \ell$ from $S_{1}$. So, if there were three points $X$ satisfying the conditions of this subcase, the composition of the three projective transforms would be the identity. But, if we apply it to $X=Q$, we successively get some point $R_{1}^{\prime}$, then $R_{2}$, and then some point different from $Q$, a contradiction.  Thus Case 2.2 also occurs for only finitely many points $X$, as desired. Step 3: We show that $f(P Q)=P$, as desired. The argument is similar to that in Step 2, with the roles of $Q$ and $X$ swapped. Again, we show that there are only finitely many possible positions for a point $X \in \ell \backslash\{P, Q\}$, which is absurd. Case 3.1: $t(Q)$ is parallel to one of the $p_{i}$; say, to $p_{1}$. Let $t(Q)$ cross $p_{2}$ at $R$; then $g(R)$ is the circle $(P R Q)$. Let $R X \operatorname{cross} g(R)$ again at $S$. Then $f(R X) \in\{R, S\} \cap g(X)$, so $g(X)$ contains one of the points $R$ and $S$. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-084.jpg?height=729&width=909&top_left_y=1126&top_left_x=568)  Subcase 3.1.1: $S=f(R X) \in g(X)$. We have $\angle(t(X), Q X)=\angle(S X, S Q)=\angle(S R, S Q)=\angle(P R, P Q)=\angle\left(p_{2}, \ell\right)$. Hence $t(X) \| p_{2}$. Now we recall Case 2.1: we let $t(X)$ cross $p_{1}$ at $R^{\prime}$, so $g\left(R^{\prime}\right)=\left(P R^{\prime} X\right)$, and let $R^{\prime} Q$ meet $g\left(R^{\prime}\right)$ again at $S^{\prime}$; notice that $S^{\prime} \neq Q$. Excluding one position of $X$, we may assume that $R^{\prime} \notin g(Q)$, so $R^{\prime} \neq f\left(R^{\prime} Q\right)$. Therefore, $S^{\prime}=f\left(R^{\prime} Q\right) \in g(Q)$. But then, as in Case 2.1, we get $\angle(t(Q), P Q)=\angle\left(Q S^{\prime}, S^{\prime} P\right)=\angle\left(R^{\prime} X, X P\right)=\angle\left(p_{2}, \ell\right)$. This means that $t(Q)$ is parallel to $p_{2}$, which is impossible. Subcase 3.1.2: $R=f(R X) \in g(X)$. In this case, we have $\angle(t(X), \ell)=\angle(R X, R Q)=\angle\left(R X, p_{1}\right)$. Again, let $R^{\prime}=t(X) \cap p_{1}$; this point exists for all but at most one position of $X$. Then $g\left(R^{\prime}\right)=\left(R^{\prime} X P\right)$; let $R^{\prime} Q$ meet $g\left(R^{\prime}\right)$ again at $S^{\prime}$. Due to $\angle\left(R^{\prime} X, X R\right)=\angle(Q X, Q R)=\angle\left(\ell, p_{1}\right), R^{\prime}$ determines $X$ in at most two ways, so for all but finitely many positions of $X$ we have $R^{\prime} \notin g(Q)$. Therefore, for those positions we have $S^{\prime}=f\left(R^{\prime} Q\right) \in g(Q)$. But then $\angle\left(R X, p_{1}\right)=\angle\left(R^{\prime} X, X P\right)=\angle\left(R^{\prime} S^{\prime}, S^{\prime} P\right)=$ $\angle\left(Q S^{\prime}, S^{\prime} P\right)=\angle(t(Q), Q P)$ is fixed, so this case can hold only for one specific position of $X$ as well.  Thus, in Case 3.1, there are only finitely many possible positions of $X$, yielding a contradiction.  Case 3.2: $t(Q)$ crosses $p_{1}$ and $p_{2}$ at $R_{1}$ and $R_{2}$, respectively. By Step $2, R_{1} \neq R_{2}$. Notice that $g\left(R_{i}\right)$ is the circle $\left(P Q R_{i}\right)$. Let $R_{i} X$ meet $g\left(R_{i}\right)$ at $S_{i}$; then $S_{i} \neq X$. Then $f\left(R_{i} X\right) \in\left\{R_{i}, S_{i}\right\}$, and we distinguish several subcases. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-085.jpg?height=854&width=866&top_left_y=424&top_left_x=589)  Subcase 3.2.1: $f\left(R_{1} X\right)=S_{1}$ and $f\left(R_{2} X\right)=S_{2}$, so $S_{1}, S_{2} \in g(X)$. As in Subcase 2.2.1, we have $0=\angle\left(R_{1} Q, Q P\right)+\angle\left(Q P, R_{2} Q\right)=\angle\left(X S_{1}, S_{1} P\right)+\angle\left(S_{2} P, S_{2} X\right)$, which shows $P \in g(X)$. But $X, Q \in g(X)$ as well, so $g(X)$ meets $\ell$ at three distinct points, which is absurd. Subcase 3.2.2: $f\left(R_{1} X\right)=R_{1}, f\left(R_{2} X\right)=R_{2}$, so $R_{1}, R_{2} \in g(X)$. Now three distinct collinear points $R_{1}, R_{2}$, and $Q$ belong to $g(X)$, which is impossible. Subcase 3.2.3: $f\left(R_{1} X\right)=S_{1}, f\left(R_{2} X\right)=R_{2}$ (the case $f\left(R_{1} X\right)=R_{1}, f\left(R_{2} X\right)=S_{2}$ is similar). We have $\angle\left(X R_{2}, R_{2} Q\right)=\angle\left(X S_{1}, S_{1} Q\right)=\angle\left(R_{1} S_{1}, S_{1} Q\right)=\angle\left(R_{1} P, P Q\right)=\angle\left(p_{1}, \ell\right)$, so this case can occur for a unique position of $X$.  Thus, in Case 3.2 , there is only a unique position of $X$, again yielding the required contradiction.


Solution: We will get an upper bound on $n$ from the speed at which $v_{2}\left(L_{n}\right)$ grows. From  $$ L_{n}=\left(2^{n}-1\right)\left(2^{n}-2\right) \cdots\left(2^{n}-2^{n-1}\right)=2^{1+2+\cdots+(n-1)}\left(2^{n}-1\right)\left(2^{n-1}-1\right) \cdots\left(2^{1}-1\right) $$  we read  $$ v_{2}\left(L_{n}\right)=1+2+\cdots+(n-1)=\frac{n(n-1)}{2} . $$  On the other hand, $v_{2}(m!)$ is expressed by the Legendre formula as  $$ v_{2}(m!)=\sum_{i=1}^{\infty}\left\lfloor\frac{m}{2^{i}}\right\rfloor . $$  As usual, by omitting the floor functions,  $$ v_{2}(m!)<\sum_{i=1}^{\infty} \frac{m}{2^{i}}=m $$  Thus, $L_{n}=m$ ! implies the inequality  $$ \frac{n(n-1)}{2}<m . $$  In order to obtain an opposite estimate, observe that  $$ L_{n}=\left(2^{n}-1\right)\left(2^{n}-2\right) \cdots\left(2^{n}-2^{n-1}\right)<\left(2^{n}\right)^{n}=2^{n^{2}} . $$  We claim that  $$ 2^{n^{2}}<\left(\frac{n(n-1)}{2}\right)!\text { for } n \geqslant 6 $$  For $n=6$ the estimate (3) is true because $2^{6^{2}}<6.9 \cdot 10^{10}$ and $\left(\frac{n(n-1)}{2}\right)!=15!>1.3 \cdot 10^{12}$. For $n \geqslant 7$ we prove (3) by the following inequalities:  $$ \begin{aligned} \left(\frac{n(n-1)}{2}\right)! & =15!\cdot 16 \cdot 17 \cdots \frac{n(n-1)}{2}>2^{36} \cdot 16^{\frac{n(n-1)}{2}-15} \\ & =2^{2 n(n-1)-24}=2^{n^{2}} \cdot 2^{n(n-2)-24}>2^{n^{2}} . \end{aligned} $$  Putting together (2) and (3), for $n \geqslant 6$ we get a contradiction, since  $$ L_{n}<2^{n^{2}}<\left(\frac{n(n-1)}{2}\right)!<m!=L_{n} $$  Hence $n \geqslant 6$ is not possible. Checking manually the cases $n \leqslant 5$ we find  $$ \begin{gathered} L_{1}=1=1!, \quad L_{2}=6=3!, \quad 5!<L_{3}=168<6!, \\ 7!<L_{4}=20160<8!\quad \text { and } \quad 10!<L_{5}=9999360<11! \end{gathered} $$  So, there are two solutions:  $$ (m, n) \in\{(1,1),(3,2)\} $$


Solution: Like in the previous solution, the cases $n=1,2,3,4$ are checked manually. We will exclude $n \geqslant 5$ by considering the exponents of 3 and 31 in (1).  For odd primes $p$ and distinct integers $a, b$, coprime to $p$, with $p \mid a-b$, the Lifting The Exponent lemma asserts that  $$ v_{p}\left(a^{k}-b^{k}\right)=v_{p}(a-b)+v_{p}(k) . $$  Notice that 3 divides $2^{k}-1$ if only if $k$ is even; moreover, by the Lifting The Exponent lemma we have  $$ v_{3}\left(2^{2 k}-1\right)=v_{3}\left(4^{k}-1\right)=1+v_{3}(k)=v_{3}(3 k) $$  Hence,  $$ v_{3}\left(L_{n}\right)=\sum_{2 k \leqslant n} v_{3}\left(4^{k}-1\right)=\sum_{k \leqslant\left\lfloor\frac{n}{2}\right\rfloor} v_{3}(3 k) $$  Notice that the last expression is precisely the exponent of 3 in the prime factorisation of (3 $\left.\left.\frac{n}{2}\right\rfloor\right)$ !. Therefore  $$ \begin{gathered} v_{3}(m!)=v_{3}\left(L_{n}\right)=v_{3}\left(\left(3\left\lfloor\frac{n}{2}\right\rfloor\right)!\right) \\ 3\left\lfloor\frac{n}{2}\right\rfloor \leqslant m \leqslant 3\left\lfloor\frac{n}{2}\right\rfloor+2 . \end{gathered} $$  Suppose that $n \geqslant 5$. Note that every fifth factor in $L_{n}$ is divisible by $31=2^{5}-1$, and hence we have $v_{31}\left(L_{n}\right) \geqslant\left\lfloor\frac{n}{5}\right\rfloor$. Then  $$ \frac{n}{10} \leqslant\left\lfloor\frac{n}{5}\right\rfloor \leqslant v_{31}\left(L_{n}\right)=v_{31}(m!)=\sum_{k=1}^{\infty}\left\lfloor\frac{m}{31^{k}}\right\rfloor<\sum_{k=1}^{\infty} \frac{m}{31^{k}}=\frac{m}{30} $$  By combining (4) and (5),  $$ 3 n<m \leqslant \frac{3 n}{2}+2 $$  so $n<\frac{4}{3}$ which is inconsistent with the inequality $n \geqslant 5$. Comment 1. There are many combinations of the ideas above; for example combining (2) and (4) also provides $n<5$. Obviously, considering the exponents of any two primes in (1), or considering one prime and the magnitude orders lead to an upper bound on $n$ and $m$.  Comment 2. This problem has a connection to group theory. Indeed, the left-hand side is the order of the group $G L_{n}\left(\mathbb{F}_{2}\right)$ of invertible $n$-by- $n$ matrices with entries modulo 2 , while the right-hand side is the order of the symmetric group $S_{m}$ on $m$ elements. The result thus shows that the only possible isomorphisms between these groups are $G L_{1}\left(\mathbb{F}_{2}\right) \cong S_{1}$ and $G L_{2}\left(\mathbb{F}_{2}\right) \cong S_{3}$, and there are in fact isomorphisms in both cases. In general, $G L_{n}\left(\mathbb{F}_{2}\right)$ is a simple group for $n \geqslant 3$, as it is isomorphic to $\operatorname{PS} L_{n}\left(\mathbb{F}_{2}\right)$.  There is also a near-solution of interest: the left-hand side for $n=4$ is half of the right-hand side when $m=8$; this turns out to correspond to an isomorphism $G L_{4}\left(\mathbb{F}_{2}\right) \cong A_{8}$ with the alternating group on eight elements.  However, while this indicates that the problem is a useful one, knowing group theory is of no use in solving it!


Find all pairs $(m, n)$ of positive integers satisfying the equation  $$ \left(2^{n}-1\right)\left(2^{n}-2\right)\left(2^{n}-4\right) \cdots\left(2^{n}-2^{n-1}\right)=m! $$  (El Salvador)


Find all triples $(a, b, c)$ of positive integers such that $a^{3}+b^{3}+c^{3}=(a b c)^{2}$. (Nigeria)

Solution: We will start by proving that $c=1$. Note that  $$ 3 a^{3} \geqslant a^{3}+b^{3}+c^{3}>a^{3} . $$  So $3 a^{3} \geqslant(a b c)^{2}>a^{3}$ and hence $3 a \geqslant b^{2} c^{2}>a$. Now $b^{3}+c^{3}=a^{2}\left(b^{2} c^{2}-a\right) \geqslant a^{2}$, and so  $$ 18 b^{3} \geqslant 9\left(b^{3}+c^{3}\right) \geqslant 9 a^{2} \geqslant b^{4} c^{4} \geqslant b^{3} c^{5} $$  so $18 \geqslant c^{5}$ which yields $c=1$. Now, note that we must have $a>b$, as otherwise we would have $2 b^{3}+1=b^{4}$ which has no positive integer solutions. So  $$ a^{3}-b^{3} \geqslant(b+1)^{3}-b^{3}>1 $$  and  $$ 2 a^{3}>1+a^{3}+b^{3}>a^{3} $$  which implies $2 a^{3}>a^{2} b^{2}>a^{3}$ and so $2 a>b^{2}>a$. Therefore  $$ 4\left(1+b^{3}\right)=4 a^{2}\left(b^{2}-a\right) \geqslant 4 a^{2}>b^{4} $$  so $4>b^{3}(b-4)$; that is, $b \leqslant 4$. Now, for each possible value of $b$ with $2 \leqslant b \leqslant 4$ we obtain a cubic equation for $a$ with constant coefficients. These are as follows:  $$ \begin{array}{rlrl} b=2: & & & a^{3}-4 a^{2}+9=0 \\ b=3: & & a^{3}-9 a^{2}+28=0 \\ b=4: & & a^{3}-16 a^{2}+65=0 . \end{array} $$  The only case with an integer solution for $a$ with $b \leqslant a$ is $b=2$, leading to $(a, b, c)=(3,2,1)$. Comment 1.1. Instead of writing down each cubic equation explicitly, we could have just observed that $a^{2} \mid b^{3}+1$, and for each choice of $b$ checked each square factor of $b^{3}+1$ for $a^{2}$.  We could also have observed that, with $c=1$, the relation $18 b^{3} \geqslant b^{4} c^{4}$ becomes $b \leqslant 18$, and we can simply check all possibilities for $b$ (instead of working to prove that $b \leqslant 4$ ). This check becomes easier after using the factorisation $b^{3}+1=(b+1)\left(b^{2}-b+1\right)$ and observing that no prime besides 3 can divide both of the factors.  Comment 1.2. Another approach to finish the problem after establishing that $c \leqslant 1$ is to set $k=b^{2} c^{2}-a$, which is clearly an integer and must be positive as it is equal to $\left(b^{3}+c^{3}\right) / a^{2}$. Then we divide into cases based on whether $k=1$ or $k \geqslant 2$; in the first case, we have $b^{3}+1=a^{2}=\left(b^{2}-1\right)^{2}$ whose only positive root is $b=2$, and in the second case we have $b^{2} \leqslant 3 a$, and so  $$ b^{4} \leqslant(3 a)^{2} \leqslant \frac{9}{2}\left(k a^{2}\right)=\frac{9}{2}\left(b^{3}+1\right), $$  which implies that $b \leqslant 4$.


Solution: Again, we will start by proving that $c=1$. Suppose otherwise that $c \geqslant 2$. We have $a^{3}+b^{3}+c^{3} \leqslant 3 a^{3}$, so $b^{2} c^{2} \leqslant 3 a$. Since $c \geqslant 2$, this tells us that $b \leqslant \sqrt{3 a / 4}$. As the right-hand side of the original equation is a multiple of $a^{2}$, we have $a^{2} \leqslant 2 b^{3} \leqslant 2(3 a / 4)^{3 / 2}$. In other words, $a \leqslant \frac{27}{16}<2$, which contradicts the assertion that $a \geqslant c \geqslant 2$. So there are no solutions in this case, and so we must have $c=1$.  Now, the original equation becomes $a^{3}+b^{3}+1=a^{2} b^{2}$. Observe that $a \geqslant 2$, since otherwise $a=b=1$ as $a \geqslant b$.  The right-hand side is a multiple of $a^{2}$, so the left-hand side must be as well. Thus, $b^{3}+1 \geqslant$ $a^{2}$. Since $a \geqslant b$, we also have  $$ b^{2}=a+\frac{b^{3}+1}{a^{2}} \leqslant 2 a+\frac{1}{a^{2}} $$  and so $b^{2} \leqslant 2 a$ since $b^{2}$ is an integer. Thus $(2 a)^{3 / 2}+1 \geqslant b^{3}+1 \geqslant a^{2}$, from which we deduce $a \leqslant 8$.  Now, for each possible value of $a$ with $2 \leqslant a \leqslant 8$ we obtain a cubic equation for $b$ with constant coefficients. These are as follows:  $$ \begin{array}{ll} a=2: & \\ b^{3}-4 b^{2}+9=0 \\ a=3: & \\ b^{3}-9 b^{2}+28=0 \\ a=4: & \\ a=5: 16 b^{2}+65=0 \\ a=6: & \\ b^{3}-25 b^{2}+126=0 \\ a=7: & \\ b^{3}-36 b^{2}+217=0 \\ a=8: & \\ b^{3}-49 b^{2}+344=0 \\ b^{3}-64 b^{2}+513=0 . \end{array} $$  The only case with an integer solution for $b$ with $a \geqslant b$ is $a=3$, leading to $(a, b, c)=(3,2,1)$. Comment 2.1. As in Solution 1, instead of writing down each cubic equation explicitly, we could have just observed that $b^{2} \mid a^{3}+1$, and for each choice of $a$ checked each square factor of $a^{3}+1$ for $b^{2}$.  Comment 2.2. This solution does not require initially proving that $c=1$, in which case the bound would become $a \leqslant 108$. The resulting cases could, in principle, be checked by a particularly industrious student.


Solution: Set $k=\left(b^{3}+c^{3}\right) / a^{2} \leqslant 2 a$, and rewrite the original equation as $a+k=(b c)^{2}$. Since $b^{3}$ and $c^{3}$ are positive integers, we have $(b c)^{3} \geqslant b^{3}+c^{3}-1=k a^{2}-1$, so  $$ a+k \geqslant\left(k a^{2}-1\right)^{2 / 3} $$  As in Comment 1.2, $k$ is a positive integer; for each value of $k \geqslant 1$, this gives us a polynomial inequality satisfied by $a$ :  $$ k^{2} a^{4}-a^{3}-5 k a^{2}-3 k^{2} a-\left(k^{3}-1\right) \leqslant 0 . $$  We now prove that $a \leqslant 3$. Indeed,  $$ 0 \geqslant \frac{k^{2} a^{4}-a^{3}-5 k a^{2}-3 k^{2} a-\left(k^{3}-1\right)}{k^{2}} \geqslant a^{4}-a^{3}-5 a^{2}-3 a-k \geqslant a^{4}-a^{3}-5 a^{2}-5 a $$  which fails when $a \geqslant 4$. This leaves ten triples with $3 \geqslant a \geqslant b \geqslant c \geqslant 1$, which may be checked manually to give $(a, b, c)=(3,2,1)$.


Solution: Again, observe that $b^{3}+c^{3}=a^{2}\left(b^{2} c^{2}-a\right)$, so $b \leqslant a \leqslant b^{2} c^{2}-1$. We consider the function $f(x)=x^{2}\left(b^{2} c^{2}-x\right)$. It can be seen that that on the interval $\left[0, b^{2} c^{2}-1\right]$ the function $f$ is increasing if $x<\frac{2}{3} b^{2} c^{2}$ and decreasing if $x>\frac{2}{3} b^{2} c^{2}$. Consequently, it must be the case that  $$ b^{3}+c^{3}=f(a) \geqslant \min \left(f(b), f\left(b^{2} c^{2}-1\right)\right) $$  First, suppose that $b^{3}+c^{3} \geqslant f\left(b^{2} c^{2}-1\right)$. This may be written $b^{3}+c^{3} \geqslant\left(b^{2} c^{2}-1\right)^{2}$, and so  $$ 2 b^{3} \geqslant b^{3}+c^{3} \geqslant\left(b^{2} c^{2}-1\right)^{2}>b^{4} c^{4}-2 b^{2} c^{2} \geqslant b^{4} c^{4}-2 b^{3} c^{4} $$  Thus, $(b-2) c^{4}<2$, and the only solutions to this inequality have $(b, c)=(2,2)$ or $b \leqslant 3$ and $c=1$. It is easy to verify that the only case giving a solution for $a \geqslant b$ is $(a, b, c)=(3,2,1)$.  Otherwise, suppose that $b^{3}+c^{3}=f(a) \geqslant f(b)$. Then, we have  $$ 2 b^{3} \geqslant b^{3}+c^{3}=a^{2}\left(b^{2} c^{2}-a\right) \geqslant b^{2}\left(b^{2} c^{2}-b\right) . $$  Consequently $b c^{2} \leqslant 3$, with strict inequality in the case that $b \neq c$. Hence $c=1$ and $b \leqslant 2$. Both of these cases have been considered already, so we are done.  Comment 4.1. Instead of considering which of $f(b)$ and $f\left(b^{2} c^{2}-1\right)$ is less than $f(a)$, we may also proceed by explicitly dividing into cases based on whether $a \geqslant \frac{2}{3} b^{2} c^{2}$ or $a<\frac{2}{3} b^{2} c^{2}$. The first case may now be dealt with as follows. We have $b^{3} c^{3}+1 \geqslant b^{3}+c^{3}$ as $b^{3}$ and $c^{3}$ are positive integers, so we have  $$ b^{3} c^{3}+1 \geqslant b^{3}+c^{3} \geqslant a^{2} \geqslant \frac{4}{9} b^{4} c^{4} $$  This implies $b c \leqslant 2$, and hence $c=1$ and $b \leqslant 2$.


We say that a set $S$ of integers is rootiful if, for any positive integer $n$ and any $a_{0}, a_{1}, \ldots, a_{n} \in S$, all integer roots of the polynomial $a_{0}+a_{1} x+\cdots+a_{n} x^{n}$ are also in $S$. Find all rootiful sets of integers that contain all numbers of the form $2^{a}-2^{b}$ for positive integers $a$ and $b$. (Czech Republic)

Solution: The set $\mathbb{Z}$ of all integers is clearly rootiful. We shall prove that any rootiful set $S$ containing all the numbers of the form $2^{a}-2^{b}$ for $a, b \in \mathbb{Z}_{>0}$ must be all of $\mathbb{Z}$.  First, note that $0=2^{1}-2^{1} \in S$ and $2=2^{2}-2^{1} \in S$. Now, $-1 \in S$, since it is a root of $2 x+2$, and $1 \in S$, since it is a root of $2 x^{2}-x-1$. Also, if $n \in S$ then $-n$ is a root of $x+n$, so it suffices to prove that all positive integers must be in $S$.  Now, we claim that any positive integer $n$ has a multiple in $S$. Indeed, suppose that $n=2^{\alpha} \cdot t$ for $\alpha \in \mathbb{Z}_{\geqslant 0}$ and $t$ odd. Then $t \mid 2^{\phi(t)}-1$, so $n \mid 2^{\alpha+\phi(t)+1}-2^{\alpha+1}$. Moreover, $2^{\alpha+\phi(t)+1}-2^{\alpha+1} \in S$, and so $S$ contains a multiple of every positive integer $n$.  We will now prove by induction that all positive integers are in $S$. Suppose that $0,1, \ldots, n-$ $1 \in S$; furthermore, let $N$ be a multiple of $n$ in $S$. Consider the base- $n$ expansion of $N$, say $N=a_{k} n^{k}+a_{k-1} n^{k-1}+\cdots+a_{1} n+a_{0}$. Since $0 \leqslant a_{i}<n$ for each $a_{i}$, we have that all the $a_{i}$ are in $S$. Furthermore, $a_{0}=0$ since $N$ is a multiple of $n$. Therefore, $a_{k} n^{k}+a_{k-1} n^{k-1}+\cdots+a_{1} n-N=0$, so $n$ is a root of a polynomial with coefficients in $S$. This tells us that $n \in S$, completing the induction.


Solution: As in the previous solution, we can prove that 0,1 and -1 must all be in any rootiful set $S$ containing all numbers of the form $2^{a}-2^{b}$ for $a, b \in \mathbb{Z}_{>0}$.  We show that, in fact, every integer $k$ with $|k|>2$ can be expressed as a root of a polynomial whose coefficients are of the form $2^{a}-2^{b}$. Observe that it suffices to consider the case where $k$ is positive, as if $k$ is a root of $a_{n} x^{n}+\cdots+a_{1} x+a_{0}=0$, then $-k$ is a root of $(-1)^{n} a_{n} x^{n}+\cdots-$ $a_{1} x+a_{0}=0$.  Note that  $$ \left(2^{a_{n}}-2^{b_{n}}\right) k^{n}+\cdots+\left(2^{a_{0}}-2^{b_{0}}\right)=0 $$  is equivalent to  $$ 2^{a_{n}} k^{n}+\cdots+2^{a_{0}}=2^{b_{n}} k^{n}+\cdots+2^{b_{0}} $$  Hence our aim is to show that two numbers of the form $2^{a_{n}} k^{n}+\cdots+2^{a_{0}}$ are equal, for a fixed value of $n$. We consider such polynomials where every term $2^{a_{i}} k^{i}$ is at most $2 k^{n}$; in other words, where $2 \leqslant 2^{a_{i}} \leqslant 2 k^{n-i}$, or, equivalently, $1 \leqslant a_{i} \leqslant 1+(n-i) \log _{2} k$. Therefore, there must be $1+\left\lfloor(n-i) \log _{2} k\right\rfloor$ possible choices for $a_{i}$ satisfying these constraints.  The number of possible polynomials is then  $$ \prod_{i=0}^{n}\left(1+\left\lfloor(n-i) \log _{2} k\right\rfloor\right) \geqslant \prod_{i=0}^{n-1}(n-i) \log _{2} k=n!\left(\log _{2} k\right)^{n} $$  where the inequality holds as $1+\lfloor x\rfloor \geqslant x$. As there are $(n+1)$ such terms in the polynomial, each at most $2 k^{n}$, such a polynomial must have value at most $2 k^{n}(n+1)$. However, for large $n$, we have $n!\left(\log _{2} k\right)^{n}>2 k^{n}(n+1)$. Therefore there are more polynomials than possible values, so some two must be equal, as required.


Let $\mathbb{Z}_{>0}$ be the set of positive integers. A positive integer constant $C$ is given. Find all functions $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ such that, for all positive integers $a$ and $b$ satisfying $a+b>C$,  $$ a+f(b) \mid a^{2}+b f(a) $$  (Croatia)

Solution: First, we show that $b \mid f(b)^{2}$ for all $b$. To do this, we choose a large positive integer $n$ so that $n b-f(b) \geqslant C$. Setting $a=n b-f(b)$ in (*) then shows that  $$ n b \mid(n b-f(b))^{2}+b f(n b-f(b)) $$  so that $b \mid f(b)^{2}$ as claimed. Now in particular we have that $p \mid f(p)$ for every prime $p$. If we write $f(p)=k(p) \cdot p$, then the bound $f(p) \leqslant f(1) \cdot p$ (valid for $p$ sufficiently large) shows that some value $k$ of $k(p)$ must be attained for infinitely many $p$. We will show that $f(a)=k a$ for all positive integers $a$. To do this, we substitute $b=p$ in (*), where $p$ is any sufficiently large prime for which $k(p)=k$, obtaining  $$ a+k p \mid\left(a^{2}+p f(a)\right)-a(a+k p)=p f(a)-p k a . $$  For suitably large $p$ we have $\operatorname{gcd}(a+k p, p)=1$, and hence we have  $$ a+k p \mid f(a)-k a $$  But the only way this can hold for arbitrarily large $p$ is if $f(a)-k a=0$. This concludes the proof.  Comment. There are other ways to obtain the divisibility $p \mid f(p)$ for primes $p$, which is all that is needed in this proof. For instance, if $f(p)$ were not divisible by $p$ then the arithmetic progression $p^{2}+b f(p)$ would attain prime values for infinitely many $b$ by Dirichlet's Theorem: hence, for these pairs p , b, we would have $p+f(b)=p^{2}+b f(p)$. Substituting $a \mapsto b$ and $b \mapsto p$ in $(*)$ then shows that $\left(f(p)^{2}-p^{2}\right)(p-1)$ is divisible by $b+f(p)$ and hence vanishes, which is impossible since $p \nmid f(p)$ by assumption.


Solution: First, we substitute $b=1$ in (*) and rearrange to find that  $$ \frac{f(a)+f(1)^{2}}{a+f(1)}=f(1)-a+\frac{a^{2}+f(a)}{a+f(1)} $$  is a positive integer for sufficiently large $a$. Since $f(a) \leqslant a f(1)$, for all sufficiently large $a$, it follows that $\frac{f(a)+f(1)^{2}}{a+f(1)} \leqslant f(1)$ also and hence there is a positive integer $k$ such that $\frac{f(a)+f(1)^{2}}{a+f(1)}=k$ for infinitely many values of $a$. In other words,  $$ f(a)=k a+f(1) \cdot(k-f(1)) $$  for infinitely many $a$. Fixing an arbitrary choice of $a$ in (*), we have that  $$ \frac{a^{2}+b f(a)}{a+k b+f(1) \cdot(k-f(1))} $$  is an integer for infinitely many $b$ (the same $b$ as above, maybe with finitely many exceptions). On the other hand, for $b$ taken sufficiently large, this quantity becomes arbitrarily close to $\frac{f(a)}{k}$; this is only possible if $\frac{f(a)}{k}$ is an integer and  $$ \frac{a^{2}+b f(a)}{a+k b+f(1) \cdot(k-f(1))}=\frac{f(a)}{k} $$  for infinitely many $b$. This rearranges to  $$ \frac{f(a)}{k} \cdot(a+f(1) \cdot(k-f(1)))=a^{2} $$  Hence $a^{2}$ is divisible by $a+f(1) \cdot(k-f(1))$, and hence so is $f(1)^{2}(k-f(1))^{2}$. The only way this can occur for all $a$ is if $k=f(1)$, in which case $(* *)$ provides that $f(a)=k a$ for all $a$, as desired.


Solution: Fix any two distinct positive integers $a$ and $b$. From (*) it follows that the two integers  $$ \left(a^{2}+c f(a)\right) \cdot(b+f(c)) \text { and }\left(b^{2}+c f(b)\right) \cdot(a+f(c)) $$  are both multiples of $(a+f(c)) \cdot(b+f(c))$ for all sufficiently large $c$. Taking an appropriate linear combination to eliminate the $c f(c)$ term, we find after expanding out that the integer  $$ \left[a^{2} f(b)-b^{2} f(a)\right] \cdot f(c)+[(b-a) f(a) f(b)] \cdot c+[a b(a f(b)-b f(a))] $$  is also a multiple of $(a+f(c)) \cdot(b+f(c))$. But as $c$ varies, $(\dagger)$ is bounded above by a positive multiple of $c$ while $(a+f(c)) \cdot(b+f(c))$ is bounded below by a positive multiple of $c^{2}$. The only way that such a divisibility can hold is if in fact  $$ \left[a^{2} f(b)-b^{2} f(a)\right] \cdot f(c)+[(b-a) f(a) f(b)] \cdot c+[a b(a f(b)-b f(a))]=0 $$  for sufficiently large $c$. Since the coefficient of $c$ in this linear relation is nonzero, it follows that there are constants $k, \ell$ such that $f(c)=k c+\ell$ for all sufficiently large $c$; the constants $k$ and $\ell$ are necessarily integers.  The value of $\ell$ satisfies  $$ \left[a^{2} f(b)-b^{2} f(a)\right] \cdot \ell+[a b(a f(b)-b f(a))]=0 $$  and hence $b \mid \ell a^{2} f(b)$ for all $a$ and $b$. Taking $b$ sufficiently large so that $f(b)=k b+\ell$, we thus have that $b \mid \ell^{2} a^{2}$ for all sufficiently large $b$; this implies that $\ell=0$. From ( $\dagger \dagger \dagger$ ) it then follows that $\frac{f(a)}{a}=\frac{f(b)}{b}$ for all $a \neq b$, so that there is a constant $k$ such that $f(a)=k a$ for all $a$ ( $k$ is equal to the constant defined earlier).


Solution: Let $\Gamma$ denote the set of all points $(a, f(a))$, so that $\Gamma$ is an infinite subset of the upper-right quadrant of the plane. For a point $A=(a, f(a))$ in $\Gamma$, we define a point $A^{\prime}=\left(-f(a),-f(a)^{2} / a\right)$ in the lower-left quadrant of the plane, and let $\Gamma^{\prime}$ denote the set of all such points $A^{\prime}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-095.jpg?height=552&width=652&top_left_y=449&top_left_x=702)  Claim. For any point $A \in \Gamma$, the set $\Gamma$ is contained in finitely many lines through the point $A^{\prime}$. Proof. Let $A=(a, f(a))$. The functional equation (with $a$ and $b$ interchanged) can be rewritten as $b+f(a) \mid a f(b)-b f(a)$, so that all but finitely many points in $\Gamma$ are contained in one of the lines with equation  $$ a y-f(a) x=m(x+f(a)) $$  for $m$ an integer. Geometrically, these are the lines through $A^{\prime}=\left(-f(a),-f(a)^{2} / a\right)$ with gradient $\frac{f(a)+m}{a}$. Since $\Gamma$ is contained, with finitely many exceptions, in the region $0 \leqslant y \leqslant$ $f(1) \cdot x$ and the point $A^{\prime}$ lies strictly in the lower-left quadrant of the plane, there are only finitely many values of $m$ for which this line meets $\Gamma$. This concludes the proof of the claim.  Now consider any distinct points $A, B \in \Gamma$. It is clear that $A^{\prime}$ and $B^{\prime}$ are distinct. A line through $A^{\prime}$ and a line through $B^{\prime}$ only meet in more than one point if these two lines are equal to the line $A^{\prime} B^{\prime}$. It then follows from the above claim that the line $A^{\prime} B^{\prime}$ must contain all but finitely many points of $\Gamma$. If $C$ is another point of $\Gamma$, then the line $A^{\prime} C^{\prime}$ also passes through all but finitely many points of $\Gamma$, which is only possible if $A^{\prime} C^{\prime}=A^{\prime} B^{\prime}$.  We have thus seen that there is a line $\ell$ passing through all points of $\Gamma^{\prime}$ and through all but finitely many points of $\Gamma$. We claim that this line passes through the origin $O$ and passes through every point of $\Gamma$. To see this, note that by construction $A, O, A^{\prime}$ are collinear for every point $A \in \Gamma$. Since $\ell=A A^{\prime}$ for all but finitely many points $A \in \Gamma$, it thus follows that $O \in \ell$. Thus any $A \in \Gamma$ lies on the line $\ell=A^{\prime} O$.  Since $\Gamma$ is contained in a line through $O$, it follows that there is a real constant $k$ (the gradient of $\ell$ ) such that $f(a)=k a$ for all $a$. The number $k$ is, of course, a positive integer.  Comment. Without the $a+b>C$ condition, this problem is approachable by much more naive methods. For instance, using the given divisibility for $a, b \in\{1,2,3\}$ one can prove by a somewhat tedious case-check that $f(2)=2 f(1)$ and $f(3)=3 f(1)$; this then forms the basis of an induction establishing that $f(n)=n f(1)$ for all $n$.


Let $a$ be a positive integer. We say that a positive integer $b$ is $a$-good if $\binom{a n}{b}-1$ is divisible by $a n+1$ for all positive integers $n$ with $a n \geqslant b$. Suppose $b$ is a positive integer such that $b$ is $a$-good, but $b+2$ is not $a$-good. Prove that $b+1$ is prime. (Netherlands)

Solution: For $p$ a prime and $n$ a nonzero integer, we write $v_{p}(n)$ for the $p$-adic valuation of $n$ : the largest integer $t$ such that $p^{t} \mid n$.  We first show that $b$ is $a$-good if and only if $b$ is even, and $p \mid a$ for all primes $p \leqslant b$. To start with, the condition that $a n+1 \left\lvert\,\binom{ a n}{b}-1\right.$ can be rewritten as saying that  $$ \frac{a n(a n-1) \cdots(a n-b+1)}{b!} \equiv 1 \quad(\bmod a n+1) $$  Suppose, on the one hand, there is a prime $p \leqslant b$ with $p \nmid a$. Take $t=v_{p}(b!)$. Then there exist positive integers $c$ such that $a c \equiv 1\left(\bmod p^{t+1}\right)$. If we take $c$ big enough, and then take $n=(p-1) c$, then $a n=a(p-1) c \equiv p-1\left(\bmod p^{t+1}\right)$ and $a n \geqslant b$. Since $p \leqslant b$, one of the terms of the numerator an $(a n-1) \cdots(a n-b+1)$ is $a n-p+1$, which is divisible by $p^{t+1}$. Hence the $p$-adic valuation of the numerator is at least $t+1$, but that of the denominator is exactly $t$. This means that $p \left\lvert\,\binom{ a n}{b}\right.$, so $p \nmid\binom{a n}{b}-1$. As $p \mid a n+1$, we get that $a n+1 \nmid\binom{a n}{b}$, so $b$ is not $a$-good.  On the other hand, if for all primes $p \leqslant b$ we have $p \mid a$, then every factor of $b$ ! is coprime to $a n+1$, and hence invertible modulo $a n+1$ : hence $b$ ! is also invertible modulo $a n+1$. Then equation (1) reduces to:  $$ a n(a n-1) \cdots(a n-b+1) \equiv b!\quad(\bmod a n+1) $$  However, we can rewrite the left-hand side as follows:  $$ a n(a n-1) \cdots(a n-b+1) \equiv(-1)(-2) \cdots(-b) \equiv(-1)^{b} b!\quad(\bmod a n+1) $$  Provided that $a n>1$, if $b$ is even we deduce $(-1)^{b} b!\equiv b$ ! as needed. On the other hand, if $b$ is odd, and we take $a n+1>2(b!)$, then we will not have $(-1)^{b} b!\equiv b$ !, so $b$ is not $a$-good. This completes the claim.  To conclude from here, suppose that $b$ is $a$-good, but $b+2$ is not. Then $b$ is even, and $p \mid a$ for all primes $p \leqslant b$, but there is a prime $q \leqslant b+2$ for which $q \nmid a$ : so $q=b+1$ or $q=b+2$. We cannot have $q=b+2$, as that is even too, so we have $q=b+1$ : in other words, $b+1$ is prime.


Solution: We show only half of the claim of the previous solution: we show that if $b$ is $a$-good, then $p \mid a$ for all primes $p \leqslant b$. We do this with Lucas' theorem.  Suppose that we have $p \leqslant b$ with $p \nmid a$. Then consider the expansion of $b$ in base $p$; there will be some digit (not the final digit) which is nonzero, because $p \leqslant b$. Suppose it is the $p^{t}$ digit for $t \geqslant 1$.  Now, as $n$ varies over the integers, an +1 runs over all residue classes modulo $p^{t+1}$; in particular, there is a choice of $n$ (with $a n>b$ ) such that the $p^{0}$ digit of $a n$ is $p-1$ (so $p \mid a n+1$ ) and the $p^{t}$ digit of $a n$ is 0 . Consequently, $p \mid a n+1$ but $p \left\lvert\,\binom{ a n}{b}\right.$ (by Lucas' theorem) so $p \nmid\binom{a n}{b}-1$. Thus $b$ is not $a$-good.  Now we show directly that if $b$ is $a$-good but $b+2$ fails to be so, then there must be a prime dividing $a n+1$ for some $n$, which also divides $(b+1)(b+2)$. Indeed, the ratio between $\binom{a n}{b+2}$ and $\binom{a n}{b}$ is $(b+1)(b+2) /(a n-b)(a n-b-1)$. We know that there must be a choice of $a n+1$ such that the former binomial coefficient is 1 modulo $a n+1$ but the latter is not, which means that the given ratio must not be $1 \bmod a n+1$. If $b+1$ and $b+2$ are both coprime to $a n+1$ then the ratio is 1 , so that must not be the case. In particular, as any prime less than $b$ divides $a$, it must be the case that either $b+1$ or $b+2$ is prime.  However, we can observe that $b$ must be even by insisting that $a n+1$ is prime (which is possible by Dirichlet's theorem) and hence $\binom{a n}{b} \equiv(-1)^{b}=1$. Thus $b+2$ cannot be prime, so $b+1$ must be prime.