Solution: First, let $\ell$ be a line in the plane such that there are $n$ points on one side and the other $n$ points on the other side. For convenience, assume $\ell$ is horizontal (otherwise, we can rotate the plane). Then we can use the terms "above", "below", "left" and "right" in the usual way. We denote the $n$ points above the line in an arbitrary order as $P_{1}, P_{2}, \ldots, P_{n}$, and the $n$ points below the line as $Q_{1}, Q_{2}, \ldots, Q_{n}$.  If we connect $P_{i}$ and $Q_{j}$ with a line segment, the line segment will intersect with the line $\ell$. Denote the intersection as $I_{i j}$. If $P_{i}$ is connected to $Q_{j}$ and $Q_{k}$, where $j<k$, then $I_{i j}$ and $I_{i k}$ are two different points, because $P_{i}, Q_{j}$ and $Q_{k}$ are not collinear.  Now we define a "sign" for each angle $\angle Q_{j} P_{i} Q_{k}$. Assume $j<k$. We specify that the sign is positive for the following two cases:  - if $i$ is odd and $I_{i j}$ is to the left of $I_{i k}$, - if $i$ is even and $I_{i j}$ is to the right of $I_{i k}$.  Otherwise the sign of the angle is negative. If $j>k$, then the sign of $\angle Q_{j} P_{i} Q_{k}$ is taken to be the same as for $\angle Q_{k} P_{i} Q_{j}$.  Similarly, we can define the sign of $\angle P_{j} Q_{i} P_{k}$ with $j<k$ (or equivalently $\angle P_{k} Q_{i} P_{j}$ ). For example, it is positive when $i$ is odd and $I_{j i}$ is to the left of $I_{k i}$.  Henceforth, whenever we use the notation $\angle Q_{j} P_{i} Q_{k}$ or $\angle P_{j} Q_{i} P_{k}$ for a numerical quantity, it is understood to denote either the (geometric) measure of the angle or the negative of this measure, depending on the sign as specified above.  We now have the following important fact for signed angle measures:  $$ \angle Q_{i_{1}} P_{k} Q_{i_{3}}=\angle Q_{i_{1}} P_{k} Q_{i_{2}}+\angle Q_{i_{2}} P_{k} Q_{i_{3}} $$  for all points $P_{k}, Q_{i_{1}}, Q_{i_{2}}$ and $Q_{i_{3}}$ with $i_{1}<i_{2}<i_{3}$. The following figure shows a "natural" arrangement of the points. Equation (1) still holds for any other arrangement, as can be easily verified. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-045.jpg?height=346&width=806&top_left_y=2163&top_left_x=631)  Similarly, we have  $$ \angle P_{i_{1}} Q_{k} P_{i_{3}}=\angle P_{i_{1}} Q_{k} P_{i_{2}}+\angle P_{i_{2}} Q_{k} P_{i_{3}}, $$  for all points $Q_{k}, P_{i_{1}}, P_{i_{2}}$ and $P_{i_{3}}$, with $i_{1}<i_{2}<i_{3}$.  We are now ready to specify the desired ordering $A_{1}, \ldots, A_{2 n}$ of the points:  - if $i \leqslant n$ is odd, put $A_{i}=P_{i}$ and $A_{2 n+1-i}=Q_{i}$; - if $i \leqslant n$ is even, put $A_{i}=Q_{i}$ and $A_{2 n+1-i}=P_{i}$.  For example, for $n=3$ this ordering is $P_{1}, Q_{2}, P_{3}, Q_{3}, P_{2}, Q_{1}$. This sequence alternates between $P$ 's and $Q$ 's, so the above conventions specify a sign for each of the angles $A_{i-1} A_{i} A_{i+1}$. We claim that the sum of these $2 n$ signed angles equals 0 . If we can show this, it would complete the proof.  We prove the claim by induction. For brevity, we use the notation $\angle P_{i}$ to denote whichever of the $2 n$ angles has its vertex at $P_{i}$, and $\angle Q_{i}$ similarly.  First let $n=2$. If the four points can be arranged to form a convex quadrilateral, then the four line segments $P_{1} Q_{1}, P_{1} Q_{2}, P_{2} Q_{1}$ and $P_{2} Q_{2}$ constitute a self-intersecting quadrilateral. We use several figures to illustrate the possible cases.  The following figure is one possible arrangement of the points. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-046.jpg?height=435&width=803&top_left_y=1047&top_left_x=632)  Then $\angle P_{1}$ and $\angle Q_{1}$ are positive, $\angle P_{2}$ and $\angle Q_{2}$ are negative, and we have  $$ \left|\angle P_{1}\right|+\left|\angle Q_{1}\right|=\left|\angle P_{2}\right|+\left|\angle Q_{2}\right| . $$  With signed measures, we have  $$ \angle P_{1}+\angle Q_{1}+\angle P_{2}+\angle Q_{2}=0 $$  If we switch the labels of $P_{1}$ and $P_{2}$, we have the following picture: ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-046.jpg?height=438&width=809&top_left_y=2031&top_left_x=629)  Switching labels $P_{1}$ and $P_{2}$ has the effect of flipping the sign of all four angles (as well as swapping the magnitudes on the relabelled points); that is, the new values of ( $\angle P_{1}, \angle P_{2}, \angle Q_{1}, \angle Q_{2}$ ) equal the old values of ( $-\angle P_{2},-\angle P_{1},-\angle Q_{1},-\angle Q_{2}$ ). Consequently, equation (3) still holds. Similarly, when switching the labels of $Q_{1}$ and $Q_{2}$, or both the $P$ 's and the $Q$ 's, equation (3) still holds.  The remaining subcase of $n=2$ is that one point lies inside the triangle formed by the other three. We have the following picture. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-047.jpg?height=495&width=820&top_left_y=358&top_left_x=615)  We have  $$ \left|\angle P_{1}\right|+\left|\angle Q_{1}\right|+\left|\angle Q_{2}\right|=\left|\angle P_{2}\right| . $$  and equation (3) holds. Again, switching the labels for $P$ 's or the $Q$ 's will not affect the validity of equation (3). Also, if the point lying inside the triangle of the other three is one of the $Q$ 's rather than the $P$ 's, the result still holds, since our sign convention is preserved when we relabel $Q$ 's as $P$ 's and vice-versa and reflect across $\ell$.  We have completed the proof of the claim for $n=2$. Assume the claim holds for $n=k$, and we wish to prove it for $n=k+1$. Suppose we are given our $2(k+1)$ points. First ignore $P_{k+1}$ and $Q_{k+1}$, and form $2 k$ angles from $P_{1}, \ldots, P_{k}$, $Q_{1}, \ldots, Q_{k}$ as in the $n=k$ case. By the induction hypothesis we have  $$ \sum_{i=1}^{k}\left(\angle P_{i}+\angle Q_{i}\right)=0 $$  When we add in the two points $P_{k+1}$ and $Q_{k+1}$, this changes our angles as follows:  - the angle at $P_{k}$ changes from $\angle Q_{k-1} P_{k} Q_{k}$ to $\angle Q_{k-1} P_{k} Q_{k+1}$; - the angle at $Q_{k}$ changes from $\angle P_{k-1} Q_{k} P_{k}$ to $\angle P_{k-1} Q_{k} P_{k+1}$; - two new angles $\angle Q_{k} P_{k+1} Q_{k+1}$ and $\angle P_{k} Q_{k+1} P_{k+1}$ are added.  We need to prove the changes have no impact on the total sum. In other words, we need to prove  $$ \left(\angle Q_{k-1} P_{k} Q_{k+1}-\angle Q_{k-1} P_{k} Q_{k}\right)+\left(\angle P_{k-1} Q_{k} P_{k+1}-\angle P_{k-1} Q_{k} P_{k}\right)+\left(\angle P_{k+1}+\angle Q_{k+1}\right)=0 $$  In fact, from equations (1) and (2), we have  $$ \angle Q_{k-1} P_{k} Q_{k+1}-\angle Q_{k-1} P_{k} Q_{k}=\angle Q_{k} P_{k} Q_{k+1} $$  and  $$ \angle P_{k-1} Q_{k} P_{k+1}-\angle P_{k-1} Q_{k} P_{k}=\angle P_{k} Q_{k} P_{k+1} $$  Therefore, the left hand side of equation (4) becomes $\angle Q_{k} P_{k} Q_{k+1}+\angle P_{k} Q_{k} P_{k+1}+\angle Q_{k} P_{k+1} Q_{k+1}+$ $\angle P_{k} Q_{k+1} P_{k+1}$, which equals 0 , simply by applying the $n=2$ case of the claim. This completes the induction.


Let $n>1$ be an integer. Suppose we are given $2 n$ points in a plane such that no three of them are collinear. The points are to be labelled $A_{1}, A_{2}, \ldots, A_{2 n}$ in some order. We then consider the $2 n$ angles $\angle A_{1} A_{2} A_{3}, \angle A_{2} A_{3} A_{4}, \ldots, \angle A_{2 n-2} A_{2 n-1} A_{2 n}, \angle A_{2 n-1} A_{2 n} A_{1}$, $\angle A_{2 n} A_{1} A_{2}$. We measure each angle in the way that gives the smallest positive value (i.e. between $0^{\circ}$ and $180^{\circ}$ ). Prove that there exists an ordering of the given points such that the resulting $2 n$ angles can be separated into two groups with the sum of one group of angles equal to the sum of the other group.  Comment. The first three solutions all use the same construction involving a line separating the points into groups of $n$ points each, but give different proofs that this construction works. Although Solution 1 is very short, the Problem Selection Committee does not believe any of the solutions is easy to find and thus rates this as a problem of medium difficulty.

Solution: We shall think instead of the problem as asking us to assign a weight $\pm 1$ to each angle, such that the weighted sum of all the angles is zero.  Given an ordering $A_{1}, \ldots, A_{2 n}$ of the points, we shall assign weights according to the following recipe: walk in order from point to point, and assign the left turns +1 and the right turns -1 . This is the same weighting as in Solution 3, and as in that solution, the weighted sum is a multiple of $360^{\circ}$.  We now aim to show the following: Lemma. Transposing any two consecutive points in the ordering changes the weighted sum by $\pm 360^{\circ}$ or 0 .  Knowing that, we can conclude quickly: if the ordering $A_{1}, \ldots, A_{2 n}$ has weighted angle sum $360 k^{\circ}$, then the ordering $A_{2 n}, \ldots, A_{1}$ has weighted angle sum $-360 k^{\circ}$ (since the angles are the same, but left turns and right turns are exchanged). We can reverse the ordering of $A_{1}$, $\ldots, A_{2 n}$ by a sequence of transpositions of consecutive points, and in doing so the weighted angle sum must become zero somewhere along the way.  We now prove that lemma: Proof. Transposing two points amounts to taking a section $A_{k} A_{k+1} A_{k+2} A_{k+3}$ as depicted, reversing the central line segment $A_{k+1} A_{k+2}$, and replacing its two neighbours with the dotted lines. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-048.jpg?height=424&width=1357&top_left_y=1156&top_left_x=354)  Figure 1: Transposing two consecutive vertices: before (left) and afterwards (right) In each triangle, we alter the sum by $\pm 180^{\circ}$. Indeed, using (anticlockwise) directed angles modulo $360^{\circ}$, we either add or subtract all three angles of each triangle.  Hence both triangles together alter the sum by $\pm 180 \pm 180^{\circ}$, which is $\pm 360^{\circ}$ or 0 .


There are 60 empty boxes $B_{1}, \ldots, B_{60}$ in a row on a table and an unlimited supply of pebbles. Given a positive integer $n$, Alice and Bob play the following game.  In the first round, Alice takes $n$ pebbles and distributes them into the 60 boxes as she wishes. Each subsequent round consists of two steps: (a) Bob chooses an integer $k$ with $1 \leqslant k \leqslant 59$ and splits the boxes into the two groups $B_{1}, \ldots, B_{k}$ and $B_{k+1}, \ldots, B_{60}$. (b) Alice picks one of these two groups, adds one pebble to each box in that group, and removes one pebble from each box in the other group.  Bob wins if, at the end of any round, some box contains no pebbles. Find the smallest $n$ such that Alice can prevent Bob from winning. (Czech Republic) Answer: $n=960$. In general, if there are $N>1$ boxes, the answer is $n=\left\lfloor\frac{N}{2}+1\right\rfloor\left\lceil\frac{N}{2}+1\right\rceil-1$. Common remarks. We present solutions for the general case of $N>1$ boxes, and write $M=\left\lfloor\frac{N}{2}+1\right\rfloor\left\lceil\frac{N}{2}+1\right\rceil-1$ for the claimed answer. For $1 \leqslant k<N$, say that Bob makes a $k$-move if he splits the boxes into a left group $\left\{B_{1}, \ldots, B_{k}\right\}$ and a right group $\left\{B_{k+1}, \ldots, B_{N}\right\}$. Say that one configuration dominates another if it has at least as many pebbles in each box, and say that it strictly dominates the other configuration if it also has more pebbles in at least one box. (Thus, if Bob wins in some configuration, he also wins in every configuration that it dominates.)  It is often convenient to consider ' V -shaped' configurations; for $1 \leqslant i \leqslant N$, let $V_{i}$ be the configuration where $B_{j}$ contains $1+|j-i|$ pebbles (i.e. where the $i^{\text {th }}$ box has a single pebble and the numbers increase by one in both directions, so the first box has $i$ pebbles and the last box has $N+1-i$ pebbles). Note that $V_{i}$ contains $\frac{1}{2} i(i+1)+\frac{1}{2}(N+1-i)(N+2-i)-1$ pebbles. If $i=\left\lceil\frac{N}{2}\right\rceil$, this number equals $M$.  Solutions split naturally into a strategy for Alice (starting with $M$ pebbles and showing she can prevent Bob from winning) and a strategy for Bob (showing he can win for any starting configuration with at most $M-1$ pebbles). The following observation is also useful to simplify the analysis of strategies for Bob. Observation A. Consider two consecutive rounds. Suppose that in the first round Bob made a $k$-move and Alice picked the left group, and then in the second round Bob makes an $\ell$-move, with $\ell>k$. We may then assume, without loss of generality, that Alice again picks the left group. Proof. Suppose Alice picks the right group in the second round. Then the combined effect of the two rounds is that each of the boxes $B_{k+1}, \ldots, B_{\ell}$ lost two pebbles (and the other boxes are unchanged). Hence this configuration is strictly dominated by that before the first round, and it suffices to consider only Alice's other response.  Solution 1 (Alice). Alice initially distributes pebbles according to $V_{\left\lceil\frac{N}{2}\right\rceil}$. Suppose the current configuration of pebbles dominates $V_{i}$. If Bob makes a $k$-move with $k \geqslant i$ then Alice picks the left group, which results in a configuration that dominates $V_{i+1}$. Likewise, if Bob makes a $k$-move with $k<i$ then Alice picks the right group, which results in a configuration that dominates $V_{i-1}$. Since none of $V_{1}, \ldots, V_{N}$ contains an empty box, Alice can prevent Bob from ever winning.  Solution 1 (Bob). The key idea in this solution is the following claim. Claim. If there exist a positive integer $k$ such that there are at least $2 k$ boxes that have at most $k$ pebbles each then Bob can force a win. Proof. We ignore the other boxes. First, Bob makes a $k$-move (splits the $2 k$ boxes into two groups of $k$ boxes each). Without loss of generality, Alice picks the left group. Then Bob makes a $(k+1)$-move, $\ldots$, a $(2 k-1)$-move. By Observation A, we may suppose Alice always picks the left group. After Bob's $(2 k-1)$-move, the rightmost box becomes empty and Bob wins.  Now, we claim that if $n<M$ then either there already exists an empty box, or there exist a positive integer $k$ and $2 k$ boxes with at most $k$ pebbles each (and thus Bob can force a win). Otherwise, assume each box contains at least 1 pebble, and for each $1 \leqslant k \leqslant\left\lfloor\frac{N}{2}\right\rfloor$, at least $N-(2 k-1)=N+1-2 k$ boxes contain at least $k+1$ pebbles. Summing, there are at least as many pebbles in total as in $V_{\left\lceil\frac{N}{2}\right\rceil}$; that is, at least $M$ pebbles, as desired.  Solution 2 (Alice). Let $K=\left\lfloor\frac{N}{2}+1\right\rfloor$. Alice starts with the boxes in the configuration $V_{K}$. For each of Bob's $N-1$ possible choices, consider the subset of rounds in which he makes that choice. In that subset of rounds, Alice alternates between picking the left group and picking the right group; the first time Bob makes that choice, Alice picks the group containing the $K^{\text {th }}$ box. Thus, at any time during the game, the number of pebbles in each box depends only on which choices Bob has made an odd number of times. This means that the number of pebbles in a box could decrease by at most the number of choices for which Alice would have started by removing a pebble from the group containing that box. These numbers are, for each box,  $$ \left\lfloor\frac{N}{2}\right\rfloor,\left\lfloor\frac{N}{2}-1\right\rfloor, \ldots, 1,0,1, \ldots,\left\lceil\frac{N}{2}-1\right\rceil $$  These are pointwise less than the numbers of pebbles the boxes started with, meaning that no box ever becomes empty with this strategy.  Solution 2 (Bob). Let $K=\left\lfloor\frac{N}{2}+1\right\rfloor$. For Bob's strategy, we consider a configuration $X$ with at most $M-1$ pebbles, and we make use of Observation A. Consider two configurations with $M$ pebbles: $V_{K}$ and $V_{N+1-K}$ (if $n$ is odd, they are the same configuration; if $n$ is even, one is the reverse of the other). The configuration $X$ has fewer pebbles than $V_{K}$ in at least one box, and fewer pebbles than $V_{N+1-K}$ in at least one box.  Suppose first that, with respect to one of those configurations (without loss of generality $V_{K}$ ), $X$ has fewer pebbles in one of the boxes in the half where they have $1,2, \ldots,\left\lceil\frac{N}{2}\right\rceil$ pebbles (the right half in $V_{K}$ if $N$ is even; if $N$ is odd, we can take it to be the right half, without loss of generality, as the configuration is symmetric). Note that the number cannot be fewer in the box with 1 pebble in $V_{K}$, because then it would have 0 pebbles. Bob then does a $K$-move. If Alice picks the right group, the total number of pebbles goes down and we restart Bob's strategy with a smaller number of pebbles. If Alice picks the left group, Bob follows with a $(K+1)$-move, a $(K+2)$-move, and so on; by Observation A we may assume Alice always picks the left group. But whichever box in the right half had fewer pebbles in $X$ than in $V_{K}$ ends up with 0 pebbles at some point in this sequence of moves.  Otherwise, $N$ is even, and for both of those configurations, there are fewer pebbles in $X$ only on the $2,3, \ldots, \frac{N}{2}+1$ side. That is, the numbers of pebbles in $X$ are at least  $$ \frac{N}{2}, \frac{N}{2}-1, \ldots, 1,1, \ldots, \frac{N}{2} $$  with equality occurring at least once on each side. Bob does an $\frac{N}{2}$-move. Whichever group Alice chooses, the total number of pebbles is unchanged, and the side from which pebbles are removed now has a box with fewer pebbles than in $(C)$, so the previous case of Bob's strategy can now be applied.  Solution 3 (Bob). For any configuration $C$, define $L(C)$ to be the greatest integer such that, for all $0 \leqslant i \leqslant N-1$, the box $B_{i+1}$ contains at least $L(C)-i$ pebbles. Similarly, define $R(C)$ to be greatest integer such that, for all $0 \leqslant i \leqslant N-1$, the box $B_{N-i}$ contains at least $R(C)-i$ pebbles. (Thus, $C$ dominates the 'left half' of $V_{L(C)}$ and the 'right half' of $V_{N+1-R(C)}$.) Then $C$ dominates a 'V-shaped' configuration if and only if $L(C)+R(C) \geqslant N+1$. Note that if $C$ dominates a V -shaped configuration, it has at least $M$ pebbles.  Now suppose that there are fewer than $M$ pebbles, so we have $L(C)+R(C) \leqslant N$. Then Bob makes an $L(C)$-move (or more generally any move with at least $L(C)$ boxes on the left and $R(C)$ boxes on the right). Let $C^{\prime}$ be the new configuration, and suppose that no box becomes empty (otherwise Bob has won). If Alice picks the left group, we have $L\left(C^{\prime}\right)=L(C)+1$ and $R\left(C^{\prime}\right)=R(C)-1$. Otherwise, we have $L\left(C^{\prime}\right)=L(C)-1$ and $R\left(C^{\prime}\right)=R(C)+1$. In either case, we have $L\left(C^{\prime}\right)+R\left(C^{\prime}\right) \leqslant N$.  Bob then repeats this strategy, until one of the boxes becomes empty. Since the condition in Observation A holds, we may assume that Alice picks a group on the same side each time. Then one of $L$ and $R$ is strictly decreasing; without loss of generality assume that $L$ strictly decreases. At some point we reach $L=1$. If $B_{2}$ is still nonempty, then $B_{1}$ must contain a single pebble. Bob makes a 1-move, and by Observation A, Alice must (eventually) pick the right group, making this box empty.


Alice has a map of Wonderland, a country consisting of $n \geqslant 2$ towns. For every  pair of towns, there is a narrow road going from one town to the other. One day, all the roads are declared to be "one way" only. Alice has no information on the direction of the roads, but the King of Hearts has offered to help her. She is allowed to ask him a number of questions. For each question in turn, Alice chooses a pair of towns and the King of Hearts tells her the direction of the road connecting those two towns.  Alice wants to know whether there is at least one town in Wonderland with at most one outgoing road. Prove that she can always find out by asking at most $4 n$ questions.  Comment. This problem could be posed with an explicit statement about points being awarded for weaker bounds $c n$ for some $c>4$, in the style of IMO 2014 Problem 6. (Thailand)

Solution: We will show Alice needs to ask at most $4 n-7$ questions. Her strategy has the following phases. In what follows, $S$ is the set of towns that Alice, so far, does not know to have more than one outgoing road (so initially $|S|=n$ ).  Phase 1. Alice chooses any two towns, say $A$ and $B$. Without loss of generality, suppose that the King of Hearts' answer is that the road goes from $A$ to $B$.  At the end of this phase, Alice has asked 1 question. Phase 2. During this phase there is a single (variable) town $T$ that is known to have at least one incoming road but not yet known to have any outgoing roads. Initially, $T$ is $B$. Alice does the following $n-2$ times: she picks a town $X$ she has not asked about before, and asks the direction of the road between $T$ and $X$. If it is from $X$ to $T, T$ is unchanged; if it is from $T$ to $X, X$ becomes the new choice of town $T$, as the previous $T$ is now known to have an outgoing road.  At the end of this phase, Alice has asked a total of $n-1$ questions. The final town $T$ is not yet known to have any outgoing roads, while every other town has exactly one outgoing road known. The undirected graph of roads whose directions are known is a tree.  Phase 3. During this phase, Alice asks about the directions of all roads between $T$ and another town she has not previously asked about, stopping if she finds two outgoing roads from $T$. This phase involves at most $n-2$ questions. If she does not find two outgoing roads from $T$, she has answered her original question with at most $2 n-3 \leqslant 4 n-7$ questions, so in what follows we suppose that she does find two outgoing roads, asking a total of $k$ questions in this phase, where $2 \leqslant k \leqslant n-2$ (and thus $n \geqslant 4$ for what follows).  For every question where the road goes towards $T$, the town at the other end is removed from $S$ (as it already had one outgoing road known), while the last question resulted in $T$ being removed from $S$. So at the end of this phase, $|S|=n-k+1$, while a total of $n+k-1$ questions have been asked. Furthermore, the undirected graph of roads within $S$ whose directions are known contains no cycles (as $T$ is no longer a member of $S$, all questions asked in this phase involved $T$ and the graph was a tree before this phase started). Every town in $S$ has exactly one outgoing road known (not necessarily to another town in $S$ ).  Phase 4. During this phase, Alice repeatedly picks any pair of towns in $S$ for which she does not know the direction of the road between them. Because every town in $S$ has exactly one outgoing road known, this always results in the removal of one of those two towns from $S$. Because there are no cycles in the graph of roads of known direction within $S$, this can continue until there are at most 2 towns left in $S$.  If it ends with $t$ towns left, $n-k+1-t$ questions were asked in this phase, so a total of $2 n-t$ questions have been asked.  Phase 5. During this phase, Alice asks about all the roads from the remaining towns in $S$ that she has not previously asked about. She has definitely already asked about any road between those towns (if $t=2$ ). She must also have asked in one of the first two phases about at least one other road involving one of those towns (as those phases resulted in a tree with $n>2$ vertices). So she asks at most $t(n-t)-1$ questions in this phase.  At the end of this phase, Alice knows whether any town has at most one outgoing road. If $t=1$, at most $3 n-3 \leqslant 4 n-7$ questions were needed in total, while if $t=2$, at most $4 n-7$ questions were needed in total.  Comment 1. The version of this problem originally submitted asked only for an upper bound of $5 n$, which is much simpler to prove. The Problem Selection Committee preferred a version with an asymptotically optimal constant. In the following comment, we will show that the constant is optimal.  Comment 2. We will show that Alice cannot always find out by asking at most $4 n-3\left(\log _{2} n\right)-$ 15 questions, if $n \geqslant 8$.  To show this, we suppose the King of Hearts is choosing the directions as he goes along, only picking the direction of a road when Alice asks about it for the first time. We provide a strategy for the King of Hearts that ensures that, after the given number of questions, the map is still consistent both with the existence of a town with at most one outgoing road, and with the nonexistence of such a town. His strategy has the following phases. When describing how the King of Hearts' answer to a question is determined below, we always assume he is being asked about a road for the first time (otherwise, he just repeats his previous answer for that road). This strategy is described throughout in graph-theoretic terms (vertices and edges rather than towns and roads).  Phase 1. In this phase, we consider the undirected graph formed by edges whose directions are known. The phase terminates when there are exactly 8 connected components whose undirected graphs are trees. The following invariant is maintained: in a component with $k$ vertices whose undirected graph is a tree, every vertex has at most $\left[\log _{2} k\right\rfloor$ edges into it.  - If the King of Hearts is asked about an edge between two vertices in the same component, or about an edge between two components at least one of which is not a tree, he chooses any direction for that edge arbitrarily. - If he is asked about an edge between a vertex in component $A$ that has $a$ vertices and is a tree and a vertex in component $B$ that has $b$ vertices and is a tree, suppose without loss of generality that $a \geqslant b$. He then chooses the edge to go from $A$ to $B$. In this case, the new number of edges into any vertex is at most $\max \left\{\left\lfloor\log _{2} a\right\rfloor,\left\lfloor\log _{2} b\right\rfloor+1\right\} \leqslant\left\lfloor\log _{2}(a+b)\right\rfloor$.  In all cases, the invariant is preserved, and the number of tree components either remains unchanged or goes down by 1. Assuming Alice does not repeat questions, the process must eventually terminate with 8 tree components, and at least $n-8$ questions having been asked.  Note that each tree component contains at least one vertex with no outgoing edges. Colour one such vertex in each tree component red.  Phase 2. Let $V_{1}, V_{2}$ and $V_{3}$ be the three of the red vertices whose components are smallest (so their components together have at most $\left\lfloor\frac{3}{8} n\right\rfloor$ vertices, with each component having at most $\left\lfloor\frac{3}{8} n-2\right\rfloor$ vertices). Let sets $C_{1}, C_{2}, \ldots$ be the connected components after removing the $V_{j}$. By construction, there are no edges with known direction between $C_{i}$ and $C_{j}$ for $i \neq j$, and there are at least five such components.  If at any point during this phase, the King of Hearts is asked about an edge within one of the $C_{i}$, he chooses an arbitrary direction. If he is asked about an edge between $C_{i}$ and $C_{j}$ for $i \neq j$, he answers so that all edges go from $C_{i}$ to $C_{i+1}$ and $C_{i+2}$, with indices taken modulo the number of components, and chooses arbitrarily for other pairs. This ensures that all vertices other than the $V_{j}$ will have more than one outgoing edge.  For edges involving one of the $V_{j}$ he answers as follows, so as to remain consistent for as long as possible with both possibilities for whether one of those vertices has at most one outgoing edge. Note that as they were red vertices, they have no outgoing edges at the start of this phase. For edges between two of the $V_{j}$, he answers that the edges go from $V_{1}$ to $V_{2}$, from $V_{2}$ to $V_{3}$ and from $V_{3}$ to $V_{1}$. For edges between $V_{j}$ and some other vertex, he always answers that the edge goes into $V_{j}$, except for the last such edge for which he is asked the question for any given $V_{j}$, for which he answers that the edge goes out of $V_{j}$. Thus, as long as at least one of the $V_{j}$ has not had the question answered for all the vertices that are not among the $V_{j}$, his answers are still compatible both with all vertices having more than one outgoing edge, and with that $V_{j}$ having only one outgoing edge.  At the start of this phase, each of the $V_{j}$ has at most $\left\lfloor\log _{2}\left\lfloor\frac{3}{8} n-2\right\rfloor\right\rfloor<\left(\log _{2} n\right)-1$ incoming edges. Thus, Alice cannot determine whether some vertex has only one outgoing edge within $3(n-$ $\left.3-\left(\left(\log _{2} n\right)-1\right)\right)-1$ questions in this phase; that is, $4 n-3\left(\log _{2} n\right)-15$ questions total.  Comment 3. We can also improve the upper bound slightly, to $4 n-2\left(\log _{2} n\right)+1$. (We do not know where the precise minimum number of questions lies between $4 n-3\left(\log _{2} n\right)+O(1)$ and $4 n-2\left(\log _{2} n\right)+$ $O(1)$.) Suppose $n \geqslant 5$ (otherwise no questions are required at all).  To do this, we replace Phases 1 and 2 of the given solution with a different strategy that also results in a spanning tree where one vertex $V$ is not known to have any outgoing edges, and all other vertices have exactly one outgoing edge known, but where there is more control over the numbers of incoming edges. In Phases 3 and 4 we then take more care about the order in which pairs of towns are chosen, to ensure that each of the remaining towns has already had a question asked about at least $\log _{2} n+O(1)$ edges.  Define trees $T_{m}$ with $2^{m}$ vertices, exactly one of which (the root) has no outgoing edges and the rest of which have exactly one outgoing edge, as follows: $T_{0}$ is a single vertex, while $T_{m}$ is constructed by joining the roots of two copies of $T_{m-1}$ with an edge in either direction. If $n=2^{m}$ we can readily ask $n-1$ questions, resulting in a tree $T_{m}$ for the edges with known direction: first ask about $2^{m-1}$ disjoint pairs of vertices, then about $2^{m-2}$ disjoint pairs of the roots of the resulting $T_{1}$ trees, and so on. For the general case, where $n$ is not a power of 2 , after $k$ stages of this process we have $\left\lfloor n / 2^{k}\right\rfloor$ trees, each of which is like $T_{k}$ but may have some extra vertices (but, however, a unique root). If there are an even number of trees, then ask about pairs of their roots. If there are an odd number (greater than 1) of trees, when a single $T_{k}$ is left over, ask about its root together with that of one of the $T_{k+1}$ trees.  Say $m=\left\lfloor\log _{2} n\right\rfloor$. The result of that process is a single $T_{m}$ tree, possibly with some extra vertices but still a unique root $V$. That root has at least $m$ incoming edges, and we may list vertices $V_{0}$, $\ldots, V_{m-1}$ with edges to $V$, such that, for all $0 \leqslant i<m$, vertex $V_{i}$ itself has at least $i$ incoming edges.  Now divide the vertices other than $V$ into two parts: $A$ has all vertices at an odd distance from $V$ and $B$ has all the vertices at an even distance from $B$. Both $A$ and $B$ are nonempty; $A$ contains the $V_{i}$, while $B$ contains a sequence of vertices with at least $0,1, \ldots, m-2$ incoming edges respectively, similar to the $V_{i}$. There are no edges with known direction within $A$ or within $B$.  In Phase 3, then ask about edges between $V$ and other vertices: first those in $B$, in order of increasing number of incoming edges to the other vertex, then those in $A$, again in order of increasing number of incoming edges, which involves asking at most $n-1-m$ questions in this phase. If two outgoing edges are not found from $V$, at most $2 n-2-m \leqslant 4 n-2\left(\log _{2} n\right)+1$ questions needed to be asked in total, so we suppose that two outgoing edges were found, with $k$ questions asked in this phase, where $2 \leqslant k \leqslant n-1-m$. The state of $S$ is as described in the solution above, with the additional property that, since $S$ must still contain all vertices with edges to $V$, it contains the vertices $V_{i}$ described above.  In Phase 4, consider the vertices left in $B$, in increasing order of number of edges incoming to a vertex. If $s$ is the least number of incoming edges to such a vertex, then, for any $s \leqslant t \leqslant m-2$, there are at least $m-t-2$ vertices with more than $t$ incoming edges. Repeatedly asking about the pair of vertices left in $B$ with the least numbers of incoming edges results in a single vertex left over (if any were in $B$ at all at the start of this phase) with at least $m-2$ incoming edges. Doing the same with $A$ (which must be nonempty) leaves a vertex with at least $m-1$ incoming edges.  Thus if only $A$ is nonempty we ask at most $n-m$ questions in Phase 5 , so in total at most $3 n-m-1$ questions, while if both are nonempty we ask at most $2 n-2 m+1$ questions in Phase 5 , so in total at most $4 n-2 m-1<4 n-2\left(\log _{2} n\right)+1$ questions.  This page is intentionally left blank


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) Answer: The maximum possible size of $\mathcal{F}$ is $2^{k}$. Common remarks. For convenience, we extend the use of the word scale: we say that the scale between two reals $x$ and $y$ is $D(x, y)$.

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.


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) ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-062.jpg?height=763&width=715&top_left_y=578&top_left_x=662)

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 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) Common remarks. Throughout the solution, $\angle(a, b)$ denotes the directed angle between lines $a$ and $b$, measured modulo $\pi$.

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) Common remarks. The condition on $f$ is equivalent to the following: There is some function $g$ that assigns to each point $X$ a circle $g(X)$ passing through $X$ such that for any line $\ell$ passing through $X$, the point $f(\ell)$ lies on $g(X)$. (The function $g$ may not be uniquely defined for all points, if some points $X$ have at most one value of $f(\ell)$ other than $X$; for such points, an arbitrary choice is made.)  If there were two points $P$ and $Q$ with the given property, $f(P Q)$ would have to be both $P$ and $Q$, so there is at most one such point, and it will suffice to show that such a point exists.

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.