a)For odd $n$ pick a regural $n$-gon and for $n=4$ pick a circle,let its center be $O$ and pick $3$ points $A,B,C$ on the circle such that $AOB$ and $BOC$ are equilateral.Now,for even $n$ to prove that it is true for $n+2$ just pick two random points $A,B$ on the circle such that $AOB$ is equilateral. b)For all odd $n$,a regural $n$-gon.Now,let $n=2k$.For any pair of points there exists a point such that $AC=BC$.Call that point good.A point can be good at most $k-1$ times,else it will be good for some $A,B$ and $A,C$ and that is a contradiction.But,we have $k*(2k-1)$ pairs and at most $2k*(k-1)$ good points,which is less than $k*(2k-1)$,contradiction.

Here's my solution: (a) For $n \ge 4$ let $\omega$ be a circle centered at a point $O$. Notice that for any $3 \le k \ge 6$, we can add onto $\omega$ the $k$ consecutive vertices of a regular hexagon, and the set will be good. (b) For odd $n$, use a regular $n$-gon. For even $n$ it's impossible because there are $\textstyle\binom n2$ pairs of points, hence some point will be selected at least $\tfrac12(n-1)$ times as a balancing point for pairs $\{A,B\}$. Since $n$ is even, there is in fact a point $P$ selected at least $\tfrac12 n$ times as a pivot. Since there are only $n-1$ other points it follows that there exists $\{A,B\}$ and $\{A,C\}$ such that $PA=PB$ and $PA=PC$. Hence the answer is odd $n$. Part (a) and the first half of (b) are good examples of ad-hoc constructions, while the latter half of part (b) is a very typical example of a (somewhat greedy) "global"-type double-counting argument. ssk9208 wrote: I also think that this is harder than recent P1s. I thought each of (a) and (b) individually was comparable to a past P1, the thing is there are two of them. I also thought (a) was somewhat harder than (b), as (a) took me a good 15 minutes while (b) took me about 15 seconds. Granted, time spent thinking about (a) helped with (b), but still. leminscate wrote: Also Q3 has been geometry for the last 3 years now, which is very surprising. Guess they're still having {1,2,4,5} be different subjects? We'll see tomorrow. math_explorer wrote: In fact, there are former IMO contestants claiming that it is harder than P3. I must be missing something obvious on P3 then

Solution found at home, (even though it is kind of a replica of the above solutions) (a) We can, for $n$ odd, take the regular $n$-gon. For $n$ even, we proceed with induction. Base Case: $n=4$ can be solved by taking a rhombus with $60-120-60-120$ angles. Induction Step: Induct from $n$ to $n+2$. In the above rhombus, take a point with the angle of $120$ and call it $X$. Let the length of each side of the rhombus $1$. Draw a equilateral triangle with length $1$ and $X$ as one of its points. It is clear that this works. (b) For $n$ odd, take the regular $n$-gon again. I claim that $n$ even doesn't work. Denote all points $P_1, P_2 ....., P_n$. Define $f(x)$ the number of $(i,j), i<j$ such that $\overline {P_xP_i} = \overline {P_xP_j}$ It is clear that $f(x) \le [\frac{n-1}{2}]$ ($[x]$ is the largest integer that is not larger than $x$) Define $g(x,y)$ the number of $i$ such that $\overline {P_xP_i} = \overline {P_yP_i}$ It is given that $g(x,y) \ge 1$ for all $x \not= y$ Also, by simple double counting, $n \times [\frac{n-1}{2}] \ge \sum_{i=1}^n f(i) = \sum_{1\le i < j \le n} g(i,j) \ge \textstyle\binom n2$ Substituting $n=2k$, we have $2k(k-1) \ge k(2k-1)$, which is $k \le 0$, contradiction. $\blacksquare$

randomusername wrote: We say that a finite set $\mathcal{S}$ of points in the plane is balanced if, for any two different points $A$ and $B$ in $\mathcal{S}$, there is a point $C$ in $\mathcal{S}$ such that $AC=BC$. We say that $\mathcal{S}$ is centre-free if for any three different points $A$, $B$ and $C$ in $\mathcal{S}$, there are no points $P$ in $\mathcal{S}$ such that $PA=PB=PC$. (a) Show that for all integers $n\ge 3$, there exists a balanced set consisting of $n$ points. (b) Determine all integers $n\ge 3$ for which there exists a balanced centre-free set consisting of $n$ points. slight grammar corrections.

a part is very easy. We make a circle with center point $P$ which depend on $S$. And it is enough to construct in cases $n=3k,3k+1,3k+2$!

Dear Mathelinkers, Q1 a) part isn't a new problem! It has been proposed before in a Chinese Junior High School contest. In fact, the 'Zu Chongzhi Cup' of 1990 contained this problem. And it's included in Dr. Shan Zun's book (数学竞赛研究教程)which is used extensively by Chinese contestants and trainers. So if the jury wants new problem for the paper, they didn't do a very great job. b) part, on the other hand, is not included in that book (not that I am aware of) and I guess is new. How to upload photo when replying with my phone, mathlinkers? Wilson

Likewise, I found (something clearly equivalent to) part A in a 2011 handout from my olympiad training, although I confess to not remembering it until I rediscovered the solution. (Somebody deleted this last time I posted it though.) (English translation:) Quote: A finite set in the plane contains $n$ points. If the perpendicular bisector of any two points in the set passes through at least one other point, then the set is good. What are the possible values of $n$?

Prof. Abderrahim Ouardini informs me that this problem also appears in his book 112 problems of mathematics competition. Ed Ellipses 2000. I suppose that's a fourth source for the problem. I suppose the problem got selected because of its (b) part, which does seem to be new.

1b) $n=2k$, Assume that there exists a balanced and center-free set $S$ consisting of n points.There are $k(2k-1)$ pairs of points in those points or there are are $k(2k-1)$ central lines of pairs of points consisting $2k$ points. Applying Pigeonhole principe, There exists a point $A$ which there are $[\frac{k(2k-1)}{2k}]+1=k$ lines passing, these lines correspond with $2k$ points. because $S$ is a center-free set, these $2k$ are distinct $\implies |S| \geq 2k+1$. This is a contradiction. For $n=2k+1$, Considering a regular polygon, it is easy to see that the set of all points of this polygons satisfying the conditions of problems

a) For $n=3$ and $n=4$, consider an equilateral triangle and a parallelogram with angles $60^{\circ}$ and $120^{\circ}$ respectively. Assume that there exists a balanced set of $k$ points such that i. $k-1$ of those points lie on the same circle, and ii. the other point is the center of that circle. As above, such balanced sets exist for $n=3$ and $n=4$. Then we add two points to the circle such that those two and the center makes an equilateral triangle. For any two points on the circle, the center is equidistant from them. If we choose one of the initial $k-1$ points on the circle and the center, there exist a point on the circle equidistant from them by assumption. If we choose one of the new points and the center, the other point is equidistant from them, so we are done by induction. b) Consider a balanced center-free set $S$ and let $f(P,Q)$ be the point equidistant from $P$ and $Q$. By the center-freeness of the set, we have that $f(P,Q)=f(P,R)$ if and only if $Q \equiv R$. So when we choose a pivot point $P$, for any other point $Q$ on the set, we have a point $R$ in $S$ such that $f(P,R)=Q$. By letting all of them be pivot points, for all $P,Q$, we have $R$ such that $f(P,R)=Q$. Define a pivot point $P_1$ in the set, and let $P_2$ be another point in $S$. We have $P_3$ in $S$ such that $f(P_2,P_3)=P_ 1$. Then consider another point $P_4$, and let $f(P_4,P_5)=P_1$. $P_5$ must be different from $P_2$ and $P_3$ because of the center-freeness of the set. Continuing likewise, for each new point we consider, we have another point on the set which corresponds to that point. Hence $n$ must be odd. For odd $n$, consider the vertices of a regular $n$-gon. It is easy to show that it is center-free and balanced, so the answer for (b) is all odd $n$.

An example of construction for even $n$: Take $A_1,A_2, \dots ,A_k$ points on some circle and center of the circle $O$. Let $\angle A_iOA_{i+1}=\frac{60}{k-1}$. Those are $k+1$ points.Rotating around $O$ for $60$ gives us more $k-1$ points because $A_1$ goes to $A_k$ so we have $2k$ points and we are finished.

randomusername wrote: We say that a finite set $\mathcal{S}$ of points in the plane is balanced if, for any two different points $A$ and $B$ in $\mathcal{S}$, there is a point $C$ in $\mathcal{S}$ such that $AC=BC$. We say that $\mathcal{S}$ is centre-free if for any three different points $A$, $B$ and $C$ in $\mathcal{S}$, there is no points $P$ in $\mathcal{S}$ such that $PA=PB=PC$. (a) Show that for all integers $n\ge 3$, there exists a balanced set consisting of $n$ points. (b) Determine all integers $n\ge 3$ for which there exists a balanced centre-free set consisting of $n$ points. Proposed by Netherlands isn't there a slight typo for #1b?

hwl0304 wrote: isn't there a slight typo for #1b? I can't see any typo...

(a) Let $O$ be the center of a circle and $A_1A_2A_3A_4A_5A_6$ be a regular hexagon inscribed in the circle. For $3 \le n \le 7$, take the points $O, A_1, A_2, A_3,...,A_{n-1}$. For $n = 8$, remove $A_6$ from $S$ and use points $B_1, B_2$ such that $OB_1B_2$ is an equilateral triangle (and such that $B_1$ and $B_2$ are on that same circle). Notice that this would form a unique regular hexagon that is inscribed in the circle. Keep adding consecutive points into $S$ as we increment $n$, until we run out of points, then follow the procedure mentioned earlier about going from $n = 8$ to $n = 9$. (b) For odd $n$ we can use a regular $n$-gon. It is clear that this is balanced and center-free. For even $n$, each pair of points $AB$ there must exist a point on the perpendicular bisector of $AB$. There are $\frac{n(n-1)}{2}$ perpendicular bisectors, and if we let $n = 2k$, $k(2k-1) = 2k^2 - k$ perpendicular bisectors. This implies that some point is going to be selected at least $k$ times as the point that is equidistant to a pair of points. Let this point be $P$. Suppose we don't have $A, B, C$ such that $PA = PB = PC$. That means the set of points that constitute the points that $P$ resolves can be paired up. Then, there must be at least $2k$ points in that set, but since we already are using $P$, there are only $2k-1$ points left. Thus, the answer is odd $n$

(a) For $n \geqslant 3$ odd, just take a regular $n$-gon. For $n>3$ even, fix a point $P$ and take $Q, R, S$ such that $PQRS$ is a $120-60-120-60$ rhombus. Now, just take pairs of points $A, A'$ on the circle centered at $P$ of radius $PQ$ such that $\triangle PAA'$ is equilateral. Adding sufficiently many such pairs we can reach the goal for any $n>3$ even. (b) The only valid $n \geqslant 3$ are the odd ones. For construction, merely the regular polygon case suffices. To see that $n$ cannot be even admit a balanced-centre-free configuration; look at pairs of vertices $\{A_i, A_j\}$ for $1 \leqslant i,j \leqslant n$ for $i \ne j$. Each such pair is assigned an index $k$ so that $A_kA_i=A_kA_j$. Note that the balanced condition ensures that each pair has an image; while the centre-free condition ensures that $(i, j)$ and $(i, j')$ are assigned different indices for $j \ne j'$. Evidently, each vertex occurs as an image exactly $\frac{n-1}{2}$ instances, so $n$ must be odd. $\blacksquare$

My feeling when I spent 90% of my time trying to find a center free construction for $n$ odd. Here's a different construction for (a). Basically glue the constructions for $n=5$ and $n=4$ together at a vertex. This gives $n = 1 + 3a + 4b$ for all $a, b$. Thus, we immediately get everything except $n = 6$, for which we can take an equilateral triangle along with its midpoints.

(a): For odd $n$, an $n$-gon works by inspection. For even $n$, consider the following construction: $$0, 1, e^{2\pi i/6}, e^{4 \pi i/6}, \underbrace{e^{2\pi i \frac{1}{3n}}, e^{2\pi i \frac{1}{3n} + 2 \pi i /6}, e^{2\pi i \frac{2}{3n}}, e^{2\pi i \frac{2}{3n} + 2 \pi i /6}, \dots,e^{2\pi i \frac{\frac{n-4}{2}}{3n}}, e^{2\pi i \frac{\frac{n-4}{2}}{3n} + 2\pi i/6}}_{n-4}.$$For every two points on the unit circle, point $0$ the origin satisfies the balanced condition. For any point on the unit circle and $O$, there exists a corresponding point $\pm \frac{2\pi}{6}$ always on the circle, which lies on the perpendicular bisector. (b): We claim that the statement holds for odd $n$ and not even $n.$ Just take a $n$-gon for odd $n$ by inspection. For even $n$, it suffices to find two adjacent edges whose perpendicular bisectors concur at a point in $S$ as the edge between the other vertices will automatically concur with the other perpendicular bisectors. There are a total of $\binom{n}{2} = n(n-1)/2$ edges so by PhP, there exists at least one vertex with $> \frac{n-1}{2}$ edges whose perpendicular bisector it lies on. Since there are $>n-1/2$ edges, $ > n-1$ counts twice for each vertex, and $n-1$ vertices to choose from, by PhP there is some vertex contained in $> 1$ edges. Hence there are two adjacent edges. Their perpendicular bisectors concur at a vertex in $S$, so the three points are not center-free. $\blacksquare$

For part (a), take a circle centered at a point $O$, and add $n-1$ additional points by adding pairs of points separated by an arc of $60^{\circ}$ or similar triples. Now we only have to consider case (b). If $n$ is odd, then taking regular $n$-gon works. Therefore, we can assume $n = 2k$. Now assume there exists a balanced center free set consisting of $2k$ point. Since there are $k(2k-1)$ ways to pick 2 points from the set and for every point $A, B$ in the set, there exists point $C$ such that $CA = CB$. Combining above facts, we get there exists a point lies on at least $\frac{k(2k-1)}{2k} \frac{2k-1}{2}$ segments perpendicular bisectors and since we have $n$ point, hence that point must be center of some three point, which contradicts the fact that the set of $2k$ point is center free. $\blacksquare$

(a) For $n$ being odd, take the regular polygon with $n$ sides. For $n$ being even, we have $n\geq 4$. So, consider equilateral triangle $XYZ$ with center $O$. Then, arbitrary take $\frac n2-2$ distinct points on this circle. Take the second set of points with each point exactly $60^\circ$ from previous one on the circle. (b) We claim that only odd $n$ work, for which the construction in (a) works. Suppose there is a proper config for some even $n$. Let $n=2k$. Consider the $\binom{n}{2}=k(2k-1)$ segments and the corresponding points which lie on their perpendicular bisectors. Then, by Pigeonhole Principle (perpendicular bisectors are pigeons and points are pigeonholes), there is one point $X$ which lies on at least $\text{ceil}\left(\frac{k(2k-1)}{2k}\right)=k$ perpendicular bisectors. Since there are only $2k-1$ points available (i.e., points other than $X$), and $k$ segments, on point lies on at least two line segments of these $k$ line segments by PHP again. Let this point be $A$ and the segments be $\overline{AB}$ and $\overline{AC}$. So, $XA=XB$ and $XA=XC$, which gives the desired contradiction as $XA=XB=XC$. $\blacksquare$

For part (a) combine triangles and rhombi to get anything except for $n=5$. Then for $n=5$ take a regular pentagon. For part (b) notice that any point can act as a center at most $\left\lfloor \frac{n-1}{2}\right\rfloor$ times, and yet there are at least $\frac{n(n-1)}{2}$ centers. Thus $n$ is odd. Use a regular polygon for the construction. $\blacksquare$

Solved with a lot of help from wzs26843545602, I wonder if I'm the only one who struggled so hopelessly with this. We first deal with part (a). For odd $n$, we simply consider a regular $n-$gon. To see why this works, consider any two vertices of the $n-$gon $A $ and $B$. Then, segment $AB$ seperates the points on the $n-$gon into two groups. Since $n$ is odd, one of these groups has an odd number of points, of which the middle point in fact lies on the perpendicular bisector of $AB$. For even $n$, we consider a regular $n-$gon with center for $n=6k$. This must work since it works for $n=6$ (easy to verify) and for larger $n$, it is clear that for any pair of vertices $A$ and $B$ on the $n-$gon, the center is equidistant. For any vertex and the center, we consider the regular hexagon starting from that vertex, which is included in the $n-$gon (pick every $k^{th}$ vertex) from which it follows that this works. For part (b), odd $n$ clearly work. For even $n$, note that since there are $\binom{n}{2}$ pairs of points, there must exist some point which lies on $\frac{n-1}{2}$ perpendicular bisectors. Since $n$ is even, we in fact must have a point which is on $\frac{n}{2}$ perpendicular bisectors. But, there are only $n-1$ points besides this chosen point, so if no 3 points have this chosen point as its balancing point, we must have that this point lies on at most $\frac{n-1}{2}$ perpendicular bisectors which is a clear contradiction.

For (a), if $n$ is odd, a regular $n-$gon suffices. For $n$ even, first make a circle centered at $O$ with radius one. Then take $A.B,C$ on $(O)$ such that $\Delta OAB$ and $\Delta OAC$ are equilateral. Then keep on putting points $X,Y$ on the circle such that $\Delta OXY$ are equilateral. For (b), I claim only odd $n$ works (the construction in (a) works). For $n$ even, FTSoC assume there exists a working set. There are $\frac{n(n-1)}{2}$ pairs of points, but since we have $n$ points, we must have a point $P$ that forms isosceles triangles with \[\ge \frac{n-1}{2} \text{ pairs of points}\]as that is the expected value of the pairs of points any point forms isosceles triangles with. Since $n$ even, it is actually $\ge \frac{n}{2}$ pairs of points. However, we have $n-1$ that are not point $P$, so there must be a point $A$ such that $P$ is the vertex of isosceles triangle $\Delta PAB$ and $\Delta PAC$, implying that $PA = PB = PC$, a contradiction.

A teeny bit hard for 1/4 just because no one likes combinatorial geometry... We solve both parts simultaneously. In particular, the answer for (b) is odd $n$ only. Center-free construction for odd $n$: Just take a regular $n$-gon. Any diagonal $d$ of the $n$-gon divides it into two polygons $\mathcal A$ and $\mathcal B$, one of which contains an odd number of sides of the original $n$-gon; then the perpendicular bisector of $d$ passes through the top vertex of that polygon. Construction for even $n$: Fix some angles $\theta_1, \theta_2, \dots, \theta_k \in [0, 2\pi)$. For each $i \geq 2$, construct the points $\exp(\theta_i)$ and $\exp(\theta_i + \pi/3)$ on the unit circle, and add the points $\exp(\theta_1)$, $\exp(\theta_1+\pi/3)$, $\exp(\theta_1+2\pi/3)$. The set $\mathcal S$ consisting of all these points and the origin works. In particular, the perpendicular bisector of any two points on the circle passes through the origin, and the perpendicular bisector of the line through $0$ and $\exp(\theta)$ passes through $\exp(\theta \pm \pi/3)$ for both choices of sign. This construction has $1+3+2k = 2k+4$ points, so it works for all even $n \geq 4$. Proof that even $n$ cannot be center-free: Suppose that there is a balanced center-free set $\mathcal S$ with $2k$ points. It follows that for any $P \in \mathcal S$ and any $A \in \mathcal S$, there is at most one $B \in \mathcal S$ such that $P$ lies on the perpendicular bisector of $\overline{AB}$. In particular, if $P$ lies on at least $k$ perpendicular bisectors of segments $\overline{AB}$, then two such $\overline{AB}$ share a vertex by Pigeonhole. So $P$ lies on at most $k-1$ such bisectors. But now, double counting pairs $\left(\{A, B\}, P\right)$ where $P$ lies on the perpendicular bisector of $\overline{AB}$, we have ${2k \choose 2} \leq 2k(k-1)$ which is a clear contradiction.

a) take an equilateral triangle and keep adding new equilateral triangles extending outside. after adding 3 new pts one for each original side consider the larger equilateral triangle and repeat. this clearly works. b) only odd n work. for odd n a regular n-gon suffices. Now we count the pairs of (segment, center). by the balanced condition there always is a center for all segments and the center free condition implies that $AB$ and $AB'$ do not have the same center. Now this shows that some point is counted $\frac{n-1}{2}$ times and hence n is odd.

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OG! a) Consider a circle with a centre point $A$. Case1 : $n$ is odd. Consider points $X_1,X_2 \cdots X_{n-1}$, lying on the circle such that $AX_1X_2$, $AX_3X_4$........$AX_{n-2}X_{n-1}$ all are equilateral triangles. Case2 : $n$ is even. Consider points $X_1,X_2 \cdots X_{n-1}$, lying on the circle such that $AX_1X_2$, $AX_3X_4$........$AX_{n-2}X_{n-1}$ all are equilateral triangles and $AX_1X_n$ is also equilateral. It is easy to see that such points exists and indeed satisfy the condition b For any odd $n$, consider a regular $n$- sided regular polygon. It is easy to this is centre-free and balanced and for $n$ =even consider #9 th post

a) If $n$ is odd, just consider a regular $n$-gon. If $n$ is even, consider a rhombus created by connecting two congruent equilateral triangles together at a side, and then from some point $A,$ a vertice of both the two original equilateral triangles, add equilateral triangles (with vertices not overlapping) congruent to the original equilateral triangles and one vertice at $A$ until we have the required number of points. b) We claim that only odd $n$ work. For the construction, just consider a regular $n$-gon. For necessity, suppose for the sake of a contradiction that we have a balanced center-free set with $n=2k$ points for some positive integer $k.$ Note that we have $\binom{2k}{2}$ pairwise points each determining a perpendicular bisector, so by the Pigeonhole Principle there is a point $P$ lying on at least $k$ bisectors. In particular, we have $k$ distinct pairs of points $\{ A_1, B_1 \}, \cdots \{ A_{k}, B_{k} \}$ such that $PA_i=PB_i$ for each $1 \leq i \leq k.$ Thus by the Pigeonhole Principle some two pairs, say $\{ A_1, B_1 \}, \{A_2, B_2 \},$ must overlap. If WLOG $A_1=A_2, B_1 \neq B_2$ we have that $$PB_1=PA_1=PA_2=PB_2,$$so $P$ is the center of distinct points $A_1, B_1, B_2,$ a contradiction. QED

Same as the above- I'll just post motivaiton for (b). Once you have the construction for $n$ odd, we can notice that in a sense, each vertex is acting as a balancing point as much as it possibly can in a centre-free balanced set, i. e. the equality configuration, in a sense, is very sharp. But this is literally impossible for $n$ even, as one vertex is always left out, so we can't have a similar balanced centre-free set.