Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

Cute problem . First of all, let's observe that for any $i,j\in\overline{1,5}$, there is a separator through $A_i,A_j$. This is because there are three circles through $A_i,A_j$, and there is at most one circle containing the other two points and at most one leaving the other two points outside, so there is at least one which is a separator. This, in turn, means that there are at least $4$ separators because if there are $\le$, then these have at most $3\cdot 3=9$ chords with endpoints $A_i$, but there are $10$ segments $A_iA_j$, a contradiction. Another observation: through a point (let's call it $A_1$) there are $2$ or $3$ separators: after the inversion of pole $A_1$ the quadrilateral $\mathcal Q=A_2'A_3'A_4'A_5'$ (where, in general, $X'$ is the image of $X$) is either convex or concave. A separator is the preimage of a line $A_l'A_k'$ which separates the other two images of points, and there are two such lines (the diagonals) if $\mathcal Q$ is convex, and three (concurrent) if $\mathcal Q$ is concave. From this we have also obtained the fact that if through $A_1$ there are $3$ separators, then there is an $i\ne 1$ s.t. all the three separators also pass through $A_i$. Assume now that through each point there are $2$ separators. It's easy to get a contradiction based only on the fact that we can't find at least $4$ subsets of $\{A_i\}$ s.t. each point belongs to exactly two sets. This means that there is at least a pair (call it $(A_1,A_2)$) of points with $3$ separators each. Then $A_1A_2A_3,A_1A_2A_4,A_1A_2A_5$ are separators, and it's pretty clear that $A_3A_4A_5$ is also a separator. Any attempt to add another separator will cause one of $A_1,A_2$ to have more than $3$ separators passing through it, and this is a contradiction. Sorry if it appears long and boring..

OFFICIAL SOLUTIONS Solution 1: Let $\{A,B,C,D\}$ be the inverses of four of the points of the set through an inversion having as pole the fifth point. Notice that a a separator which passes through the pole is transformed into a straight line which passes through two of the points $A,B,C,D$ and separates the other two. Notice also that a separator which does not pass through the pole is transformed into a circle which passes through three of the points $A,B,C,D$ and contains the fourth point inside. Considering K - the convex hull of the set $\{A,B,C,D\}$, two cases may appear. - case 1: K = quadrilateral (for instance $ABCD$). In this case we have as separators the two diagonals of the quadrilateral and one circle from each pair $\{(ABC),(ABD)\},\{(CDA),(CDB)\}$ (the corresponding to the smaller angle from $\{\widehat{ACB},\widehat{ADB}\}$ and $\{\widehat{CAD},\widehat{CBD}\}$ respectively). - case 2: K = triangle (for instance $\triangle ABC$). In this case we have as separators, with obvious arguments, the straight lines $DA,DB,DC$ and the circle $(ABC)$. Solution 2: Consider coordinates and the points $A_{i}(x_{i},y_{i}),i=1,2,3,4,5$. Let $d_{ijkl}$ be the value of the determinant $|x_{p}^{2}+y_{p}^{2}\hspace{0.3cm}x_{p} \hspace{0.3cm} y_{p} \hspace{0.3cm} 1 \hspace{0.3cm} 1|_{P=i,j,k,l}$. The circle $(A_{1}A_{2}A_{3})$ is separator iff $d_{1234}$ and $d_{1235}$ have different signs. Let us look at ten pairs $(d_{1234},d_{1235})$ and $(d_{1243},d_{1245})$, etc. corresponding to the ten circles which pass through three of the five given points. Denoting by $a_{n}$ the number $d_{ijkl} (i<j<k<l,\{i,j,k,l\}=\{1,2,3,4,5\})$, the ten pairs are $(a_{5},a_{4}),(-a_{5},a_{3}),(-a_{4},-a_{3}),(a_{5},a_{2}), (a_{4},-a_{2}),(a_{3},a_{2}),(-a_{5},a_{1}), (-a_{4},-a_{1}),(-a_{3},a_{1}),(-a_{2},-a_{1}).$ We notice that the number of separators is equal to the number of pairs of terms with the same sign from the sequence $S = (a_{1},-a_{2},a_{3},-a_{4},a_{5})$. We also remark (using the determinant $|x_{p}^{2}+y_{p}^{2}\hspace{0.3cm}x_{p} \hspace{0.3cm} y_{p} \hspace{0.3cm} 1 \hspace{0.3cm} 1|_{P=1,2,3,4}$) that $a_{1}-a_{2}+a_{3}-a_{4}+a_{5}=0.$ This shows that $S$ cannot have all its terms of the same sign. We claim also that $S$ cannot have four terms of the same sign. Indeed, if four terms have the same sign then all the six circles passing through one point of the given five are separators. Taking this point as origin, the $Ox$ axis through an other of the given points and passing to polar coordinates we would get, for instance $ \begin{vmatrix} r_{1} \quad \cos a \quad \sin a\\ r_{2} \quad \cos b \quad \sin b\\ r_{3} \quad \cos c \quad \sin c\\ \end{vmatrix} > 0 \quad \begin{vmatrix} r_{1} \quad \cos a \quad \sin a\\ r_{2} \quad \cos b \quad \sin b\\ r_{4} \quad 1 \quad 0\\ \end{vmatrix} < 0 \\ \begin{vmatrix} r_{1} \quad \cos a \quad \sin a\\ r_{3} \quad \cos b \quad \sin b\\ r_{4} \quad 1 \quad 0\\ \end{vmatrix} > 0, \quad \textnormal{and} \quad \begin{vmatrix} r_{2} \quad \cos a \quad \sin a\\ r_{3} \quad \cos b \quad \sin b\\ r_{4} \quad 1 \quad 0 \end{vmatrix} < 0. \\ $

No inversion, but this is pretty long and boring. Case 1: Convex hull is a pentagon, ABCDE. Then, it's easy to see that AB, BC, CD, DE, EA are each part of exactly one separator triangle. Then, AC, BD, CE, DA, EB are each part of at least one. From here, we get 10 edges that are part of separator triangles, so we'll have at least 4 of them. We claim that the number of separator triangles containing AC is either 1 or 3. Consider the angles, ABC, ADC, AEC. If ABC+AEC, ABC+ADC < 180, and, WLOG, AEC < ADC, we have that the only separator triangle among triangles ABC, AEC, ADC is AEC. Similar stuff happens if ABC+AEC, ABC+ADC > 180. The only other case to consider is if, say ABC+AEC<180<ABC+ADC. Then, we see that all three triangles are separator triangles. So of AC, BD, CE, DA, EB, each is part of either one separator triangle or three. Call it super if it's part of 3. Note that given a super diagonal, the triangle directly above it, containing two of the sides of ABCDE, must be a separator triangle. We see that thus there can be at most 2 super diagonals, or else we have 6 sides of the pentagon used in a separator triangle, contradiction. Now, adding, we get at most 10+2(2)=14 sides of separator triangles, so at most [14/3]=4 separators. We know there are at least 4, so there are exactly 4. Case 2: Convex hull is a quadrilateral, ABCD, with E inside. As in case 1, we get exactly one separator triangle that contains each of AB, BC, CD, DA. Then, we have six other lengths that are part of either 1 or 3 separator triangles. First, consider what happens when AC (or BD) is super. Then, ACE, ACB, ACD are all separator triangles. The only thing left that can be a separator triangle is BED because all of the others contain one of AB, BC, CD, DA, but we see that their separator triangles have already been found. Now, we claim that BED is a separator as well. Note that AECD is a convex quadrilateral. WLOG, BEDA is a convex quadrilateral (that is, E and A are on opposite sides of BD). First, ACD is a separator, so its circumcircle contains E and not B. Thus, ADC+ABC<180. It follows that BAD+BCD>180. Now, BED+BAD>BCD+BAD>180, so the circumcircle of BED contains A. Clearly, it does not contain C, so BED is a separator. This gives us 4 in total. Now, assume AC and BD are not super. Then, the only diagonals that can be super are EA, EB, EC, ED. Say that x of them are super. Then, we have exactly 10+2x edges of separators. This has to be a multiple of 3, so x=1, 4. If x=1, we're done. We show that x=4 is impossible. That is, all of EA, EB, EC, ED are super. This gives us that EAB, EBC, ECD, EDA, EAC, EBD are super, and none ABC, BCD, CDA, DAB can be separators because the separator triangles for AB, BC, CD, DA have already been used up. But let AC and BD meet at P, and WLOG let E be inside PAB. Then, AEB>ACB,ADB, so one of ACB and ADB is a separator triangle, contradiction. Thus, x=1, and we have 4 separators. Case 3: Convex hull is a triangle, ABC, with D and E inside. WLOG, let BCDE be a convex quadrilateral. DE only forms a convex quadrilateral with one side of ABC, because, if say, BCDE and ACDE are convex quadrilaterals, we have that angle CDE in both quad CDEB and quad CDEA are <180, which is impossible, because these two angles CDE have to sum to 360. Now, for each side AB, BC, CA, there is exactly one separator triangle, and note that ABC is not a separator triangle. Now, the only three triangles left to consider are ADE, BDE, CDE, and we wish to show that exactly one of these is a separator triangle. Note that DAE+DBE<BAC+ABC=180, and similarly DAE+DCE<180. Thus, ADE is not a separator. Furthermore, the circumcircles of BDE and CDE do not contain A. Now, for X={B,C}, exactly one of the circumcircles of XDE contains the other of B and C, and it doesn't contain A, so we have the fourth separator.

For the sake of completeness, I'll mention two solutions to the generalized problem with a set $S$ of $2n+1$ points ($n\ge1$). The first is Federico Ardila's smoothing argument, which someone linked to in another thread. The second (which is not as motivated, but perhaps more beautiful) is based on ideas from a paper on the connection between balancing lines and halving triangles (in particular, Theorem 2.2 from here, which is essentially a subtler application of the invariant used in IMO 2011.2). Here we use (a special case of) the inverted form of Theorem 2.2, i.e. the fact that for a fixed point $P\in S$ and circle $C$ passing through $P$ but no other points of $S$, there are exactly $\min(m,2n-m)$ ordered pairs of points $(Q,R)\in S^2$ with $(PQR)$ a separator of $S$, $Q$ inside $C$, and $R$ outside $C$, where $m$ is the number of points in $S$ lying inside $C$. Take a point $O$ such that $OP\ne OQ$ for any two distinct points $P,Q\in S$, and label the points $P_i$ ($1\le i\le 2n+1$) of $S$ so that $OP_1<\cdots OP_{2n+1}$ (this may be more natural with stereographic projection, interpreting this as "decreasing $z$-coordinates" instead); let $C_i$ denote the circle centered at $O$ passing through $P_i$. Then we can identify each separator $(P_i P_j P_k)$ (where $i<j<k$) with $C_j$, the unique circle $C_l$ passing through one of $P_i,P_j,P_k$ and "separating" the other two. Furthermore, $C_i$ contains exactly $i-1$ points in its interior for every $i$, so by the previous paragraph, there must be exactly \[\sum_{i=1}^{2n+1}\min(i-1,2n-i+1)=n^2\]separators, as desired. And as noted in Ardila's paper, we can just as easily count the total number of separators of type $(a,b)$ or $(b,a)$ (where there are either exactly $a$ inside and $b$ outside or $a$ outside and $b$ inside) whenever $a+b=2n-2$ (the original problem is just $(n-1,n-1)$). But the general inverted form of Theorem 2.2 still applies, except we have to sum up $2\min(i-1,2n-i+1,a+1,b+1)$ instead of $\min(i,2n-i)$, where the extra factor of 2 accounts for our inability (with this method) to distinguish between types $(a,b)$ and $(b,a)$ whenever $a\ne b$.

orl wrote: OFFICIAL SOLUTIONS Solution 1: Let $\{A,B,C,D\}$ be the inverses of four of the points of the set through an inversion having as pole the fifth point. Notice that a a separator which passes through the pole is transformed into a straight line which passes through two of the points $A,B,C,D$ and separates the other two. Notice also that a separator which does not pass through the pole is transformed into a circle which passes through three of the points $A,B,C,D$ and contains the fourth point inside. Considering K - the convex hull of the set $\{A,B,C,D\}$, two cases may appear. Why is it true that a separator which passes through the pole is transformed into a straight line which passes through two of the points $A,B,C,D$ and separates the other two and that a separator which does not pass through the pole is transformed into a circle which passes through three of the points $A,B,C,D$ and contains the fourth point inside? Thanks a lot!

Anyone? I would really appreciate the help!

orl wrote: OFFICIAL SOLUTIONS Solution 1: Let $\{A,B,C,D\}$ be the inverses of four of the points of the set through an inversion having as pole the fifth point. Notice that a a separator which passes through the pole is transformed into a straight line which passes through two of the points $A,B,C,D$ and separates the other two. Notice also that a separator which does not pass through the pole is transformed into a circle which passes through three of the points $A,B,C,D$ and contains the fourth point inside. Considering K - the convex hull of the set $\{A,B,C,D\}$, two cases may appear. - case 1: K = quadrilateral (for instance $ABCD$). In this case we have as separators the two diagonals of the quadrilateral and one circle from each pair $\{(ABC),(ABD)\},\{(CDA),(CDB)\}$ (the corresponding to the smaller angle from $\{\widehat{ACB},\widehat{ADB}\}$ and $\{\widehat{CAD},\widehat{CBD}\}$ respectively). - case 2: K = triangle (for instance $\triangle ABC$). In this case we have as separators, with obvious arguments, the straight lines $DA,DB,DC$ and the circle $(ABC)$. Solution 2: Consider coordinates and the points $A_{i}(x_{i},y_{i}),i=1,2,3,4,5$. Let $d_{ijkl}$ be the value of the determinant $|x_{p}^{2}+y_{p}^{2}\hspace{0.3cm}x_{p} \hspace{0.3cm} y_{p} \hspace{0.3cm} 1 \hspace{0.3cm} 1|_{P=i,j,k,l}$. The circle $(A_{1}A_{2}A_{3})$ is separator iff $d_{1234}$ and $d_{1235}$ have different signs. Let us look at ten pairs $(d_{1234},d_{1235})$ and $(d_{1243},d_{1245})$, etc. corresponding to the ten circles which pass through three of the five given points. Denoting by $a_{n}$ the number $d_{ijkl} (i<j<k<l,\{i,j,k,l\}=\{1,2,3,4,5\})$, the ten pairs are $(a_{5},a_{4}),(-a_{5},a_{3}),(-a_{4},-a_{3}),(a_{5},a_{2}), (a_{4},-a_{2}),(a_{3},a_{2}),(-a_{5},a_{1}), (-a_{4},-a_{1}),(-a_{3},a_{1}),(-a_{2},-a_{1}).$ We notice that the number of separators is equal to the number of pairs of terms with the same sign from the sequence $S = (a_{1},-a_{2},a_{3},-a_{4},a_{5})$. We also remark (using the determinant $|x_{p}^{2}+y_{p}^{2}\hspace{0.3cm}x_{p} \hspace{0.3cm} y_{p} \hspace{0.3cm} 1 \hspace{0.3cm} 1|_{P=1,2,3,4}$) that $a_{1}-a_{2}+a_{3}-a_{4}+a_{5}=0.$ This shows that $S$ cannot have all its terms of the same sign. We claim also that $S$ cannot have four terms of the same sign. Indeed, if four terms have the same sign then all the six circles passing through one point of the given five are separators. Taking this point as origin, the $Ox$ axis through an other of the given points and passing to polar coordinates we would get, for instance $ \begin{vmatrix} r_{1} \quad \cos a \quad \sin a\\ r_{2} \quad \cos b \quad \sin b\\ r_{3} \quad \cos c \quad \sin c\\ \end{vmatrix} > 0 \quad \begin{vmatrix} r_{1} \quad \cos a \quad \sin a\\ r_{2} \quad \cos b \quad \sin b\\ r_{4} \quad 1 \quad 0\\ \end{vmatrix} < 0 \\ \begin{vmatrix} r_{1} \quad \cos a \quad \sin a\\ r_{3} \quad \cos b \quad \sin b\\ r_{4} \quad 1 \quad 0\\ \end{vmatrix} > 0, \quad \textnormal{and} \quad \begin{vmatrix} r_{2} \quad \cos a \quad \sin a\\ r_{3} \quad \cos b \quad \sin b\\ r_{4} \quad 1 \quad 0 \end{vmatrix} < 0. \\ $ Can someone help me understand the first solution, case II? How do we know that $(A'B'C')$ does not contain the center of inversion $E$ in its interior? Because that would imply that $(ABC)$ contains both $D$ and $E$, so it is not a separator, and the solution would have a hole.

Let $V,W,X,Y,Z$ be the five points, and consider the separators that pass through $V$. Consider an inversion $\Psi$ with centre $V$ and some arbitrary positive radius. A separator gets mapped to a line joining some two of the remaining four points such that the other two of these four points lie on opposite sides of this line. If $W',X',Y',Z'$ are the images of $W,X,Y,Z$ respectively under $\Psi$, then we have two cases. The first case is when $W',X',Y',Z'$ form a convex quadrilateral, in which case the diagonals are the only separators. The second case is one of the four points, say Z', lies inside the triangle formed by the other three points, which is $\Delta X'Y'W'$ in this case. Here, $X'Z' , Y'Z' , W'Z'$ are the separators. Thus we see that there are either two or three separators that pass through a specified point. Furthermore, if a point $V$ has three separators passing through it, then there a unique point, different from $V$, through which all these separators pass, which was $Z$ in the previous example. This also means that there are three separators through $Z$, and these pass through $V$ too. This means there are an even number of points that lie on three separators. If $2i$ points lie on three separators and $5-2i$ points lie on exactly two separators, where $0 \leq i \leq 2$, then the total number of separators , $S$, can be found as $$3S = 3(2i) + 2(5-2i) = 10+2i.$$The only $i$ for which $3 \mid 10+2i$ is $i=1$, and thus $S=4$. Therefore there are $4$ separators.

Let $P_j$, $j=1,2,3,4,5$ be the five points in the plane such that no three of them are colinear and no four of them are concyclic. Then there exists at least one circle through three of then which contains the other two points on its interior. To see this, take the convex hall and pick an edge, say $P_1P_2$. Choose $P_j$ so that the $\angle P_1P_jP_2$ becomes minimal among $j=3,4,5$. WLOG let the circle through $P_1P_2P_3$ encloses the other two. Let $z_j$, $j=1,2,3,4,5$ be the complex numbers which corresponds to those given five points. Let $\mu$ be the Mobius transform which maps $z_j$ to $w_j$ where $w_1=0$, $w_2=1$, $w_3=\infty$. Note that being separator is preserved under $\mu$. Also note that $w_4$, $w_5$ are in the upper half or lower half plane simultaneously. WLOG assume that they are on the upper half plane. Let $W_j$ be the point corresponding to $w_j$ where $j=1,2,3,4,5$. From now on, by the circle, we mean generalized circle which includes lines. Note that circle through $W_1(0), W_2(1)$ never encloses $W_3(\infty)$, there exists exactly one separator through $W_1,W_2$; choose $W_j$ which makes angle $W_1W_jW_2$ smallest among $j=4,5$. Similarly, there exists exactly one separator through $W_2W_3$; choose the line $W_2W_j$ which separates $W_1$, $W_k$ where $\{ j,k \} = \{4,5\}$. Similarly, there exists exactly one separator through $W_1, W_3$. Therefore it remains to prove that exactly one separator exists through $W_4$, $W_5$. If the line $W_4W_5$ separates $W_1$ and $W_2$, then the circle through $W_4W_5W_3$ separates; that is the line $W_4W_5$; no other circles separates, i.e. $W_4W_5W_j$, $j=1,2$ does not separates. Finally, if the line $W_4W_5$ does not separates $W_1$, $W_2$, the circle through $W_jW_4W_5$ with angle $W_4W_jW_5$ minimum among $j=1,2$ separates which completes the proof.

Consider a segment $XY$ between two of these points, $X$ and $Y$. We claim that there is either 1 or 3 separators passing through $X$ and $Y$. Denote the other points $A$, $B$, and $C$. Let $a$, $b$, $c$ denote the angle measures of $\angle XAY$, $\angle XBY$, $\angle XCY$, respectively. Consider the case where $A$, $B$, and $C$ lie on the same side of line $XY$. In this case, no two of $a, b, c$ can be equal, since no four points are concyclic. So, WLOG, let $a>b>c$. The circle $(AXY)$ does not contain $B$ or $C$, so it is not a seperator. The circle $(BXY)$ contains $A$ but not $C$, so it is a separator. The circle $(CXY)$ contains both $A$ and $B$, so it is not a separator. Thus, there is exactly one separator passing through $X$ and $Y$. Consider the case where $A$ and $B$ lie on one side of $XY$, and $C$ lies on the other side. WLOG let $a>b$. Then there are 3 cases. Case 1: $180-c>a>b$: The circle $(AXY)$ does not contain $B$ or $C$, so it is not a separator. The circle $(BXY)$ contains $A$ but not $C$, so it is a separator. The circle $(CXY)$ does not contain $A$ or $B$, so it is not a separator. Thus, exactly 1 separator passes through $X$ and $Y$. Case 2: $a>180-c>b$: The circle $(AXY)$ contains $C$ but not $B$, so it is a separator. The circle $(BXY)$ contains $A$ but not $C$, so it is a separator. The circle $(CXY)$ contains $A$ but not $B$, so it is a separator. Thus, exactly 3 separators pass through $X$ and $Y$. Case 3: $a>b>180-c$: The circle $(AXY)$ contains $C$ but not $B$, so it is a separator. The circle $(BXY)$ contains $A$ and $C$, so it is not a separator. The circle $(CXY)$ contains $A$ and $B$, so it is not a separator. Thus, exactly 1 separator passes through $X$ and $Y$. There are 10 segments that contain 2 points from the set. Let a segment's "separator score" be the number of separators that pass through both vertices. Each segment's separator score is 1 or 3. When we add all the separator scores, we will get $N$, which is an even number greater than 10. Note that $\frac{N}{3}$ is the number of separators. So, the number of separators is even and at least $4$. We will now show the number of separators is at most $4$, which completes the proof. We proceed by casework on the number of segments on the convex hull of the set. Note that a segment on the convex hull has a separator score of 1. If there are 5 segments on the convex hull, then $N$ is at most $5\cdot 1+5\cdot 3=15$, so there are at most $5$ separators. However, the number of separators is even, so there are at most $4$ separators. If there are 4 segments on the convex hull, then there is 1 point not on the convex hull. Let $A, B, C, D$ be on the convex hull and $X$ not be. Denote by $a$ the angle measure of $\angle BAC$ and define $b, c, d$ similarly. Since $ABCD$ is not cyclic, WLOG let $a+c<180$. $X$ is inside either $\triangle ABC$ or $\triangle ADC$. WLOG let $X$ be inside $\triangle ABC$. Then, $(ABC)$ is a separator because $X$ is inside it and $D$ is not. Because every segment of the convex hull has a separator score of 1, and this separator has 2 segments of the convex hull, there are at most 3 separators with any segment from the convex hull. After removing the segments of the convex hull, there are only 2 triangles left. So, there are at most 2 separators that don't include segments from the convex hull. So, there are at most 5 separators total, and since the number of separators is even, there are at most 4. If there are 3 segments on the convex hull, let $A, B, C$ be on the convex hull, and $X$ and $Y$ are not. Denote $a=\angle BAC$ and define $b, c$ similarly. We claim segment $XY$ has a separator score of $1$. Assume WLOG that $A$ is on one side of line $XY$ and $B$ and $C$ are on the other. We know $\angle XAY+\angle XBY+\angle XCY< a+b+c=180$. So, $\angle XBY+angle XCY<180-\angle XAY$. So, WLOG, $\angle XCY<\angle XBY<180-\angle XAY$. This corresponds to "Case 1" earlier in the proof, so $XY$ has a separator score of 1. Every triangle without a segment from the convex hull includes $XY$. So, there is only 1 triangle that does not include a segment from the convex hull. There are at most 3 separators that include segments from the convex hull, since there are 3 segments on the convex hull. So, there are at most 4 separators. So, in any case, there are at least 4 and at most 4 separators, so there are exactly 4 separators.

Let the points be $A_1$, $A_2$, $A_3$, $A_4$, and $A_5$. Invert about $A_1$ to get the images $A_i\to B_i$. Then, $A_1A_iA_j$ is a separator if and only if $B_iB_j$ separates the other two $B$s, which is equivalent to it not being on the convex hull of $B_2$, $B_3$, $B_4$, $B_5$. $~$ If the $B_2B_3B_4B_5$ is convex then there's two separators about $A_1$, and three otherwise. Clearly, any $A_iA_jA_k$ is a separator with $i,j,k\ge 2$ if and only if $(B_iB_jB_k)$ contains the other point. In this case, if it was convex then the two with angle at least $180^\circ$ are the separators and otherwise it is just the convex hull. Thus, there's always four separators.