My solution: let ri be the radis of Ci let di be the distance from Oi to l d=the distance of the midpoint of AB,CD then first prove for a point X on l ,if \[\sqrt{XA*XB}-\sqrt{XC*XD}\]=t fixed then X fixed LET X1,2 be the meet point of the common e-tanget of C1,C4;C2,C3 so we have \[\sqrt{X1A*X1B}-\sqrt{X1C*X1D}\]=\[\sqrt{d^2+(d4-d1)^2-(r1-r4)^2}\] so X1=X2 iff \[ -(r4-r1)^2+(d4-d1)^2=-(r3-r2)^2+(d3-d2)^2\] BUT \[ r1^2-r3^2=d1^2-d3^2 r2^2-r4^2=d2^2-d4^2\] only to prove \[ d1d4-r1r4=d2d3-r2r3\] first notice the both side have the same plus-minus only to prove \[ d1^2d4^2+d2^2d3^2-2d1d2d3d4=r1^2r4^2+r2^2r3^2-2r1r2r3r4\] ...(*) and we have \[ (r1+r3)^2=(d1-d3)^2+d^2\] \[ (r2+r4)^2=(d2-d4)^2+d^2\] so \[ r1r3+d1d3=r2r3+d2d3\] then \[ r1^2r3^2+r2^2r4^2-2r1r2r3r4=d1^2d3^2+d2^2d4^2-2d1d2d3d4\] ...(1) and \[ (r1^2-r2^2)(r3^2-r4^2)=(d1^2-d2^2)(d3^2-d4^2) \] ...(2) from1,2 we get *

This solution is so short that it is probably wrong but I am not able to find a flaw. If you can find one, please notify me. Let the common tangent to $C_1, C_4$ meet $ABCD$ at $P$. By an inversion with pole $P$ and power $PA\cdot PD$, $C_4$ maps to $C_1$ and due to tangency, $C_3$ maps to $C_1$ and so we have that the common external tangent for $C_2, C_3$ to pass through $P$. Edit: Wrong solution as pointed out.

You are wrong C1not map toC4

Dear Mathlinkers, 1. through A and D passes a circle which goes through the points of contact of the common exterior tangent to C1 and C4 (on the side of l) 2. through B and C passes a circle which goes through the points of contact of the common exterior tangent to C2 and C3 (on the side of l) 3. according to the three chords theorem (of Monge) we are done... Sincerely Jean-Louis

Could you explain in detail? And Where do you use C1C3tangent to each other

Dear Mathlinkers, its work also with none tangent circles... Sincerely Jean-Louis

duanby wrote: My solution: let ri be the radis of Ci let di be the distance from Oi to l d=the distance of the midpoint of AB,CD then first prove for a point X on l ,if \[\sqrt{XA*XB}-\sqrt{XC*XD}\]=t fixed then X fixed LET X1,2 be the meet point of the common e-tanget of C1,C4;C2,C3 so we have \[\sqrt{X1A*X1B}-\sqrt{X1C*X1D}\]=\[\sqrt{d^2+(d4-d1)^2-(r1-r4)^2}\] so X1=X2 iff \[ -(r4-r1)^2+(d4-d1)^2=-(r3-r2)^2+(d3-d2)^2\] BUT \[ r1^2-r3^2=d1^2-d3^2 r2^2-r4^2=d2^2-d4^2\] only to prove \[ d1d4-r1r4=d2d3-r2r3\] first notice the both side have the same plus-minus only to prove \[ d1^2d4^2+d2^2d3^2-2d1d2d3d4=r1^2r4^2+r2^2r3^2-2r1r2r3r4\] ...(*) and we have \[ (r1+r3)^2=(d1-d3)^2+d^2\] \[ (r2+r4)^2=(d2-d4)^2+d^2\] so \[ r1r3+d1d3=r2r3+d2d3\] then \[ r1^2r3^2+r2^2r4^2-2r1r2r3r4=d1^2d3^2+d2^2d4^2-2d1d2d3d4\] ...(1) and \[ (r1^2-r2^2)(r3^2-r4^2)=(d1^2-d2^2)(d3^2-d4^2) \] ...(2) from1,2 we get * how do you prove the first part \[\sqrt{XA*XB}-\sqrt{XC*XD}\]=t is fixed means $X$ is fixed?????

Oh,I'm sorry ,YOU ARE RIGHT,that it's not ture,X can be one of the meetpoint of the external tangent of circle C1,C4 and I may expain more, IF not ture there is two point P1,P2 satified then P1,P2 is the meet point of two external tanget of circle C1,C4,and C2,C3 the same.then only need to prove the external Homothetic Center of C1,C4;C2,C3 on the same side of l that is (d1r4-d4r1)*(r4-r1)*(d2r3-d3r2)*(r3-r2)>=0 that is (d1/r1-d4/r4)*(r4-r1)*(d2/r2-d3/r3)*(r3-r2)>=0 whitch is not hard to prove and i hope it's OK now

i still don't get why could't all of the external common tangents of C1,C4 or C3,C2 meet $l$ at $4$ different points.

leader wrote: i still don't get why could't all of the external common tangents of C1,C4 or C3,C2 meet $l$ at $4$ different points.

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Because the pointX satisfy \[sqrt{XA*XB}-\sqrt{XC*XD}=t or -t\] is at most 2

that's what confuses me. yunxiu i know the problem is true what i meant was. if $X\in l$ andv$X,D$ are on opposite sides of $A$ then there can be 2 points $X_{1},X_{2}$ that give the same value for that function $f(X)=\sqrt {XA*XB} - \sqrt {XC*XD}$. i am sorry if this are stupid questions(if this is trivial an i'm not seeing it) but i really want to understand this solution because when i started solving the problem i got to that function in about 10 mins and i knew that if i could prove that it is a bijection when $X$ is on the side of $A$ where $D$ is not i would finish the problem quickly. But i couldn't after a few hours. I mean the function becomes $\sqrt {x(x+a)} -\sqrt {(x+c)*(x+d)}$ for $x=XA>0$ and $a=AB<b=AC<c=AD$ which can be proven that it is not a bijection.(there are 2 numbers which give the same value $t$ of $f$) and on the other side there can be $2$ numbers that give that value $t$) as well a total of $4$ for $t$ or $-t$ if i'm not missing something.

It can at most 2 solution For t or-t

how do you prove that?

Only square it twice Or in geometry \[\sqrt{XA*XB}-\sqrt{XC*XD}=t or-t \] X must on the external tangent of C1C4

thanks finally i got how it works with geometry. for a fixed difference in tangents if the tangent to C4 is of length $p$ it belongs to a circle concentric with C4(call it k). if their difference is $t$ then then it's easy to prove that all the possibilities for such points are some of the points that belong to both the external common tangents of C1,C4 and k.

Here's a possible approach, seeing that the only synthetic solution appears to be wrong (drawing the diagram in Geogebra yields that the 4 points are not concyclic). We use the half-plane model for hyperbolic geometry (by using $l$ as the line at infinity). Thus $C_1$ and $C_2$ are hypercycles (to the h-line through $A$ and $B$), and similarly $C_3, C_4$ are hypercycles (to the h-line through $C,D$). But $C_1, C_3$ are tangent to each other, and so are $C_2, C_4$. Thus we have that the distance between these two hypercycles are the same[!] Now we let $l_1$ be the tangent to $C_1,C_4$, and $l_2$ be the tangent to $C_2,C_3$. But due to our previous points, $l_1,l_2$ are hypercycles to the same vertical h-line (since the distance between them are equal at two points). Thus we conclude that they must intersect each other on $l$. Can someone more well-versed in hyperbolic geometry validate (or otherwise correct) this? Especially the [!] bit, which I am rather unsure of. This is merely a guess that I pieced together...

But $C_1, C_3$ are tangent to each other, and so are $C_2, C_4$. Thus we have that the distance between these two hypercycles are the same[!] explain, please

DVDthe1st wrote: Here's a possible approach, seeing that the only synthetic solution appears to be wrong (drawing the diagram in Geogebra yields that the 4 points are not concyclic). We use the half-plane model for hyperbolic geometry (by using $l$ as the line at infinity). Thus $C_1$ and $C_2$ are hypercycles (to the h-line through $A$ and $B$), and similarly $C_3, C_4$ are hypercycles (to the h-line through $C,D$). But $C_1, C_3$ are tangent to each other, and so are $C_2, C_4$. Thus we have that the distance between these two hypercycles are the same[!] Now we let $l_1$ be the tangent to $C_1,C_4$, and $l_2$ be the tangent to $C_2,C_3$. But due to our previous points, $l_1,l_2$ are hypercycles to the same vertical h-line (since the distance between them are equal at two points). Thus we conclude that they must intersect each other on $l$. Can someone more well-versed in hyperbolic geometry validate (or otherwise correct) this? Especially the [!] bit, which I am rather unsure of. This is merely a guess that I pieced together... uuuh,,, I understood,,, it is vary interesting solution... applaud...

I've attached the official solution(s) of this problem.

Attachments:
Iran MO - Team Selection 2013 - Problem 6.pdf (332kb)

Problem. Points $A, B, C$ and $D$ lie on line $l$ in this order. Two circular arcs $\omega_1$ and $\omega_2$, which both lie on one side of line $l$, pass through points $A$ and $B$ and two circular arcs $\omega_3$ and $\omega_4$ pass through points $C$ and $D$ such that $C_1$ is tangent to $C_3$ and $C_2$ is tangent to $C_4$. Prove that the common external tangent of $\omega_2$ and $\omega_3$ and the common external tangent of $\omega_1$ and $\omega_4$ meet each other on line $l$. Solution 1. Lemma 1. Given two arcs $\gamma_1$ and $\gamma_2$ intersect at $A$ and $C$ such that they lie on the same half plane wrt $AC$. Let $C_1, C_2$ be two circles that touch $\gamma_1$ and $\gamma_2$. Then the exsimilicenter of $C_1$ and $C_2$ lie on $AC$. We can easily prove this lemma by using Monge D'Alembert theorem and some facts about radical axis. Lemma 2. Two arcs $(O_1)$ and $(O_2)$ intersect r at A and B such that they lie on the same half plane wrt AB. Let C,D be two arbitrary points on AB and outside the segment AB. Prove that 4 tangents from C,D to that arcs form a circumscribed quadrilateral. This lemma is also trivial so I leave to AoPSers. (See also: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=573502) Back to our problem. Let $(O_1), (O_2), (O_3), (O_4)$ be the circles wrt $\omega_1, \omega_2,\omega_3,\omega_4$ respectively; $P,Q$ be the tangencies of $(O_1)$ and $(O_3)$, $(O_2)$ and $(O_4)$, respectively. Let the in-common tangent of $(O_1)$ and $(O_3)$ which passes through P intersects $d$ at $X.$ We have $XP^2=XA.XB=XC.XD$ therefore $X$ lies on the radical axis of $(O_2)$ and $(O_4)$, in other words, $XQ$ is the in-common tangent of $(O_2)$ and $(O_4).$ Assume the ex-common tangent $ME$ of $\omega_2$ and $\omega_3$ intersects $d$ at $Y$. Let $YN$ be the tangent of $\omega_1$, $YF$ be the tangent $\omega_2.$ Applying lemma 2 for two circles $(O_1)$ and $(O_2)$ with two points $X,Y$ we conclude that $YM,YN,XP,XQ$ form a circumscribed quadrilateral with its incircle $(I).$ Applying lemma 2 again for two circles $(O_3)$ and $(O_4)$ with two points $X,Y$ we have $YF$ touches the incircle of triangle formed by 3 lines $XP,XQ,YM$ or $(I)$. Thus $YN, YF$ are both tangent to $(I)$, this means $Y,N,F$ are collinear. We are done. Solution 2. The tangent $ME$ of $\omega_2$ and $\omega_3$ cuts $l$ at $Y$. Construct two tangents $YN, YF$ of $\omega_2$, $\omega_4$. We need to prove that $Y,N,F$ are collinear. Since $XP=XQ, YM=YN, YE=YF$, there exists a circle $C_1$ that is tangent to $(O_1)$ and $(O_3)$ at $P$, tangent to $(O_2)$ and $(O_4)$ at $Q$; a circle $C_2$ that is tangent to $(O_2)$ at $M$ and $(O_1)$ at $N$, a circle $C_3$ that is tangent to $(O_3)$ at $E$ and $(O_4)$ at $F.$ According to lemma 1, $MQ$ cuts $NP$ at the exsimilicenter $G$ of $C_1$ and $C_2$ and $G\in l$, $EP$ cuts $FQ$ at the exsimilicenter $H$ of $C_1$ and $C_3$ and $H\in l.$ Applying Monge-D'Alembert theorem for 3 circles $C_1, C_2, C_3$ we conclude that $ME$ intersects $d$ at the exsimilicenter of $C_2$ and $C_3$, or $Y$ is the exsimilicenter of $C_2$ and $C_3$. Therefore $Y,N,F$ are collinear. We are done. Solution 3. Lemma 3. (Apollonius circles chain). Given 3 circles $(O_1), (O_2), (O_3)$ in the plane. Let $C_{12}$ be the circle tangent to $(O_1)$ and $(O_2)$, $C_{23}$ be the circle tangent to $C_{12}$, $(O_2)$, $(O_3)$; $C_{34}$ be the circle tangent to $C_{23}$, $(O_3)$, $(O_1)$; $C_{45}$ be the circle tangent to $C_{34}$, $(O_1)$, $(O_2)$; $C_{56}$ be the circle tangent to $C_{45}$, $(O_2)$, $(O_3)$, $C_{61}$ be the circle tangent to $C_5$, $(O_3)$, $(O_1)$. Then $C_{12}, C_{23},...,C_{61}$ is a closed chain, in other word $C_{61}$ is tangent to $C_{12}$. I don't have time so I will not post the solution of this lemma here. Back to our problem. The common tangent $ME$ of $(O_2)$ and $(O_3)$ intersect $l$ at $Y$. Construct the tangent $YN$ of $\omega_1$, a circle $C_3$ tangent to $YM,YN$ at $E,F'$, respectively. The circles $C_1$ and $C_2$ are defined the same as solution 2. Applying lemma 3 for the circle $C_1$ and two lines $YM$, $YN$ we have: $(O_2)$ is tangent to $C_1, YM$. $C_2$ is tangent to $(O_2), YM, YN.$ $(O_1)$ is tangent to $C_2, YN, C_1.$ $(O_3)$ is tangent to $(O_1), C_1, YM.$ $C_3$ is tangent to $(O_3), YM, YN.$ $(O'_4)$ is tangent to $C_3, YN, C_1.$ Since $(O_2), C_2, (O_1), (O_3), C_3, (O'_4)$ is a closed chain then $(O'_4)$ is tangent to $(O_2), C_1, YN.$ This means $(O'_4)$ is tangent to $C_1$ at $P$ and tangent to $YN$ at $F'.$ On the other hand, the tangent through $Q$ of $(O_2)$ intersect the tangent through $P$ of $(O_1)$ at $X$ and $XP=XQ$ then $X$ lies on the radical axis of $(O'_4)$ and $(O_3).$ But $YE=YF'$ hence $Y$ also lies on the radical axis of $(O'_4)$ and $(O_3)$. Therefore $XY$ is the radical axis of $(O'_4)$ and $(O_3)$ or $(O'_4)$ passes through $C,D.$ Then $(O'_4)\equiv (O_4)$. We are done.

Attachments:
Iran2.pdf (20kb)
Iran3.pdf (20kb)
Iran5.pdf (19kb)

Generalization. Given $2n$ points $A_1,B_1,A_2,B_2,...,A_n,B_n$ $(n\geq 2)$ in that order lie on the line $l$. Two arcs $\omega_k$ and $\gamma_k$ are constructed base on $A_kB_k$ $(k=\overline{1,n})$ such that $2n$ they lie on the same half plane wrt $l$ and $\omega_k$ is tangent to $\omega_{k-1}$ ; $\omega_{k+1}$, $\gamma_k$ is tangent to $\gamma_{k-1}$ and $\gamma_{k+1}$. then the intersection of the common external tangents of $\omega_1$ and $\omega_n$, $\gamma_1$ and $\gamma_n$ lies on $l$. Solution. Lemma 4. Given 6 points $A_1,B_1,A_2,B_2,A_3,B_3$ lie on the line $l$ such that the segments $A_1B_1, A_2B_2, A_3B_3$ are separated. Two circles $\omega_i$ and $\gamma_i$ are constructed base on $A_iB_i$ $(i=\overline{1,3})$ such that the lie on the same half plane wrt $l.$ Let $d_{ij}$ be the ex-common tangent of $\omega_i$ and $\omega_j$, $k_{ij}$ be the ex-common tangent of $\gamma_i$ and $\gamma_j$. Assume that the intersections of $d_{12}$ and $k_{12}$, $d_{23}$ and $k_{23}$ are on $l$. Then the intersection of $d_{13}$ and $k_{13}$ are also on $l$. Solution for lemma 4. Let $X,Y$ be the intersections of $d_{12}$ and $k_{12}$, $d_{23}$ and $k_{23}$; $Z$ be the intersection of $d_{13}$ and $l$. Denote $k'_{13}$ the tangent through $Z$ of $\gamma_1.$ Applying lemma 2 for two circles $\omega_1, \gamma_1$ and two points $X,Z$ then there exists a circle $C_1$ inscribes the quadrilateral formed by the intersections of $d_{12}, k_{12}, d_{13}, k'_{13}$. Similarly, there exists a circle $C_2$ inscribes the quadrilateral formed by the intersections of $d_{12}, k_{12}, d_{23}, k_{23}$. Let $C_3$ be the incircle of the triangle formed by $d_{13}, d_{23}, k_{23}$. Applying Monge-D'Alembert theorem for 3 circles $C_1, C_2, C_3$ we have $X$ is the exsimilicenter of $C_1$ and $C_2$; $Y$ is the exsimilicenter of $C_2$ and $C_3$ hence the exsimilicenter of $C_1$ and $C_3$ lies on $XY$. Because $d_{13}$ is the common tangent of $C_1$ and $C_3$ then $d_{13}$ cuts $l$ at the exsimilicenter $Z$ of $C_1$ and $C_3$. But $k'_{13}$ is tangent to $C_1$ then $k'_{13}$ is tangent to $C_3.$ Let $k''_{13}$ be the tangent through $Z$ of $\gamma_3$, applying lemma 2 again for 2 circles $\omega_3$ and $\gamma_3$ with 2 points $Y,Z$ we deduce that $C_3$ is tangent to $k''_{13}$. Therefore $k'_{13}\equiv k''_{13}\equiv k_{13}$. We are done. Back to our problem. Define $d_{ij}$ and $k_{ij}$ the same as lemma 4. Let $X_{ij}$ be the intersection of $d_{ij}$ and $k_{ij}.$ According to P6 Iran TST 2013 we have $X_{12}$ and $X_{23}$ lie on $l$. Applying lemma 4 we conclude that $X_{13}$ lies on $l$. Moreover, $X_{34}$ lies on $l$ then applying lemma 4 again we have $X_{14}$ lies on $l$. Similarly, $X_{1n}$ lies on $l$.

Attachments:
Iran6.pdf (12kb)
Iran7.pdf (15kb)

Hello, Nice proofs, my idea how to get another solution to Generalization (sorry if it's already posted). We only use livetolove212 wrote: Lemma 1. Given two arcs $\gamma_1$ and $\gamma_2$ intersect at $A$ and $C$ such that they lie on the same half plane wrt $AC$. Let $C_1, C_2$ be two circles that touch $\gamma_1$ and $\gamma_2$. Then the exsimilicenter of $C_1$ and $C_2$ lie on $AC$. We can easily prove this lemma by using Monge D'Alembert theorem and some facts about radical axis. and from Monge D'Alembert theorem we get that: Lemma 1': Let for line $l$ given circles $C_i, i=1,...,4$, such that $C_1, C_2$ intersects on $l$, $C_3, C_4$ intersects on $l$, let circle $\omega_1$ is tangent to $C_1, C_2$ and circle $\omega_2$ is tangent to $C_3C_4$, then if common external tangent of $\omega_1$ and $\omega_2$ is on $l$, then for every other circles $\omega_1'$, $\omega_2'$ were $\omega_1'$ is tangent to $C_1, C_2$ and $\omega_2'$ is tangent to $C_3C_4$ common external tangent of $\omega_1'$ and $\omega_2'$ is on $l$. From Lemma 1' we can move from pare $C_i, C_{i+1}$, to pare $C_{i+1}, C_{i+2}$ (look on picture).

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Another Generalization see on picture

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DVDthe1st wrote: But due to our previous points, $l_1,l_2$ are hypercycles to the same vertical h-line (since the distance between them are equal at two points). I don't understand this, even with a comment in brackets. Can you explain?

Amir Hossein wrote: I've attached the official solution(s) of this problem. Is it public? Or could you tell me how to get this archive?