n=4,a very long discuss.

shobber wrote: Find the greatest integer $n$, such that there are $n+4$ points $A$, $B$, $C$, $D$, $X_1,\dots,~X_n$ in the plane with $AB\ne CD$ that satisfy the following condition: for each $i=1,2,\dots,n$ triangles $ABX_i$ and $CDX_i$ are equal. See here: http://mks.mff.cuni.cz/kalva/apmo/asoln/asol015.html

Here is the solution from our yearbook containing all our training material from 2001. This solution is probably copied from the official solution. One of the sides $AX_i$ or $BX_i$ is equal to $CD$, thus $X_i$ is on one of the circles of radius $CD$ and centre $A$ or $B$. In the same way $X_i$ is on one of the circles of radius $AB$ with centre $C$ or $D$. The intersection of these four circles has no more than 8 points so that $n\le 8$. Suppose that circle $S_B$ with centre $B$ and radius $CD$ intersects circle $S_C$ with centre $C$ and radius $AB$ in two points $X_1$ and $X_2$ which satisfy the conditions of the problem. Then in triangles $ABX_1$ and $CDX_1$ we have $BX_1=CD$ and $CX_1=AB$. Since these triangles are congruent then $AX_1=DX_1$, therefore $X_1$ and $X_2$ are on the perpendicular bisector of $AD$. On the other hand $X_1X_2$ is perpendicular to segment $BC$. Then $BC\parallel AD$ and $AB$ and $CD$ are the diagonals or non-parallel sides of a trapezoid. Suppose that $AB<CD$. Then $BX_1=CD>AB=CX_1$. It follows that the distance from $A$ to the perpendicular bisector of $BC$ must be less than the distance from $D$ to this line otherwise we obtain a contradiction to the condition $AB<CD$. Then for any point $X$ in the perpendicular bisector of $BC$ we have $AX<DX$ and it is not impossible to have $AX=CD$, $DX=AB$. Thus if the circle with centre $A$ and radius $CD$ intersects the circle with centre $D$ and radius $AB$, then the points of intersection of $S_B$ with $S_C$ satisfy the condition of congruence, then the points of intersection of $S_B$ with $S_D$ do not. Thus no more than half of the 8 points of intersection of these circles can satisfy the condition of congruence, i.e. $n\le 4$. If $n=4$ we have the following example where $ABDX_3X_2C$ is a regular hexagon, and $X_1$ and $X_4$ are the intersections of $X_2X_3$ with respectively $AC$ and $BD$. [asy][asy] pair A; pair B; pair C; pair D; pair X1; pair X2; pair X3; pair X4; D=(1,0); C=(-1,0); B=(0.5,sqrt(3)/2); A=B+C; X2=-B; X3=-A; X4=X3+D; X1=X2+C; draw(A--X1--X4--B--A); draw(X2--C--D--X3); draw(circumcircle(A,B,C)); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,W); label("$D$",D,E); label("$X_1$",X1,S); label("$X_2$",X2,S); label("$X_3$",X3,S); label("$X_4$",X4,S); [/asy][/asy]

What motivated the solver to get that configuration? While it does have nice symmetric properties, the motivation seems a bit obscure besides that bit about AB, CD being a trapezoid but even that contradicts the diagram.