IMO 2006 Shortlisted Problems

We will work with oriented angles between lines. For two straight lines $\ell ,m$ in the plane, $\angle (\ell ,m)$ denotes the angle of counterclockwise rotation which transforms line $\ell$ into a line parallel to $m$ (the choice of the rotation centre is irrelevant). This is a signed quantity; values differing by a multiple of $\pi$ are identified, so that $$\angle (\ell ,m)=-\angle (m,\ell ),\angle (\ell ,m)+\angle (m,n)=\angle (\ell ,n).$$ If $\ell$ is the line through points $K,L$ and $m$ is the line through $M,N$, one writes $\angle (KL,MN)$ for $\angle (\ell ,m)$; the characters $K,L$ are freely interchangeable; and so are $M,N$. The counterpart of the classical theorem about cyclic quadrilaterals is the following: If $K,L,M,N$ are four noncollinear points in the plane then $$K,L,M,N\text{ are concyclic if and only if }\angle (KM,LM)=\angle (KN,LN).\text{\hspace{1em}(1)}$$ Passing to the solution proper, we first show that the three circles $(A{B}_1{C}_1)$, $(B{C}_1{A}_1)$, $(C{A}_1{B}_1)$ have a common point. So, let $(A{B}_1{C}_1)$ and $(B{C}_1{A}_1)$ intersect at the points $C_1$ and $P$. Then by (1) $$\begin{array}{rl}& \angle (P{A}_1,C{A}_1)=\angle (P{A}_1,B{A}_1)=\angle (P{C}_1,B{C}_1)\\ =& \angle (P{C}_1,A{C}_1)=\angle (P{B}_1,A{B}_1)=\angle (P{B}_1,C{B}_1)\end{array}$$ Denote this angle by $\phi$. The equality between the outer terms shows, again by (1), that the points $A_1,B_1,P,C$ are concyclic. Thus $P$ is the common point of the three mentioned circles. From now on the basic property (1) will be used without explicit reference. We have $$\phi =\angle (P{A}_1,BC)=\angle (P{B}_1,CA)=\angle (P{C}_1,AB).\text{\hspace{1em}(2)}$$ Let lines $A_{2}P$, $B_{2}P$, $C_{2}P$ meet the circle $(ABC)$ again at $A_4$, $B_4$, $C_4$, respectively. As $$\angle (A_4{A}_2,A{A}_2)=\angle (P{A}_2,A{A}_2)=\angle (P{C}_1,A{C}_1)=\angle (P{C}_1,AB)=\phi ,$$ we see that line $A_{2}A$ is the image of line $A_2{A}_4$ under rotation about $A_2$ by the angle $\phi$. Hence the point $A$ is the image of $A_4$ under rotation by $2\phi$ about $O$, the centre of $(ABC)$. The same rotation sends $B_4$ to $B$ and $C_4$ to $C$. Triangle $ABC$ is the image of $A_4{B}_4{C}_4$ in this map. Thus $$\angle (A_4{B}_4,AB)=\angle (B_4{C}_4,BC)=\angle (C_4{A}_4,CA)=2\phi .\text{\hspace{1em}(3)}$$ Since the rotation by $2\phi$ about $O$ takes $B_4$ to $B$, we have $\angle (A{B}_4,AB)=\phi$. Hence by (2) $$\angle (A{B}_4,P{C}_1)=\angle (A{B}_4,AB)+\angle (AB,P{C}_1)=\phi +(-\phi )=0,$$ which means that $A{B}_4\parallel P{C}_1$. Let $C_5$ be the intersection of lines $P{C}_1$ and $A_4{B}_4$; define $A_5,B_5$ analogously. So $A{B}_4\parallel C_1{C}_5$ and, by (3) and (2), $$\angle (A_4{B}_4,P{C}_1)=\angle (A_4{B}_4,AB)+\angle (AB,P{C}_1)=2\phi +(-\phi )=\phi ;\text{\hspace{1em}(4)}$$ i.e., $\angle (B_4{C}_5,C_5{C}_1)=\phi$. This combined with $\angle (C_5{C}_1,C_{1}A)=\angle (P{C}_1,AB)=\phi$ (see (2)) proves that the quadrilateral $A{B}_4{C}_5{C}_1$ is an isosceles trapezoid with $A{C}_1=B_4{C}_5$. Interchanging the roles of $A$ and $B$ we infer that also $B{C}_1=A_4{C}_5$. And since $A{C}_1+B{C}_1=AB=A_4{B}_4$, it follows that the point $C_5$ lies on the line segment $A_4{B}_4$ and partitions it into segments $A_4{C}_5$, $B_4{C}_5$ of lengths $B{C}_1(=A{C}_3)$ and $A{C}_1(=B{C}_3)$. In other words, the rotation which maps triangle $A_4{B}_4{C}_4$ onto $ABC$ carries $C_5$ onto $C_3$. Likewise, it sends $A_5$ to $A_3$ and $B_5$ to $B_3$. So the triangles $A_3{B}_3{C}_3$ and $A_5{B}_5{C}_5$ are congruent. It now suffices to show that the latter is similar to $A_2{B}_2{C}_2$. Lines $B_4{C}_5$ and $P{C}_5$ coincide respectively with $A_4{B}_4$ and $P{C}_1$. Thus by (4) $$\angle (B_4{C}_5,P{C}_5)=\phi .$$ Analogously (by cyclic shift) $\phi =\angle (C_4{A}_5,P{A}_5)$, which rewrites as $$\phi =\angle (B_4{A}_5,P{A}_5).$$ These relations imply that the points $P,B_4,C_5,A_5$ are concyclic. Analogously, $P,C_4,A_5,B_5$ and $P,A_4,B_5,C_5$ are concyclic quadruples. Therefore $$\angle (A_5{B}_5,C_5{B}_5)=\angle (A_5{B}_5,P{B}_5)+\angle (P{B}_5,C_5{B}_5)=\angle (A_5{C}_4,P{C}_4)+\angle (P{A}_4,C_5{A}_4).\text{\hspace{1em}(5)}$$ On the other hand, since the points $A_2,B_2,C_2,A_4,B_4,C_4$ all lie on the circle $(ABC)$, we have $$\angle (A_2{B}_2,C_2{B}_2)=\angle (A_2{B}_2,B_4{B}_2)+\angle (B_4{B}_2,C_2{B}_2)=\angle (A_2{A}_4,B_4{A}_4)+\angle (B_4{C}_4,C_2{C}_4).\text{\hspace{1em}(6)}$$ But the lines $A_2{A}_4,B_4{A}_4,B_4{C}_4,C_2{C}_4$ coincide respectively with $P{A}_4,C_5{A}_4,A_5{C}_4,P{C}_4$. So the sums on the right-hand sides of (5) and (6) are equal, leading to equality between their left-hand sides: $\angle (A_5{B}_5,C_5{B}_5)=\angle (A_2{B}_2,C_2{B}_2)$. Hence (by cyclic shift, once more) also $\angle (B_5{C}_5,A_5{C}_5)=\angle (B_2{C}_2,A_2{C}_2)$ and $\angle (C_5{A}_5,B_5{A}_5)=\angle (C_2{A}_2,B_2{A}_2)$. This means that the triangles $A_5{B}_5{C}_5$ and $A_2{B}_2{C}_2$ have their corresponding angles equal, and consequently they are similar.