31st Turkish Mathematical Olympiad

From the condition $BD/DC=A{B}^2/A{C}^2$ we know that $AD$ is the $A$-symmedian of the triangle $ABC$. Let $a,b,c$ be the side lengths of the triangle, then we get $AF/FB=CD/DB=b^2/c^2$ hence $AF=b^{2}c/(b^2+c^2)$ and similarly $AE=c^{2}b/(b^2+c^2)$. Using these we get $AE\cdot AC=AF\cdot AB$ hence $B,E,C,F$ are concyclic. Let the circumcircle of the triangle $ABC$ be $\Gamma$ and let the circumcircle of the triangle $AEF$ be $\omega$. Let $AD$ meet $\Gamma$ for the second time at $Q$ and let $AS\cap BC=P$. Let the intersection of $\Gamma$ and $\omega$ other than $A$ be $M$. The radical axis of the circles $\Gamma$, $\omega$ and $(BCEF)$ are concurrent, let this point be $AM\cap EF\cap BC=L$. Let the line passing through $A$ and parallel to $BC$ intersect $\Gamma$ for the second time at $A^{\prime }$. Since $\angle C=\angle AFE=\angle EA{A}^{\prime }$ we find that $A{A}^{\prime }$ and $\omega$ are tangent to each other.

Let us consider the spiral similarity $\sigma$ centred at $M$ and satisfying $F\to B,E\to C$. It is in fact the spiral similarity that sends the circle $\omega$ to the circle $\Gamma$, thus $\sigma$ also satisfies $S\to A,A\to A^{\prime }$ and $R\to Q$. Since $AQ$ is a symmedian, the tangent lines to $\Gamma$ passing through $Q$ and $A$ intersect on the line $BC$. Therefore, under $\sigma$, the tangent lines to $\omega$ passing through $R$ and $S$ intersect on the line $EF$. In addition, simple angle chasing gives that $\angle BDF=\angle C$ and $\angle AEF=\angle B,\angle DEC=\angle A$ which means $\angle FED=\angle C$ hence, circumcircle of the triangle $DEF$ and the line $BC$ are tangent to each other. From here we find $L{D}^2=LE\cdot LF=LB\cdot LC$, also since we know $(P,D;B,C)=-1$ we obtain that $L$ is the midpoint of the segment $PD$. We found that the lines $A{A}^{\prime }$ and $BC$ are parallel, therefore we have $(AP;AD;AL,A{A}^{\prime })=-1$ and projecting these lines to $\omega$ we find that $(S,R;M,A)=-1$, which means the tangent lines to $\omega$ passing through $S,R$ intersect on the line $AM$. In this case, this intersection point was also on the line $EF$, it must be $L$.

So, we find that the polar line of $L$ with respect to $\omega$ is $SR$, and since $AM\cap EF=L$, from Brokard's Theorem in quadrilateral $AMEF$, the lines $AF$ and $EM$ intersect on the polar line $SR$. Hence $AB$, $EM$, $SR$ are concurrent, and we also have $(T,L;F,E)=-1$. Furthermore, on the circle $\omega$ we have $(S,Q;A,M)=-1$ and under $\sigma$ its image is $(A,Q;M,A^{\prime })=1$, hence the points $A^{\prime },M,P$ are collinear. Let $A^{\prime }M\cap AQ=V$, then we obtain $(P,V;M,A^{\prime })=-1$. Finally, by definition we have $A{A}^{\prime }\parallel BC$ and under $\sigma$ it becomes

Now, we will work on the problem statement to finish the solution. We will prove that both conditions are equivalent to $\angle B=2\angle C$ or $180^{\circ }+\angle B=2\angle C$. Solution to both cases are identical, thus we only study the case where $\angle B>\angle C$. First, we know that $\angle M{A}^{\prime }=\angle B-\angle C$ and under $\sigma$ we have $\angle SMA=\angle B-\angle C$. Also, we have $\angle AME=\angle AFE=\angle C$. $\angle B=2\angle C$ is equivalent to the following statements: $$(1)\angle B=2\angle C\iff \angle SMA+\angle AMB=\angle B-\angle C+(180^{\circ }-\angle C)=180^{\circ }\iff B,M,W,S\text{ are collinear.}$$ $$(2)\angle B=2\angle C\iff \angle M{A}^{\prime }=\angle B-\angle C=\angle C=\angle AME\iff P,M,E,A^{\prime }\text{ are collinear.}$$ From the first condition, $BS$ bisecting $[TL]$ is equivalent to $\angle B=2\angle C$. From the concurrency $AB,SR,EM$ we find that $SR$ bisecting $[AB]$ is equivalent to $EM$ bisecting $[AB]$. $EM$ bisecting $[AB]$ is equivalent to the condition $(EA,EB;EM,ED)=-1$, and reflecting these lines to the line $BC$ shows that it is in fact equivalent to $E,M,P$ being collinear. From the second condition this is equivalent to $\angle B=2\angle C$ hence the required statement is proven.