Team Selection Test for IMO

Lemma. Let $ABC$ be a triangle with $|AB{|}^2+|AB|\cdot |AC|=|BC{|}^2$. Then $m(\hat{A})=2m(\hat{C})$. Proof: Let $D$ be an intersection of interior angle bisector of $\hat{A}$ with $BC$. Then $|BD|=|AB|\cdot k$, $|CD|=|AC|\cdot k$. $|AB|\cdot (|AB|+|BC|)\cdot k=|BC{|}^2\cdot k\Rightarrow |AB|\cdot |BC|=|BC{|}^2\cdot k\Rightarrow |AB|=|BC|$. Since $|BD|=|AB|\cdot k$, $△ABD\sim △CBA\Rightarrow m(\widehat{BCA})=m(\widehat{BAD})=\frac{1}{2}m(\widehat{BAC})$. Done.

By the lemma, $|AD{|}^2+|AD|\cdot |AB|=|BD{|}^2\Rightarrow m(\widehat{BAD})=2m(\hat{B})$ and $|AD{|}^2+|AD|\cdot |AC|=|CD{|}^2\Rightarrow m(\widehat{CAD})=2m(\hat{C})$. Let $m(\hat{B})=\beta$ and $m(\hat{C})=\alpha$. Then $m(\hat{A})=120^{\circ }$.

$|DE{|}^2=|CD{|}^2+|CD|\cdot |CE|\Rightarrow m(\widehat{DCE})=2m(\widehat{DEC})$ and $m(\widehat{DCE})+m(\widehat{DEC})=m(\widehat{ADC})=3\beta \Rightarrow m(\widehat{DCE})=2\beta$ and $m(\widehat{DEC})=\beta$. Thus, $A$, $B$, $E$, $C$ are concyclic.

Since $\alpha =60^{\circ }-\beta$, the measure of the arc $\widehat{ACE}$ is equal to $240^{\circ }-2\beta$. Let us take a point $F$ on the arc $\widehat{BEC}$ satisfying $m(\widehat{BAF})=m(\widehat{CAF})=60^{\circ }$. The measure of the arc $\widehat{ABF}$ is equal to $240^{\circ }$. Therefore, $|AF|=|AE|$.

$m(\widehat{BCF})=m(\widehat{BAF})=60^{\circ }$, $m(\widehat{CBF})=m(\widehat{CAF})=60^{\circ }$, $m(\widehat{BFC})=180^{\circ }-m(\widehat{BAC})=60^{\circ }\Rightarrow △BCF$ is equilateral.

Ptolemy's cyclic quadrilateral theorem applied to $ABFC$ yields: $|AB|\cdot |CF|+|AC|\cdot |BF|=|AF|\cdot |BC|$. Therefore, $|AB|+|AC|=|AF|$ (since $|BF|=|CF|=|BC|$). Thus, $|AB|+|AC|=|AE|$. Done.