Problema 1, Finala Sharygin 2012, clasa a 8-a

[Laurențiu Ploscaru]

Fie $M$ mijlocul bazei $[AC]$ a $\triangle ABC$ ascuțitunghic isoscel. Fie $N$ simetricul lui $M$ față de $[BC]$. Paralela la $AC$ care trece prin $N$ intersectează $[AB]$ în $K$.
Aflați măsura unghiului $\widehat{AKC}$.

[Stefan Tudose]

Fie $\widehat{ABC}=2\alpha$ si $CM=MA=a$

Aplicand T. sin. in $\triangle{BMC}$,

$\dfrac{BM}{\cos{\alpha}}=\dfrac{a}{\sin{\alpha}} \Rightarrow BM=\dfrac{a\cdot \cos{\alpha}}{\sin{\alpha}}$

Aplicand T. sin. in $\triangle{NBM}$,

$\dfrac{MN}{\sin{2\cdot \alpha}}=\dfrac{BM}{\cos{\alpha}} \Rightarrow MN=\dfrac{BM\cdot \sin{(2\alpha)}}{\cos{\alpha}}\Leftrightarrow MN=2\cdot a \cdot \cos{\alpha}$

Patrulaterul NMKB ortodiagonal $\Rightarrow NM^2+BK^2=MK^2+NB^2 \Leftrightarrow MK=\sqrt{NM^2+BK^2-BM^2}$

$MK=\sqrt{4\cdot a^2\cdot \cos^2{\alpha}+\dfrac{a^2\cdot \cos^2{(2\alpha)}}{\sin^2{\alpha}}-\dfrac{a^2\cdot \cos^2{\alpha}}{\sin^2{\alpha}}}$


$MK=a\cdot \sqrt{\dfrac{4\cdot \sin^2{\alpha} \cdot \cos^2{\alpha} + \cos^2{(2\alpha)}-\cos^2{\alpha}}{\sin^2{\alpha}}$

$MK=a\cdot\sqrt{\dfrac{\sin^2{(2\alpha)}+ \cos^2{(2\alpha)} - \cos^2{\alpha}}{\sin^2{\alpha}}$

$MK=a\cdot\sqrt{\dfrac{1- \cos^2{(\alpha)}}{\sin^2{\alpha}}$

$MK=a \Rightarrow MK=MA$

Conchidem că $\widehat{AKC}=90^\circ$

[mihai miculita]

1). Notam cu $P$ proiectia lui $M$ pe $BC$ si cu $\{Q\}=NK\cap BC.$ Cum $|MP|=|PN|$ si $MC\parallel NK\Rightarrow MCNQ\mbox{-romb}\Rightarrow |CP|=|PQ|.\,\,(1)$
2). $M$ fiind mijlocul laturii $[AC]$, tinand seama de (1)$\Rightarrow [MP]\mbox{-linie mijlocie in }\triangle{ACQ}\Rightarrow MP\parallel AQ$ si intrucat $MP\perp BC\Rightarrow AQ\perp BC.\,\,(2)$
3). In ipoteza $|AB|=|BC|\Rightarrow ACQK\mbox{-trapez isoscel}$ si tinand acum seama de (2) $\Rightarrow CK\perp AB.$

[sunken rock]

Little bit shorter, $CMQN$ kite with $NQ\parallel CM\implies CMQN$ rhombus, so $QM=CM$. But $BQMK$ is a kite as well, so $MQ=MK=CM=AM$, from $MK=CM=AM\implies\triangle ACK$ right angled at $K$.

Best regards,
sunken rock