Solution of task "IMO Shortlist 2006 problem G2" by user "binkret"

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May 26, 2022, 10:22 p.m. (4 years, 3 months)

User: binkret
Task: IMO Shortlist 2006 problem G2 (Hide problem text)

Let $ABC$ be a trapezoid with parallel sides $AB > CD$ . Points $K$ and $L$ lie on the line segments $AB$ and $CD$ , respectively, so that $\frac {AK}{KB} = \frac {DL}{LC}$ . Suppose that there are points $P$ and $Q$ on the line segment $KL$ satisfying $\angle{APB} = \angle{BCD}$ and $\angle{CQD} = \angle{ABC}$ . Prove that the points $P$ , $Q$ , $B$ and $C$ are concylic.

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Po obratu Talesa, $AD$ , $BC$ , i $KL$ sijeku se u jednoj tocki $X$ . Zelimo $XC*XB=XQ*XP$

$\angle BPA=\angle BCD$ i $\angle BPA + \angle DQC =180^{\circ}$ se mogu upotrijebiti istovremeno uvodenjem tocke $R$ kao tocka na duzini ${LX}$ takva da je $\angle {CRD}= \angle {BPA}$ . Ovo je zgodno jer je onda ${DQCR}$ tetivan i $\angle{CRQ}=\angle{BCQ}$ pa je $XC^2=XR*XQ$ . Zbog homoteticnosti je jasno da je ${DR}||{AP}$ i ${CR}||{BP}$ , iz druge od te dvije paralelnosti zajedno s prijasnjom jednakosti slijedi $XC*XB=XQ*XP$ .

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