IGO 2019 - Nivel Medio - P5

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IGO 2019 - Nivel Medio - P5

Mensaje sin leer por BrunZo » Lun 30 Sep, 2019 7:51 pm

Sea $ABC$ un triángulo con $\angle A=60^{\circ}$. Los puntos $E$ y $F$ son los pies de las bisectrices de los vértices $B$ y $C$. Los puntos $P$ y $Q$ son tales que los cuadriláteros $BFPE$ y $CEQF$ son paralelogramos. Probar que $\angle PAQ>150^{\circ}$. (Considerar el ángulo $PAQ$ que no contiene al lado $AB$ del triángulo.)

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Re: IGO 2019 - Nivel Medio - P5

Mensaje sin leer por caioh_br » Jue 24 Oct, 2019 9:30 am

La solución oficial del problema:
Spoiler: mostrar
Let I and be the intersection point of lines BE and CF, and let R be the intersection point of lines QE and PF. It is easy to see that ∠BIC = 120◦. Thus AEIF is a cyclic quadrilateral and so CE · CA = CI · CF (1)

Also ∠PRQ = ∠BIC = 120◦, therefore it suffices to show that at least one of the angles ∠APR or ∠AQR is greater than or equal to 30◦.

Assume the contrary, meaning both of these angles are less than 30◦. Hence there exists a point K on the extension of ray CA such that
∠KQE = 30◦. Since ∠IAC = 30◦ and ∠ACI = ∠KEQ, it is deduced that AIC ∼ QKE. This implies

CI / CA = KE / QE > AE / CF ⇒ AE < (CF · CI) / CA = CE. (1)

Similarly, it is obtained that AF < BF. On the other hand at least on of the angles ∠ABC or ∠ACB are not less than 60◦. Without loss of generality one can assume that ∠ABC ≥ 60◦ thus AC ≥ BC and according to angle bisector theorem it is obtained that AF ≥ BF, which is a contradiction. Hence the claim of the problem.