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A. 495. In the acute triangle ABC we have \angleBAC=\alpha. The point D lies in the interior of the triangle, on the bisector of \angleBAC, and points E and F lie on the sides AB and BC, respectively, such that \angleBDC=2\alpha, \angleAED=90o+\frac{\alpha}{2}, and \angleBEF=\angleEBD. Determine the ratio BF:FC.

(5 points)

Deadline expired on 11 January 2010.


Solution. Let G be the reflection of E through the bisector AD. Reflect point D through the lines AB and AC; denote the reflections by U and V, respectively. We have UE=DE=DG=VG.

From the deltoid AEDG we obtain AEG\angle=AGE\angle=90^\circ-\frac\alpha2 and DEG\angle=DGE\angle=AED\angle-AEG\angle=
\bigg(90^\circ+\frac\alpha2\bigg)-
\bigg(90^\circ-\frac\alpha2\bigg) =\alpha.

Moreover, BEU\angle=BED\angle=180^\circ-AED\angle=90^\circ-\frac\alpha2 and CGU\angle=CGD\angle=180^\circ-AGD\angle=90^\circ-\frac\alpha2.

Since BEU\angle=AEG\angle and CGV\angle=AGE\angle, the ponts U, E, G and V are collinear.

Leet BEF\angle=BDE\angle=EBU\angle=x and CGF\angle=CDG\angle=GCV\angle=z. From the quadrilateral ABDC and the triangle ABC, we have x+y=360o-\alpha-(360o-2\alpha)=\alpha and

UBC\angle+BCV\angle=(ABC\angle+x)+(ACB\angle+y)=180o.

Therefore, the lines BU and CV are parallel. Due to EBU\angle=VEF\angle, also the line EF is parallel with them.

Hence,


\frac{BF}{FC} = \frac{UE}{EV} =
\frac{1}{2\cos\alpha + 1}.


Statistics on problem A. 495.
5 students sent a solution.
5 points:Bodor Bertalan, Éles András, Nagy 235 János, Nagy 648 Donát.
2 points:1 student.


  • Problems in Mathematics of KöMaL, December 2009

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