IMO 2018 1first 's problem solution

Let acute-angled triangle ABC, and let \Gamma  be the circumscribed circle. Points Dand ISin the parts FROMand AMcorresponding order AD AE.

The perpendicular bisector of BDand THIS intersecting the small arches FROMand AM the \Gamma in points Fand G respectively.

Prove that the lines FROMand FG are either parallel or are identical.



Solution




IMO 2018 P1.png


Suppose that FDintersects the circumscribed circle of ABCa point The.

We observe that \widehat{ADL}=\widehat{FDB}=\widehat{FBD}=\widehat{FBA}=\widehat{FLA}=\widehat{ALD}.

Thus the triangle LADis isosceles with AL = AD = AE.

Similarly if GIVEintersects circumcircle of ABCat K, then AK AL = AE = AD =.

We will prove that FG // THE, ie \widehat{LFG}=\widehat{LDE}.

We note that \widehat{LFG}=\widehat{LAG}, while \widehat{LDE}=\dfrac{\widehat{LAE}}{2} \widehat{LAE}is the core of \widehat{LDE} the circle with center A and radius AK AL = AE = AD =.

So it suffices to prove that \widehat{LAG}=\widehat{CAG}, that GL=GC.

But we know that GC = GE, by definition G.

So long GL=GE.

We observe that the quadrilateral KALGis recordable and AND AL =therefore the AGis the bisector \widehat{LGE}.

If it LGEwas not equilateral, then since Ait belongs to the bisector \widehat{LGE} and AE = AL, should it glaerbe writable (south pole theorem) inappropriate, since the circumscribed circle Alargis the circumscribed circle ABCand ISdoes not belong to this circle.

So it LGEis isosceles with GL=GEand the requested follow.

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