Nice Geometry Problem

The incenter of ABC\triangle ABC touches its sides at D,E,FD,E,F respectively, and BE,CFBE, CF intersect the incircle second time at M,NM, N respectively. Extend MDMD to PP so that DP=2MDDP=2MD. Prove that DEDE is tangent to the circle (DPN)\odot (DPN).

#Geometry

Note by Alan Yan
5 years, 6 months ago

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Comments

More incircle properties!

Property1: DNEN=DFEF\frac {DN}{EN}=\frac {DF}{EF}.

Proof: Since CD,CECD,CE are tangent to the incircle, CDNCFD    DN2DF2=CNCF\triangle CDN\sim \triangle CFD\implies \frac {DN^2}{DF^2}=\frac {CN}{CF}. Symmetrically we have EN2EF2=CNCF\frac {EN^2}{EF^2}=\frac {CN}{CF}, thus DNDF=ENEF\frac {DN}{DF}=\frac {EN}{EF} and switching DF,ENDF,EN gives us our desired equality.

Property2: Let LL denote the midpoint of EMEM, then DMLDEF\triangle DML\sim \triangle DEF

Proof: Note that ILB=IFB=IDB=90\angle ILB=\angle IFB=\angle IDB=90^{\circ}, therefore D,I,L,F,BD,I,L,F,B are concyclic where II is the incenter of ABCABC. Hence DLM=DFB=DEF\angle DLM=\angle DFB=\angle DEF. Since DML=DFE\angle DML=\angle DFE by cyclic, we have our desired similarity by AA

Correlary: DNEN=DFEF=DMML=2DMEM\frac {DN}{EN}=\frac {DF}{EF}=\frac {DM}{ML}=\frac {2DM}{EM}

Main proof: It suffices to show DPN=NDE=NME\angle DPN=\angle NDE=\angle NME, since PDN=NEM\angle PDN=\angle NEM by cyclic, we will prove that NDPNEM\triangle NDP\sim \triangle NEM. This is equivalent to DNEN=DPEM=2DMEM\frac {DN}{EN}=\frac {DP}{EM}=\frac {2DM}{EM} by SAS similarity, which follows from the correlary above.

Note: A cyclic quadrilateral that satisfies property 1 is called harmonic quadrilateral, and it is intimately related to concepts such as harmonic division and poles and polar.

Xuming Liang - 5 years, 6 months ago

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BTW did u get this from AoPS? I just went on and saw the exact same problem

Xuming Liang - 5 years, 6 months ago

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Yes I did :)

Alan Yan - 5 years, 6 months ago

Btw do you have a document that goes over some incircle properties? I am not that familiar with them and they will be very useful.

Alan Yan - 5 years, 6 months ago
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