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Nice Geometry Problem

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

Note by Alan Yan
1 year ago

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More incircle properties!

Property1: \(\frac {DN}{EN}=\frac {DF}{EF}\).

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

Property2: Let \(L\) denote the midpoint of \(EM\), then \(\triangle DML\sim \triangle DEF\)

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

Correlary: \(\frac {DN}{EN}=\frac {DF}{EF}=\frac {DM}{ML}=\frac {2DM}{EM}\)

Main proof: It suffices to show \(\angle DPN=\angle NDE=\angle NME\), since \(\angle PDN=\angle NEM\) by cyclic, we will prove that \(\triangle NDP\sim \triangle NEM\). This is equivalent to \(\frac {DN}{EN}=\frac {DP}{EM}=\frac {2DM}{EM}\) 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 · 1 year ago

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@Xuming Liang 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 · 1 year ago

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@Xuming Liang BTW did u get this from AoPS? I just went on and saw the exact same problem Xuming Liang · 1 year ago

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@Xuming Liang Yes I did :) Alan Yan · 1 year ago

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