Let $P$ be a point in the interior of a triangle $ABC$. The lines $AP, BP$ and $CP$ intersect again the circumcircles of the triangles $PBC, PCA$ and $PAB$ at $D, E$ and $F$ respectively. Prove that $P$ is the orthocenter of the triangle $DEF$ if and only if $P$ is the incenter of the triangle $ABC$. Proposed by Romania
Problem
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Tags: geometry
cadaeibf
12.09.2020 15:09
By inversion with respect to $P$ we get that $D^*=A^*P\cap B^*C^*$ and similarly for $E^*$ $F^*$. From hypothesis we have $P$ being the Incenter of $D^*E^*F^*$, and so, since $D^*E^*F^*$ is the orthic triangle of $A^*B^*C^*$, $P$ Is the orthocenter of $A^*B^*C^*$. Finally, this implies that $P$ is the Incenter of the original triangle $ABC$.
Mahdi_Mashayekhi
17.01.2022 12:35
Assume P is incenter : ∠PAF = ∠PAB + ∠BAF = ∠PAB + ∠BPF = 90 and ∠PAE = ∠PAC + ∠CAE = ∠PAC + ∠CPE = 90 so F,A,E are collinear and DP is perpendicular to EF. with same approach we can prove P is orthocenter of DEF. Assume P is not incenter : PA is perpendicular to EF and from last part we know it's not perpendicular at A so we have to cases: 1 - ∠PAE and ∠PAD > 90. so ∠PBE and ∠PCD < 90 so ∠PBE + ∠PCD = ∠PBA + ∠APE + ∠PCA + ∠APD = 180 and it's contradiction. 2 - ∠PAE and ∠PAD < 90. so ∠PBE and ∠PCD > 90 so ∠PBE + ∠PCD = ∠PBA + ∠APE + ∠PCA + ∠APD = 180 and it's contradiction. we're Done.