H is the orthocenter of a non-isosceles acute triangle △ABC. M is the midpoint of BC and BB1,CC1 are two altitudes of △ABC. N is the midpoint of B1C1. Prove that AH is tangent to the circumcircle of △MNH.
Problem
Source: Turkey EGMO TST 2020 P1
Tags: geometry, circumcircle
06.02.2020 10:31
Let P be midpoint of AH. Then P,N,M lie on the Newton Gauss line of quadrilateral BB1CC1. Also, by shooting lemma (since PB1MC1 is cyclic with PB1=PC1), we get PB21=PN⋅PM. Combined with PH=PB1, we have the result.
06.02.2020 10:33
Let U be midpoint of AH then: UH2=UB21=UC21=¯UM.¯UN of AH tangents (HMN) at H
17.07.2020 18:39
Let AA1 the third altitude. We now that M is the circumcenter of BC1B1C. Then MN⊥B1C1 at N. ∠HNM=90−∠C1NH∠MHA1=90−∠HMBSince △HB1C1 is similar to △HCB we have ∠HMB=∠HNC1⇒∠HNM=∠MHA1
04.01.2022 11:47
Gauss theorem for BB1CC1 M,N,AH2 collinear. AH2=K MC1=MB1=BC2 MN perpendicular B1C1 AH and B1C1 intersect at P (A,H,P,A1)=−1 KP⋅KA1=KH2=KN⋅KM
04.01.2022 12:44
ISL 2009 G4 wrote: Given a cyclic quadrilateral ABCD, let the diagonals AC and BD meet at E and the lines AD and BC meet at F. The midpoints of AB and CD are G and H, respectively. Show that EF is tangent at E to the circle through the points E, G and H. The end ◼ This config was also present in USA TSTST 2012 #7
31.12.2024 14:50
Firstly we note that <C₁CB=α and <B₁BC=β.We get <CAH=β=<B₁C₁H; <C₁AH=α=<C₁B₁H; <C₁BB₁=<C₁CB₁=90-α-β.HN and HM is median and we get with median sine <C₁HN=<BHM.AH intersect BC at D.We note that <DHM=θ.We get <HMB=90-θ,<AHN=α-β+θ.We want to find <HNM=θ.It is easy. Because MC₁=MB₁=MB=MC .We are done.
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