$X$ and $Y$ are two points lying on or on the extensions of side $BC$ of $\triangle{ABC}$ such that $\widehat{XAY} = 90$. Let $H$ be the orthocenter of $\triangle{ABC}$. Take $X'$ and $Y'$ as the intersection points of $(BH,AX)$ and $(CH,AY)$ respectively. Prove that circumcircle of $\triangle{CYY'}$,circumcircle of $\triangle{BXX'}$ and $X'Y'$ are concurrent.
Problem
Source: Iranian 3rd round Geometry exam P5 - 2014
Tags: geometry, circumcircle, geometry proposed
jayme
28.09.2014 16:39
Dear Mathlinkers, see for example http://jl.ayme.pagesperso-orange.fr/ vol. 4 A new mixtilinear incircle adventure III p. 44-45. Sincerely Jean-Louis
yunxiu
28.09.2014 18:20
Denote $\left( {AXC} \right) \cap \left( {ABY} \right) = \left\{ {A,D} \right\}$， and $l \bot AD$, $l \cap AX = U$, $l \cap AY = V$. Then we have $\angle DBY = \angle DAY = \angle AUD$, so $BXUD$ is cyclic, hence $\angle AUB = \angle BUD + \angle AUD = \angle BXD + \angle VAD = \angle CAD + \angle VAD = 90^\circ - \angle CAU$ so $BU \bot AC$, and we have $U = X'$, $D \in \left( {BXX'} \right)$. Same we have $V = Y'$ and $D \in \left( {CYY'} \right)$, so we done.
saturzo
02.10.2014 18:04
Let $O$ be the foot of perpendicular from $A$ to $X'Y'$. Then simple angle-chasing just kills the problem, i.e., it implies that $O \in \odot BXX' \cap \odot CYY'$. Done!
sualehasif996
31.01.2016 16:50
Here is a sketch of another solution: If the circles $(BXX')$ and $(CYY')$ intersect at T. Then consider the Miquel point of the triangle $AXC$ with reference to the points $X',B,B'$ where $B'$ is the foot of perpendiculars from B. Similarly consider the Miquel point of the triangle $AYB$ with reference to the points $Y',C,C'$ where $C'$ is the foot of perpendicular from C. Since $B,C',B',C$ are concyclic we have that the miquel points are common in both cases, I.e. T. $\implies \angle BTC = 90^{\circ}$ and simple angle chasing yields our result.
sualehasif996
31.01.2016 16:56
K.N
05.07.2016 08:42
saturzo wrote: Let $O$ be the foot of perpendicular from $A$ to $X'Y'$. Then simple angle-chasing just kills the problem, i.e., it implies that $O \in \odot BXX' \cap \odot CYY'$. Done! Could someone explain more this soloution?!
GMOH
23.02.2018 15:44
K.N wrote: saturzo wrote: Let $O$ be the foot of perpendicular from $A$ to $X'Y'$. Then simple angle-chasing just kills the problem, i.e., it implies that $O \in \odot BXX' \cap \odot CYY'$. Done! Could someone explain more this soloution?! notice that $\angle Y'C'O =\angle Y'AO = \angle OX'A = \angle OB'A$ ($B'$ and $C'$ are the feet of perpendiculars from $B$ and $C$ to $HC$ and $HB$ ) so we get that $BC'OB'C$ is cyclic and with a bit more angle chasing we can prove that $BXOX'$ and $CYOY'$ are cyclic.