In an acute $\Delta ABC$, $AH_a$ and $BH_b$ are altitudes and $M$ is the middle point of $AB$. The circumscribed circles of $\Delta AMH_a$ and $\Delta BMH_b$ intersect for a second time in $P$. Prove that point $P$ lies on the circumscribed circle of $\Delta ABC$.
Problem
Source: V International Festival of Young Mathematicians Sozopol 2014, Theme for 10-12 grade
Tags: geometry, circumscribed
Ricochet
10.10.2019 00:31
Please check if im right: Let $H$ be the orthocenter and let $Xc=$$CM$$\cap$$(AHB)$. Clearly by contruction this is the $C$-humpty point of $\triangle$$ABC$, this implies $\angle$$MBXc=$$\angle$$BCXc$ hence $\triangle$$MBXc$ $\cong$ $MCB$ so ${MB}^2=MXc$ $\cdot$ $MC$. Now consider the inversion with center $M$ radius $MA=MHb=MHa=MB$ (this are equal cause $M$ is midpoint of the hipotenuse in triangles $\triangle$$ABHb$ and $\triangle$$ABHa$, so $AHa$ $\mapsto$ $(AMHa)$ , $BHb$ $\mapsto$ $(BMHb)$ and so $H=$$AHa$$\cap$$BHb$ $\mapsto$ $P=$$(AMHa)$$\cap$$(BMHb)$ also since ${MB}^2=MXc$ $\cdot$ $MC$ we have that $Xc$ $\mapsto$ $C$ , but $A$ $\mapsto$ $A$ and $B$ $\mapsto$ $B$ , so $(AHXcB)$ must map to circunference that passes through the inverses of $A, H, Xc, B$ wich are $A, P, C, B$ and we are done
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LKira
10.10.2019 01:01
Ricochet wrote:
Please check if im right: Let $H$ be the orthocenter and let $Xc=$$CM$$\cap$$(AHB)$. Clearly by contruction this is the $C$-humpty point of $\triangle$$ABC$, this implies $\angle$$MBXc=$$\angle$$BCXc$ hence $\triangle$$MBXc$ $\cong$ $MCB$ so ${MB}^2=MXc$ $\cdot$ $MC$. Now consider the inversion with center M radius $MA=MHb=MHa=MB$ (this are equal cause $M$ is midpoint of the hipotenuse in triangles $\triangle$$ABHb$ and $\triangle$$ABHa$, so $AHa$ $\mapsto$ $(AMHa)$ , $BHb$ $\mapsto$ $(BMHb)$ and so $H=$$AHa$$\cap$$BHb$ $\mapsto$ $P=$$(AMHa)$$\cap$$(BMHb)$ also since ${MB}^2=MXc$ $\cdot$ $MC$ we have that $Xc$ $\mapsto$ $C$ , but $A$ $\mapsto$ $A$ and $B$ $\mapsto$ $B$ , so $(AHXcB)$ must map to circunference that passes through the inverses of $A, H, Xc, B$ wich are $A, P, C, B$ and we are done
I think your solution is correct Btw, we can just use the inversion center $H$ swap $A$ and $H_a$
Pluto1708
10.10.2019 01:08
Pinko wrote: In an acute $\Delta ABC$, $AH_a$ and $BH_b$ are altitudes and $M$ is the middle point of $AB$. The circumscribed circles of $\Delta AMH_a$ and $\Delta BMH_b$ intersect for a second time in $P$. Prove that point $P$ lies on the circumscribed circle of $\Delta ABC$. First of all note that power of $H$ wrt both the circles is same hence $H, M, P$ collinear. Now since $MH_a=MH_b=MA=MB$ hence $MP$ bisects $\angle APH_a$ and $\angle BPH_b$.Hence $\dfrac{PA}{PH_a}=\dfrac{PB}{PH_b} \implies \dfrac{PA}{PB}=\dfrac{PH_a}{PH_b}$ hence since $ABH_aH_b$ is cyclic we obtain by some trivial calculation that $P$ is center of spiral similarity taking $H_aH_b \to BA$ hence $P\in \odot{ABC}$ as desired.
sarjinius
10.10.2019 03:04
We let $D$ be the reflection of the orthocenter $H$ across $H_a$, $E$ be the reflection of $H$ across $H_b$, and $F$ be the reflection of $H$ across $M$. Note that $D, E, F$ are on $(ABC)$. Claim: $P$ lies on line $MH$. Proof: $MP$ is the radical axis of $(AMH_a)$ and $(BMH_b)$. So we just have to prove $H$ lies on the radical axis, i.e. has equal power wrt both circles. $Pow_{(AMH_a)}H = HA*HH_a$ $Pow_{(BMH_b)}H = HB*HH_b$ But since $\angle AH_aB = \angle AH_bB = 90^{\circ}$, $ABH_aH_b$ is cyclic. So $HA*HH_a = HB*HH_b$, so $H$ lies on the radical axis and the claim is proved. $HP*HF = 2HM*HP = 2HA*HH_a =HA*HD$ So $AFDP$ is cyclic, but $(AFD)$ is $(ABC)$, so $P$ lies on $(ABC)$.