Let $\Delta ABC$ be an acute triangle and $\Omega$ its circumscribed circle, with diameter $AP$. Points $E$ and $F$ are the orthogonal projections from $B$ on $AC$ and $AP$, points $M$ and $N$ are the midpoints of segments $EF$ and $CP$. Prove that $\angle BMN=90$.
Problem
Source: Moldova TST 2020
Tags: geometry
achen29
07.03.2020 22:26
What are E, F?
EmirhanYagcioglu
07.03.2020 23:07
Let $O$ be the center of the $\Omega$, and $L$ be the midpoint of the segment $AB$. Then the quadrilaterals $BLEF$ and $BOCP$ are similar.
fastlikearabbit
07.03.2020 23:16
Notice that the quadrilateral $ABFE$ is cyclic as $\angle{BFA} = \angle{BEA} = 90^{\circ}$. Hence$$\angle{EBF} = \angle{EAF} = \angle{PAC} = \angle{CBP}.$$Also,$$\angle{BEF} = \angle{BAF} = \angle{BAP} = \angle{BCP}.$$Therefore, we can conclude that $\triangle{BEF} \sim \triangle{BCP}$ due to AA similarity. The aforementioned similarity basically tells us that there's a spiral similarity centered at $B$ which sends $\overline{EF}$ to $\overline{CP}$. Now, as $M$ and $N$ are the midpoints of $\overline{EF}$ and $\overline{CP}$, respectively, we can conclude that $M \to N$, which in particular means that $\triangle{BMN} \sim \triangle{BEC}$, and we're done.
achen29
08.03.2020 00:04
A complex bash would do too
kirillnaval
08.03.2020 00:28
achen29 wrote: What are E, F? Perpendiculars on $AC$ and $AP$
WolfusA
08.03.2020 00:34
achen29 wrote: A complex bash would do too It wouldn't be even so bashy
Com10atorics
10.09.2020 15:57
Holds for any $P$ in arc $BC$ which does not contain $A$ not just the antipode.
Let $D=EF\cap CP$.
Since $\angle AEB=90=\angle AFB$ we have that $ABEF$ is cyclic. Now,
$\angle BFE=180-\angle A=\angle BPC$
$\angle BEF=\angle BAP=\angle BCP$.
So $\triangle BEF \sim \triangle BCP$ and in particular $\triangle BMF\sim \triangle BNP$.
So $\angle BMD=\angle BMF=\angle BNP=\angle BND$.
Which means that $DBMN$ is cyclic.
Now it is clear that $EF$ is the Simson line of $\triangle ACP$ from $B$. And therefore
$\angle BMN=180-\angle BDN=90$
We can let $a=1$ and $p=-1$. So now we have
$e=\tfrac{1}{2}(1+c+b-\tfrac{ac}{b})=\tfrac{b+bc+b^2-c}{2b}$
$f=\tfrac{1}{2}(1-1+b+\overline{b})=\tfrac{b^2+1}{2b}$
$n=\tfrac {-1+c}{2}$
$m=\tfrac {e+f}{2}=\tfrac {b-c+2b^2+bc+1}{2}.$
Now it is easy to check that $BM$ is perpendicular to $MN$
i3435
10.09.2020 16:05
Com10atorics wrote:
Holds for any $P$ in arc $BC$ which does not contain $A$ not just the antipode.
?
Math00954
10.09.2020 16:09
The condition augustin_p wrote: diameter $AP$ is redundant. Even @above's condition is redundant, I think. The same end result follows even if we are given only that $ABCP$ is cyclic.