In the triangle $ABC$ let $B'$ and $C'$ be the midpoints of the sides $AC$ and $AB$ respectively and $H$ the foot of the altitude passing through the vertex $A$. Prove that the circumcircles of the triangles $AB'C'$,$BC'H$, and $B'CH$ have a common point $I$ and that the line $HI$ passes through the midpoint of the segment $B'C'.$
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Tags: geometry, circumcircle, Euler, power of a point, IMO Shortlist
Amir Hossein
23.09.2010 13:32
Let $F$ be the midpoint of $B'C'$, $A'$ the midpoint of $BC$, and $I$ the intersection point of the line $HF$ and the circle circumscribed about $\triangle BHC'.$
Denote by $M$ the intersection point of the line $AA'$ with the circumscribed circle about the triangle $ABC$. Triangles $HB'C'$ and $ABC$ are similar. Since $ \angle C'IF = \angle ABC = \angle A'MC, \angle C'FI = \angle AA'B = \angle MA'C$, we have $2C'F = C'B'$, and $2A'C = CB$, it follows that $\triangle C'IB' \sim \triangle CMB$, hence $ \angle FIB' = \angle A'MB = \angle ACB$. Now one concludes that $I$ belongs to the circumscribed circles of $\triangle AB'C'$ (since $\angle C'IB' = 180^\circ -\angle C'AB'$) and $ \triangle HCB'.$
thecmd999
01.01.2014 09:36
Sigh...
By Miquel, $I$ exists. Since $HI$ is the radical axis of $\odot(BC'H)$, $\odot(B'CH)$, it suffices to show that $M$ has the same power wrt them. But angle chasing yields $MC'$, $MB'$ are tangents to them, so we're done. $\blacksquare$
jayme
01.01.2014 10:13
Dear Mathlinkers, 1. yes, with the nice theorem of Miquel... 2. consider the Euler circle of ABC and the Reim's theorem... we obtain that B'C' is tangent to (BC'H) and (B'CH) 3. By a converse of a theorem concernant the median of a triangle, we are done... Sincerely Jean-Louis
sunken rock
01.01.2014 21:24
Additional requirement: $I$ lies onto the $A-$symmedian of $\triangle ABC$. Best regards, sunken rock
thecmd999
01.01.2014 23:45
Indeed, $I$ is the same point as $F$ in 2008 USAMO #2
TheMaskedMagician
02.01.2014 00:06
Wow, thecmd999's solution was so short and simple.
jayme
17.09.2020 13:52
Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Quickies%209.pdf p. 13-15. Sincerely Jean-Louis
CrazyInMath
13.11.2022 09:14
Solution involving Humpty Point The first part can be easily done with Miquel Theorem. For the second part, let the intersection of $AH$ and $B'C'$ be $Y$, then we have $\triangle AYC'=\triangle HYC'$ and $\triangle AYB'=\triangle HYB'$, so we would get $\triangle AB'C'=\triangle HB'C'$ (here $=$ means congruence since I don't know how to type congruent symbol). Now we have $\measuredangle AB'C'=\measuredangle C'B'H=\measuredangle B'CH$, so $BC$ is tangent to $(B'HC)$. Now we have $\measuredangle PB'I=\measuredangle B'CI=\measuredangle B'HI$, and similarly $\measuredangle PC'I=\measuredangle C'HI$. So $I$ is the $H$-Humpty Point of $\triangle HB'C'$, the conclusion follows.
IAmTheHazard
18.09.2023 03:24
part of this shows up in usa tst 2023/2. The first part is just Miquel's Theorem. Now let $M$ be the midpoint of $\overline{B'C'}$ and let $I'=\overline{HM} \cap (ABC)$, so we wish to show that $B'CHI'$ is cyclic. Indeed note that $\overline{AI'}$ is a symmedian in $\triangle AB'C'$ (we REFLECT OVER the perpendicular bisector of $\overline{B'C'}$, or $\sqrt{bc}$ invert and angle chase if you're silly like me). But then $\triangle IB'M \sim \triangle I'AC'$, hence $\measuredangle HI'B'=\measuredangle MI'B'=\measuredangle C'I'A=\measuredangle C'B'A=\measuredangle HCB'$. $\blacksquare$
Jishnu4414l
01.01.2024 17:55
Let $X=AH\cap B'C'$. The first part is obvious as $I$ is a miquel point of $\triangle ABC$. Now for the second part, we have $\triangle AC'X \cong \triangle HC'X$ (by SAS) congruency. So $\angle ABH=\angle AC'X=\angle HC'X$. But $\angle HC'X=\angle HBC' \implies$ $B'C'$ is tangent to $(BC'H)$. by analogous angle chasing, we have that $B'C'$ is tangent to $(B'CH)$ Now notice that $HI$ is the radical axis of the two circles $(BC'H)$ and $(B'CH)$. Thus it must have equal powers to both of these circles. Since $B'C'$ is tangent to both of these circles, we have that $HI$ bisects $B'C'$. Our proof is thus complete.