Točno
11. travnja 2022. 11:47 (2 godine, 3 mjeseci)
Let
![H](/media/m/4/c/0/4c0872a89da410a25f00b86366efece7.png)
be the orthocenter of an acute-angled triangle
![ABC](/media/m/a/c/7/ac75dca5ddb22ad70f492e2e0a153f95.png)
. The circle
![\Gamma_{A}](/media/m/5/d/a/5da953383f4b346be9912f2e7bec0ab2.png)
centered at the midpoint of
![BC](/media/m/5/0/0/5005d4d5eac1b420fbabb76c83fc63ad.png)
and passing through
![H](/media/m/4/c/0/4c0872a89da410a25f00b86366efece7.png)
intersects the sideline
![BC](/media/m/5/0/0/5005d4d5eac1b420fbabb76c83fc63ad.png)
at points
![A_{1}](/media/m/9/7/4/9742b2655cd943b758073e1f1d090c23.png)
and
![A_{2}](/media/m/a/2/5/a25771e5c2a6c9c6113eab3c46cf63d8.png)
. Similarly, define the points
![B_{1}](/media/m/f/a/5/fa55cf39f6736c287bf64ee9471f00f1.png)
,
![B_{2}](/media/m/5/f/0/5f0ace33ca787cd757fffa052940b6a9.png)
,
![C_{1}](/media/m/e/4/6/e46111370b6102ad343bcdc7190d9ff9.png)
and
![C_{2}](/media/m/1/3/c/13ce248ee56b45a1b9e032736b8100a1.png)
.
Prove that six points
![A_{1}](/media/m/9/7/4/9742b2655cd943b758073e1f1d090c23.png)
,
![A_{2}](/media/m/a/2/5/a25771e5c2a6c9c6113eab3c46cf63d8.png)
,
![B_{1}](/media/m/f/a/5/fa55cf39f6736c287bf64ee9471f00f1.png)
,
![B_{2}](/media/m/5/f/0/5f0ace33ca787cd757fffa052940b6a9.png)
,
![C_{1}](/media/m/e/4/6/e46111370b6102ad343bcdc7190d9ff9.png)
and
![C_{2}](/media/m/1/3/c/13ce248ee56b45a1b9e032736b8100a1.png)
are concyclic.
Author: Andrey Gavrilyuk, Russia
%V0
Let $H$ be the orthocenter of an acute-angled triangle $ABC$. The circle $\Gamma_{A}$ centered at the midpoint of $BC$ and passing through $H$ intersects the sideline $BC$ at points $A_{1}$ and $A_{2}$. Similarly, define the points $B_{1}$, $B_{2}$, $C_{1}$ and $C_{2}$.
Prove that six points $A_{1}$ , $A_{2}$, $B_{1}$, $B_{2}$, $C_{1}$ and $C_{2}$ are concyclic.
Author: Andrey Gavrilyuk, Russia
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Postavimo $ABC$ na jediničnu kružnicu kompleksne ravnine. Neka je $M$ polovište $BC$, dokazat ćemo da je centar tražene kružnice centar opisane kružnice $ABC$. Uvjeti na $a_1$ su:
\[|a_1-m|=|h-m|\]
\[b+c=a_1+bc\overline{a_1}\]
Kvadriranjem prvog uvjeta i izmnažanjem dobiva se:
\[(a_1-\frac{b+c}{2})(\overline{a_1}-\frac{b+c}{2bc})=\frac{2a+b+c}{2}\cdot \frac{2bc+ab+ac}{2abc}\]
Ponovnim izmnažanjem imamo:
\[RHS=a_1\overline{a_1}-\frac{b+c}{2bc}(b+c-bc\overline{a_1})-\frac{b+c}{2}\overline{a_1}+\frac{(b+c)^2}{4bc}=a_1\overline{a_1}-\frac{(b+c)^2}{4bc}\]
Prebacivanjem konstante imamo konačno:
\[|a_1|^2=\frac{(2a+b+c)(2bc+ab+ac)+a(b+c)^2}{4abc}=\frac{8abc+4\sum_{sym}a^2b}{4abc}\], što je simetrično u $a,b,c$, što dokazuje našu tvrdnju, stoga sve tražene točke leže na kružnici s centrom $O$