Triangle $ABC$ is given with its centroid $G$ and cicumcentre $O$ is such that $GO$ is perpendicular to $AG$. Let $A'$ be the second intersection of $AG$ with circumcircle of triangle $ABC$. Let $D$ be the intersection of lines $CA'$ and $AB$ and $E$ the intersection of lines $BA'$ and $AC$. Prove that the circumcentre of triangle $ADE$ is on the circumcircle of triangle $ABC$.
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Tags: geometry, circumcircle, ratio, geometric transformation, homothety, geometry unsolved
Luis González
10.04.2011 20:49
Since $A'$ is a point on the A-median, it is clear that $ED \parallel BC.$ If $M$ is the midpoint of $BC,$ then $AM$ is also the A-median of $\triangle ADE,$ i.e. $P \equiv AM \cap DE$ is the midpoint of $DE.$ From the harmonic cross ratio $(A,A',M,P),$ keeping in mind that $\overline{GA}=-\overline{GA'},$ we have $\frac{_{\overline{PA}}}{^{\overline{PA'}}}=-\frac{_{\overline{MA}}}{^{\overline{MA'}}}=\frac{_{\overline{MA}}}{^{\overline{MG}}}=3$ $ \Longrightarrow$ $ A'$ is centroid of $\triangle ADE,$ i.e. $B,C$ are midpoints of $AD,AE$ $\Longrightarrow$ circumcenter of $\triangle ADE$ is the antipode of $A$ WRT $(O).$
FantasyLover
16.04.2011 01:16
First of all, since $\angle AGO=\angle AMO=90^{\circ}$, points $A, M, G, O, N$ are concyclic.
As a result, $\angle BCA'=\angle BAA'=\angle MAG=\angle MNG=\angle MNB=\angle GBC$, and therefore $BN\parallel CD$. Analogously, $CM\parallel BE$.
However, $BN\parallel CD$ along with the fact that $N$ is the midpoint of $AC$ implies that $B$ is the midpoint of $AD$, and analogously $C$ is the midpoint of $AE$.
Hence, there exists a homothety centered at $A$ with magnitude $2$ that sends $\triangle ABC$ to $\triangle ADE$. On the other hand, this homothety maps $O$ to a certain point on $(ABC)$, but this point is also the circumcenter of $\triangle ADE$. Thus, circumcenter of $\triangle ADE$ lies on the circumcircle of $\triangle ABC$, as desired. $\blacksquare$
littletush
22.10.2011 13:50
it's trivial that $G$ is the migpointg of $AA'$,let $T$ be the midpoint of $DE$ then by Newton's line,$G,M,T$ collinear so $A,G,M,A'.T$ collinear hence$BC$∥$DE$ so it suffices to prove $DE=2BC$ which is trivial
Virgil Nicula
22.10.2011 22:11
Remark. $GO\perp GA\iff 2a^2=b^2+c^2$ .
bojler
20.11.2014 21:37
Let $X$ be intersection of $AO$ and circumcircle of $ABC$ and $A_1,B_1,C_1$ be the midpoints of $BC,AC,AB$.$H_{A,2}$ maps $G$ to $A'$ so $CA'||GB_1$ so $B,C$ are midpoints of $DA,AE$ and thus $X$ is the circumcenter of $ADE$.