Let $ABC$ be an acute-angled triangle with $AB < AC$. Tangent to its circumcircle $\Omega$ at $A$ intersects the line $BC$ at $D$. Let $G$ be the centroid of $\triangle ABC$ and let $AG$ meet $\Omega$ again at $H \neq A$. Suppose the line $DG$ intersects the lines $AB$ and $AC$ at $E$ and $F$, respectively. Prove that $\angle EHG = \angle GHF$.(Slovakia)
Problem
Source: Czech-Polish-Slovak Match 2016,P3,day 2
Tags: geometry proposed, geometry, circumcircle, Centroid
jayme
12.07.2016 14:18
See http://www.artofproblemsolving.com/community/c6t48f6h1265773_geometry Sincerely Jean-Louis
jayme
12.07.2016 14:33
Dear Mathlikers, what is the oficial solution of this problem? Sincerely Jean-Louis
j___d
12.07.2016 14:57
Let $p$ be the line parallel to $BC$ that passes through $A$ and let $P = p \cap DG$. We denote the midpoint of $BC$ by $T$.
We project the harmonic ratio $(BTC\infty)=-1$ from $A$ onto the line $DG$ and learn that $(EGFP)=-1$. Therefore by a well-known Apollonian property of harmonic ratios it suffices to prove $\angle PHA = 90^{\circ}$.
Now let $Q$ be the orthogonal projection of $D$ onto $AH$. The homothety centered at $G$ with factor $-\dfrac{1}{2}$ maps the line $p$ onto the line $BC$ and hence it maps $P$ to $D$. Moreover, it leaves the line $AH$ intact, so we just need to prove that it maps $H$ to $Q$.
As $H$ lies on the circumcircle of $ABC$, which is mapped to the nine-point circle $\omega$ of $ABC$ by the considered homothety, we just need to verify that $Q$ lies on $\omega$ and that $Q$ is not the “wrong” intersection point of $\omega$ and $AH$. But this other point is the midpoint of $BC$, which does not coincide with $Q$ when $AB \neq AC$.
So our current claim is just that $Q$ lies on $\omega$. To verify it, we denote the orthogonal projection of $A$ to $BC$ by $R$ and the midpoint of $AB$ by $S$. It is known that $R$, $S$, and $T$ lie on $\omega$. Further, the points $Q$ and $R$ lie on the circle with diameter $AD$. Hence
$\angle RQT=\angle BDA=\beta - \angle BAD=\beta - \gamma = \alpha - (180^{\circ}-2\beta)=\\\angle BST-\angle BSR=\angle RST$
and we may conclude.
FabrizioFelen
20.07.2016 17:52
Let $\ell$ the line parallel to $BC$ through $A$, let $Y$ the point in $\ell$ such that $YA=YH$ and let $X=DG\cap \ell$. Define $M$ the midpoint of $BC$ and $N$ the midpoint of $AH$ and $N'$ is the projection of $D$ in $AM$. Define $BM=MC=a$, $BD=b$, $HM=y$ and $AN=NH=x+y$. Since $AD^2$ $=$ $DB.DC$ and $BM$ $.$ $MC$ $=$ $MH$ $.$ $MA$ we get $AD^2$ $=$ $b(b+2a)$ and $a^2$ $=$ $y(2x+y)$ $\Longrightarrow$ since $AN'$ $\perp$ $AM$ we get $MN'^2$ $-$ $N'A^2$ $=$ $DM^2$ $-$ $AD^2$ $=$ $(a+b)^2$ $-$ $b(b+2a)$ $=$ $a^2$ $...(1)$. Furthermore $AN^2$ $-$ $NM^2$ $=$ $(x+y)^2$ $-$ $x^2$ $=$ $y(2x+y)$ $=$ $a^2$ $...(2)$ $\Longrightarrow$ by $(1)$ and $(2)$ we get $MN'^2$ $-$ $N'A^2$ $=$ $AN^2$ $-$ $NM^2$ $\Longrightarrow$ $AN'$ $=$ $MN$ and $AN$ $=$ $MN'$. Furthermore $\measuredangle N'MD$ $=$ $\measuredangle NAY$ and $AN$ $=$ $MN'$ we get $\triangle N'MD$ $\cong$ $\triangle NAY$ $\Longrightarrow$ $YA$ $=$ $MD$ $=$ $a+b$. On the other hand since $\triangle GDM$ $\sim$ $\triangle GXA$ we get $\tfrac{AX}{DM}$ $=$ $\tfrac{AG}{GM}$ $=$ $2$ $\Longrightarrow$ $AX$ $=$ $2(a+b)$ $\Longrightarrow$ $AY$ $=$ $YX$ $=$ $YH$ $=$ $a+b$ $\Longrightarrow$ $\measuredangle XHA$ $=$ $90^{\circ}...(3)$, also that $A(B,C,M,\infty)$ $=$ $H(E,F,G,X)=-1$ combining this result with $(3)$ we get $\measuredangle EHG$ $=$ $\measuredangle GHF$
EthanWYX2009
04.03.2023 13:17
Another link to this(sorry for posting it again).