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Symmedian line
April   91
N an hour ago by BS2012
Source: All Russian Olympiad - Problem 9.2, 10.2
Let be given a triangle $ ABC$ and its internal angle bisector $ BD$ $ (D\in BC)$. The line $ BD$ intersects the circumcircle $ \Omega$ of triangle $ ABC$ at $ B$ and $ E$. Circle $ \omega$ with diameter $ DE$ cuts $ \Omega$ again at $ F$. Prove that $ BF$ is the symmedian line of triangle $ ABC$.
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Geometry from germany
shinhue   3
N Jul 1, 2023 by aqwxderf
Source: German TST 2022 extension
Let $ABC$ be a triangle with orthocenter $H$ and circumcenter $O$. Let $P$ be a point in the plane such that $AP$ passes through the Euler center. Let $Q$ and $R$ be the reflections of $P$ in the lines $CA$ and $AB$, respectively. Let $Y$ be the orthogonal projection of $R$ onto $CA$. Let $Z$ be the orthogonal projection of $Q$ onto $AB$. Assume that $H \neq O$ and $Y \neq Z$. Prove that $YZ||BC$.
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shinhue
Jun 29, 2023
aqwxderf
Jul 1, 2023
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Source: German TST 2022 extension
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shinhue
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shinhue
#1 h  Jun 29, 2023, 2:11 AM
Y by
Let $ABC$ be a triangle with orthocenter $H$ and circumcenter $O$. Let $P$ be a point in the plane such that $AP$ passes through the Euler center. Let $Q$ and $R$ be the reflections of $P$ in the lines $CA$ and $AB$, respectively. Let $Y$ be the orthogonal projection of $R$ onto $CA$. Let $Z$ be the orthogonal projection of $Q$ onto $AB$. Assume that $H \neq O$ and $Y \neq Z$. Prove that $YZ||BC$.
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pi_quadrat_sechstel
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pi_quadrat_sechstel
#2 h  Jun 30, 2023, 8:03 PM • 1 Y
Y by AndresSchwepps
shinhue wrote:
Let $ABC$ be a triangle with orthocenter $H$ and circumcenter $O$. Let $P$ be a point in the plane such that $AP$ passes through the Euler center. Let $Q$ and $R$ be the reflections of $P$ in the lines $CA$ and $AB$, respectively. Let $Y$ be the orthogonal projection of $R$ onto $CA$. Let $Z$ be the orthogonal projection of $Q$ onto $AB$. Assume that $H \neq O$ and $Y \neq Z$. Prove that $YZ||BC$.

Use complex coordinates with the excircle of $ABC$ as the unit circle. We have $o=0,h=a+b+c,p=\frac{a+b+c}{2}$. Let $R',Q'$ be the reflection of $R,Q$ in $AC,AB$ respectivly. We have
\begin{align*} r=a+b-ab\overline{p}=\frac{a+b-ab\overline{c}}{2}\\ r'=a+c-ac\overline{r}=\frac{2a+c-ac\overline{b}+c^2\overline{b}}{2}\\ y=\frac{r+r'}{2}=\frac{3a+b+c-ab\overline{c}-ac\overline{b}+c^2\overline{b}}{4} \end{align*}Analogously $z=\frac{3a+b+c-ab\overline{c}-ac\overline{b}+b^2\overline{c}}{4}$ So
\[ \frac{y-z}{c-b}=\frac{c^2\overline{b}-b^2\overline{b}}{4(c-b)}=\frac{b^2+bc+c^2}{4bc}=\frac{b\overline{c}+1+c\overline{b}}{4}\in\mathbb{R} \]So $YZ||BC$.
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AndresSchwepps
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AndresSchwepps
#3 h  Jul 1, 2023, 10:21 AM
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Well, after such a short and clean complex bash, it is difficult to offer something better. I will illustrate an application of how to use the "co-tangent trick".
Let $a,b,c$ be the lenghts of the side of $ABC$, and $\alpha,\beta,\gamma$ the corresponding angles. If $X$ is the Euler center, let $\lambda=\angle QAC=\angle CAX$ and $\mu=\angle XAB=\angle BAR$. We have to prove that $\frac{AY}{AZ}=\frac{AC}{AB}$. Note that:
\begin{eqnarray*} \frac{AY}{AZ}&=&\frac{AR\cos(2\mu+\lambda)}{AQ\cos(2\lambda+\mu)}=\frac{\cos(2\alpha-\lambda)}{\cos(\alpha+\lambda)}=\\ &=&\frac{\cos(2\alpha)\cos(\lambda)+\sin(2\alpha)\sin(\lambda)}{\cos(\alpha)\cos(\lambda)-\sin(\alpha)\sin(\lambda)}=\\ &=&\frac{\cos(2\alpha)\cot(\lambda)+\sin(2\alpha)}{\cos(\alpha)\cot(\lambda)-\sin(\alpha)}. \end{eqnarray*}The problem will be solved if we prove that the above expression is equal to $\frac{b}{c}$. We need some more information about $\lambda,\mu$. Let $M,N$ be the midpoints of $AC,AB$; let $E,F$ be the feet of the altitudes of $B,C$, and let $K,N$ be the projections of $X$ onto $AC,AB$.
Let us compute $\frac{AK}{AL}$ in two different ways:
\begin{eqnarray*} \frac{AK}{AL}&=&\frac{(AM+AE)/2}{(AN+AF)/2}=\frac{b+2c\cdot \cos(\alpha)}{c+2b\cdot \cos(\alpha)},\\ \frac{AK}{AL}&=&\frac{AX\cos(\lambda)}{AX\cos(\mu)}=\frac{\cos(\lambda)}{\cos(\alpha-\lambda)}=\\ &=&\frac{\cos(\lambda)}{\cos(\alpha)\cos(\lambda)+\sin(\alpha)\sin(\lambda)}=\\ &=&\frac{\cot(\lambda)}{\cos(\alpha)\cot(\lambda)+\sin(\alpha)}. \end{eqnarray*}Equating both ratios we obtain:
\begin{eqnarray*} \cot(\lambda)(c+2b\cdot \cos(\alpha))&=&(b+2c\cdot \cos(\alpha))(\cos(\alpha)\cot(\lambda)+\sin(\alpha))\\ \cot(\lambda)(c+2b\cdot \cos(\alpha)-b\cos(\alpha)-2c\cdot \cos^2(\alpha))&=&b\sin(\alpha)+2c\cdot \cos(\alpha)\sin(\alpha)\\ \cot(\lambda)(b\cos(\alpha)-c \cos(2\alpha))&=&b\sin(\alpha)+c\sin(2\alpha)\\ b(\cot(\lambda)\cos(\alpha)-\sin(\alpha))&=&c(\cos(2\alpha)\cot(\lambda)+\sin(2\alpha)), \end{eqnarray*}as we wanted to prove.
We are still waiting for a synthetic solution.
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aqwxderf
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aqwxderf
#4 h  Jul 1, 2023, 4:36 PM • 1 Y
Y by AndresSchwepps
Let $M $ and $N $ be the midpoints of $[PR]$ and $[PQ]$.

$AMNP $- cyclic $\Rightarrow \measuredangle ANM = \measuredangle APM$.
$[AP] = [AR] \Rightarrow \measuredangle APM = \measuredangle PRA$.
$AMRY $- cyclic $\Rightarrow \measuredangle PRA = \measuredangle MYA$.
So $[MN] = [MY]$ and similarly $[MN] = [NZ]$.

Let $I $ be the orthocenter of $\triangle MAN \Rightarrow \overline{AI}$, $\overline{AP}$ - isogonal in $\measuredangle BAC$.
$\overline{MI} \perp \overline{YN}$, $[MY] = [MN] \Rightarrow \measuredangle IYN = \measuredangle INY = 90^{\circ} - \measuredangle BAC$.
Similarly $\measuredangle IZM = 90^{\circ} - \measuredangle BAC$.

Let $K $ be the reflection of $O $ over $\overline{BC}$. Its known that $K \in \overline{AP}$.
Let $L $ be the isogonal conjugate of $K $ with respect to $\triangle ABC $.
$\overline{AI}$, $\overline{AP}$ - isogonal in $\measuredangle BAC$, $K \in \overline{AP} \Rightarrow L \in \overline{AI}$.

$\measuredangle LCA = \measuredangle BCK = \measuredangle OCB = 90^{\circ} - \measuredangle BAC \Rightarrow \measuredangle LCA = \measuredangle IYN \Rightarrow IY \parallel LC$.
Similarly $IZ \parallel LB$.
$IY \parallel LC$, $IZ \parallel LB \Rightarrow YZ \parallel BC$.
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