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Source: German TST 2022 extension

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#1 h Jun 29, 2023, 2:11 AM

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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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#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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168 posts

#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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