We claim that $XY$ is the Steiner Line of $M$ WRT $(ABC)$, where $A \neq M= (ABC) \cap (AEF)$. Let $X_1$ the reflection of $X$ WRT $AB$ and $C \neq H_C= CH \cap (ABC)$. Define $Y_1,H_B$ similarly. In order to prove that $XY$ is the Steiner Line of $M$ WRT $(ABC)$, it suffices to prove that $X_1,H_C,M$ are collinear, as well as $Y_1,H_B,M$, since it's well known that $H_B,H_C$ are the reflections of $H$ WRT $AC,AB$, respectively. Now, observe that since $M= (ABC) \cap (AEF) \implies M$ is the center of the spiral similarity mapping $FE \mapsto BC (\star)$. Since $X_1$ is the reflection of $X$ WRT $AB \implies$ $\Delta FX_1B \sim \Delta FXB \sim \Delta EJC \implies$ $X_1 \mapsto J$ in this spiral similarity. Thus, $\angle X_1MJ= \angle BMC = \angle BAC (\star \star)$. Furthermore, from $(\star)$ and since $BCEF$ is cyclic with center $J$, it's well known that $\angle AMJ= 90º$. Let $M \neq A'= MJ \cap (ABC) \implies A'$ is the antipode of $A$ WRT $(ABC)$. Then, $\angle H_CMJ= \angle H_CMA' = \angle H_CAA'= \angle H_CAB+ \angle BAA'= 90º- \angle ABC + 90º - \angle ACB= \angle BAC$. Thus, from $(\star \star)$, we have that $X_1,H_C,M$ are collinear. Similarly, $Y_1,H_B,M$ are collinear, implying the desired result. Hence, $XY$ is the Steiner Line of $M$ WRT $(ABC)$. This implies that $H,X,Y$ are collinear, as desired. $\blacksquare$

I think the Moving Points Method works. We will prove the statement is true for all points $J$ in the perpendicular bisector $m$ of $BC$. Let $H$ be the orthocenter of $ABC$, and animate $J$. Consider the following cases: If $J=m\cap CH$, then $X\equiv H$ and the statement is true. If $J=m\cap BH$, then $Y\equiv H$, and the statement is true. If $J$ is the circumcenter of $ABC$, then $E\equiv A\equiv F$, so $\angle BAX=\angle CAJ=\angle BAH$ and $\angle CAY=\angle BAJ=\angle CAH$, hence $A,X,H,Y$ are collinear.

GianDR wrote: I think the Moving Points Method works. We will prove the statement is true for all points $J$ in the perpendicular bisector $m$ of $BC$. Let $H$ be the orthocenter of $ABC$, and animate $J$. Consider the following cases: If $J=m\cap CH$, then $X\equiv H$ and the statement is true. If $J=m\cap BH$, then $Y\equiv H$, and the statement is true. If $J$ is the circumcenter of $ABC$, then $E\equiv A\equiv F$, so $\angle BAX=\angle CAJ=\angle BAH$ and $\angle CAY=\angle BAJ=\angle CAH$, hence $A,X,H,Y$ are collinear. You just proved the problem for $3$ choices of $J$. In order to use the Moving Points Method, you must define some projective maps or the degree of the Moving Points.

rcorreaa wrote: You just proved the problem for $3$ choices of $J$. In order to use the Moving Points Method, you must define some projective maps or the degree of the Moving Points. Points $E$ has degree $1$ because it's the reflection of $C$ through the projection of $J$ in $CA$. Analogously, $F$ has degree $1$. Now $X=f(J)$ where $f$ is the composition of the spiral similarity centered at $A$ and mapping $CE\mapsto BF$ and the reflection through line $AB$, hence $X$ has degree $1$. Analogously $Y$ has degree $1$. So it's enough to prove the statement for $3$ positions of $J$. Is it complete now? BTW How would you write this? I am not very familiar with this method

You cannot use the spiral similarity as a projective map, since its center is not fixed when you move $J$. You can learn more about the Moving Points Method here: https://artofproblemsolving.com/community/q1h1884540p12835147

The center of spiral similarity is $A$, and it's fixed since it's independent of $J$. Shouldn't it work then?

One solution for this problem.

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GianDR wrote: The center of spiral similarity is $A$, and it's fixed since it's independent of $J$. Shouldn't it work then? The spiral similarity mapping $CE \mapsto BF$ is $(ABC) \cap (AEF)$.

Yes of course, you are right. Is there any way to find $\text{deg}(X)$ avoiding tedious calculations?

Nice problem! Let: $X_b$ and $X_c$ be the foot of the perpendicular from $X$ to $AB$ and $AC$, $Y_b$ and $Y_c$ be the foot from $Y$ to $AB$ and $AC$, $H_b$ and $H_c$ be the foot from $H$ to $AB$ and $AC$, $R$ be the radius of $(BCEF)$. And finally $\angle JFB = \theta$, $\angle JEC = \alpha$, $\angle CAO = \beta$, $\angle OAB = A - \beta$. It's enough to prove that $\dfrac{X_cH_c}{X_bH_b} = \dfrac{Y_cH_c}{Y_bH_b}$. So, let's Compute $X_cH_c$ and $Y_cH_c$: (we can assume $WLOG$ that $AX_b < AY_b < AH_b$ and $AX_c < AY_c < AH_c$) 1) $X_cH_c = AH_c$ $-$ $AX_c$. Verify that $AX_c = AX_bcos\angle A + XX_bsin\angle A$ (law of sines on $(AX_bXX_c)$). 2) $AH_c = ABcos\angle A$ 3) $BX_b = Rcos\angle \theta$ 4) $XX_b = \dfrac{Rcos\angle \theta sin\angle \alpha}{cos\angle \alpha}$ 5) Then: $X_cH_c = AH_c$ $-$ $AX_c = ABcos \angle A - AX_bcos\angle A - XX_bsin\angle A$ = $BX_bcos\angle A$ $-$ $XX_bsin\angle A$ $ = \dfrac{Rcos\angle \theta(cos\angle \alpha cos\angle A - sin\angle \alpha sin\angle A) }{cos\angle \alpha} = \dfrac{Rcos\angle \theta cos\angle (\alpha + A)}{cos\angle \alpha}$ Now, to compute $Y_cH_c$: 1) $AH_c = ABcos \angle A = \dfrac{AJ(cos\ A)(sen \angle (A - \beta + \theta))}{sin \angle \theta}$ 2) $AY_c = AJcos \angle \beta$ 3) Then: $Y_cH_c = AH_c - AY_c = \dfrac{AJ(cos \angle A sen \angle (A - \beta + \theta) - sin \angle \theta cos \angle \beta)}{sin \angle \theta} = \dfrac{AJ(cos \angle (A + \theta))(sin \angle (A - \beta)) }{sin \angle \theta} = R(cos \angle (A + \theta)) $ As the problem is symmetric in $B$ and $C$, we have $Y_bH_b = \dfrac{Rcos\angle \alpha cos\angle (A + \theta)}{cos\angle \theta}$ and $X_bH_b = R(cos \angle (A + \alpha))$. Now we just have to verify that $\dfrac{X_cH_c}{X_bH_b} = \dfrac{Y_cH_c}{Y_bH_b} = \frac{cos\angle \theta}{cos \angle \alpha}$ So we are done. $\blacksquare$

There is a complex solution too

alinazarboland wrote: There is a complex solution too Can anyone post a complex solution I wasn't able to do it using complex numbers?

My general problem and proof.

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geometry6 wrote: Can anyone post a complex solution I wasn't able to do it using complex numbers? You're welcome! Let $H$ be the orthocenter of $\triangle ABC$. We use complex numbers with $(BCEF)$ as the unit circle. $\triangle FXB \sim \triangle EJC\implies \frac{x-f}{x-b} = \frac{c}{e}\implies x = \frac{bc-ef}{c-e}$. Let the second intersection of $BH,CH$ with $(BCEF)$ be $H_b$ and $H_c$ respectively. $h_b = -\frac{ec}{b}, h_c = -\frac{bf}{c}$. As $H = CH_c\cap BH_b\implies$ $$h = \frac{e(c^2-bf)-f(b^2-ce)}{ce-bf}$$\begin{align*} x-h = \frac{(ec^2 - bef - b^2f+cef)(e-c) + (bc-ef)(ce-bf)}{(ce-bf)(c-e)} =\\ = \frac{e^2c^2 - be^2f -b^2ef + ce^2f - ec^3 + bcef + b^2cf - c^2ef + bc^2e + bef^2 - ce^2f - b^2cf} {(ce-fb)(c-e)} = \\ = \frac{e^2c^2 - be^2f -b^2ef - ec^3 + bcef - c^2ef + bc^2e + bef^2} {(ce-fb)(c-e)} = \\ = \frac{e(c^2(e-f)-fb(b-c)+c^2(b-c)-fb(e-f))} {(ce-fb)(c-e)} = \\ = \frac{e(c^2-fb)(b+e-c-f)} {(ce-fb)(c-e)}\\ \end{align*}and by symmetry $y-h = \frac{f(b^2-ec)(c+f-b-e)}{(bf-ce)(b-f)}$. $$ \frac{\overline{x-h}}{\overline{y-h}} = \overline{\frac{e}{f}\cdot \frac{c^2-bf}{b^2-ec}\cdot \frac{b-f}{c-e}} = \frac{e}{f}\cdot \frac{c^2-bf}{b^2-ec}\cdot \frac{b-f}{c-e}\blacksquare$$Remark: One can also show that $Z=BE\cap CF$ is on that line using the same method.

buratinogigle wrote: My general problem and proof. Can you send a pdf or something this way? The graphics resolution of your picture is quite bad and it is illegible.

dungnguyentien wrote: One solution for this problem. The same idea as here; let $\beta ,\gamma$ be circles with diameters $BE,CF$, then well-known that orthocenter lies on radical axis of these circles. Denote by $M,N$ midpoints of $CF,BE$. By angle chasing $\angle NBX=\angle FCJ$ and $$|FJ|^{2}=|JC|^{2}+4|FM|^{2}-4|JC|\cdot |FM|\cdot \cos \angle FCJ,$$i.e. $|FM|=|JC|\cdot \cos \angle NBX$. Analogously $|BN|=|JB|\cdot \cos \angle MFX$, and thus $$|BN|^{2}-|XN|^{2}=|BX|\cdot (2|BN| \cdot \cos \angle NBX-|BX|)=|FX|\cdot (2|FM| \cdot \cos \angle MFX-|FX|)=|CM|^{2}-|XM|^{2}.$$Hence $X$ lies on radical axis of $\beta ,\gamma,$ and analogously $Y$ lies implying desired.

GianDR wrote: rcorreaa wrote: hence $X$ has degree $1$. Analogously $Y$ has degree $1$

I don't know much about MMP

dungnguyentien wrote: One solution for this problem. How did you get the idea to take the circles $(BE),(CF)$?