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A beautiful collinearity regarding three wonderful points
math_pi_rate   10
N 6 minutes ago by alexanderchew
Source: Own
Let $\triangle DEF$ be the medial triangle of an acute-angle triangle $\triangle ABC$. Suppose the line through $A$ perpendicular to $AB$ meet $EF$ at $A_B$. Define $A_C,B_A,B_C,C_A,C_B$ analogously. Let $B_CC_B \cap BC=X_A$. Similarly define $X_B$ and $X_C$. Suppose the circle with diameter $BC$ meet the $A$-altitude at $A'$, where $A'$ lies inside $\triangle ABC$. Define $B'$ and $C'$ similarly. Let $N$ be the circumcenter of $\triangle DEF$, and let $\omega_A$ be the circle with diameter $X_AN$, which meets $\odot (X_A,A')$ at $A_1,A_2$. Similarly define $\omega_B,B_1,B_2$ and $\omega_C,C_1,C_2$.
1) Show that $X_A,X_B,X_C$ are collinear.
2) Prove that $A_1,A_2,B_1,B_2,C_1,C_2$ lie on a circle centered at $N$.
3) Prove that $\omega_A,\omega_B,\omega_C$ are coaxial.
4) Show that the line joining $X_A,X_B,X_C$ is perpendicular to the radical axis of $\omega_A,\omega_B,\omega_C$.
10 replies
math_pi_rate
Nov 8, 2018
alexanderchew
6 minutes ago
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big-picture ??
geomath111   2
N Oct 28, 2023 by geomath111
Let ABC be a triangle with incenter I. Points M and N are the midpoints of side AB
and AC, respectively. Points D and E lie on lines AB and AC, respectively, such that BD = CE = BC. Line L_1 pass through D and is perpendicular to line IM. Line L_2 passes through E and is perpendicular to line IN. Let P be the intersection of lines L_1 and L_2. Prove that AP is perpendicular BC.
2 replies
geomath111
Oct 22, 2023
geomath111
Oct 28, 2023
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big-picture ??
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geomath111
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geomath111
#1 h  Oct 22, 2023, 6:40 AM
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Let ABC be a triangle with incenter I. Points M and N are the midpoints of side AB
and AC, respectively. Points D and E lie on lines AB and AC, respectively, such that BD = CE = BC. Line L_1 pass through D and is perpendicular to line IM. Line L_2 passes through E and is perpendicular to line IN. Let P be the intersection of lines L_1 and L_2. Prove that AP is perpendicular BC.
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v4913
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v4913
#2 h  Oct 27, 2023, 6:00 PM
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Let $X$ and $Y$ be the intouch points on $AB$ and $AC$, and $r$ be the inradius, and let the $A$-altitude intersect $\ell_1$ and the perpendicular from $D$ to $AD$ at $L, P_1$ respectively; then by the Ratio Lemma on $\triangle{ADL}$, $$\frac{AP_1}{P_1L} = \tan \angle{B} \tan \angle{MIX} = \frac{MX \tan \angle{B}}{r}$$$$\implies AP_1 = \frac{AD}{\sin \angle{B}} \cdot \frac{MX \tan \angle{B}}{MX \tan \angle{B} + r}.$$
Similarly, if $P_2$ is the intersection of $\ell_2$ with the $A$-altitude, then $AP_2 = \frac{AE}{\sin \angle{C}} \cdot \frac{NY \tan \angle{C}}{NY \tan \angle{C} + r}$.

Since $MX = \frac{b-a}{2}$ and $AE = b-c$, $AE = 2 \cdot MX$ and similarly $AD = 2 \cdot NY$, so $$AP_1 = AP_2 \iff \frac{2 \cdot MX}{AB} \cdot \frac{NY \tan \angle{C}}{NY \tan \angle{C} + r} = \frac{2 \cdot NY}{AC} \cdot \frac{MX \tan \angle{B}}{MX \tan \angle{B} + r}$$$$\iff AB\left(NY + \frac{r}{\tan \angle{C}}\right) = AC\left(MX + \frac{r}{\tan \angle{B}}\right)$$which is equivalent to $AB(NY + KC_1) = AC(MX + KB_1)$ if $K$ is the intouch point on $AB$ and $B_1, C_1 \in BC$ such that $IB_1 \parallel AB, IC_1 \parallel AC$. And this evaluates to $$c\left(\frac{c-a}{2} + s-c - \frac{r}{\sin \angle{C}}\right) = b\left(\frac{b-a}{2} + s-b - \frac{r}{\sin \angle{B}}\right)$$$$\frac{bc}{2} - \frac{rc}{\sin \angle{C}} = \frac{bc}{2} - \frac{rb}{\sin \angle{B}}$$which is clearly true by the Law of Sines in $\triangle{ABC}$. $\square$
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geomath111
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geomath111
#3 h  Oct 28, 2023, 12:55 PM
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thank u so much.
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