concurrency wanted,3 equilateral triangle on sides of a triangle

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Source: 2007 Ukraine MO grade VIII P8

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#1 h Jul 7, 2020, 11:14 AM

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On each side of the triangle $ABC$ on the outside are constructed equilateral triangles: $AB {{C} _ {1}}$ , $A {{B} _ {1}} C$ and ${ {A} _ {1}} BC$ . Through the midpoints of the segments ${{A} _ {1}} {{B} _ {1}}$ , ${{B} _ {1}} {{C} _ {1}}$ and ${{C} _ {1}} {{A} _ {1}}$ held lines perpendicular to the sides $AB$ , $BC$ and $AC$ , respectively. Prove that the drawn lines intersect at one point.

(Alexey Chubenko)

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#2 h Jul 12, 2020, 4:14 PM • 1 Y

Y by Mango247

Seems like I've been in eighth grade too long ago, since only I see here complex numbers or cartesian coordinates.

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#3 h Jul 12, 2020, 4:21 PM • 1 Y

Y by Mango247

how old is eighth grade?

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#4 h Jul 12, 2020, 4:37 PM

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it depends on the country, whether they have 11 or 12 years in secondary education, its either 15->16 or 14->15 years old respectively (I think)

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#5 h Jul 13, 2020, 4:49 PM

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Cartesian return. We can place $\triangle ABC$ on a plane so that there exist $a\in \mathbb{R}, c\in \mathbb{R}_+$ such that $B=(0,0), A=(2a,2), C=(2c,0)$ . Let me remind two important theorems.
$C_1=\left( ,\right)$ Theorem 1.
Given lines $m:\ B_1y=A_1x+C_1,\ n:\ B_2y=A_2x+C_2$ . Then $\tan\measuredangle(m,n)=\frac{A_1B_2-A_2B_1}{B_1B_2+A_1A_2}.$
remark on notation
Theorem 2.
Line passing through points $(x_1,y_1),(x_2,y_2)$ is given by equation $(x_2-x_1)(y-y_1)=(x-x_1)(y_2-y_1)$ .

Since $\triangle A_1BC$ is equilateral we have $A_1=(c,-c\sqrt3)$ We have
$AB:\ y=\frac{x}{a},\ AC: (a-c)y=(x-2c).$ Let $p(XY)$ be the slope of line $XY$ on a plane. By theorem 1
$\sqrt3=\tan60^\circ=\frac{p(BC_1)-p(AB)}{1+p(BC_1)p(AB)}\iff p(BC_1)=\frac{\sqrt3+p(AB)}{1-p(AB)\sqrt3}\implies p(BC_1)=\frac{1+a\sqrt{3}}{a-\sqrt3}.$ You may say "oops, if $a-\sqrt3=0$ we have a problem. But actually we don't because whether it's equal to $0$ or not you get the following:
$\exists_{d\in \mathbb{R}}\ C_1=\left( (a-\sqrt3)d,(a\sqrt3+1)d\right).$ Because $AC_1=BC_1$ and $(a,1)$ is the midpoint of segment $AB$ we have perpendicularity implying the following equation concerning scalar product $0=\left[ (a-\sqrt3)d-a,(a\sqrt3+1)d-1\right]\cdot[a,1]\iff d(a^2+1)=(a^2+1)\iff d=1\iff C_1=\left( a-\sqrt3,a\sqrt3+1\right).$ No comment for the following calculations as they are analogous.
$p(AB_1)=\frac{\sqrt3+p(AC)}{1-p(AC)\sqrt3}\implies p(AB_1)=\frac{1+(a-c)\sqrt{3}}{a-c-\sqrt3}$ $\exists_{z\in \mathbb{R}}\ B_1=\left( 2a+(a-c-\sqrt3)z,1+\left((a-c)\sqrt3+1\right)z\right)$ Midpoint of segment $AC$ is $(a+c,1)$ .
$\left[ a-c+(a-c-\sqrt3)z,\left((a-c)\sqrt3+1\right)z\right]\cdot[a+c,1]=0\iff z=-1$ $B_1=\left( a+c+\sqrt3,(c-a)\sqrt3+1\right)$ .
It's time to get intersection of lines $AA_1$ , ${{B} _ {1}} {{C} _ {1}}$ and ${{C} _ {1}} {{A} _ {1}}$ .
$AA_1:\ (2a-c)(y+c\sqrt3)=(x-c)(2+c\sqrt3)$ $CC_1:\ (a-\sqrt3-2c)y=(x-2c)(a\sqrt3+1)$ $BB_1:\ (a+c+\sqrt3)y=x\left((c-a)\sqrt3+1\right)$ Let's findout the intersection of $BB_1, CC_1$ (call it $P$ , does it always exist?) and then we'll prove it belongs to $AA_1$ .
$(a-\sqrt3-2c)x_P\left((c-a)\sqrt3+1\right)=(a-\sqrt3-2c)(a+c+\sqrt3)y_P=(a+c+\sqrt3)(x_P-2c)(a\sqrt3+1)$ $x_P=\frac{2c(a+c+\sqrt3)(a\sqrt3+1)}{(a+c+\sqrt3)(a\sqrt3+1)-(a-\sqrt3-2c)\left((c-a)\sqrt3+1\right)}$ $x_P=\frac{c(a+c+\sqrt{3})(a\sqrt{3}+1)}{ \sqrt3 a^2 - \sqrt3 a c + \sqrt3 c^2 + 3c + \sqrt3}$ From $BB_1$ equation conclude
$y_P=\frac{c\left((c-a)\sqrt3+1\right)(a\sqrt{3}+1)}{ \sqrt3 a^2 - \sqrt3 a c + \sqrt3 c^2 + 3c + \sqrt3}$ And now the cherry on top
$x_P-c=\frac{c(2a-c)(2+c\sqrt3)}{ \sqrt3 a^2 - \sqrt3 a c + \sqrt3 c^2 + 3c + \sqrt3}$ $y_P+c\sqrt{3}=\frac{c(2+c\sqrt3)^2}{ \sqrt3 a^2 - \sqrt3 a c + \sqrt3 c^2 + 3c + \sqrt3}$ Thus equality holds $(2a-c)(y_P+c\sqrt3)=\frac{c(2a-c)(2+c\sqrt3)^2}{ \sqrt3 a^2 - \sqrt3 a c + \sqrt3 c^2 + 3c + \sqrt3}=(x_P-c)(2+c\sqrt3)$ and $P\in AA_1$ QED

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#6 h Jul 13, 2020, 6:20 PM • 3 Y

Y by GeoMetrix, Bumblebee60, Aritra12

parmenides51 wrote:

On each side of the triangle $ABC$ on the outside are constructed equilateral triangles: $AB {{C} _ {1}}$ , $A {{B} _ {1}} C$ and ${ {A} _ {1}} BC$ . Through the midpoints of the segments ${{A} _ {1}} {{B} _ {1}}$ , ${{B} _ {1}} {{C} _ {1}}$ and ${{C} _ {1}} {{A} _ {1}}$ held lines perpendicular to the sides $AB$ , $BC$ and $AC$ , respectively. Prove that the drawn lines intersect at one point.

Clearly notice that the Perpendiculars from $A_1,B_1,C_1$ to $\overline{BC},\overline{CA},\overline{AB}$ are concurrent at the Circumcenter of $\Delta ABC$ . So, $\{\Delta ABC,\Delta A_1B_1C_1\}$ are Orthologic triangles, so by Sondat's Theorem the Perpendicular from $A,B,C$ to $\overline{B_1C_1},\overline{C_1A_1},\overline{A_1B_1}$ are concurrent $(\bigstar)$ . Let the midpoints of segments $B_1C_1,C_1A_1,A_1B_1$ be $A^*,B^*,C^*$ respectively. So as $A_1B_1\|A^*B^*, B_1C_1\|B^*C^* , C_1A_1\|C^*A^*$ , so the Perpendiculars from $A,B,C$ to $\overline{B^*C^*},\overline{C^*A^*},\overline{A^*B^*}$ are concurrent from $(\bigstar)$ . So, $\{\Delta ABC,\Delta A^*B^*C^*\}$ are Orthologic Triangles, so again by Sondat we get the desired conclusion. $\blacksquare$

This post has been edited 1 time. Last edited by amar_04, Jul 13, 2020, 6:23 PM

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