perpendicular wanted, circumcenter, circumcircle 2016 Estonia Open Senior 2.5

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perpendicular wanted, circumcenter, circumcircle 2016 Estonia Open Senior 2.5

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#1 h Oct 15, 2020, 6:40 PM

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The circumcentre of an acute triangle $ABC$ is $O$ . Line $AC$ intersects the circumcircle of $AOB$ at a point $X$ , in addition to the vertex $A$ . Prove that the line $XO$ is perpendicular to the line $BC$ .

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#2 h Feb 24, 2021, 6:09 PM

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Let $A = (a,n), B = (0,0),$ and $C = (c,0)$
Perpendicular bisector of $BC$ is $x = \frac{c}{2}$ . Perpendicular bisector of $AB$ is $y = -\frac{a}{n} x + \frac{a^2+n^2}{2n}$
$O$ is the intersection of these $2$ lines, so we get
$y = -\frac{a}{n} (\frac{c}{2}) + \frac{a^2+n^2}{2n} \Rrightarrow y = \frac{a^2-ac+n^2}{2n}$
and
$x = \frac{c}{2} \Rrightarrow$
$O = (\frac{c}{2}, \frac{a^2-ac+n^2}{2n})$ Let the circumcenter of $AOB$ be $P$ .
As before, perpendicular bisector of $AB = -\frac{a}{n}x + \frac{a^2+n^2}{2n}$ , and we can find that the perpendicular bisector of $BO$ is given by $y = \frac{-cn}{a^2-ac+n^2} x + \frac{(a^2+n^2)(a^2-2ac+c^2+n^2)}{4n(a^2-ac+n^2)}$
$P$ is the intersection of these $2$ lines, so setting them equal and solving, we get
$P = (\frac{a^2-c^2+n^2}{4(a-c)}, \frac{a^3-2a^2c+ac^2+an^2-2cn^2}{4n(a-c)}$ $X$ is the point of intersection, other than $A$ , of the circle centered at $P$ and the line $AC$ . We see the circle centered at $P$ is given by the equation
$(x- \frac{a^2-c^2+n^2}{4a-4c})^2 + (y - \frac{a^3-2a^2c+ac^2+an^2-2cn^2}{4n(a-c)})^2 = (\frac{a^2-c^2+n^2}{4a-4c})^2 + (\frac{a^3-2a^2c+ac^2+an^2-2cn^2}{4n(a-c)})^2$ Line $AC$ is given by the equation $y = \frac{-n}{c-a}x + \frac{cn}{c-a}$ . Substituting this value of $y$ into the equation of the circle, and simplifying, we get
$x^2 + 2x(\frac{a^2-c^2+n^2}{4a-4c}) + y^2 - 2y(\frac{a^3-2a^2c+ac^2+an^2-2cn^2}{4n(a-c)}) = 0$ $\Rrightarrow x^2 + 2x(\frac{a^2-c^2+n^2}{4a-4c}) + (\frac{-n}{c-a}x + \frac{cn}{c-a})^2 - 2(\frac{-n}{c-a}x + \frac{cn}{c-a})(\frac{a^3-2a^2c+ac^2+an^2-2cn^2}{4n(a-c)}) = 0$ Solving for $x$ in the above equation, we get $x = a, y = n$ , which are the coordinates of point $A$ , and $x = \frac{c}{2}, y = \frac{cn}{2(c-a)}$ . Thus, $XO \perp BC$ . $\blacksquare$

This post has been edited 2 times. Last edited by JustinLee2017, Feb 24, 2021, 6:15 PM

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#3 h Feb 24, 2021, 6:12 PM • 5 Y

Y by fukano_2, JustinLee2017, IAmTheHazard, vsamc, mathleticguyyy

Hello JustinLee2017,

You have killed a dolphin because of your unethical methods to approaching this problem.

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#4 h Feb 24, 2021, 6:16 PM

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@tree_4 what-

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