Point that is orthocenter, incenter, and centroid

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Point that is orthocenter, incenter, and centroid

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Source: 2017 Peru Southern Cone TST P9

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#1 h Apr 5, 2025, 7:50 PM • 1 Y

Y by PikaPika999

Let $BXC$ be a triangle and $A_1, A_2, A_3$ points in the same plane such that $X$ is the orthocenter of triangle $A_1BC$ , $X$ is the incenter of triangle $A_2BC$ , and $X$ is the centroid of triangle $A_3BC$ . If line $A_1A_3$ is parallel to $BC$ , prove that $A_2$ is the midpoint of segment $A_1A_3$ .

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#2 h Apr 5, 2025, 9:48 PM • 1 Y

Y by PikaPika999

Not bad to complex bash. Let $\triangle BXC$ be inscribed in the unit circle so that
$|b|=|c|=|x|=1$ Since $A_1$ is the orthocenter of $\triangle BXC$ , we have
$a_1 = b+c+x$ Since $X$ is the centroid of $\triangle A_3BC$ , we have $x = \frac{a_3+b+c}3$ and so
$a_3 = 3x-b-c$ And since $BX$ bisects $\angle A_2BC$ and $CX$ bisects $\angle A_2CB$ , we have
$a_2 = \frac{b\left(\frac{x^2}c\right)\left(c+\frac{x^2}b\right) - c\left(\frac{x^2}b\right)\left(b+\frac{x^2}c\right)}{b\left(\frac{x^2}c\right) - c\left(\frac{x^2}b\right)} = \frac{bc+x^2}{b+c}$ Since $A_1A_3\parallel BC$ , we have
$\frac{a_1-a_3}{b-c} \in \mathbb{R}$ $\frac{2b+2c-2x}{b-c} = \frac{2bx+2cx-2bc}{cx-bx}$ $bx+cx-x^2 = bc-bx-cx$ $2bx+2cx = bc+x^2$ Then rearranging gives
$a_2 = \frac{bc+x^2}{b+c} = 2x = \frac{(b+c+x) + (3x-b-c)}2 = \frac{a_1+a_3}2$ as desired. $\blacksquare$

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