the triangles have the same centroid

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the triangles have the same centroid

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V-217

#1 h Jul 20, 2023, 5:39 PM

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Let $ABC$ an acute triangle inscribed in a circle with its center $O$ . Let the lines $B_{a}C_{a}, C_{b}A_{b}$ and $A_{c}B_{c}$ the perpendiculars in $O$ to $AO, BO, CO$ respectively, where $A_{b},A_{c}\in BC$ , $B_{a},B_{c}\in AC$ and $C_{a},C_{b}\in AB$ . Let $O_{a}, O_{b}, O_{c}$ be the circumcircles of triangles $AC_{a}B_{a}, BA_{b}C_{b}, CA_{c}B_{c}$ .
Prove that the triangles $O_{a}O_{b}O_{c}$ and $ABC$ have the same centroid.

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Ianis

#2 h Jul 20, 2023, 10:04 PM

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V-217 wrote:

Let $ABC$ an acute triangle inscribed in a circle with its center $O$ . Let the lines $B_{a}C_{a}, C_{b}A_{b}$ and $A_{c}B_{c}$ the perpendiculars in $O$ to $AO, BO, CO$ respectively, where $A_{b},A_{c}\in BC$ , $B_{a},B_{c}\in AC$ and $C_{a},C_{b}\in AB$ . Let $O_{a}, O_{b}, O_{c}$ be the circumcircles of triangles $AC_{a}B_{a}, BA_{b}C_{b}, CA_{c}B_{c}$ .
Prove that the triangles $O_{a}O_{b}O_{c}$ and $ABC$ have the same centroid.

Use complex numbers with $(ABC)=\mathbb{S}^1$ . Define $a'=ia$ and $a''=-ia$ , then $B_a=A'A''\cap CA$ , so $\begin{align*}b_a & =\frac{a'a''(c+a)-ca(a'+a'')}{a'a''-ca} \\ & =\frac{(ia)(-ia)(c+a)-ca((ia)+(-ia))}{(ia)(-ia)-ca} \\ & =\frac{a^2(c+a)}{a^2-ca} \\ & =a\frac{a+c}{a-c}. \end{align*}$ Analogously we find $c_a=a\frac{a+b}{a-b}$ . We have $o_a=a+\frac{(b_a-a)(c_a-a)\overline{(b_a-c_a)}}{\overline{(b_a-a)}(c_a-a)-(b_a-a)\overline{(c_a-a)}}.$ Now compute

$b_a-a=a\frac{c+a}{a-c}-a=\frac{2ac}{a-c}$ ,
$c_a-a=a\frac{a+b}{a-b}-a=\frac{2ab}{a-b}$ ,
$b_a-c_a=a\frac{c+a}{a-c}-a\frac{a+b}{a-b}=\frac{2a^2(c-b)}{(a-b)(a-c)}$ ,
$\overline{(b_a-a)}=\frac{\frac{2}{ac}}{\frac{1}{a}-\frac{1}{c}}=\frac{2}{c-a}$ ,
$\overline{(c_a-a)}=\frac{\frac{2}{ab}}{\frac{1}{a}-\frac{1}{b}}=\frac{2}{b-a}$ ,
$\overline{(b_a-c_a)}=\frac{\frac{2}{a^2}\left (\frac{1}{c}-\frac{1}{b}\right )}{\left (\frac{1}{a}-\frac{1}{b}\right )\left (\frac{1}{a}-\frac{1}{c}\right )}=\frac{2(b-c)}{(a-b)(a-c)}$ .

Then $o_a=a+\frac{\dfrac{2ac}{a-c}\dfrac{2ab}{a-b}\dfrac{2(b-c)}{(a-b)(a-c)}}{\dfrac{2}{c-a}\dfrac{2ab}{a-b}-\dfrac{2ac}{a-c}\dfrac{2}{b-a}}=a+\frac{2abc}{(c-a)(a-b)}.$ Analogously we find $o_b=b+\frac{2abc}{(a-b)(b-c)}$ and $o_c=c+\frac{2abc}{(b-c)(c-a)}$ . Then $\frac{o_a+o_b+o_c}{3}=\frac{a+b+c}{3}+\frac{2}{3}abc\left (\frac{1}{(c-a)(a-b)}+\frac{1}{(a-b)(b-c)}+\frac{1}{(b-c)(c-a)}\right )=\frac{a+b+c}{3},$ so both triangles have the same centroid.

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