RMO 2016 ,Q1

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Source: Oct 23,2016

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sqing

#1 h Oct 25, 2016, 6:51 AM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be an isosceles triangle with $AB=AC.$ Let $\Gamma$ be its circumcircle and let $O$ be the centre of $\Gamma$ . let $CO$ meet $\Gamma$ in $D .$ Draw a line parallel to $AC$ thrugh $D.$ Let it intersect $AB$ at $E.$ Suppose $AE : EB=2:1$ .Prove that $ABC$ is an equilateral triangle.

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#2 h Oct 25, 2016, 9:55 AM • 4 Y

Y by svatejas, dumbduck, Adventure10, mqoi_KOLA

sqing wrote:

Let $ABC$ be an isosceles triangle with $AB=AC.$ Let $\Gamma$ be its circumcircle and let $O$ be the centre of $\Gamma$ . let $CO$ meet $\Gamma$ in $D .$ Draw a line parallel to $AC$ thrugh $D.$ Let it intersect $AB$ at $E.$ Suppose $AE : EB=2:1$ .Prove that $ABC$ is an equilateral triangle.

Let $AH$ be the altitude of the triangle. Because of $DE||AC$ we have $E\widehat DA = D\widehat AC = {90^0}$ and $D\widehat EA = \widehat A$ .

But, $A\widehat OD = O\widehat AC + O\widehat CA = \widehat A = D\widehat EA$ , thus $AOED$ is cyclic and $EO||BC$ , so $\frac{{AO}}{{OH}} = \frac{{AE}}{{EB}} = \frac{2}{1}$ , that means $O$ is centroid.

In the isosceles triangle $ABC$ , circumcenter $O$ is also centroid, hence the triangle is equilateral.

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41379 posts

sqing

#3 h Oct 26, 2016, 1:20 AM • 1 Y

Y by Adventure10

george_54 wrote:

sqing wrote:

Let $ABC$ be an isosceles triangle with $AB=AC.$ Let $\Gamma$ be its circumcircle and let $O$ be the centre of $\Gamma$ . let $CO$ meet $\Gamma$ in $D .$ Draw a line parallel to $AC$ thrugh $D.$ Let it intersect $AB$ at $E.$ Suppose $AE : EB=2:1$ .Prove that $ABC$ is an equilateral triangle.

Let $AH$ be the altitude of the triangle. Because of $DE||AC$ we have $E\widehat DA = D\widehat AC = {90^0}$ and $D\widehat EA = \widehat A$ .

But, $A\widehat OD = O\widehat AC + O\widehat CA = \widehat A = D\widehat EA$ , thus $AOED$ is cyclic and $EO||BC$ , so $\frac{{AO}}{{OH}} = \frac{{AE}}{{EB}} = \frac{2}{1}$ , that means $O$ is centroid.

In the isosceles triangle $ABC$ , circumcenter $O$ is also centroid, hence the triangle is equilateral.

Wonderful.
Thanks.

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#4 h Feb 10, 2019, 1:10 PM • 1 Y

Y by Adventure10

Easy to see, $\Delta AOB \cong \Delta AOC$ $\implies$ $\angle ABO=\angle ACO$ , Since, $\angle DAC=90^{\circ}$ $\implies$ $\angle EDA=90^{\circ}$ , hence, if $A'$ is the $A-$ antipode in $\odot (ABC)$ , then, $A' \in ED$ , $\angle DA'A=\angle DCA=\angle EBO$ $\implies$ $EBA'O$ is cyclic, $\implies$ $EO||BC$ hence, if $G$ is the centroid of $\Delta ABC$ , then, $O \equiv G$ $\implies$ $\Delta ABC$ is equilateral

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#5 h Sep 30, 2019, 1:13 PM • 1 Y

Y by Adventure10

At last solved a RMO geometery problem...I am a geo noob :oops:

:oops:

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