Bosnia and Herzegovina 2011

by franzliszt, Jun 26, 2024, 8:52 PM

Bosnia and Herzegovina 2011 wrote:

In triangle $ABC$ , $AB+AC=2BC$ . Let $M$ and $N$ be midpoints of $AB$ and $AC$ , and let $I$ be the incenter of $ABC$ . Prove that $A, M, I, N$ are concyclic.

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Proof. Clearly $M,N$ , the midpoints of $AB,AC,$ in addition to the circumcenter $O$ and $A$ are concyclic with diameter $OA.$ Thus, it suffices to show that $I$ lies on $(AMN)$ .

The general equation of a circle is $-a^2yz-b^2zx-c^2xy+(x+y+z)(ux+vy+wz)=0$ for some constants $u,v,w$ . To find the constants $u,v,w$ which determine $\Omega$ , we can plug in each of $A,M,N$ in the general form and solve for $u,v,w$ . Note that the equation of a circle is homogenous so we can use homogenized coordinates here.

Plugging in $A=(1,0,0)$ gives $u=0$ .
Plugging in $M=(1:0:1)$ gives $u+w=\frac{b^2}2$ .
Plugging in $N=(1:1:0)$ gives $u+v=\frac{c^2}2$ .

Plugging in these constants in the general equation, we find that $\Omega$ has equation $-a^2yz-b^2zx-c^2xy+(x+y+z)\left(\frac{c^2}2\cdot y+\frac{b^2}2\cdot z\right)=0.$ Hence, using Barycentric Power of a Point, we can compute $\text{Pow}_{\Omega}(I)=-(a+b+c)(abc)+(a+b+c)\left(\frac{c^2}2\cdot b+\frac{b^2}2\cdot c\right)$ since $I=(a:b:c)$ . But this is $0$ since $a=\frac{b+c}2$ . Done. $\square$

This post has been edited 1 time. Last edited by franzliszt, Jun 26, 2024, 8:55 PM

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Geometry through Barycentric Coordinates

Bosnia and Herzegovina 2011

by franzliszt, Jun 26, 2024, 8:52 PM

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