Team Selection Test for IMO

Let $H$ be the orthocenter of the triangle $ABC$. Let the points $K,M,L$ be the intersection of the lines $AI$ and $OD$, $AI$ and $OH$, $AH$ and $OI$. We know that $AH=2OD$. By a simple angle chasing we see that $AI$ is an angle bisector of $HAO$. Therefore $$\frac{AO}{OD}=2\frac{AO}{AH}=2\frac{OM}{MH}$$ (1).

That is easy to see that $K$ is on the circumcircle of $ABC$. Let $\angle A=2\alpha$. Then we have $BC=2R\sin (2\alpha )$ and $AT=r\cdot \cot \alpha$. Then $$\frac{BC}{AT}=2\frac{2R\sin \alpha }{r/\sin \alpha }=2\frac{BK}{AI}=2\frac{KI}{AI}=2\frac{OI}{IL}$$ (2) since $BK=KI$ and $AH\parallel OK$.

Let $HM=x$, $MP=y$, $IL=k$, $IQ=s$. Then $PO=x+y$ by the very well known fact that $P$ is the midpoint of $[HO]$, $OQ=s$, $OM=x+2y$, $OI=2s$, $QL=k+s$. Applying Menelaus Theorem on the triangle $ALI$ with respect to the collinear points $H,M,O$ gives $$\frac{AH}{AL}=\frac{OIHM}{ILMO}=\frac{2sx}{k(x+2y)}.$$ Applying Menelaus Theorem on the triangle $ALQ$ with respect to the collinear points $H,P,O$ gives $$\frac{AH}{AL}=\frac{OQHP}{QLPO}=\frac{s}{k+s}.$$ Therefore $2x(k+s)=k(x+2y)$, that is $$\frac{y}{x}-\frac{s}{k}=\frac{1}{2}$$ (3).

Finally (1), (2) and (3) result $$\frac{AO}{OD}-\frac{BC}{AT}=2\frac{x+2y}{x}-2\frac{2s}{k}=2+4\left(\frac{y}{x}-\frac{s}{k}\right)=2+2=4.$$