## IMO 2001 Problem1

Discussion on International Mathematical Olympiad (IMO)
Ananya Promi
Posts: 36
Joined: Sun Jan 10, 2016 4:07 pm

### IMO 2001 Problem1

Consider an acute angled triangle $ABC$. Let $P$ be the foot of the altitude of triangle $ABC$ issuing from the vertex $A$, and let $O$ be the circumvented of triangle $ABC$. Assume that $$\angle{C}\geq\angle{B}+30°$$. Prove that $$\angle{A}+\angle{COP}<90°$$

Ananya Promi
Posts: 36
Joined: Sun Jan 10, 2016 4:07 pm

### Re: IMO 2001 Problem1

$$\angle{A}+\angle{COP} <90° \Rightarrow \angle{COP} <90°-\angle{COD}$$
$$\Rightarrow \angle{COP} <\angle{OCP} \Rightarrow CP<PO$$
is enough to prove.
$\angle C -\angle B \geq 30°$ as both $(\angle{C}-\angle{B})$ and 30° are less than $90°$,
\begin{align*}&\sin(\angle{C}-\angle{B})\geq \sin30°\\ \Rightarrow &\sin\angle{C}\cos\angle{B}-\sin\angle{B}\cos\angle{C}\geq \sin30°= \frac{1}{2}\\ \Rightarrow &\frac{AP*BP}{AC*AB}- \frac{AP*CP}{AC*AB}\geq \frac{1}{2}\\ \Rightarrow &\frac{AP}{AC*AB}(BP-CP) \geq \frac{1}{2}\\ \Rightarrow &\frac{1}{2R}(BP-CP) \geq \frac{1}{2}\\ \Rightarrow &BP-CP \geq R\\ \Rightarrow &BP-R \geq CP\end{align*}
Again, \begin{align*}PC+PO>R \Rightarrow PC>R-PO\\ \Rightarrow \frac{PC*BP}{R-PO}>BP\end{align*}
Again, \begin{align*}&PO^2=PC^2+CO^2-2PC*CO*cos\angle{OCP}\\ \Rightarrow &PO^2=PC^2+CO^2-2PC*CO\frac{CD}{OC}\\ &\Rightarrow PO^2=PC^2+CO^2-PC*BC\\ &\Rightarrow PO^2=CO^2-PC(BC-PC)= R^2-PC*PB\\ &\Rightarrow PO^2-R^2=-PC*PB\\ &\Rightarrow PO+R=\frac{PC*PB}{R-PO}>BP\\ &\Rightarrow PO+R>BP\\ &\Rightarrow BP-R<PO\end{align*}
Again, we have got $$BP-R\geq CP$$
So, $PO>CP$
So, $CP<PO$
So, $$\angle{A}+\angle{COP} <90°$$
We are done.

Atonu Roy Chowdhury
Posts: 63
Joined: Fri Aug 05, 2016 7:57 pm