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IMO 2001 Problem1

Posted: Fri Jun 02, 2017 3:01 pm
by Ananya Promi
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 $$\angle{A}+\angle{COP}<90°$$

Re: IMO 2001 Problem1

Posted: Fri Jun 02, 2017 3:04 pm
by Ananya Promi
$$\angle{A}+\angle{COP} <90°
\Rightarrow \angle{COP} <90°-\angle{COD}$$
$$\Rightarrow \angle{COP} <\angle{OCP}
\Rightarrow CP<PO$$
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. :roll:

Re: IMO 2001 Problem1

Posted: Mon Jun 05, 2017 12:02 am
by Atonu Roy Chowdhury
Our problem is equivalent to $OP>CP$.

Lemma 1: $BP \ge R+PC$ where $R$ is the circumradius of $\triangle ABC$.
Proof of lemma: $BP-PC=AB \cos \angle B - AC \cos \angle C = 2R(\sin \angle C \cos \angle B - \sin \angle B \cos \angle C ) \ge R$ because $\angle C- \angle B \ge 30$.

Now we back to our problem.
$R+OP=BO+OP>BP \ge R+PC$