In an acute angled $\triangle ABC$, $AP \perp BC$and $O$ is its circumcenter. If $\angle C \ge \angle B + 30^\circ$, then prove that $$\angle A + \angle COP < 90^\circ$$
My Attempt:
Extending the line $AP$ to the circumferential point $D$, I connected $O,D$ and $C,D$ and got two lines $OD$ and $CD$ as such below
Given that, $\angle C \ge \angle B + 30^\circ$ and so from that I got
$\angle C +\angle B+ \angle A \ge \angle B + \angle B +\angle A + 30^\circ \implies 180^\circ \ge 2\angle B + \angle A + 30^\circ \implies 150^\circ - \angle A \ge 2\angle B$
$2\angle B \le 150^\circ - \angle A$.....(1)
In right angled $\triangle APC, \angle APC = 90^\circ$
So, $\angle PAC = 90^\circ - \angle C$
After that, we know that $\angle COD = 2\angle DAC$
$\angle COD = 2(90^\circ - \angle C) \implies 180^\circ - \angle COD = 2\angle C$
$180^\circ - \angle COD \ge 2\angle B + 60^\circ \implies 180^\circ - 60^\circ - \angle COD \ge 2\angle B$
$120^\circ - \angle COD \ge 2\angle B$.....(2)
Notice that, both ($150^\circ - \angle A$) and ($120^\circ - \angle COD$) are greater than or equal to $2\angle B$.
So, how could I show the relation between ($150^\circ - \angle A$) and ($120^\circ - \angle COD$). Or, any other way to prove for the desired inequality? Thanks in advance.
Well, afterall this is an IMO 2001 geometry problem and you can find two solutions, along with a solutions of the other problems from that year, here:
https://sms.math.nus.edu.sg/Simo/IMO_Problems/01.pdf