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.
Best Answer
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