# Proving the inequality $\angle A+\angle COP < 90^\circ$ in $\triangle ABC$

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.

• this is very hard, whic contest is it? Mar 3, 2019 at 10:40
• @greedoid I know that you are doing fun with me. You are a very expert mathematician. I thought that it would be easy for any other professional like you. And your answer is BANGLADESH MATH OLYMPIAD. Mar 3, 2019 at 10:43
• I'm not doing fun out of anybody and thus neither with you. At first I thought that it was some IMO problem. Mar 3, 2019 at 10:45
• @greedoid Oh my God!! Forgive me. I can never think of that. Mar 3, 2019 at 10:48
• I thought equality in the hypothesis would imply equaility in the thesis, i.e. $\angle C - \angle B = 30° \Rightarrow \angle COP + \angle A = 90°$. But it doesn't look like it is correct... By the way, it could be useful to observe that $\angle OAP = \angle C-\angle B$.
– dfnu
Mar 3, 2019 at 13:37