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.

  • $\begingroup$ this is very hard, whic contest is it? $\endgroup$
    – nonuser
    Mar 3, 2019 at 10:40
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    $\begingroup$ @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. $\endgroup$ Mar 3, 2019 at 10:43
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    $\begingroup$ I'm not doing fun out of anybody and thus neither with you. At first I thought that it was some IMO problem. $\endgroup$
    – nonuser
    Mar 3, 2019 at 10:45
  • $\begingroup$ @greedoid Oh my God!! Forgive me. I can never think of that. $\endgroup$ Mar 3, 2019 at 10:48
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    $\begingroup$ 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 $\angle OAP = \angle C-\angle B$. $\endgroup$
    – dfnu
    Mar 3, 2019 at 13:37

1 Answer 1


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:


  • $\begingroup$ no, but thank you $\endgroup$
    – nonuser
    Mar 5, 2019 at 7:02

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