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P.S: I only want a hint,not the whole solution.

BdMO 2009 Problem 5 Secondary

In triangle ABC, $\angle A = 90$. M is the midpoint of BC. Choose D on AC such that AD = AM. The circumcircles of triangles AMC and BDC intersect at C and at a point P. What is the ratio of angles: $(\angle ACB)/( \angle PCB)$?

The diagram is extremely cluttered.I think the answer is $2:1$ based on some poorly drawn figures.

My try:

For our purposes,let the circle intersect $AB$ at Q.Then $QC=QA$ is the diameter of one of the circles.We can also get some angle equalities by angle chasing.The penultimate step seems to be showing that the arcs AP and PM are equal.I have listed out some of the possibilities down below,in increasing order of sophistication:

0)Show that BP=PD

1)Solve the question purely using angle-chasing.

2)Show that the arcs AP and PM are equal.

3)Construct an angle equal to $\angle ACP$ and then show it to be equal to $\angle PCB$.

4)Converse of Angle Bisector Theorem

None of the above have yielded anything good to work with.I am NOT looking for the whole solution,but merely a hint.Any help will be appreciated.

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up vote 1 down vote accepted

enter image description here

I hope that the solution is clear from the picture:

  1. $\angle PAD=\angle PMB$ because $P$ lies on the circumcircle of $\triangle ACM$,

  2. similar for $\angle ADP=\angle PBM$

  3. $AD=BM(=AM)$ as $\angle A=90^\circ$.

Therefore $\triangle APD$ and $\triangle MPB$ are congruent.

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Nice solution.Looks like my $0$th idea worked.A question though.What motivated you to look for congruent triangles in this messy figure?I mean,the figure was cluttered enough and trying to look for congruent triangles seemed like a mad thing to do.Your idea was quite ingenious.But may I know how you came up with it? – rah4927 Aug 2 '14 at 16:53
It's sorta backward reasoning, based on $AP=PM$ and $DP=PB$. Plus, I almost always try to use the fact that $BC=2AM$. – Quang Hoang Aug 2 '14 at 17:13

After giving this more thought, I see that the answer is in fact 2, and that this is the correct answer for any right triangle. The diagram provided by Quang Hoang is very well done, and Quang's observation about the congruent triangles is insightful, but I don't see how it settles the issue. Here is my most recent response to the original question:

First, there is a wonderful circle theorem that goes something like this: In a circle, if an inscribed $\angle A$ subtends arc $K$, then $A = \frac{K}{2}$. Let's call this the circle theorem.

After having drawn the specified figure, we can state that point $P$ must lie at the intersection of the following two lines: The perpendicular bisector of chord $AM$, and the perpendicular bisector of chord $DB$. The perpendicular bisector of chord $AM$ clearly intersects arc $AM$ into two equal arcs, which are the arc $PM$ and the arc $PA$. Note that inscribed $\angle PCA$ subtends arc $PA$. The circle theorem tells us that $\angle PCA$ is equal to half of arc $PA$. Similarly, inscribed $\angle PCM$ subtends arc $PM$, and so $\angle PCM$ is equal to half of arc $PM$. Because arcs $PM$ and $PA$ are equal, the angles that subtend them, specifically angles $\angle PCA$ and $\angle PCB$, are equal to each other. It follows immediately that $\angle ACB$ divided by $\angle PCB$ equals $2$.

My earlier response to the question, which follows, is a brute-force approach. Nothing wrong with that, but it lacks elegance.

One approach is based on the assumption that you can state your answer in terms of various distances. For example, the distance between point $C$ and point $P$, which can be stated as $\overline{CP}$. Furthermore, I also assume that you can use a trigonometry formula, in particular the law of cosines. If so, then you can simply forget about the circles and simply consider a point $P$ inside $\triangle{ACB}$, which makes another triangle, which is $\triangle{PCB}$. You can also forget about right triangles, because the answer I have in mind works for any type of triangle.

The law of cosines can be stated as: $$c^2 = a^2 + b^2 - 2ab\cos C$$ Solving this for $\cos C$ gives: $$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$$ Solving for $C$ gives: $$C =\cos^{-1}\left( \frac{a^2 + b^2 - c^2}{2ab} \right) $$ You need to show the ratio of two angles, correct? You asked for a hint. This is it: Apply the third version of the law of cosines to your problem. You will need to apply it twice -- once for each angle as specified by the problem.

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Use your eyes. Thanks to Quang Hoang's beautifully accurate drawing, anyone can see that DP and PM are not equal in length! In fact, DP is longer than PM – O.C.Riley Aug 17 '14 at 8:09
Ignore my last comment.I made a typo. It follows from his observations that $DP=PB$,which implies that $DB$ arc is bisected by $CP$. This implies in turn that ∠DCP=∠PCB and the conclusion follows. – rah4927 Aug 18 '14 at 3:23
In Quang Hoang's hint to you, he deliberately does not present prima facie evidence that DP = PB. You were supposed to fill in the details. – O.C.Riley Aug 20 '14 at 17:27
Yes I knew that but it was your comment ". . .but I don't see how that settles the issue" that made me write up the last comment. – rah4927 Aug 24 '14 at 17:08

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