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Let $\triangle ABC$ be any triangle. Suppose the angle bisector of $\angle BAC$ intersects $BC$ at $D$. Let $\Gamma$ be a circle tangent to $BC$ at $D$ and so that $A$ belongs to the circumference of $\Gamma$. If $M$ is the (second) intersection point of $AC$ and $\Gamma$, and if $BM$ intersects $\Gamma$ at $P$, then prove that $AP$ must be a median of $\triangle ABD$.

I need some help with this problem. It was taken from an Iran Math Olympiad (from 1999 I believe).

I've mainly tried proving that $BD \over JD$$=2$ (here $J$ is $AP \cap BC$), by various methods. First, using power of points and the angle bisector theorem, no success. Then I've tried some angle chasing to find some similar triangles, just to find the same relations I had found using power of points... Finally I went full trigonometry over it, but then again I was never quite good at trigonometry anyway, so I couldn't get very far.

I would love some hints, because I'm pretty much stuck. Thanks in advance.

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  • $\begingroup$ Is this how it's supposed to be constructed? i.imgur.com/UzPldD1.png I think I must have made an error somewhere because segment AP is certainly not a median of △ABD. $\endgroup$
    – Mathemert
    Jun 18, 2014 at 1:20
  • $\begingroup$ Yup, that's how it's constructed. It's not that easy to construct it on Geogebra since there isn't a premade function that lets you construct that particular circle (tangent to a line at a point, and at the same time passing through another point). I see your figure a little bit off (the circle doesn't contains A, it's just pretty close), so when you actually measure the segments on Geogebra you won't get a median. But if you do a larger diagram and make that circle a little more precise, you'll see that it approximates a median. (also, the line AP is a median, not the segment) $\endgroup$
    – Deathkamp Drone
    Jun 18, 2014 at 3:27
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    $\begingroup$ i.imgur.com/Kd1JThg.png. Are you a student in the US? Have you participated in IMO? $\endgroup$
    – Mathemert
    Jun 18, 2014 at 4:44
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    $\begingroup$ How have you been studying? I haven't studied problem solving techniques at all so have never been able to solve even a level 4+ olympiad problem. Why do you want to qualify for IMO? $\endgroup$
    – Mathemert
    Jun 18, 2014 at 4:55
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    $\begingroup$ Ah, cool. But why do you want to do IMO? $\endgroup$
    – Mathemert
    Jun 18, 2014 at 5:07

3 Answers 3

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P is not a nice point, so let's try and avoid it.

Hint: show that $JB^2=JP \times JA = JD^2$.

Hint: Show that 2 triangles are similar to prove the first.

Hint: define N in a similar manner to M

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  • $\begingroup$ Oh, I was so close to the solution, the first hint is really the key. Thank you, I can go on from there! $\endgroup$
    – Deathkamp Drone
    Jun 18, 2014 at 3:51
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    $\begingroup$ @DeathkampDrone This proof was motivated from "P is not a nice point", and doing my best to not use it at all. You should try and deconstruct it, to understand how to approach this problem in future. $\endgroup$
    – Calvin Lin
    Jun 20, 2014 at 2:32
  • $\begingroup$ I was going to ask you that, but I forgot. I don't understand what you mean by "P is not a nice point". Is it some kind of well-known olympiad math heuristic? $\endgroup$
    – Deathkamp Drone
    Jun 20, 2014 at 2:57
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    $\begingroup$ @DeathkampDrone The definition of $P$ is strange, and not easily understood. In fact, $BM$ itself is a weird line, and it's hard to relate it to anything else, which makes $P$ somewhat worse. As such, it would be nice to avoid $P$ where possible. For example, when analyzing $\angle PBJ$, we are actually working with $\angle MBJ$ instead. And even then, we're working with $\angle NMB$. This is a 'nice angle' since the point $ P, B, N$ lie on a circle, which is thus equal to $\angle NAP = \alpha - \beta$. $\endgroup$
    – Calvin Lin
    Jun 20, 2014 at 5:23
  • $\begingroup$ I see. Thank you for your insight! $\endgroup$
    – Deathkamp Drone
    Jun 20, 2014 at 6:19
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For completeness, I'll post my solution here.

By looking at the power of $J$ with respect to $\Gamma$, we see that: $|JD|^2 = |JP|\cdot|JA|$.

On the other hand, let $\angle BAD = \alpha$ and $\angle PAD = \beta$. From this we can see that $\angle BAJ = \alpha - \beta$. It is also true that $\angle DAM = \alpha$ (since $AD$ is the angle bisector of $\angle BAM$). So we have $\angle MBD = \angle DAM - \angle PAD = \alpha - \beta$. It follows that $\triangle BJP \sim \triangle AJB$, whence: $$\frac{|JB|}{|JA|}=\frac{|JP|}{|JB|} \Rightarrow |JB|^2 = |JA| \cdot |JP|$$

So $|JD|^2 = |JP| \cdot |JA| = |JB|^2 \Rightarrow |JD| = |JB|$, which is what we wanted to prove. $\blacksquare$

enter image description here

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See the Deathkamp Drone's picture.

$$\measuredangle MBC=\measuredangle AMB-\measuredangle C=\measuredangle AMD-\measuredangle PMD-\measuredangle C=\measuredangle ADB-\measuredangle PAD-\measuredangle C=$$ $$=\measuredangle DAC-\measuredangle PAD=\measuredangle BAD-\measuredangle PAD=\measuredangle BAP,$$ which says $$\Delta BPJ\sim\Delta ABJ,$$ which gives $$BJ^2=JP\cdot JA=JD^2$$ and we are done!

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