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I'm trying to find an easy way to solve the problem below:

basic geometry problem

Of course you could solve it by "brute force", example:
- numerical means (vectors and dot product), or
- long algebra calculations (law of sines/cosines and trigonometric identities).

However, considering this was a question in a high school exam (3~5 min per question), most likely there is a shortcut. Any idea?

Mark Messa
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1 Answers1

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Introduce isosceles $\triangle PAB$, then chase some angles:

enter image description here

Blue
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  • Impressive! Mind to explain a little bit how did you come up with that? And how long did you take to 'see' this solution? – Mark Messa Jun 11 '18 at 04:10
  • Also, how hard did you consider this problem to be? Do you think it is reasonable to be in a high school exam designed for 3~5 minutes per question? – Mark Messa Jun 11 '18 at 04:21
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    Well, I have to admit, it took a bit longer than $3$-$5$ minutes, but I was exploring multiple trigonometric approaches to see if I could distill them to something self-evident. I had a sense that $\angle ADC = 60^\circ$ (which is easy to determine) was key, and I fairly quickly introduced $30^\circ$-$60^\circ$-$90^\circ$ $\triangle PDC$. I could then show that $\overline{AP}\cong\overline{CP}$ with trig, which solved the problem. Starting a little longer at the diagram made me realize that $P$ was actually the circumcenter, and the self-evident decomposition fell into place. (continued) – Blue Jun 11 '18 at 04:32

  • "P was actually the circumcenter" This was the main point I was hoping to hear ... whether you started considering the circumcenter was key or if there was something else that trigger this ...

     – Mark Messa Jun 11 '18 at 04:38
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    (continuing) I certainly wouldn't expect someone to derive my angle-chasing solution on a time-constrained exam. I'll note that my trig approach, upon creating $\triangle PDC$, was roughly: Let $\triangle ABP$ have circumdiameter $1$, then $|BD|=|PD|=\frac12|CD| = \sin 15^\circ$, and $|AD| = \sin 45^\circ$, so that $$|AP| = \sin 45^\circ - \sin 15^\circ = 2\cos 30^\circ \sin 15^\circ = |CD|\cos 30^\circ = |CP|$$ That might still require a little more inspiration than exam conditions allow. You should ask the instructor what the "expected" solution was supposed to be ... and report back here! – Blue Jun 11 '18 at 04:48
  • To answer your specific question: I didn't actually think about the fact that $P$ was the circumcenter until I needed a concise way to describe it in these limited-character comments. :) – Blue Jun 11 '18 at 04:53
  • "until I needed a concise way to describe it" Ok, the first time I've watched your answer with the circumcenter I've thought you were some kind of Will Hunting. However, even without realizing it at first, you are a genius nonetheless! :-) – Mark Messa Jun 11 '18 at 04:58
  • Whoops. Typo in my trigonometric solution; it should read: Let $\triangle ABD$ have circumdiameter $1$, then ... – Blue Jun 11 '18 at 05:10