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Let a point $P$ lie in a triangle $\triangle ABC$ such that $\angle BCP = \angle PCA = 13^\circ$, $\angle CAP = 30^\circ$, and $\angle BAP = 73^\circ$. Compute $\angle BPC$.

what

I have an ugly trig solution that looks something like this:

Let $\angle PBC = \theta$. It follows that $\angle PBA = 51-\theta$. From trig Ceva, we see that: $$\frac{\sin(30)}{\sin(73)}*\frac{\sin(51-\theta)}{\sin(\theta)}*\frac{\sin(13)}{\sin(13)} = 1$$ Observe that $90-73=17$, and conveniently $17*3=51$. This inspires the following manipulations: $$\frac{1}{2\sin(73)} * \frac{\sin(51-\theta)}{\sin(\theta)} = 1$$ $$\frac{1}{2\cos(17)} * \frac{\sin(51)\cos(\theta)-\cos(51)\sin(\theta)}{\sin(\theta)} = 1$$ $$\sin(51)\cos(\theta)-\cos(51)\sin(\theta)= 2\cos(17)\sin(\theta) $$ $$\sin(51)\cos(\theta) = 2\cos(17)\sin(\theta) + \cos(51)\sin(\theta)$$ $$\sin(51)\cos(\theta) = \sin(\theta)(2\cos(17) + \cos(51))$$ $$\tan(\theta) = \frac{\sin(51)}{2\cos(17) + \cos(51)}$$ Proceeding with triple-angle formulae: $$\tan(\theta) = \frac{3\sin(17)-4\sin^3(17)}{2\cos(17) + 4\cos^3(17)-3\cos(17)}$$ $$\tan(\theta) = \frac{\sin(17)}{\cos(17)} * \frac{3-4\sin^2(17)}{4\cos^2(17)-1}$$ $$\tan(\theta) = \tan(17) * \frac{3-4(1-\cos^2(17))}{4\cos^2(17)-1}$$ $$\tan(\theta) = \tan(17) * \frac{4\cos^2(17)-1}{4\cos^2(17)-1}$$ $$\tan(\theta) = \tan(17)$$ We conclude that $\theta = 17$ and $\boxed{\angle BPC = 150}$.

This is simply horrific. Is there a more elegant method? I notice that $73 = 13 + 60$, but I don't see where I would put an equilateral triangle.

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2 Answers 2

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Using only elementary geometry:

Make $CD=CA$, and join $PD$. Then by SAS$$\triangle APC\cong\triangle DPC$$and$$\angle DPC=\angle APC=137^o$$Drawing $PE\perp AC$, since $\angle EAP=30^o$, then$$PE=PF=AF$$And joining $FB$, since $FB\perp AP$ then by SAS$$\triangle AFB\cong\triangle PFB$$and triangle $ABP$ is isosceles. compute angle BPC

Therefore$$\angle ABP=34^o$$and$$\angle PBD=51^o-34^o=17^o$$And since$$\angle BDP=180^o-30^o=150^o$$then$$\angle BPD=180^o-17^o-150^o=13^o$$But$$\angle DPC=137^o$$Therefore$$\angle BPC=137^o+13^o=150^o$$

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  • $\begingroup$ Where did $FB \perp AP$ come from? $\endgroup$ Oct 10, 2018 at 12:02
  • $\begingroup$ Fair question. I'm afraid I've assumed without proof that $BF$ is tangent to the circle at $F$, and so perpendicular to $PA$. This needs some work. $\endgroup$ Oct 10, 2018 at 17:54
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Hint. Reflect $AC$ along $PC$ onto $BC$ to get the line $A'C$. To show that $\angle BPC=150^\circ$, it suffices to show that $\triangle BA'P$ is similar to $\triangle BPC$.

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  • $\begingroup$ I'm not seeing where to follow through with this. Can you elaborate a bit? $\endgroup$ Oct 11, 2018 at 17:42

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