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enter image description here

This image was doing the rounds on a popular text messaging application, so I decided to give it a try.

From sine rule in $\triangle ABP$: $$\frac{AB}{\sin(150^\circ)} = \frac{AP}{\sin(10^\circ} \\ \implies AP = 2AB \sin(10^\circ)$$

Applying sine rule again in $\triangle APC$: $$\frac{AP}{\sin(60^\circ + x)} = \frac{AC}{\sin(x)}$$ Manipulating the equation and using some properties gives us $$x = \arctan\left(\frac{\sqrt 3}{4\sin(10^\circ) - 1}\right)$$ This gives $x = -80^\circ$, but since it's an arctan, $x = 100^\circ$. Also, since $\sin(x) = \sin(\pi - x)$, $x = 80^\circ$ as well.

My question is: Is there a way to solve this problem that does not require a calculator? I tried to chase angles but that did not work out in this case. This solution requires computing $\sin(10^\circ)$ as well as the $\arctan$ of that expression, which needs a calculator.

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  • $\begingroup$ Apologies for my earlier bad answer. I should comment that there should only be one solution, since the information in the picture does not allow any degrees of freedom (besides global scaling), so we must use one more piece of information to eliminate one of your answers. $\endgroup$ – angryavian Jun 18 '20 at 5:08
  • $\begingroup$ Note that it is not always true that $\tan x=y\implies x=\arctan y$. One can see that $x=100^\circ$ satisfies $\tan x=\frac{\sqrt 3}{4\sin(10^\circ)-1}$, but $x=80^\circ$ doesn't. $\endgroup$ – mathlove Jun 18 '20 at 6:02
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Let $D$ be the circumcenter of $\triangle APB$. Since $\angle BPA=150^\circ$, we have $\angle ADB = 360^\circ - 2\angle BPA = 60^\circ$, and since $DA=DB$ it follows that $\triangle ABD$ is equilateral. So, $AD=AB=AC$ and therefore $A$ is the circumcenter of $\triangle DBC$.

Now, $\angle PDB = 2\angle PAB = 40^\circ$ and $\angle CDB = \frac 12 \angle CAB = 40^\circ$. Hence $\angle PDB = \angle CDB$. It follows that $D,P,C$ are collinear. Now it is easy to find $\angle DPA = 90^\circ - \frac 12 \angle ADP = 90^\circ - \angle ABP = 80^\circ$. Thus $\angle APC = 180^\circ - \angle DPA = 100^\circ$.

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Let $\angle ACP =z $. By trigonometric form of Ceva's theorem we have: $$\frac {\sin60^\circ}{\sin20^\circ}\frac {\sin10^\circ}{\sin40^\circ}\frac {\sin(50^\circ-z)}{\sin z}=1.\tag1 $$

Further we have the following property for product of sines: $$ \prod_{k=1}^{n-1}2\sin\frac{k\pi}n=n. $$ Particularly for $n=9$ it gives $$ (2^4\sin20^\circ\sin40^\circ\sin60^\circ\sin80^\circ)^2=9 \implies\sin20^\circ\sin40^\circ\sin60^\circ\sin80^\circ=\frac3{16}.\tag2 $$ Combining (1) and (2) we obtain: $$ \frac {\sin z}{\sin(50^\circ-z)}=\frac{\sin^2 60^\circ\sin10^\circ\sin80^\circ}{\frac3{16}} =4\sin10^\circ\cos10^\circ=\frac{\sin20^\circ}{\sin30^\circ}\implies z=20^\circ, $$ and finally $$x=180^\circ-60^\circ-z=100^\circ.$$

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