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$\triangle{ABC}$ is isosceles with $AB=AC$, $P$ is a point inside $\triangle {ABC}$ such that $\angle{BCP} = 30^{o}$, $\angle{APB} = 150 ^{o}$, $\angle{CAP}=39^{o}$. Find $\angle{BAP}$
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This problem is from BdMO 2016 Nationals ( Secondary ), worth of $20$ points. I've tried it after the contest many many times. But couldn't find any solution. Any Hint/Full Solution will be helpful.

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  • $\begingroup$ i have found an equation for the unknown angle $x$ which containes trigonometric functions $\endgroup$ – Dr. Sonnhard Graubner Dec 29 '16 at 17:26
  • $\begingroup$ @Dr.SonnhardGraubner But in MO we are not supposed to use calculator. But if all the functions cancel out in the end then it is possible to use trigonometric functions. $\endgroup$ – Rezwan Arefin Dec 29 '16 at 17:29
  • $\begingroup$ i have note used any calculators $\endgroup$ – Dr. Sonnhard Graubner Dec 29 '16 at 17:30
  • $\begingroup$ Is the equation solvable without using calculator? ! $\endgroup$ – Rezwan Arefin Dec 29 '16 at 17:32
  • $\begingroup$ i will try it to solve the equation $\endgroup$ – Dr. Sonnhard Graubner Dec 29 '16 at 17:33
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Let $Q$ be the point such that $\triangle{PQB}$ is equilateral and $C$ and $Q$ are on the same side of $PB$. Then $Q$ is the center of $\odot BPC$ and so $AQ$ is the perpendicular bisector of $BC$. Finally, as $\angle{APB} = 150^{\circ}$ we have $AP \perp QB$ and so $AP$ bisects $\angle{QAB}$. So $4\angle{BAP} = \angle{BAC}$ which implies that $3\angle{BAP} = \angle{CAP}$ which implies that $\angle{BAP} = 13^{\circ}$.

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  • 1
    $\begingroup$ AND WE HAVE A WINNER! How did you figure out that you needed to construct $Q$? $\endgroup$ – 1-___- Dec 30 '16 at 4:05
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    $\begingroup$ added illustration $\endgroup$ – Joffan Dec 30 '16 at 6:36
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    $\begingroup$ Could you explain a bit more why $Q$ is the center of $\odot BPC$? $\endgroup$ – yurnero Dec 30 '16 at 6:47
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    $\begingroup$ @yurnero angle (60) at the centre of the circle is half the angle at the circumference (30, at $\angle BCP$), and $Q$ lies on the bisector of $BP$ $\endgroup$ – Joffan Dec 30 '16 at 7:07
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Angle-chasing with notations found on the figure below gives all angles as functions of a unique unknown angle $s$, as follows

enter image description here $$\left\{\begin{aligned}t&=&141-s\\u&=&210-s\\z&=&s-40\\y&=&231-2s \end{aligned}\right.\tag{1}$$

and

$$x=2s-210\tag{2}$$

Remarks: 1) the last equation in (1) takes into account the fact that angle in $B$ = angle in $C$. 2) In this last expression, as the considered angle is acute, we must have $0 \leq 231-2s<90 \ \iff \ $

$$70.5 < s <115.5\tag{3}$$

It remains to find $s$. Thus, we have to find another relationship that could be called a contiguity constraint for triangles $ABP$ and $APC$ ; this constraint will come from the law of sines in each triangle :

$$\dfrac{a}{\sin(231-2v)}=\dfrac{b}{\sin(150)} \ \ \text{and} \ \ \dfrac{a}{\sin(141-s)}=\dfrac{b}{\sin(s)}\tag{4}$$

Expressing ratio $a/b$ in two ways and taking into account the value $\sin 150 =\tfrac12$, we get :

$$\underbrace{\sin(141-s)-2\sin(s)\sin(231-2s)}_{= \ \sin(3s+39)}=0 \tag{4}$$

(We don't enter here into computational details)

(4) is possible iff

$$3s+39=0 \ \text{mod.} \ 180 \tag{5}$$

which, taking into account the range given by (3) enforces $3s+39=360 \ \iff s=107$. Plugging this value into (2) gives

$$x=13°$$

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