# BdMO 2016 National Secondary Problem 3.

$\triangle{ABC}$ 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 $\angle{BAP}$ 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.

• i have found an equation for the unknown angle $x$ which containes trigonometric functions Dec 29, 2016 at 17:26
• @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. Dec 29, 2016 at 17:29
• i have note used any calculators Dec 29, 2016 at 17:30
• Is the equation solvable without using calculator? ! Dec 29, 2016 at 17:32
• i will try it to solve the equation Dec 29, 2016 at 17:33

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 $\angle{BAP} = 13^{\circ}$. • AND WE HAVE A WINNER! How did you figure out that you needed to construct $Q$? Dec 30, 2016 at 4:05
• added illustration Dec 30, 2016 at 6:36
• Could you explain a bit more why $Q$ is the center of $\odot BPC$? Dec 30, 2016 at 6:47
• @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 $BP$ Dec 30, 2016 at 7:07

Angle-chasing with notations found on the figure below gives all angles as functions of a unique unknown angle $$s$$, as follows \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

$$x=13°$$