Just an angle 2

I'm usually a fan of trig, but it is absolutely unnecessary here. As in the above picture, extend $\overline{AC}$ and $\overline{PB}$ until they meet at Q. Since $m\angle APB=60$ , then $m\angle AQP=30$ and $m\angle QCB=60$ .

You now have two 30-60-90 triangles, which means if you know one side you can know all sides (if you are unfamiliar, the side opposite the 30 = a, the side opposite the 60 = a $\sqrt{3}$ , the side opposite the 90 = 2a). So:

1) $BC=3 \color{#D61F06}{\Rightarrow} \color{#333333} CQ=6$ and $BQ=3\sqrt{3}$

2) $AQ=8 \color{#D61F06}{\Rightarrow} \color{#3D99F6}{AP=\frac{8}{\sqrt{3}}}$ and $PQ=\frac{16}{\sqrt{3}}$

3) $\color{#3D99F6}{PB}\color{#333333}=PQ-BQ=\frac{16}{\sqrt{3}}-3\sqrt{3}\color{#3D99F6}{=\color{#3D99F6}\frac{7}{\sqrt{3}}}$

Finally, $3AP+2PB=\frac{38}{\sqrt{3}}=\sqrt{\frac{1444}{3}}=\sqrt{\frac{E}{F}} \color{#D61F06}{\Rightarrow}\color{#333333} E+F=\color{#D61F06}{1447}\color{#333333}.\space\space\space\Box$

Note: Scout's honor, I did not see the hint from Akshat Sharda when I came up with this :-)

Let $PC = x$ and $\angle APC = \theta$ , then using sine rule, we have:

$\begin{aligned} \frac{x}{\sin 90^\circ} = x & = \frac{2}{\sin{\theta}} = \frac{3}{\sin{(60^\circ - \theta)}} \\ \Rightarrow 2(\sin{(60^\circ - \theta)}) & = 3\sin{\theta} \\ 2\left(\frac{\sqrt{3}}{2}\cos \theta - \frac{1}{2} \sin \theta\right) & = 3 \sin \theta \\ \sqrt{3}\cos \theta - \sin \theta & = 3 \sin \theta \\ \sqrt{3}\cos \theta & = 4 \sin \theta \\ \tan \theta & = \frac{\sqrt{3}}{4} \\ \\ AP & = \frac{2}{\tan \theta} = \frac{8}{\sqrt{3}} \\ \\ \angle CPB & = 60^\circ - \theta \\ \tan (60^\circ - \theta) & = \frac{\sqrt{3} - \frac{\sqrt{3}}{4} }{1+\frac{3}{4}} = \frac{3 \sqrt{3}}{7} \\ \\ PB & = \frac{2}{\tan (60^\circ - \theta)} = \frac{7}{\sqrt{3}} \\ \\ \Rightarrow 2AP + 3PB & = \frac{24}{\sqrt{3}} + \frac{14}{\sqrt{3}} = \frac{38}{\sqrt{3}} = \sqrt{\frac{1444}{3}} \\ \\ \Rightarrow E + F & = \boxed{1447} \end{aligned}$

Hint :

$\rightarrow$ Produce $\overline{BC}$ and $\overline{PA}$ to meet at a point(say $D$ ) and then the question is all yours.

$\rightarrow$ Just apply a bit of Trigonometry and Pytagoras Theorem and solve it.

$Easy!!$ , isn't it ?

Firstly drop a perpendicular line from A which intersects PB at u and then draw a line from C that is perpendicular to the line segment AU and intersects AU at V. Now $\angle PAU = 30 \rightarrow \angle VAC = 60$ . Also, notice that BCVU is a rectangle as it contains four 90 degree angles. Let AP = x and PB = y then using special triangle (as APU and AVC are both half an equilateral triangle): $\ AU = \frac{x \sqrt{3}}{2} = AV + VU = 2 sin30 +3 = 1+3 = 4$ $\Rightarrow\ x = \frac{8}{\sqrt{3}}$ Similarly: $\ PB = PU +UB = PU +VC = \frac{x}{2} + 2sin60 = \frac{4}{\sqrt{3}} + \sqrt{3} = \frac{7}{\sqrt{3}}$ $\Rightarrow\ 3AP +2PB = 3x+2y = \frac{38}{\sqrt{3}} = \sqrt{\frac{1444}{3}}$ $\therefore\ E+F = 1447$

Note that PACB is cyclic so PA CB+AC PB = PC AB. Using the law of cosines, AB^2 = AC^2+BC^2 - 2AC BC cos(angle ACB). Because PACB is cyclic we know that angle ACB = 180 - angle APB = 120. So AB^2 = 2^2+3^2-2 2 3 (-1/2) = 19. We have that AB = sqrt(19). Since PACB is cyclic, we also know that P is on the circumcircle of ACB which is the circumcircle of PACB. Since P is on the circumcircle of ACB and angle PAC is right, PC is the diameter of the circumcircle of ACB. We know that for a triangle ABC with side lengths a, b, and c we have (AB BC AC)/4[ABC] = R where R is the circumradius of ACB. We know that [ACB] = 1/2 AC BC sin(angle ACB) = 1/2 * 2 3 * sqrt(3) / 2 = (3sqrt(3))/2. So R = (sqrt(19) * 2* 3) / (4 * (3sqrt(3))/2 ) = sqrt(19)/sqrt(3). So PC = 2R = 2sqrt(19)/sqrt(3). Now finally we have that the answer is PC * AB = 2sqrt(19)/sqrt(3) * sqrt(19) = sqrt(1444/3) and thus E+F = 1447.

I used only ptolemy's theorem and sin rule but not trig.

let $x = |AP|$ and $y = |PB|$ :

From pythagorean theorem (forming two triangles $\bigtriangleup APC$ and $\bigtriangleup CPB$ sharing the same hypotenuse $|PC|$ ):

$x^2 + 2^2 = y^2 + 3^2$

$x^2 - y^2 = 5$

Let $\theta_1 = \angle APC$ and $\theta_2 = \angle CPB$ such that $\theta_1 + \theta_2 = 60^\circ$ :

$\frac{3}{2} = \frac{ sin(60 - \theta_1)}{sin \theta_1} = \frac{\sqrt{3}}{2} cot \theta_1 - \frac{1}{2}$

Also:

$\frac{y}{x} = \frac{cos(60 - \theta_1)}{cos \theta_1} = \frac{\sqrt{3}}{2} tan \theta_1 + \frac{1}{2}$

From these, we can get:

$tan \theta_1 = \frac{\sqrt{3}}{4}$

and

$y = \frac{7}{8} x$

which yields to

$x = \frac{8\sqrt{3}}{3}$

$y = \frac{7\sqrt{3}}{3}$

However, we are to solve $3x + 2y$ , which is $\sqrt{\frac{1444}{3}}$ .