Fun with 2018 #9

Here's a solution with the least use of trignometry.

In this solution, sides opposite to vertex $B$ and $C$ are $b$ and $c$ respectively

We find that all triangles with side $a=2018$ and opposite angle $\dfrac{19\pi}{2018}$ can be formed with point A on the arc $BC$ .

We know that the area of a triangle is given by-

$\Delta=\dfrac{1}{2}bc\sin A$

Now, since $\sin A$ is constant, for the maximum area, $b\times c$ should be maximum

and according to $AM\geq GM$ , the maximum value of $b\times c$ can be obtained when $b+c$ is maximum.

In the circle we find the maximum value of $b+c$ is when the triangle is isosceles.

So, in the triangle $\angle C=\angle B$ and according to the sum angle property -

$\angle A+\angle B+\angle C=\pi$

$\Rightarrow \dfrac{19\pi}{2018} + 2\angle B=\pi$

$\Rightarrow \angle B=\dfrac{1999\pi}{4036} >\angle A=\dfrac{19\pi}{2018}$

Therefore the largest angle is $\dfrac{1999}{2}\cdot\dfrac{\pi}{2018}$

So $P=1999$ and $Q=2$

$\Rightarrow P+10Q-1=\boxed{2018}$

Let $\angle C = x$ .

Then $\angle B = \pi - x - \frac{19\pi}{2018}$ , and by the law of sines, $b = \frac{2018}{\sin (\frac{19 \pi}{2018})} \sin (\pi - x - \frac{19\pi}{2018})$ .

The area $A$ of the triangle is $A = \frac{1}{2}ab \sin C = \frac{2018^2}{2\sin (\frac{19 \pi}{2018})} \sin (\pi - x - \frac{19\pi}{2018}) \sin x = \frac{2018^2}{2\sin (\frac{19 \pi}{2018})} \sin (x + \frac{19\pi}{2018}) \sin x$ .

This means $A' = \frac{2018^2}{2\sin (\frac{19 \pi}{2018})}(\sin(x + \frac{19 \pi}{2018} \cos x) + \sin x \cos (x + \frac{19 \pi}{2018})) = \frac{2018^2}{2\sin (\frac{19 \pi}{2018})}\sin(2x + \frac{19 \pi}{2018})$ and $A'' = \frac{2018^2}{\sin (\frac{19 \pi}{2018})}\cos(2x + \frac{19 \pi}{2018})$ .

The largest area is when $A' = 0$ and $A'' < 0$ , which in the context of this problem means $2x + \frac{19 \pi}{2018} = \pi$ , and so $x = \frac{1999}{2}\frac{\pi}{2018}$ .

This makes $P = 1999$ , $Q = 2$ , and $P + 10Q - 1 = \boxed{2018}$ .

The area of the triangle is given by:

$\begin{aligned} A_\triangle & = \frac 12 a{\color{#3D99F6}b} \sin C & \small \color{#3D99F6} \text{By sine rule: } \frac a{\sin A} = \frac b{\sin B} \\ & = \frac {a^2 {\color{#3D99F6}\sin B} \sin C}{2\sin A} & \small \color{#3D99F6} \text{Note that } B = \pi - A - C \\ & = \frac {a^2 {\color{#3D99F6}\sin (\pi - A - C)} \sin C}{2\sin A} & \small \color{#3D99F6} \text{that } \sin (\pi - \theta) = \sin \theta \\ & = \frac {a^2 {\color{#3D99F6}\sin (A + C)} \sin C}{2\sin A} & \small \color{#3D99F6} \text{and that } \sin \alpha \sin \beta = \frac 12 (\cos (\alpha - \beta) - \cos (\alpha + \beta)) \\ & = \frac {a^2 (\cos A - \cos (A+2C))}{4\sin A} \end{aligned}$

We note that $A_\triangle$ is maximum, when $\cos (A+2C) = -1$ or $A+2C = \pi$ , $\implies C = \pi 2 - \dfrac {19\pi}{4036} = \dfrac {1999 \pi}{4036}$ . From $B= \pi - A - C = \dfrac {1999 \pi}{4036}$ . Therefore, $B=C= \dfrac {1999 \pi}{4036} = \frac {1999}2 \cdot \dfrac \pi{2018}$ is the largest angle. And $P+10Q-1 = \boxed{2018}$ .