Find The Length

So, really as Ronak said this question is actually a beast. Looks difficult but is very simple. It can be solved using cosine rule and Menelaus only.

In $ABC$ , using Cosine rule for $\angle A$

${AB}^{2} + {AC}^{2} - 2*AB*AC* \cos(BAC) = {BC}^{2}$

Substituting values,

$cos A = \frac{19}{96}$

Now, we must know that $CY = CX = \frac{11}{2}, AY = AZ = \frac{5}{2}, BZ = BX = \frac{7}{2}$ (These values can be found by just taking $BC = x + y, CA = y + z, AB = z + x$ and then solving for x,y,z)

Again using cosine rule in $ABY$ for $\angle A$ , and substituting $\cos(BAY) = \frac{19}{96}$

we get, ${BY}^{2} = \frac{581}{16} \Rightarrow BY = \frac{\sqrt{581}}{4}$

Now, look at the triangle $ABY$ again and the transversal line ZC. Applying Menelaus theorem here, we get,

$\frac{AC}{CY}\cdot \frac{GY}{BG}\cdot \frac{BZ}{AZ} = -1$

Plugging values, $\frac{GY}{BG} = \frac{55}{112}$

Now, $GY = 55a, BG = 112a$

We know that $GY + BG = BY$

Hence, $a = \frac{1}{167} \frac{\sqrt{581}}{4} \Rightarrow GY = \frac{55}{167} \frac{\sqrt{581}}{4}$

In $BYR$ , we know $RY = BC = 9, BR = CY = \frac{11}{2}$ (BCYR is a parallelogram)

Hence, now using cosine rule in this triangle for $\angle Y$ , we get

$cos(Y) = \frac{199\sqrt{\frac{7}{83}}}{72}$

Finally, join $GR$ and use cosine rule in $GRY$ for $\angle Y$ ,

${GR}^{2} = {GY}^{2} + {RY}^{2} - 2*GY*GR*\cos(Y)$

Plugging values for each on the RHS, we get,

${GR}^{2} = \frac{6276741}{111556}$

which finally gives us the answer -

$GR = \boxed{\sqrt{\frac{6276741}{111556}}}$

This question is a beast

I am shocked to see that the answer is :

$\displaystyle GR=\sqrt{\frac{6276741}{111556}}$

Oh no !!!!