I've Definitely Seen it Before

img

As you see in this diagram, if in any $\triangle PQR$ , if points $E$ and $F$ are on sides $PQ$ and $PR$ respectively, then the ratio of areas

$\dfrac{A(\triangle PEF)}{A( \triangle PQR)} = \dfrac{PE\times PF}{PQ\times PR}$ .... you can easily prove this using similarity of triangles.

Thus , in our actual question, which is as follows,

img2 , we have

$\dfrac{A(\triangle AYZ)}{A(\triangle ABC)} = \dfrac{AY}{AC}\times \dfrac{AZ}{AB} = \dfrac{6}{11}\times \dfrac{1}{3} = \dfrac{2}{11}$

$\dfrac{A(\triangle BXZ)}{A(\triangle ABC)} = \dfrac{BZ}{BA} \times \dfrac{BX}{BC} = \dfrac{2}{3}\times \dfrac{3}{7} = \dfrac{2}{7}$

$\dfrac{A(\triangle CXY)}{A(\triangle ABC)} = \dfrac{CY}{CA} \times \dfrac{CX}{BC} = \dfrac{5}{11}\times \dfrac{4}{7} = \dfrac{20}{77}$

Thus we have

$\dfrac{A(\triangle AYZ) + A(\triangle BXZ) +A(\triangle CXY)}{\triangle ABC }= \dfrac{2}{11}+\dfrac{2}{7}+ \dfrac{20}{77} = \dfrac{14+22+20}{77} = \dfrac{56}{77} = \dfrac{8}{11}$

Subtracting from the ratio $\dfrac{A(\triangle ABC)}{A(\triangle ABC)} = 1$ , we get the answer,

$\dfrac{A(\triangle XYZ)}{A(\triangle ABC)} = \dfrac{3}{11}$ giving the answer as $\color{#3D99F6}{3+11} = \boxed{\color{#20A900}{14}}$

Imgur

In this problem, I use the ratio of the areas of small triangles. (Note: I base on the picture shown in this problem because point Z on segment AC, not on AB as in the problem)

First, let's determine the ratio between the area of triangle $CZX$ and the area of triangle $ABC$ .

Draw segment $BZ$ . We have:

$\frac { { S }_{ CZX } }{ { S }_{ CZB } } =\frac { XC }{ BC } =\frac { XC }{ XB+XC } =\frac { 4 }{ 4+3 }$

Moreover, the ratio between the area of triangle $CZB$ and $ABC$ is

$\frac { { S }_{ CZB } }{ { S }_{ ABC } } =\frac { ZC }{ AC } =\frac { ZC }{ ZA+ZC } =\frac { 5 }{ 5+6 }$

So we got the ratio between the area of triangle $CZX$ and $ABC$ is:

$\frac { { S }_{ CZX } }{ { S }_{ ABC } } =\frac { 5 }{ 5+6 } \times \frac { 4 }{ 4+3 }$

Keep following on this way, we have:

$\frac { { S }_{ XYB } }{ { S }_{ ABC } } =\frac { 2 }{ 2+1 } \times \frac { 3 }{ 4+3 }$

$\frac { { S }_{ AZY } }{ { S }_{ ABC } } =\frac { 1 }{ 2+1 } \times \frac { 6 }{ 5+6 }$

Sum all the above results, we have

$\frac { { S }_{ AZY }+{ S }_{ BXY }+{ S }_{ CXZ } }{ { S }_{ ABC } } =\frac { 8 }{ 11 }$

So, the ratio between the area of triangle $XYZ$ and $ABC$ is

$1-\frac { 8 }{ 11 } =\frac { 3 }{ 11 }$

Finally, the sum of the $numerator$ and the $denominator$ is $3+11=\boxed { 14 }$ !

Satvik, this is your age (you know that, right?). Please correct the problem as I wrote above!