$\color{#D61F06}{\textbf{FullStop}}$ to the disputes- JOMO 7

First of all, when at first the question file of $\textbf{JOMO 7}$ was uploaded, there was a little flaw in question $9$ (short) , and so we uploaded the new file with an addition in the question (length of AC)

But later, when the results were out and solution file was uploaded, the solution was as per the previous (wrong) version of the question, so results will be posted once again.

To put a fullstop to all the disputes and clarifications, I am writing this note.

Question (JOMO 7, short 9) :- In $\triangle ABC$ ,$I$ is the incenter, and $AI,BI,CI$ meet the sides $BC,AC,AB$ at points $D,E,F$ respectively. $AB=20 , BC=14 , AC=\dfrac{438}{53}$. Then the ratio $\dfrac{ID}{IF}$ can be written as $\dfrac{a}{b}$ where $b$ and $a$ are coprime positive integers, find $a+b$

Solution :-

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For convenience, $AB=c,BC=a,AC=b$

We will use the angle bisector property in $\triangle ABC$ ,

See that $\dfrac{DC}{BD} = \dfrac{b}{c} \implies \dfrac{BD}{BC}=\dfrac{c}{b+c}$

Thus we have $BD=\dfrac{ac}{b+c}$

Now, we use angle bisector property for $\triangle BAD$ ,

$\dfrac{BD}{AB} = \dfrac{ID}{AI} = \dfrac{\frac{ac}{b+c}}{c} = \dfrac{a}{b+c}$

Thus we get $\dfrac{ID}{AD} = \dfrac{a}{a+b+c}$

Now, $AD = \dfrac{\sqrt{bc}}{b+c} \sqrt{(b+c)^2-a^2}$ (length of the angle bisector)

So we get the value of $ID= \dfrac{a}{a+b+c}\times \dfrac{\sqrt{bc}}{b+c} \sqrt{(b+c)^2-a^2}$

Similarly, we get $IF= \dfrac{c}{a+b+c} \times \dfrac{\sqrt{ab}}{a+b} \sqrt{(b+a)^2-c^2}$

Then , after taking their ratio, answer comes as

$\dfrac{ID}{IF} = \dfrac{(b+a) \sqrt{[(b+c)^2-a^2]a}}{(b+c) \sqrt{[(b+a)^2-c^2]c}}$

$\dfrac{ID}{IF}= \dfrac{(b+a) \sqrt{(b+c-a)a}}{(b+c) \sqrt{(a+b-c)c}}$

And, finally, putting values

$\dfrac{ID}{IF} = \dfrac{(b+14)\sqrt{(b+6)14}}{(b+20)\sqrt{(b-6)20}}$

And when you put $b=\dfrac{438}{53}$, answer is $\dfrac{177}{107}$, hence the asked value is $177+107=\boxed{284}$.