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** :-

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 \(177+107=\boxed{284}\).

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TopNewestThanks, but don't you think you should instead mail it to all the JOMO users, since they're all not necessarily on Brilliant.

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We did that too, and this one is for making sure no more disputes come to us :P a fullstop

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How can you derive angle-bisector theorem?

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I won't write it all here.... see it here , i know it is clear enough... ;)

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Thanks!! Can't we use tangent-secant theorem here? The answer will not be satisfactory, BTW!!

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@Aditya Raut Since u' luv b'uutifiing Brilliant, u' may like ter b'uutify yurr image as well. Try givin' yurr image the same background as Brilliant pages 'ave, 'twill look b'uutifull.

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Img

(If you wanna see actual size,the \(\color{Red}{\textbf{bigger}}\) one it's here )

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