# $\color{#D61F06}{\textbf{FullStop}}$ 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 :- img

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 $177+107=\boxed{284}$. 6 years, 2 months ago

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How can you derive angle-bisector theorem?

- 6 years, 2 months ago

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

- 6 years, 2 months ago

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

- 6 years, 2 months ago

Good.use what you think is more elegant ;) gud luck problem solving....

- 6 years, 2 months ago

@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.

- 6 years, 2 months ago

And lol, what a new invention of language !

- 6 years, 2 months ago

Wait wait wait.... What exactly ? This image is an HD one, even my fb and gmail pro pic, it's cool already! Why make it brilliant color ? Img

(If you wanna see actual size,the $\color{#D61F06}{\textbf{bigger}}$ one it's here )

- 6 years, 2 months ago

Noah. 'was talkin' abou' the image in 'dis post, the bagg-round 'iffers from the Brilliant bagg-round.

- 6 years, 2 months ago

I made that diagram in paint, so i didn't choose it that way... anyway, no matter, it doesn't make a big difference.

- 6 years, 2 months ago

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

- 6 years, 2 months ago

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

- 6 years, 2 months ago