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

First of all, when at first the question file of JOMO 7\textbf{JOMO 7} was uploaded, there was a little flaw in question 99 (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 ABC\triangle ABC ,II is the incenter, and AI,BI,CIAI,BI,CI meet the sides BC,AC,ABBC,AC,AB at points D,E,FD,E,F respectively. AB=20,BC=14,AC=43853AB=20 , BC=14 , AC=\dfrac{438}{53}. Then the ratio IDIF\dfrac{ID}{IF} can be written as ab\dfrac{a}{b} where bb and aa are coprime positive integers, find a+ba+b

Solution :-

img img

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

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

See that DCBD=bc    BDBC=cb+c\dfrac{DC}{BD} = \dfrac{b}{c} \implies \dfrac{BD}{BC}=\dfrac{c}{b+c}

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

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

BDAB=IDAI=acb+cc=ab+c\dfrac{BD}{AB} = \dfrac{ID}{AI} = \dfrac{\frac{ac}{b+c}}{c} = \dfrac{a}{b+c}

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

Now, AD=bcb+c(b+c)2a2AD = \dfrac{\sqrt{bc}}{b+c} \sqrt{(b+c)^2-a^2} (length of the angle bisector)

So we get the value of ID=aa+b+c×bcb+c(b+c)2a2ID= \dfrac{a}{a+b+c}\times \dfrac{\sqrt{bc}}{b+c} \sqrt{(b+c)^2-a^2}

Similarly, we get IF=ca+b+c×aba+b(b+a)2c2IF= \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

IDIF=(b+a)[(b+c)2a2]a(b+c)[(b+a)2c2]c\dfrac{ID}{IF} = \dfrac{(b+a) \sqrt{[(b+c)^2-a^2]a}}{(b+c) \sqrt{[(b+a)^2-c^2]c}}

IDIF=(b+a)(b+ca)a(b+c)(a+bc)c\dfrac{ID}{IF}= \dfrac{(b+a) \sqrt{(b+c-a)a}}{(b+c) \sqrt{(a+b-c)c}}

And, finally, putting values

IDIF=(b+14)(b+6)14(b+20)(b6)20\dfrac{ID}{IF} = \dfrac{(b+14)\sqrt{(b+6)14}}{(b+20)\sqrt{(b-6)20}}

And when you put b=43853b=\dfrac{438}{53}, answer is 177107\dfrac{177}{107}, hence the asked value is 177+107=284177+107=\boxed{284}.

Note by Aditya Raut
5 years ago

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

Satvik Golechha - 5 years ago

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

Aditya Raut - 5 years ago

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How can you derive angle-bisector theorem?

Kartik Sharma - 5 years ago

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

Aditya Raut - 5 years ago

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

Kartik Sharma - 5 years ago

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

Aditya Raut - 5 years ago

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@Aditya Raut @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.

Satvik Golechha - 5 years ago

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@Satvik Golechha 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 Img

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

Aditya Raut - 5 years ago

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

Satvik Golechha - 5 years ago

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

Aditya Raut - 5 years ago

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@Satvik Golechha And lol, what a new invention of language !

Aditya Raut - 5 years ago

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