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\(\color{Red}{\textbf{FullStop}}\) 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 \(177+107=\boxed{284}\).

Note by Aditya Raut
2 years, 9 months 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 · 2 years, 9 months ago

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@Satvik Golechha We did that too, and this one is for making sure no more disputes come to us :P a fullstop Aditya Raut · 2 years, 9 months ago

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How can you derive angle-bisector theorem? Kartik Sharma · 2 years, 9 months ago

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@Kartik Sharma I won't write it all here.... see it here , i know it is clear enough... ;) Aditya Raut · 2 years, 9 months ago

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@Aditya Raut Thanks!! Can't we use tangent-secant theorem here? The answer will not be satisfactory, BTW!! Kartik Sharma · 2 years, 9 months ago

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@Kartik Sharma Good.use what you think is more elegant ;) gud luck problem solving.... Aditya Raut · 2 years, 9 months 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 · 2 years, 9 months ago

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@Satvik Golechha And lol, what a new invention of language ! Aditya Raut · 2 years, 9 months 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 ?

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(If you wanna see actual size,the \(\color{Red}{\textbf{bigger}}\) one it's here ) Aditya Raut · 2 years, 9 months 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 · 2 years, 9 months 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 · 2 years, 9 months ago

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