geometry question

Might be simple for some of you but pls type solution as I couldnt solve

1) In ABC, AD,BE,CF\triangle ABC,\ AD,BE,CF are concurrent cevians and ADAD is altitude.

Then prove that ADAD bisects FDE\angle FDE

2) Inmo 1999 question 1

Note by Megh Parikh
5 years, 10 months ago

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1 vote

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Problem 1 is really simple if u know the concept of harmonic bundles. The official solution is completely unmotivated

Shrihari B - 3 years, 8 months ago

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by the way megha did you qualified for INMO?

aditya dokhale - 5 years, 10 months ago

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yes and not megha, "Megh"

Megh Parikh - 5 years, 10 months ago

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i solved the question no. 2.. please see the solution and tell me whether it is correct or not :

in triangle ABC , FM = 2 and MC = 4 thus FM/MC = 1/2 also BE which is the median passes through M. Thus we conclude that M is the centroid of triangle ABC. thus as AD is perpendicular to BC and CF is angle bisector of triangle ABC, therefore triangle ABC is equilateral. thus in triangle ACF ,by Pythagoras theorem we have AB^2 = 36 + AB^2/4 or AB = 12/3^1/2 THUS PERIMETER OF TRIANGLE ABC = 12 TIMES ROOT OF 3

aditya dokhale - 5 years, 10 months ago

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very clever. Better than mine solution.

Eloy Machado - 5 years, 9 months ago

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correct!

Sagnik Saha - 5 years, 10 months ago

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My new problem I created due to my misread of Q2

Megh Parikh - 5 years, 10 months ago

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Solution to #1 We drop FXBCFX \perp BC and EYBCEY \perp BC. We need to show that

FDA=ADE\angle FDA = \angle ADE

    \iff FDX=EDY\angle FDX = \angle EDY

    \iff tanFDX=tanEDY\tan \angle FDX = \tan \angle EDY

    \iff FXDX=EYDY\boxed{\dfrac{FX}{DX} = \dfrac{ EY}{DY}} ... (i)(i) . Thus if we can prove that (i) is true, we can claim that ADAD bisects FDE\angle FDE

Now, Ceva's theorem in ABC\triangle ABC we have

BDDC×CEEA×AFFB=1\dfrac{BD}{DC} \times \dfrac{CE}{EA} \times \dfrac{AF}{FB} = 1.... (ii)(ii)

Now, FBXABD\triangle FBX \sim \triangle ABD and ECYACD\triangle ECY \sim \triangle ACD and so we have after some computation, FX=BXBD×ADFX = \dfrac{BX}{BD} \times AD and EY=CYCD×ADEY = \dfrac{CY}{CD} \times AD.

Therefore, FXEY=BX×CDDB×CY\dfrac{FX}{EY} = \dfrac{BX \times CD}{DB \times CY} ... (iii)(iii) Coming back to ii)ii) and putting CEEA=CYYD\dfrac{CE}{EA} = \dfrac{CY}{YD} and AFFB=DXBX\dfrac{AF}{FB} = \dfrac{DX}{BX} we have

BDDC×CEEA×AFFB=1\dfrac{BD}{DC} \times \dfrac{CE}{EA} \times \dfrac{AF}{FB} = 1

    \implies BDDC×CYYD×DXXB=1\dfrac{BD}{DC} \times \dfrac{CY}{YD} \times \dfrac{DX}{XB} = 1

    \implies DXDY=BX×DCCY×BD \dfrac{DX}{DY} = \dfrac{BX \times DC}{CY \times BD} ....(iv)(iv). Equating (iii)(iii) and (iv)(iv), we have

DXDY=FXEY \dfrac{DX}{DY} = \dfrac{FX}{EY} and thus we have FXDX=EYDY\boxed{\dfrac{FX}{DX} = \dfrac{ EY}{DY}}. Thus we proved (i)(i) and hence we are done! :)

Sagnik Saha - 5 years, 10 months ago

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Well, problem 2 has many solutions i think as i got 2 solutions. The first one requires to use the formula of the length of the angle bisector repeatedly. This will finally fetch you that ABC\triangle ABC is equilateral. That is the aim of the problem actually. Prove that ABC\triangle ABC is equilateral. If u cannot do just post a comment, Il type out the detailed solution for you.

Sagnik Saha - 5 years, 10 months ago

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One more hint prove triangle QAP AND PAD similar By using cevians theorem and parallel lines

Saurav Ray - 5 years, 10 months ago

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Hint for (1): Draw the line \ell parallel to BCBC through AA. Suposse DEDE and DFDF meet \ell at PP and QQ.

Jorge Tipe - 5 years, 10 months ago

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thanks

Megh Parikh - 5 years, 10 months ago

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