Synthetic Geometry Group-Sualeh's Proposal (Pakistan Round 1)

These problems are my first submissions to Xuming's Geometry group.

However they are for the purpose of sharing the Geometry problems from Pakistan's First Round (for 2016)

Q1.The tangents at A,BA,B to the circumcircleω\omega of Triangle ABCABC meet at TT. The line through TACT||AC meets BCBC at DD. Prove that AD=CDAD=CD.

Q2. In a right angled triangle ABC,C^=90,CDABABC, \hat{C}=90^{\circ}, CD\perp AB at DD and the angle bisector of B^\hat{B} intersects CD,ACCD,AC at O,EO,E respectively. Through OO introduce FGABFG \parallel AB such that FGFG intersects AC,BCAC,BC at F,GF,G respectively. Prove that AF=CEAF=CE.

Geometry-Pakistan's First Round (for IMO 2016).

Note by Sualeh Asif
4 years ago

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Due to lack of time I have uploaded pics of my solution to problem 1.

Nihar Mahajan - 4 years ago

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Good job Nihar ! The crux move was to identify that TBDA is cyclic, by wishful thinking.

Karthik Venkata - 4 years ago

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absolutely!

Sualeh Asif - 4 years ago

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@Calvin Lin @Xuming Liang Here is my submission!

@Nihar Mahajan ,@Sharky Kesa I thought you would be interested!

Sualeh Asif - 4 years ago

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Ill add in the solutions when requested. (Not before a week though)

Sualeh Asif - 4 years ago

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1st one is just some simple angle chase using alternate segment theorem as proved in the solution by @Nihar Mahajan.

2nd one is also just trivial: Solution to II.

ΔDBCΔCBA\Delta DBC \sim \Delta CBA

=> DBCB=CBAB\dfrac{DB}{CB} = \dfrac{CB}{AB}

=> DOOC=CEAE\dfrac{DO}{OC} = \dfrac{CE}{AE} --- (Int. Angle Bisector Theorem)

=> AFCF=CEAE\dfrac{AF}{CF} = \dfrac{CE}{AE}

=> AFAC=CEAC\dfrac{AF}{AC} = \dfrac{CE}{AC} ---- (By rule of Dividendo)

=> AF=ACAF = AC

K.I.P.K.I.G

The almighty knows it all. - 2 years, 6 months ago

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