Waste less time on Facebook — follow Brilliant.
×

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 \(A,B\) to the circumcircle\(\omega\) of Triangle \(ABC\) meet at \(T\). The line through \(T||AC\) meets \(BC\) at \(D\). Prove that \(AD=CD\).

Q2. In a right angled triangle \(ABC, \hat{C}=90^{\circ}, CD\perp AB \) at \(D\) and the angle bisector of \(\hat{B}\) intersects \(CD,AC\) at \(O,E\) respectively. Through \(O\) introduce \(FG \parallel AB\) such that \(FG\) intersects \(AC,BC\) at \(F,G\) respectively. Prove that \(AF=CE\).

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

Note by Sualeh Asif
1 year, 4 months ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

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

Nihar Mahajan · 1 year, 4 months ago

Log in to reply

@Nihar Mahajan Good job Nihar ! The crux move was to identify that TBDA is cyclic, by wishful thinking. Karthik Venkata · 1 year, 4 months ago

Log in to reply

@Karthik Venkata absolutely! Sualeh Asif · 1 year, 4 months ago

Log in to reply

@Calvin Lin @Xuming Liang Here is my submission!

@Nihar Mahajan ,@Sharky Kesa I thought you would be interested! Sualeh Asif · 1 year, 4 months ago

Log in to reply

@Sualeh Asif Ill add in the solutions when requested. (Not before a week though) Sualeh Asif · 1 year, 4 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...