Home > Geometry, IMO, Problem Solving > IMO 2014 Problem 3

IMO 2014 Problem 3

Convex quadrilateral {ABCD} has {\angle ABC = \angle CDA = 90^\circ}. Point {H} is the foot of the perpendicular from {A} to {BD}. Points {S} and {T} lie on sides {AB} and {AD}, respectively, such that {H} lies inside triangle {SCT} and

\displaystyle \angle CHS -\angle CSB = 90^\circ,\ \angle THC-\angle DTC = 90^\circ.

Prove that the line {BD} is tangent to the circumcircle of triangle {TSH}.

IMO 2014 Problem 3 (Day 1)

Categories: Geometry, IMO, Problem Solving Tags: ,
  1. No comments yet.
  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: