China TST 2016 P1

For discussing Olympiad level Geometry Problems
User avatar
Kazi_Zareer
Posts:86
Joined:Thu Aug 20, 2015 7:11 pm
Location:Malibagh,Dhaka-1217
China TST 2016 P1

Unread post by Kazi_Zareer » Sat May 07, 2016 8:25 pm

$ABCDEF$ is a cyclic hexagon with $AB=BC=CD=DE$. $K$ is a point on segment $AE$ satisfying $\angle BKC=\angle KFE$, $\angle CKD=\angle KFA.$

Prove that $KC=KF$
We cannot solve our problems with the same thinking we used when we create them.

User avatar
nahin munkar
Posts:81
Joined:Mon Aug 17, 2015 6:51 pm
Location:banasree,dhaka

Re: China TST 2016 P1

Unread post by nahin munkar » Fri May 20, 2016 7:54 pm

Use $ homotety$.It may be a useful way to get the solution. :geek:
# Mathematicians stand on each other's shoulders. ~ Carl Friedrich Gauss

User avatar
Kazi_Zareer
Posts:86
Joined:Thu Aug 20, 2015 7:11 pm
Location:Malibagh,Dhaka-1217

Re: China TST 2016 P1

Unread post by Kazi_Zareer » Fri May 27, 2016 2:17 am

nahin munkar wrote:Use $ homotety$.It may be a useful way to get the solution. :geek:
Did you solve it? :?:
We cannot solve our problems with the same thinking we used when we create them.

User avatar
nahin munkar
Posts:81
Joined:Mon Aug 17, 2015 6:51 pm
Location:banasree,dhaka

Re: China TST 2016 P1

Unread post by nahin munkar » Sat May 28, 2016 12:41 am

Kazi_Zareer wrote:
nahin munkar wrote:Use $ homotety$.It may be a useful way to get the solution. :geek:
Did you solve it? :?:
Yeah, solved. :ugeek:
# Mathematicians stand on each other's shoulders. ~ Carl Friedrich Gauss

Post Reply