Olympiad Geometry Problem #6: Isosceles Triangle, Parallels, Perpendicular

Here is a very fun problem from the 2005 Iran Team Selection Test, used to select the team for the International Math Olympiad. Enjoy! Link below.

[fdf]

X, Y have equal power w.r.t. the circle, so MO⊥XY where M is the midpoint of XY. We need to show PT∥MO. But since AXPY is a parallelogram, M is the midpoint of AP, and clearly O is the midpoint of AT, so MO is the A-midline of △APT and we're done.

[franciscochavez]

Very nice solution

Here is my solution:
Let A' be the reflexion of A with respect point X.
Claim: Triangle TAA' is similar to triangle TCP
Proof:
Easy angle chasing shows that <TAA'= <TCP.
Now :
TA/TC = 2AC/BC = 2AB/BC= 2AX/CP = AA'/CP
This implies Triangle TAA' is similar to triangle TCP
Sipiral Symetry implies TAC is similar to TA'P
Thus <TPA' = <TCA = 90
On the other hand we have XAY is congruent to A'XP thus A'P is parallel to XY, and thus XY perpendicular to TP

[kerimbendov]

Here is my solution:


<(XY, TP) = 90 if and only if
XT^2 - TY^2 = PX^2 - PY^2
But XT^2 - TY^2 = BX^2+BT^2-CT^2-CY^2
= BX^2 - CY^2 and we also know that
BX = PX and CY = YP and from here, the result follows.

[김동욱-wtz]

very nice!

(my solution very similar to yours)
show X is circumcenter of BDP so XD=XP and YD=YP therefore XDY=XPY