In a convex quadrilateral , the diagonal bisects neither the angle nor the angle . The point lies inside and satisfies Prove that is a cyclic quadrilateral if and only if .
concyclic wanted, PQ = BP, cyclic quadrilateral and 2 parallelograms related
parmenides512
N3 hours ago
by SuperBarsh
Source: 2011 Italy TST 2.2
Let be a cyclic quadrilateral in which the lines and meet at a point . Let be the point of the line , different from , such that . We construct the parallelograms and . Prove that the points lie on the same circle.
Determine all functions defined on the set of all positive integers and taking non-negative integer values, satisfying the three conditions:
[list]
[*] for at least one ;
[*] for every positive integers and ;
[*] there are infinitely many positive integers such that for all .
[/list]
Perpendicular passes from the intersection of diagonals, \angle AEB = \angle CED
NO_SQUARES1
N4 hours ago
by mathuz
Source: 239 MO 2025 10-11 p3
Inside of convex quadrilateral point was chosen such that and . Prove that if perpendicular from to passes from the intersection of diagonals of , then .
Initially, Aslı distributes balls to boxes as she wishes. After that, Aslı and Zehra make alternated moves which consists of taking a ball in any wanted box starting with Aslı. One who takes the last ball from any box takes that box to herself. What is the maximum number of boxes can Aslı guarantee to take herself regardless of Zehra's moves?
is a triangle with incentre . The feet of the altitudes from to are respectively, and the line through parallel to intersects and at and respectively. Prove that the circles with diameters and have a common point on the circumcircle of
is a triangle with incentre . The feet of the altitudes from to are respectively, and the line through parallel to intersects and at and respectively. Prove that the circles with diameters and have a common point on the circumcircle of
Notice that Define and See that and swap under incircle inversion. Also, swaps with the nine-point circle of . Hence, (sharky-devil)
Now let be the -Antipode in . Applying Pascal's Theorem on we get , since . Thus, we have . Similarly, so we are done.