Let ,, be fixed positive integers. There are ducks sitting in a
circle, one behind the other. Each duck picks either rock, paper, or scissors, with ducks
picking rock, ducks picking paper, and ducks picking scissors.
A move consists of an operation of one of the following three forms:
[list]
[*] If a duck picking rock sits behind a duck picking scissors, they switch places.
[*] If a duck picking paper sits behind a duck picking rock, they switch places.
[*] If a duck picking scissors sits behind a duck picking paper, they switch places.
[/list]
Determine, in terms of ,, and , the maximum number of moves which could take
place, over all possible initial configurations.
Let be the circumcenter of triangle . Line intersects the altitude from at point . Let be the midpoints of , respectively. If intersects at , and the circumcircle of triangle meets at , prove that is cyclic.
Source: Own, Entrance Exam for Grade 10 Admission, HSGS 2025
Given the rhombus with its incircle . Let and be the points of tangency of with and respectively. On the edges and , take points and such that is tangent to at . Suppose is the intersection point of the lines and . Prove that two lines and are parallel or coincide.
Let be a triangle with orthocenter and let be the midpoint of Suppose that and are distinct points on the circle with diameter different from such that lies on line Prove that the orthocenter of lies on the circumcircle of
Let be a scalene oblique triangle with circumcenter and orthocenter , and (,) a point in the plane.
Let , be the circumcevian triangles of ,, respectively.
Let be the pedal triangle of with respect to .
Let be the reflection in of . Define , cyclically.
Let be the reflection in of . Define , cyclically.
Let , be the circumcenters of ,, respectively.
Prove that:
1) ,, are collinear if and only if lies on the Jerabek hyperbola of .
2) ,, are collinear if and only if lies on the Lemoine cubic (= ) of .
and permutations. Note that if , then by AM-GM, with equality at , which is the only solution for . Also note that if or all of are negative, then there are no solutions by the Trivial inequality. Also note that when of are negative, then this case is symmetric to . Therefore, there are solutions