Search results

  • ...ms/Problem 1 | Problem 1]] proposed by Unknown Author, Romania (taken from a Romanian problem book) * [[1969 IMO Problems/Problem 5 | Problem 5]] proposed by A. Mekei, Mongolia
    35 KB (4,009 words) - 20:25, 21 February 2024
  • ...nce at nationals can be described as spastic. Reaching a high of 2nd, with a people coming in second or making the semifinals of the countdown, to team * 2006 - George Yu (11), Amadeus Zhu, Michael Zhao, Owen Hill, Coach: Mike Rowson
    4 KB (624 words) - 09:40, 29 April 2024
  • *The 2020 State Competition marks the first time that a Chicago Public School has won in recent history. ...as Sun (2nd (on written), Q), Elbert Du (13th), Allen Chen (48th), Haoyang Yu, Coach: Nicholas Titus
    5 KB (741 words) - 10:25, 1 April 2024
  • ...lowed by lunch. After lunch, the team round takes place, and then there is a brief waiting period until the countdown round competitors are announced. A The winners progress to a second round, and the losers are eliminated. In this second round, the lose
    9 KB (1,234 words) - 08:31, 2 May 2024
  • ...guess can cause the recursion to cycle or diverge instead of converging to a root. ...exactly one root, as long as <math>f'(x_i) \neq 0</math> we can construct a [[Taylor polynomial#Tangent-line approximation|tangent-line approximation]]
    13 KB (2,298 words) - 23:34, 28 May 2023
  • ...at the Euler line of a given triangle together with two of its sides forms a triangle whose Euler line is in parallel with the third side of the given t ...d by the Euler line and the sides, taken by two, of a given triangle, form a triangle which is perspective with the given triangle and has the same Eule
    18 KB (3,046 words) - 06:44, 19 January 2023
  • ...\( (f(x) + f(y))(f(u) + f(v)) = f(xu - yv) + f(xv - yu) \), we aim to find a function that satisfies it. Using parity and taking \( a = u \) and \( b = -v \) in the original equation, we get \( f(u^2 + v^2) =
    2 KB (369 words) - 16:12, 26 March 2024
  • ...State University named after M.V. Lomonosov. The first four problems have a standard level. A right circular cylinder is located so that the circle of its upper base tou
    32 KB (5,375 words) - 11:43, 5 May 2024