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• 2019 AIME I Problems/Problem 10
  <cmath> (x-z_1)^3(x-z_2)^3 \cdots (x-z_{673})^3 </cmath>can be expressed as <math>x^{2019} + 20x^{2018} + 19x^{20 https://www.youtube.com/watch?v=6-kcdBsmCmc
  10 KB (1,689 words) - 17:02, 8 February 2024
• 2019 AIME I Problems/Problem 11
  ...However, the sum of the radius of <math>\omega</math> and <math>\frac{rs}{s-b}</math> is equivalent to the distance between the incenter and the the <ma ...\alpha=b(\alpha+IB)\to \alpha=\frac{b*IB}{a}</cmath> Again, by the incenter-excenter lemma, <math>II_B=2IM</math> so <math>II_b=\frac{2b*IB}{a}</math> a
  21 KB (3,915 words) - 19:55, 10 October 2023
• 2019 AIME I Problems/Problem 13
  dot("$A$", A, dir(A-O1)); dot("$E$", E, dir(E-O2));
  10 KB (1,620 words) - 20:44, 20 December 2023
• 2017 IMO Problems/Problem 4
  We use the tangent-chord theorem: the angle formed between a chord and a tangent line to a circ <math>KT</math> is tangent to <math>\Gamma</math> by the inverse of tangent-chord theorem.
  4 KB (681 words) - 01:41, 19 November 2023
• 2016 IMO Problems/Problem 2
  ...of this second type consists all cells <math>(i,j)</math> for which <math>i-j</math> is constant. {{IMO box|year=2016|num-b=1|num-a=3}}
  1 KB (225 words) - 01:35, 19 November 2023
• 2006 Indonesia MO Problems/Problem 6
  if number[len(number)-1-j] == 10: number[len(number)-1-j] = 0
  5 KB (819 words) - 16:21, 20 December 2019
• 2006 iTest Problems/Problem 20
  {{iTest box|year=2006|num-b=19|num-a=21|ver=[[2006 iTest Problems/Problem U1|U1]] '''•''' [[2006 iTest Proble
  4 KB (334 words) - 12:49, 15 January 2020
• 2006 iTest Problems/Problem 19
  {{iTest box|year=2006|num-b=18|num-a=20|ver=[[2006 iTest Problems/Problem U1|U1]] '''•''' [[2006 iTest Proble
  4 KB (332 words) - 00:10, 27 January 2023
• 2003 Pan African MO Problems/Problem 2
  |num-b=1 |num-a=3
  1 KB (192 words) - 16:25, 28 January 2020
• 2002 Pan African MO Problems/Problem 5
  ...= 2a</math> and <math>\angle FOB = 2b</math>, so <math>\angle EOF = 180-2a-2b</math>. Since the sum of the angles in a quadrilateral is 360 degrees, < |num-b=4
  7 KB (1,158 words) - 13:47, 27 May 2024
• 2019 USAJMO Problems/Problem 5
  ...total ways for all <math>S_{n-1, j}</math>, so doing the same thing <math>n-1</math> more times yields a final answer of <math>(2n)!\cdot 2^{n^2}</math> ...ke <math>S_{1,1}</math> and add 2 new elements, getting you <math>\binom{2n-2}{2}</math> ways to generate <math>S_{2,2}</math>. And you can keep going d
  4 KB (747 words) - 02:29, 14 May 2020
• 2016 IMO Problems
  ...of this second type consists all cells <math>(i,j)</math> for which <math>i-j</math> is constant. ...sible positive integer value of <math>b</math> such that there exists a non-negative integer <math>a</math> for which the set <math>\{P(a+1),P(a+2),\ldo
  4 KB (755 words) - 09:48, 20 December 2023
• GMAAS
  9: KK-Butter is Glass's Son and Human Host, and is the fourth deity of the Almight ...orem was proved by Almighty Gmaas. EDIT: Almighty Gmaas's Theorem has real-world applications: because Almighty Gmaas knows the answer to any math pro
  85 KB (13,971 words) - 18:02, 16 May 2024
• 2019 AMC 8 Problems
  ...to a race. The hare eagerly agrees and quickly runs ahead, leaving the slow-moving tortoise behind. Confident that he will win, the hare stops to take a fill((0,r-t)--(0,r+t)--(num[i],r+t)--(num[i],r-t)--cycle,gray);
  17 KB (2,585 words) - 15:27, 4 May 2024
• 2002 Pan African MO Problems/Problem 6
  <cmath>a_1^2+3a_2^2+5a_3^2+ \cdots +(2n-1)a_n^2 \leq 1</cmath> ...\cdots + a_n)^2 = \sum_{i=1}^{n} (a_i^2) + 2a_1a_2 + 2a_1a_3 + \cdots 2a_{n-1}a_n</math>. Additionally, if <math>i \le j</math>, then <math>a_i \ge a_j
  852 bytes (151 words) - 01:45, 7 January 2020
• 2007 Indonesia MO Problems/Problem 2
  ...nitely many positive integers <math> n</math> which satisfy <math> b(n)=k^2-k+1</math>. ...nitely many positive integers <math> n</math> which satisfy <math> p(n)=k^2-k+1</math>.
  2 KB (418 words) - 15:10, 14 March 2020
• 2004 Pan African MO Problems
  Is <math>4\sqrt{4-2\sqrt{3}}+\sqrt{97-56\sqrt{3}}</math> an integer?
  2 KB (294 words) - 12:56, 5 December 2019
• 2020 AMC 10A Problems/Problem 19
  ...st (since every upper-ring face is adjacent to exactly <math>2</math> lower-ring faces) and then once again <math>9</math> choices for how to travel aro {{AMC10 box|year=2020|ab=A|num-b=18|num-a=20}}
  8 KB (1,269 words) - 05:36, 4 November 2022
• 2020 AMC 12A Problems/Problem 2
  ...s have length <math>\sqrt{2}</math> (this can be deducted using <math>45-45-90</math>, pythag, trig, or just sense) {{AMC12 box|year=2020|ab=A|num-b=1|num-a=3}}
  2 KB (254 words) - 04:19, 7 October 2022
• 2020 AMC 12A Problems/Problem 6
  <ol style="margin-left: 1.5em;"> ...d <math>C</math> are all distinct, so we need to shade at least <math>4+4+2-3=\boxed{\textbf{(D) } 7}</math> unit squares in addition, as shown below:
  5 KB (646 words) - 06:14, 10 September 2021

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