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- size(200); MC(90,"\mbox{semiminor axis}",7,D((0,0)--(0,3),green+linewidth(1)),E);5 KB (892 words) - 21:52, 1 May 2021
- ...s for any given Diophantine equations. This is known as [[Hilbert's tenth problem]]. The answer, however, is no. ...ally proven by [[Andrew Wiles]] after he spent over 7 years working on the 200-page proof, and another year fixing an error in the original proof.9 KB (1,434 words) - 01:15, 4 July 2024
- == Problem == pathpen = black + linewidth(0.7);4 KB (693 words) - 13:03, 28 December 2021
- == Problem == size(200);13 KB (2,080 words) - 13:14, 23 July 2024
- {{AIME Problems|year=2004|n=I}} == Problem 1 ==9 KB (1,434 words) - 13:34, 29 December 2021
- == Problem == ...a [[base]] with [[radius]] <math>600</math> and [[height]] <math> 200\sqrt{7}. </math> A fly starts at a point on the surface of the cone whose distance2 KB (268 words) - 22:20, 23 March 2023
- == Problem == size(200);9 KB (1,500 words) - 20:06, 8 October 2024
- {{AIME Problems|year=2004|n=II}} == Problem 1 ==9 KB (1,410 words) - 05:05, 20 February 2019
- {{AIME Problems|year=2002|n=II}} == Problem 1 ==7 KB (1,177 words) - 15:42, 11 August 2023
- == Problem == ...other cases yield non-convex and/or degenerate hexagons, which violate the problem statement.9 KB (1,472 words) - 15:24, 29 December 2024
- == Problem == ...math> such that <math>1^2+2^2+3^2+\ldots+k^2</math> is a multiple of <math>200</math>.3 KB (403 words) - 12:10, 9 September 2023
- == Problem == ...sqrt{161}}{40}</math>, and so the final answer is <math>-1+161+40 = \boxed{200}</math>.7 KB (1,098 words) - 00:33, 21 January 2025
- == Problem == size(200); pathpen = linewidth(0.7);4 KB (589 words) - 15:33, 20 January 2025
- == Problem == ...ath>. Thus, there are <math>0 + 0 + 2 + 3 + 8 + 18 + 23 + 48 + 98 = \boxed{200}</math> solutions of <math>(a,b,c)</math>.1 KB (228 words) - 08:41, 4 November 2022
- {{AIME Problems|year=2007|n=II}} == Problem 1 ==9 KB (1,435 words) - 01:45, 6 December 2021
- == Problem == ...s 195\cdot 197\cdot 199=\frac{1\cdot 2\cdots200}{2\cdot4\cdots200} = \frac{200!}{2^{100}\cdot 100!}</math>4 KB (562 words) - 18:37, 30 October 2020
- {{AIME Problems|year=2008|n=I}} == Problem 1 ==9 KB (1,536 words) - 00:46, 26 August 2023
- == Problem == size(200);6 KB (1,092 words) - 22:22, 18 August 2024
- {{AIME Problems|year=2011|n=II}} == Problem 1 ==8 KB (1,301 words) - 08:43, 11 October 2020
- {{AIME Problems|year=2015|n=I}} ==Problem 1==10 KB (1,615 words) - 17:03, 9 October 2024