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- == Problem == ...h> and <math>(0,y)</math>. Because the segment has length 2, <math>x^2+y^2=4</math>. Using the midpoint formula, we find that the midpoint of the segmen4 KB (647 words) - 21:51, 12 January 2025
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- ...lity, with equality holding when <math>A^*, B^*, C^*</math> are collinear, i.e. when <math>A,B,C</math> lie on a circle containing <math>D.</math> Addit ===2023 AIME I Problem 5===6 KB (922 words) - 17:34, 13 January 2025
- An example of a classic problem is as follows: ...a coincidence that 67 is close to two-thirds of 100! We can approach this problem in a constructive way, building the set based on the remainders when divide4 KB (635 words) - 12:19, 2 January 2022
- This is a problem where constructive counting is not the simplest way to proceed. This next e ...proceed with the construction. If we were to go like before and break the problem down by each box, we'd get a fairly messy solution.13 KB (2,018 words) - 15:31, 10 January 2025
- ...="text-align:right">([[2000 AMC 12 Problems/Problem 4|2000 AMC 12, Problem 4]])</div> ...a triangle. <div style="text-align:right">(Manhattan Mathematical Olympiad 2004)</div>7 KB (1,111 words) - 14:57, 24 June 2024
- ...ver, it is possible to define a number, <math> i </math>, such that <math> i = \sqrt{-1} </math>. If we add this new number to the reals, we will have ...rm <math> a + bi </math> where <math> a,b\in \mathbb{R} </math> and <math> i = \sqrt{-1} </math> is the [[imaginary unit]]. The set of complex numbers i5 KB (860 words) - 15:36, 10 December 2023
- ...on difference <math>-8</math>; however, <math>7, 0, 7, 14</math> and <math>4, 12, 36, 108, \ldots</math> are not arithmetic sequences, as the difference * [[2005_AMC_10A_Problems/Problem_17 | 2005 AMC 10A Problem 17]]4 KB (736 words) - 02:00, 7 March 2024
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2004 AIME I Problems]]1 KB (135 words) - 18:15, 19 April 2021
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2004 AIME II Problems]]1 KB (135 words) - 12:24, 22 March 2011
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2005 AIME I Problems]]1 KB (154 words) - 12:30, 22 March 2011
- ...A number of '''Mock AMC''' competitions have been hosted on the [[Art of Problem Solving]] message boards. They are generally made by one community member ...amous theorems and formulas and see if there's any way you can make a good problem out of them.51 KB (6,175 words) - 21:41, 27 November 2024
- {{AIME Problems|year=2005|n=I}} == Problem 1 ==6 KB (983 words) - 05:06, 20 February 2019
- == Problem == ...h> k</math>. For example, <math> S_3 </math> is the [[sequence]] <math> 1,4,7,10,\ldots. </math> For how many values of <math> k </math> does <math> S_2 KB (303 words) - 01:31, 5 December 2022
- == Problem == We approach the problem by [[recursion]]. We [[partition]] the positive integers into the sets9 KB (1,491 words) - 01:23, 26 December 2022
- == Problem == ...> feet. The unicorn has pulled the rope taut, the end of the rope is <math>4</math> feet from the nearest point on the tower, and the length of the rope4 KB (729 words) - 01:00, 27 November 2022
- == Problem == ...<math>34</math> complex roots of the form <math> z_k = r_k[\cos(2\pi a_k)+i\sin(2\pi a_k)], k=1, 2, 3,\ldots, 34, </math> with <math> 0 < a_1 \le a_2 \2 KB (298 words) - 20:02, 4 July 2013
- == Problem == <cmath>x \in \left(\frac{1}{2},1\right) \cup \left(\frac{1}{8},\frac{1}{4}\right) \cup \left(\frac{1}{32},\frac{1}{16}\right) \cup \cdots</cmath>2 KB (303 words) - 18:43, 16 October 2024
- == Problem == A [[solid]] in the shape of a right circular [[cone]] is 4 inches tall and its base has a 3-inch radius. The entire [[surface]] of the5 KB (839 words) - 22:12, 16 December 2015
- == Problem == Let <math> ABC </math> be a [[triangle]] with sides 3, 4, and 5, and <math> DEFG </math> be a 6-by-7 [[rectangle]]. A segment is dra4 KB (647 words) - 17:43, 23 November 2024
- == Problem == ...+ 10(n + 1) + n = 3210 + 1111n</math>, for <math>n \in \lbrace0, 1, 2, 3, 4, 5, 6\rbrace</math>.2 KB (250 words) - 23:56, 2 December 2024
- == Problem == ...ath> consecutive integers that sum to <math>2m</math>, the middle integer (i.e., the median) must be <math>2</math>. Therefore, the largest element in8 KB (1,431 words) - 17:50, 29 December 2024