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- {{AIME Problems|year=2007|n=I}} [[2007 AIME I Problems/Problem 1|Solution]]7 KB (1,218 words) - 14:28, 11 July 2022
- {{AIME box|year=2007|n=I|before=First Question|num-a=2}}1 KB (204 words) - 12:56, 7 February 2023
- {{AIME box|year=2007|n=I|num-b=1|num-a=3}}2 KB (301 words) - 23:52, 9 January 2025
- ...i</math>, where <math>b</math> is a [[positive]] [[real number]] and <math>i^{2}=-1</math>. Given that the imaginary parts of <math>z^{2}</math> and <m {{AIME box|year=2007|n=I|num-b=2|num-a=4}}1,003 bytes (163 words) - 14:34, 18 February 2017
- {{AIME box|year=2007|n=I|num-b=4|num-a=6}}7 KB (1,076 words) - 07:46, 10 January 2025
- {{AIME box|year=2007|n=I|num-b=6|num-a=8}}2 KB (242 words) - 19:26, 20 April 2023
- {{AIME box|year=2007|n=I|num-b=7|num-a=9}}4 KB (728 words) - 11:07, 2 December 2024
- [[Image:AIME I 2007-10.png]] ...ath>(0,0,0,0)</math> to <math>(3,3,3,3)</math> as follows: if in the <math>i</math>th row, the <math>j</math>th and <math>k</math>th columns are shaded,14 KB (2,337 words) - 08:37, 10 January 2025
- I believe this is an easier way of organizing the solution to reduce the poss Here is my diagram to help you see what I did: https://drive.google.com/file/d/1Gk_cziYvoeg--uVTap5FGOds3SEfiDAW/view10 KB (1,561 words) - 13:27, 21 July 2024
- [[Image:AIME I 2007-9.png]] [[Image:AIME I 2007-9b.png|left]]11 KB (1,853 words) - 19:10, 21 July 2024
- {{AIME box|year=2007|n=I|num-b=3|num-a=5}}1 KB (228 words) - 17:46, 11 March 2021
- {{AIME box|year=2007|n=I|num-b=10|num-a=12}}3 KB (562 words) - 19:02, 30 December 2023
- [[Image:I-07-12.png|250px|left]] {{AIME box|year=2007|n=I|num-b=11|num-a=13}}10 KB (1,458 words) - 19:50, 3 November 2023
- The '''2007 AIME I''' was held on March 13, 2007. The first link contains the full set of test * [[2007 AIME I Problems]]1 KB (135 words) - 11:32, 22 March 2011
- i.e. ...riant]]. Defining <math>b_{i}= \frac{a_{i}}{a_{i-1}}</math> for each <math>i \ge 2</math>, the above equation means13 KB (2,214 words) - 16:39, 28 November 2024
- [[Image:AIME I 2007-13.png]] ...Q=(B+C)/2, R=(C+D)/2, Y=(-3/2,-3/2,2^.5/2),X=(3/2,3/2,2^.5/2), H=(4,2,0), I=(-2,-4,0);7 KB (1,034 words) - 22:30, 18 June 2024
- [[Image:AIME I 2007-15.png]] {{AIME box|year=2007|n=I|num-b=14|after=Last Question}}4 KB (673 words) - 21:14, 6 August 2022
Page text matches
- * [[2006_AIME_I_Problems/Problem_1 | 2006 AIME I Problem 1]]6 KB (943 words) - 09:44, 17 January 2025
- ...AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]]. == AIME Qualification ==19 KB (2,024 words) - 20:13, 22 February 2025
- ...are invited to take the [[American Invitational Mathematics Examination]] (AIME). .... Students that are in the top ~6% of the AMC 10 are invited to take the [[AIME]].4 KB (596 words) - 03:53, 3 February 2025
- ...tive integer]]s as <math>n!=n \cdot (n-1) \cdots 2 \cdot 1 = \prod_{i=1}^n i</math>. Alternatively, a [[recursion|recursive definition]] for the factor <cmath>\sum_{i=1}^{100}(i!)^{2}</cmath>10 KB (809 words) - 15:40, 17 March 2024
- '''2006 AIME II''' problems and solutions. The first link contains the full set of test * [[2006 AIME II Problems]]1 KB (133 words) - 11:32, 22 March 2011
- * [[Mock_AIME_2_2006-2007_Problems#Problem_8 | Mock AIME 2 2006-2007 Problem 8]] ([[number theory]]) *[[1994_AIME_Problems/Problem 9|1994 AIME Problem 9]]2 KB (316 words) - 15:03, 1 January 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) - 14:36, 10 December 2023
- ! scope="row" | '''Mock AMC I''' * [[Mock AIME]]51 KB (6,175 words) - 20:41, 27 November 2024
- A '''Mock AIME''' is a contest that is intended to mimic the [[AIME]] competition. (In more recent years, recurring competitions will be listed ...xOGY2Y2QwOTc3NWZiYjY0LnBkZg==&rn=TWlsZG9yZiBNb2NrIEFJTUUucGRm Mildorf Mock AIME 1]8 KB (901 words) - 19:45, 13 January 2025
- This is a list of all [[AIME]] exams in the AoPSWiki. Many of these problems and solutions are also ava ! Year || Test I || Test II3 KB (400 words) - 19:21, 13 February 2025
- Suppose <math>b_{i} = \frac {x_{i}}3</math>. ...= \sum_{i = 0}^{2005}(b_{i} + 1)^{2} = \sum_{i = 0}^{2005}(b_{i}^{2} + 2b_{i} + 1)8 KB (1,334 words) - 16:37, 15 December 2024
- ...explanations, see related [[2007 AIME I Problems/Problem 10|problem (AIME I 2007, 10)]].'' *[[2007 AIME I Problems/Problem 10]] - the same problem, but with a <math>4\times 6</math>4 KB (638 words) - 15:41, 22 January 2024
- pair A=(0,0),B=(4,0),C=(1.5,2),I=incenter(A,B,C),F=foot(I,A,B); ...(F--I--foot(I,B,C)--I--foot(I,C,A)); D(rightanglemark(I,F,B,5)); MP("r",(F+I)/2,E);2 KB (321 words) - 21:54, 20 June 2024
- ...2\right)^i = 1024 \sum_{i = 10}^\infty 2^{-i} \sum_{\omega^{10}=1} \omega^i</math>. Now, recall that if <math>z_1, z_2, \ldots, z_n</math> are the <ma *[[Mock AIME 1 2006-2007 Problems/Problem 8 | Previous Problem]]5 KB (744 words) - 18:46, 20 October 2020
- ...et circles <math>A''</math>, <math>B''</math>, <math>C''</math>, and <math>I</math> have radii <math>a</math>, <math>b</math>, <math>c</math>, and <math *[[Mock AIME 1 2006-2007 Problems/Problem 9 | Previous Problem]]1 KB (236 words) - 22:58, 24 April 2013
- Case (i) <math>A<90^\circ</math> *[[Mock AIME 1 2006-2007 Problems/Problem 13 | Previous Problem]]3 KB (541 words) - 16:32, 22 November 2023
- [[Mock AIME 1 2006-2007 Problems/Problem 1|Solution]] [[Mock AIME 1 2006-2007 Problems/Problem 2|Solution]]8 KB (1,355 words) - 13:54, 21 August 2020
- ...(a - bi)^n</math>, where <math>n</math> is as small as possible and <math>i = \sqrt{-1}</math>. Compute <math>\frac{b^2}{a^2}</math>. {{Mock AIME box|year=2006-2007|n=2|num-b=3|num-a=5}}1 KB (240 words) - 09:50, 4 April 2012
- ...3}{z^2}+\frac{4}{z ^3}+\frac{5}{z^4}+\cdots</math> and <math>z=n\pm \sqrt{-i},</math> find <math> \lfloor 100n \rfloor</math>. ...r><math>iz^2(z-1)^2=z^2\iff (z-1)^2=\frac{1}{i}=-i\Rightarrow z=1\pm\sqrt{-i}</math>.</center>912 bytes (145 words) - 09:51, 4 April 2012
- ...3}{z^2}+\frac{4}{z ^3}+\frac{5}{z^4}+\cdots</math> and <math>z=n\pm \sqrt{-i},</math> find <math> \lfloor 100n \rfloor.</math> [[Image:Mock AIME 2 2007 Problem14.jpg]]5 KB (848 words) - 22:49, 25 February 2017