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- == Problem == ...nded by the coordinate axes and the graph of <math>ax+by=6</math> has area 6, then <math>ab=</math>911 bytes (141 words) - 21:12, 8 September 2023
- == Problem == By the given condition in the problem, all the equalities in the above discussion must hold, that is, <math>AI =4 KB (700 words) - 23:18, 28 November 2014
- == Problem == ...h> and <math>y</math> are, so we can decide what they are. Let <math>x = 1.6</math> and <math>y = 1.4</math>. We round <math>x</math> to <math>2</math>1 KB (246 words) - 07:32, 29 June 2023
- == Problem == The problem statement tells us that Xiaoli performed the following computation:1 KB (187 words) - 16:07, 18 January 2020
- == Problem ==736 bytes (114 words) - 21:31, 24 March 2022
- ==Problem == ...at <math>71</math> is prime and <math>k<71</math> by the conditions of the problem, so the denominator in the argument of this value will always be <math>71.<1 KB (233 words) - 17:15, 30 July 2022
- == Problem == {{IMO box|year=1968|num-b=5|after=Last Problem}}3 KB (427 words) - 12:49, 5 December 2023
- == Problem 6 == ...o <math>z_{1}^4</math>(if you want to know why, reread what we want in the problem!)3 KB (428 words) - 02:34, 31 December 2020
- #REDIRECT [[Mock AIME 2 2006-2007 Problems/Problem 6]]54 bytes (6 words) - 15:29, 3 April 2012
- #REDIRECT [[Mock AIME 1 2006-2007 Problems/Problem 6]]54 bytes (6 words) - 15:49, 3 April 2012
- ==Problem==3 KB (478 words) - 03:06, 5 April 2012
- ==Problem==2 KB (266 words) - 17:36, 7 April 2012
- ==Problem==3 KB (556 words) - 15:08, 15 July 2021
- == Problem == ...> however, we may use some logic to first layout a plan. Since for <math>n=6,n=4,</math> and <math>n=2</math>, <math>2^{n} - 2^{\frac{n}{2}} = n2^{n-3}6 KB (1,081 words) - 13:37, 21 June 2023
- ==Problem== ...he area using the first method: <math>\frac{1}{2}\cdot 5 \cdot 12 = 5\cdot 6 = 30</math>. Therefore, we have <math>\frac{1}{2}L \cdot 13 = 30</math>. Mu915 bytes (135 words) - 00:52, 22 June 2018
- ==Problem==1 KB (258 words) - 12:42, 5 July 2013
- ==Problem== ...50\cdot51}{2}=1275 </math>. Therefore, <math> 10n+9\le1275\implies n\le126.6 </math>, so the largest possible winning score is <math> 126 </math>. Notic1 KB (221 words) - 21:14, 27 May 2012
- ==Problem== <math>c-1 = \tfrac{720}{120} = 6 \leadsto c = 7</math>2 KB (386 words) - 01:24, 31 May 2012
- ==Problem==506 bytes (75 words) - 15:40, 20 March 2018
- #redirect [[2012 USAMO Problems/Problem 5]]43 bytes (4 words) - 15:15, 25 August 2020
Page text matches
- == Problem == <cmath>(y^{12}+y^8+y^4+1)(y^2+1)=(y^{14}+y^{12}+y^{10}+y^8+y^6+y^4+y^2+1)</cmath>2 KB (279 words) - 12:33, 27 October 2019
- == Problem == ..._3O_3'</math> is similar to <math>O_1O_2O_2'</math> so <math>O_3O_3'=\frac{6}{14} \cdot 10=\frac{30}{7}</math>. From rectangles, <math>O_3'T=O_1T_1=4</m4 KB (693 words) - 13:03, 28 December 2021
- == Problem == This problem begs us to use the familiar identity <math>e^{it} = \cos(t) + i \sin(t)</ma6 KB (1,154 words) - 03:30, 11 January 2024
- == Problem == ...e(10)); pair A=(0,9), B=(9,9), C=(9,0), D=(0,0), E=(2.5-0.5*sqrt(7),9), F=(6.5-0.5*sqrt(7),9), G=(4.5,9), O=(4.5,4.5); draw(A--B--C--D--A);draw(E--O--F)13 KB (2,080 words) - 21:20, 11 December 2022
- #REDIRECT [[2006 AMC 12A Problems/Problem 6]]45 bytes (5 words) - 10:59, 20 February 2016
- #REDIRECT [[2006 AMC 12A Problems/Problem 6]]45 bytes (5 words) - 11:01, 20 February 2016
- ...cate|[[2006 AMC 12A Problems|2006 AMC 12A #4]] and [[2006 AMC 10A Problems/Problem 4|2006 AMC 10A #4]]}} == Problem ==2 KB (257 words) - 11:20, 2 January 2022
- == Problem 1 == [[2005 AIME I Problems/Problem 1|Solution]]6 KB (983 words) - 05:06, 20 February 2019
- == Problem == [[Image:2005 AIME I Problem 1.png]]1 KB (213 words) - 13:17, 22 July 2017
- == Problem == ...th>(6,334)</math>, <math>(12,167)</math>, <math>(167,12)</math>,<math>(334,6)</math>, <math>(501,4)</math>, <math>(668,3)</math>, <math>(1002,2)</math>2 KB (303 words) - 01:31, 5 December 2022
- == Problem == ==Solution 6 (NO ALGEBRA)==8 KB (1,248 words) - 11:43, 16 August 2022
- == Problem == There are two separate parts to this problem: one is the color (gold vs silver), and the other is the orientation.5 KB (830 words) - 01:51, 1 March 2023
- == Problem == ==Solution 6 (De Moivre's Theorem)==4 KB (686 words) - 12:52, 13 June 2024
- == Problem == draw((5,8.66)--(16.87,6.928));4 KB (567 words) - 20:20, 3 March 2020
- == Problem == ...ns, so from these cubes we gain a factor of <math>\left(\frac{2}{3}\right)^6</math>.4 KB (600 words) - 21:44, 20 November 2023
- == Problem == ...d <math> n. </math> For example, <math> \tau (1)=1 </math> and <math> \tau(6) =4. </math> Define <math> S(n) </math> by <math> S(n)=\tau(1)+ \tau(2) + \4 KB (647 words) - 02:29, 4 May 2021
- == Problem == ...>R</math> and <math>U</math> stay together, then there are <math>3 \cdot 2=6</math> ways.5 KB (897 words) - 00:21, 29 July 2022
- == Problem == ...n from the second gives <cmath>20a=240-40x\rightarrow a=12-2x\rightarrow x=6-\frac a2</cmath>12 KB (2,000 words) - 13:17, 28 December 2020
- == Problem == ...> Point <math> D </math> is on <math> \overline{BC} </math> with <math> CD=6. </math> Point <math> E </math> is on <math> \overline{BC} </math> such tha13 KB (2,129 words) - 18:56, 1 January 2024
- == Problem == We approach the problem by [[recursion]]. We [[partition]] the positive integers into the sets9 KB (1,491 words) - 01:23, 26 December 2022
