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- ==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
- ==Problem==900 bytes (126 words) - 18:16, 15 October 2023
- == Problem ==5 KB (871 words) - 18:59, 10 May 2023
- == Problem ==2 KB (324 words) - 20:45, 2 January 2018
- #REDIRECT [[2003 AMC 10B Problems/Problem 8]]45 bytes (5 words) - 00:17, 5 January 2014
- ==Problem==1 KB (190 words) - 11:26, 13 June 2022
- ==Problem == ...extbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 10 </math>518 bytes (78 words) - 18:23, 24 September 2016
- ==Problem==2 KB (277 words) - 13:08, 1 July 2023
- == Problem== Since the problem doesn't specify the number of 3-point shots she attempted, it can be assume3 KB (419 words) - 11:39, 10 March 2024
- #REDIRECT [[2013 AMC 12B Problems/Problem 5]]45 bytes (5 words) - 12:10, 7 April 2013
- {{duplicate|[[2013 AMC 12B Problems|2013 AMC 12B #6]] and [[2013 AMC 10B Problems|2013 AMC 10B #11]]}} ==Problem==2 KB (291 words) - 18:04, 14 July 2021
Page text matches
- == 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
- == Problem == real x = 20 - ((750)^.5)/3, CE = 8*(6^.5) - 4*(5^.5), CD = 8*(6^.5), h = 4*CE/CD;4 KB (729 words) - 01:00, 27 November 2022
- == Problem == ...19}+ \frac3{19}+ \frac 1{17}+ \frac2{17}= \frac6{19} + \frac 3{17} = \frac{6\cdot17 + 3\cdot19}{17\cdot19} = \frac{159}{323}</math> and so the answer is2 KB (298 words) - 20:02, 4 July 2013
- == Problem == ...left(\frac{\frac{4}{5}}{1-\frac{1}{25}}\right)= \frac{2}{3} \cdot \frac{5}{6} = \frac{5}{9},</cmath> and the answer is <math>m+n = 5 + 9 = \boxed{014}</2 KB (303 words) - 22:28, 11 September 2020
- == Problem == ...th>r_{ABC} = \frac{[ABC]}{s_{ABC}} = \frac{15 \cdot 36 /2}{(15+36+39)/2} = 6</math>. Thus <math>r_{A'B'C'} = r_{ABC} - 1 = 5</math>, and since the ratio5 KB (836 words) - 07:53, 15 October 2023
- == Problem == ...th> be a [[triangle]] with sides 3, 4, and 5, and <math> DEFG </math> be a 6-by-7 [[rectangle]]. A segment is drawn to divide triangle <math> ABC </math4 KB (618 words) - 20:01, 4 July 2013
- == Problem == ...n + 1) + n = 3210 + 1111n</math>, for <math>n \in \lbrace0, 1, 2, 3, 4, 5, 6\rbrace</math>.2 KB (374 words) - 14:53, 27 December 2019
- == Problem == The thing about this problem is, you have some "choices" that you can make freely when you get to a cert8 KB (1,437 words) - 21:53, 19 May 2023
- == Problem == Alpha and Beta both took part in a two-day problem-solving competition. At the end of the second day, each had attempted quest3 KB (436 words) - 18:31, 9 January 2024
- == Problem == ...\le i \neq j}^{15} S_iS_j\right)\\ (-8)^2 &= \frac{15(15+1)(2\cdot 15+1)}{6} + 2C\end{align*}</cmath>6 KB (941 words) - 11:37, 27 May 2024
- == Problem == There are no regular 3-pointed, 4-pointed, or 6-pointed stars. All regular 5-pointed stars are similar, but there are two n4 KB (620 words) - 21:26, 5 June 2021
