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2006 AIME I Problems/Problem 15

6 KB (910 words) - 19:31, 24 October 2023
2006 AMC 12B Problems/Problem 15

3 KB (458 words) - 16:40, 6 October 2019
2006 AMC 12A Problems/Problem 15

919 bytes (138 words) - 12:45, 4 August 2017
2005 AMC 12B Problems/Problem 15

2 KB (411 words) - 21:02, 21 December 2020
2006 AMC 10A Problems/Problem 15

3 KB (532 words) - 17:49, 13 August 2023
2005 AIME I Problems/Problem 15
</asy></center>

5 KB (906 words) - 23:15, 6 January 2024
2005 AIME II Problems/Problem 15
D((0,30)--(0,-10),Arrows(4));D((15,0)--(-25,0),Arrows(4));D((0,0)--MP("y=ax",(14,14 * (69/100)^.5),E),EndArrow [[Image:2005_AIME_II_-15.png||center|800px]]

12 KB (2,000 words) - 13:17, 28 December 2020
2004 AIME I Problems/Problem 15
15&2^{13}-4\cdot 7\\ 17&2^{15}-16\cdot 9\\

9 KB (1,491 words) - 01:23, 26 December 2022
2004 AIME II Problems/Problem 15
<cmath>s_{13, 4} = 2s_{15 - 13, 3} + 1 = 2s_{2, 3}+1</cmath> (46,15)&64&16\\

6 KB (899 words) - 20:58, 12 May 2022
1983 AIME Problems/Problem 15
 <cmath>0 \geq f(3) = 25 - 15\cos \alpha - 20 \sin \alpha</cmath>

19 KB (3,221 words) - 01:05, 7 February 2023
1984 AIME Problems/Problem 15
\frac{x^2}{15}+\frac{y^2}{7}-\frac{z^2}{9}-\frac{w^2}{33}=1\\ \frac{x^2}{63}+\frac{y^2}{55}+\frac{z^2}{39}+\frac{w^2}{15}=1\\

6 KB (1,051 words) - 04:52, 8 May 2024
1985 AIME Problems/Problem 15
[[Image:AIME 1985 Problem 15.png]]

2 KB (245 words) - 22:44, 4 March 2024
1986 AIME Problems/Problem 15
<math>2a^2 + 5b^2 = - \frac {15}{2}ab \ \ \ \ (2)</math></div>

11 KB (1,722 words) - 09:49, 13 September 2023
1987 AIME Problems/Problem 15

5 KB (838 words) - 18:05, 19 February 2022
1988 AIME Problems/Problem 15

7 KB (1,186 words) - 10:16, 4 June 2023
1989 AIME Problems/Problem 15
...ac{[APC] + [BPC]}{[APB]} = 3</math>, so <math>CP = \dfrac{3}{4} \cdot CF = 15</math>. ...ed out, so <math>w_C = 1</math> and <math>w_F = 3</math>. Thus, <math>CP = 15</math> and <math>PF = 5</math>.

13 KB (2,091 words) - 00:20, 26 October 2023
1990 AIME Problems/Problem 15

4 KB (644 words) - 16:24, 28 May 2023
1991 AIME Problems/Problem 15

4 KB (658 words) - 16:58, 10 November 2023
2006 AMC 10B Problems/Problem 15

3 KB (445 words) - 22:01, 20 August 2022
1992 AIME Problems/Problem 15

2 KB (358 words) - 01:54, 2 October 2020
1993 AIME Problems/Problem 15

3 KB (449 words) - 21:39, 21 September 2023
1994 AIME Problems/Problem 15
.../2,Fb,12,-35); D(La,bluedots);D(Lb,bluedots);D(endptproject((C+P)/2,Fc,18,-15),bluedots);D(IP(La,Lb),blue);

4 KB (717 words) - 22:20, 3 June 2021
1995 AIME Problems/Problem 15
...ve. This is an infinite geometric series whose sum is <math>\frac{3/64}{1-(15/32)}=\frac{3}{34}</math>, so the answer is <math>\boxed{037}</math>. ...e roll <math>4</math> consecutive <tt>H</tt>'s, and there is a <math>\frac{15}{16}</math> probability we roll a <tt>T</tt>. Thus,

6 KB (979 words) - 13:20, 11 April 2022
1996 AIME Problems/Problem 15
...\Longrightarrow \cos 2\theta = \frac{\sqrt{3}}{2} \Longrightarrow \theta = 15^{\circ}</math>. The answer is <math>\lfloor 1000r \rfloor = \left\lfloor 10 ...cdot 4}}{8} = \frac {\sqrt {6} \pm \sqrt {2}}{4}</math>, so <math>\theta = 15^{\circ}</math> (as the other roots are too large to make sense in context).

5 KB (710 words) - 21:04, 14 September 2020
1997 AIME Problems/Problem 15

4 KB (609 words) - 22:49, 17 July 2023
1998 AIME Problems/Problem 15

9 KB (1,671 words) - 22:10, 15 March 2024
1999 AIME Problems/Problem 15
...has pairwise parallel planar and oppositely equal length (<math>4\sqrt{13},15,17</math>) edges and can be inscribed in a parallelepiped (rectangular box) <math>q^2+r^2=15^2</math>

7 KB (1,169 words) - 15:28, 13 May 2024
2000 AIME I Problems/Problem 15
Round 7: <math>b</math> to <math>b</math>, <math>15</math> to right, <math>16</math> left in deck, <math>n = -2 + 8k</math>, be ...ans our sieving process will return to normal after Round 7, with <math>31-15=16</math> cards remaining. Since <math>16</math> is a perfect power of <mat

15 KB (2,673 words) - 19:16, 6 January 2024
2001 AIME I Problems/Problem 15

11 KB (1,837 words) - 18:53, 22 January 2024
2002 AIME I Problems/Problem 15

7 KB (1,181 words) - 20:32, 8 January 2024
2003 AIME I Problems/Problem 15

8 KB (1,382 words) - 14:23, 29 December 2022
2003 AIME II Problems/Problem 15

4 KB (675 words) - 17:23, 30 July 2022
2002 AIME II Problems/Problem 15

7 KB (1,182 words) - 09:56, 7 February 2022
2001 AIME II Problems/Problem 15

4 KB (518 words) - 15:01, 31 December 2021
2000 AIME II Problems/Problem 15

3 KB (469 words) - 21:14, 7 July 2022
2005 AMC 10B Problems/Problem 15

45 bytes (5 words) - 16:48, 16 July 2011
Mock AIME 1 2006-2007 Problems/Problem 15

3 KB (414 words) - 13:45, 19 February 2016
Mock AIME 2 2006-2007 Problems/Problem 15

4 KB (739 words) - 17:04, 24 November 2023
2005 AMC 10A Problems/Problem 15

2 KB (306 words) - 19:29, 13 December 2021
2004 AMC 10A Problems/Problem 15

3 KB (457 words) - 02:16, 5 July 2021
2003 AMC 10A Problems/Problem 15

2 KB (261 words) - 14:34, 17 August 2023
2004 AMC 12A Problems/Problem 15
{{duplicate|[[2004 AMC 12A Problems|2004 AMC 12A #15]] and [[2004 AMC 10A Problems/Problem 17|2004 AMC 10A #17]]}}

2 KB (309 words) - 22:27, 15 August 2023
2003 AMC 12A Problems/Problem 15
{{duplicate|[[2003 AMC 12A Problems|2003 AMC 12A #15]] and [[2003 AMC 10A Problems|2003 AMC 10A #19]]}}

3 KB (380 words) - 21:53, 19 March 2022
Mock AIME 4 2006-2007 Problems/Problem 15

3 KB (532 words) - 20:29, 31 August 2020
Mock AIME 3 Pre 2005 Problems/Problem 15

2 KB (312 words) - 10:38, 4 April 2012
2007 AIME I Problems/Problem 15
[[Image:AIME I 2007-15.png]]

4 KB (673 words) - 22:14, 6 August 2022
2007 AIME II Problems/Problem 15
...triangle <math>ABC</math> are <math>13,</math> <math>14,</math> and <math>15,</math> the radius of <math>\omega</math> can be represented in the form <m B=origin; C=15*right; A=IP(CR(B,13),CR(C,14)); P=incenter(A,B,C);

11 KB (2,099 words) - 17:51, 4 January 2024
2007 AMC 12A Problems/Problem 15

1 KB (152 words) - 14:52, 7 August 2017
2005 AMC 12A Problems/Problem 15

14 KB (1,970 words) - 17:02, 18 August 2023
2006 AIME II Problems/Problem 15
Thus, <math>A = \frac{1}{15\sqrt{7}}</math> and <math>x + y + z = 30A = \frac2{\sqrt{7}}</math> and the

4 KB (725 words) - 17:18, 27 June 2021
2000 AMC 12 Problems/Problem 15
{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #15]] and [[2000 AMC 10 Problems|2000 AMC 10 #24]]}}

3 KB (441 words) - 21:11, 29 April 2023
2002 AMC 12B Problems/Problem 15

2 KB (395 words) - 15:50, 3 April 2022
2007 AMC 10A Problems/Problem 15

2 KB (326 words) - 10:29, 4 June 2021
2008 AMC 12A Problems/Problem 15
{{duplicate|[[2008 AMC 12A Problems|2008 AMC 12A #15]] and [[2008 AMC 10A Problems/Problem 24|2008 AMC 10A #24]]}}

4 KB (547 words) - 04:19, 30 September 2023
2007 AMC 12B Problems/Problem 15

2 KB (250 words) - 15:41, 27 July 2021
2008 AIME I Problems/Problem 15

6 KB (1,041 words) - 00:54, 1 February 2024
Mock AIME 1 2007-2008 Problems/Problem 15

2 KB (276 words) - 16:27, 26 December 2015
2008 AIME II Problems/Problem 15

4 KB (628 words) - 16:23, 2 January 2024
2008 AMC 10B Problems/Problem 15

1,016 bytes (166 words) - 11:39, 8 October 2023
2008 AMC 10A Problems/Problem 15

6 KB (1,033 words) - 15:19, 1 July 2021
1995 AHSME Problems/Problem 15

1 KB (183 words) - 21:54, 2 August 2016
2002 AMC 10A Problems/Problem 15

993 bytes (147 words) - 01:50, 26 January 2018
2002 AMC 10B Problems/Problem 15

45 bytes (5 words) - 18:14, 28 July 2011
2000 AMC 10 Problems/Problem 15

45 bytes (4 words) - 23:54, 26 November 2011
1985 AJHSME Problems/Problem 15

3 KB (473 words) - 08:14, 13 January 2023
1986 AJHSME Problems/Problem 15

1 KB (155 words) - 15:32, 9 November 2017
2002 AMC 12A Problems/Problem 15
{{duplicate|[[2002 AMC 12A Problems|2002 AMC 12A #15]] and [[2002 AMC 10A Problems|2002 AMC 10A #21]]}} \text{(E) }15

3 KB (519 words) - 21:34, 23 June 2023
2004 AMC 12B Problems/Problem 15
{{duplicate|[[2004 AMC 12B Problems|2004 AMC 12B #15]] and [[2004 AMC 10B Problems|2004 AMC 10B #17]]}}

2 KB (326 words) - 09:47, 17 October 2020
2004 AMC 10B Problems/Problem 15
<math> \textbf{(A)}\ \textdollar 1.15\qquad\textbf{(B)}\ \textdollar 1.20\qquad\textbf{(C)}\ \textdollar 1.25\qqu .../math>. Her coins are worth <math>200-5n=\boxed{\mathrm{(A)\ }\textdollar1.15}</math>.

1 KB (220 words) - 00:53, 24 July 2014
2009 AMC 12A Problems/Problem 15

4 KB (634 words) - 16:34, 3 December 2020
2001 AMC 12 Problems/Problem 15

1 KB (225 words) - 15:26, 30 November 2021
2009 AMC 10B Problems/Problem 15
{{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #15]] and [[2009 AMC 12B Problems|2009 AMC 12B #8]]}}

3 KB (545 words) - 13:37, 3 September 2023
2009 AMC 12B Problems/Problem 15

2 KB (336 words) - 16:58, 27 December 2020
1987 AJHSME Problems/Problem 15

731 bytes (110 words) - 16:40, 30 July 2023
2009 AIME I Problems/Problem 15
.../math> and <math>COP</math>, with <math>BO=CO=7</math> and <math>OP=7 \tan 15=7(2-\sqrt{3})=14-7\sqrt3</math>. Then, the area of [<math>\triangle BPC</ma

6 KB (1,048 words) - 19:35, 2 January 2023
1988 AJHSME Problems/Problem 15

721 bytes (100 words) - 23:55, 4 July 2013
2009 AIME II Problems/Problem 15

11 KB (1,849 words) - 19:43, 2 January 2023
1989 AJHSME Problems/Problem 15

1 KB (194 words) - 00:03, 5 July 2013
1990 AJHSME Problems/Problem 15

863 bytes (124 words) - 00:06, 5 July 2013
1991 AJHSME Problems/Problem 15
draw((1,0)--(1+9*sqrt(3)/2,9/2)--(1+9*sqrt(3)/2,15/2)--(1+5*sqrt(3)/2,11/2)--(1+5*sqrt(3)/2,9/2)--(1+2*sqrt(3),4)--(1+2*sqrt(3 draw((1+9*sqrt(3)/2,15/2)--(9*sqrt(3)/2,15/2)--(5*sqrt(3)/2,11/2)--(5*sqrt(3)/2,5));

1 KB (197 words) - 00:07, 5 July 2013
2010 AMC 12A Problems/Problem 15
<math>\textbf{(A)}\ \frac{\sqrt{15}-3}{6} \qquad \textbf{(B)}\ \frac{6-\sqrt{6\sqrt{6}+2}}{12} \qquad \textbf{

2 KB (255 words) - 21:47, 3 July 2013
2010 AIME I Problems/Problem 15
...BC}</math> with <math>AB = 12</math>, <math>BC = 13</math>, and <math>AC = 15</math>, let <math>M</math> be a point on <math>\overline{AC}</math> such th draw((0,0)--(15,0));

14 KB (2,210 words) - 13:14, 11 January 2024
2007 AMC 8 Problems/Problem 15

1 KB (187 words) - 17:18, 3 November 2023
2010 AMC 10A Problems/Problem 15

46 bytes (5 words) - 13:25, 26 May 2020
Northeastern WOOTers Mock AIME I Problems/Problem 15
== Problem 15 ==

1 KB (226 words) - 21:35, 7 August 2021
2010 AIME II Problems/Problem 15
== Problem 15 == ...math>ABC</math>, <math>AC = 13</math>, <math>BC = 14</math>, and <math>AB=15</math>. Points <math>M</math> and <math>D</math> lie on <math>AC</math> wit

9 KB (1,523 words) - 15:24, 21 November 2023
2010 AMC 12B Problems/Problem 15
...ath>i^x</math> only cycles as <math>1, i, -1, -i</math>). So we have <math>15\cdot 1\cdot 1+20\cdot 1\cdot 1=35</math> ordered triples.

2 KB (360 words) - 17:29, 26 May 2023
1999 AMC 8 Problems/Problem 15

3 KB (516 words) - 14:50, 21 December 2022
2010 AMC 10B Problems/Problem 15

1 KB (210 words) - 02:44, 26 September 2020
2011 AMC 12A Problems/Problem 15

2 KB (376 words) - 23:14, 5 January 2024
2011 AMC 10A Problems/Problem 15
== Problem 15 ==

2 KB (264 words) - 15:31, 3 September 2022
2001 AMC 10 Problems/Problem 15
...e street at an angle. The length of the curb between the stripes is <math> 15 </math> feet and each stripe is <math> 50 </math> feet long. Find the dista ...)}\ 9 \qquad \textbf{(B)}\ 10 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 25 </math>

2 KB (253 words) - 11:57, 20 October 2020
2011 AIME I Problems/Problem 15

8 KB (1,302 words) - 04:07, 24 July 2023
2011 AIME II Problems/Problem 15
...ber <math>x</math> is chosen at random from the interval <math>5 \le x \le 15</math>. The probability that <math>\lfloor\sqrt{P(x)}\rfloor = \sqrt{P(\lfl P(15) &= 171 \\

8 KB (1,273 words) - 14:03, 7 January 2023
2001 AMC 8 Problems/Problem 15

922 bytes (135 words) - 00:38, 5 July 2013
2011 AMC 10B Problems/Problem 15

2 KB (287 words) - 01:48, 26 June 2022
2008 AMC 12B Problems/Problem 15

3 KB (505 words) - 22:58, 7 October 2021
1999 AHSME Problems/Problem 15

931 bytes (144 words) - 19:36, 1 May 2023
2011 AMC 12B Problems/Problem 15
==Problem 15==

2 KB (284 words) - 20:54, 29 May 2023
2007 AMC 10B Problems/Problem 15
...12B Problems|2007 AMC 12B #11]] and [[2007 AMC 10B Problems|2007 AMC 10B #15]]}}

997 bytes (137 words) - 12:22, 4 July 2013
1998 AHSME Problems/Problem 15

764 bytes (123 words) - 14:29, 5 July 2013
1998 AJHSME Problems/Problem 15
Problems 15, 16, and 17 all refer to the following: ==Problem 15==

1 KB (173 words) - 20:55, 11 May 2021
2003 AMC 10B Problems/Problem 15

2 KB (267 words) - 23:23, 7 September 2023
AoPSWik:Problem of the Day/June 15, 2011
#REDIRECT [[AoPSWiki:Problem of the Day/June 15, 2011]]

55 bytes (7 words) - 18:20, 14 June 2011
1984 AHSME Problems/Problem 15
==Problem 15==

3 KB (493 words) - 18:16, 4 June 2021
2002 AMC 8 Problems/Problem 15

1 KB (197 words) - 04:47, 25 November 2019
2000 AMC 8 Problems/Problem 15
<math> \text{(A)}\ 12\qquad\text{(B)}\ 13\qquad\text{(C)}\ 15\qquad\text{(D)}\ 18\qquad\text{(E)}\ 21 </math> <math>15</math>, thus the answer is <math>\boxed{C}</math>

2 KB (325 words) - 20:56, 6 November 2013
1997 AJHSME Problems/Problem 15

3 KB (473 words) - 10:58, 27 June 2023
1996 AJHSME Problems/Problem 15

1 KB (200 words) - 11:29, 27 June 2023
1997 AHSME Problems/Problem 15

2 KB (406 words) - 20:44, 15 February 2024
2008 AMC 8 Problems/Problem 15

1 KB (190 words) - 19:24, 8 August 2021
2009 AMC 8 Problems/Problem 15

1 KB (162 words) - 19:53, 20 August 2020
1994 AJHSME Problems/Problem 15
label("$\cdots$",(6.5,0.15),S);

2 KB (282 words) - 11:52, 27 June 2023
1996 AHSME Problems/Problem 15

3 KB (537 words) - 14:07, 5 July 2013
1992 AJHSME Problems/Problem 15

634 bytes (95 words) - 00:09, 5 July 2013
1993 AJHSME Problems/Problem 15

732 bytes (114 words) - 00:11, 5 July 2013
1995 AJHSME Problems/Problem 15

577 bytes (81 words) - 00:23, 5 July 2013
2007 Alabama ARML TST Problems/Problem 15

2 KB (389 words) - 18:12, 21 March 2018
1985 AHSME Problems/Problem 15

1,012 bytes (163 words) - 21:35, 19 March 2024
1950 AHSME Problems/Problem 15

2 KB (255 words) - 08:55, 1 September 2021
2003 AMC 8 Problems/Problem 15
currentprojection=orthographic(3/4,8/15,7/15);

1 KB (201 words) - 19:56, 15 April 2023
2006 AMC 8 Problems/Problem 15
Problems 14, 15 and 16 involve Mrs. Reed's English assignment. <math>75x = 45(760)</math> Divide both sides by <math>15</math> to make it easier to solve

2 KB (323 words) - 22:54, 6 January 2023
2011 AMC 8 Problems/Problem 15

846 bytes (112 words) - 15:09, 17 December 2023
2012 AMC 10A Problems/Problem 15

6 KB (935 words) - 23:41, 13 September 2023
2012 AMC 12A Problems/Problem 15

46 bytes (5 words) - 14:35, 12 February 2012
1991 AHSME Problems/Problem 15
<math>\text{(A) } 15\quad

879 bytes (143 words) - 17:38, 23 February 2018
1988 AHSME Problems/Problem 15

759 bytes (128 words) - 18:23, 26 February 2018
1989 AHSME Problems/Problem 15

2 KB (275 words) - 20:37, 24 March 2023
2012 AMC 12B Problems/Problem 15

2 KB (259 words) - 13:42, 2 September 2020
2012 AMC 10B Problems/Problem 15
...s 2.5 wins, so the maximum number of tied teams is five. Here's a chart of 15 games where five teams each have 3 wins: Note that the total number of matches is 15, and if 4 teams tie for the most wins then they can tie for 3 wins each, bu

2 KB (307 words) - 20:04, 13 July 2021
2012 AIME I Problems/Problem 15

3 KB (591 words) - 20:41, 24 January 2021
2007 iTest Problems/Problem 15

3 KB (539 words) - 21:01, 29 July 2018
2012 AIME II Problems/Problem 15
== Problem 15 == ...BD=\tfrac{35}{8}</math>, and use Stewart's Theorem to find <math>AD=\tfrac{15}{8}</math>. Use Power of Point <math>D</math> to find <math>DE=\tfrac{49}{8

13 KB (2,298 words) - 12:56, 10 September 2023
Mock AIME 2 2006-2007/Problem 15
#REDIRECT [[Mock AIME 2 2006-2007 Problems/Problem 15]]

55 bytes (6 words) - 15:31, 3 April 2012
Mock AIME 1 2006-2007/Problem 15
#REDIRECT [[Mock AIME 1 2006-2007 Problems/Problem 15]]

55 bytes (6 words) - 15:50, 3 April 2012
Mock AIME 3 Pre 2005/Problem 15
#REDIRECT [[Mock AIME 3 Pre 2005 Problems/Problem 15]]

54 bytes (6 words) - 19:42, 9 April 2012
Mock AIME II 2012 Problems/Problem 15
...h>. Now, substitute this into <math>5b_2+b_1+6=0</math> to give us <math>-15+6+b_1=0</math> or <math>b_1=9</math>. Therefore <math>f(x)=-3x^2+9x</math> ...ly. Note that <math>155^2=\underline{15*16},2,5=24025</math> (where <math>15*16, 2</math>, and <math>5</math> are digits of <math>155^2</math>), therefo

4 KB (660 words) - 15:55, 8 March 2015
Mock AIME I 2012 Problems/Problem 15
15 & 3 & 8\\

3 KB (488 words) - 20:05, 10 March 2015
KGS math club/solution 11 15

630 bytes (106 words) - 06:41, 8 April 2012
1951 AHSME Problems/Problem 15

1 KB (191 words) - 06:08, 6 April 2024
1966 AHSME Problems/Problem 15

498 bytes (87 words) - 22:11, 14 January 2018
1974 AHSME Problems/Problem 15

651 bytes (101 words) - 12:43, 5 July 2013
2010 AMC 8 Problems/Problem 15
...f gumdrops. <math>30\%</math> are blue, <math>20\%</math> are brown, <math>15\%</math> are red, <math>10\%</math> are yellow, and other <math>30</math> g ...s. If we replace half of the blue gumdrops with brown gumdrops, then <math>15\%</math> of the jar's gumdrops are brown. <math>\dfrac{35}{100} \cdot 120=4

990 bytes (148 words) - 20:55, 22 January 2023
2012 AMC 8 Problems/Problem 15

1 KB (180 words) - 19:00, 15 April 2023
2004 AMC 8 Problems/Problem 15

1 KB (215 words) - 20:20, 12 October 2020
2005 AMC 8 Problems/Problem 15

952 bytes (155 words) - 00:11, 2 July 2023
2013 AMC 12A Problems/Problem 15

3 KB (441 words) - 10:10, 4 August 2020
2013 AMC 10A Problems/Problem 15
Two sides of a triangle have lengths <math>10</math> and <math>15</math>. The length of the altitude to the third side is the average of the ...e perpendicular to that altitude will be between <math>10</math> and <math>15</math>. The only answer choice that meets this requirement is <math>\boxed{

2 KB (407 words) - 01:12, 22 September 2022
2013 AMC 10B Problems/Problem 15

2 KB (240 words) - 12:18, 18 October 2022
2013 AMC 12B Problems/Problem 15
{{duplicate|[[2013 AMC 12B Problems|2013 AMC 12B #15]] and [[2013 AMC 10B Problems|2013 AMC 10B #20]]}}

2 KB (280 words) - 16:28, 25 December 2023
2013 AIME I Problems/Problem 15
...ution. <math>B=99</math> gives no solution. Thus, <math>N=1+2+\cdots+16+16+15+\cdots+1=2\cdot\frac{16(17)}{2}=16\cdot 17=\boxed{272}</math>. When <math>x=5</math>, <math>B=6, 9, ..., 93\Rightarrow 15</math> triples..

4 KB (661 words) - 23:14, 26 May 2023
2013 AIME II Problems/Problem 15
==Problem 15== \cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ and} \\

8 KB (1,411 words) - 23:48, 11 January 2023
2003 AMC 12B Problems/Problem 15

46 bytes (5 words) - 01:01, 5 January 2014
2014 AIME II Problems/Problem 15
<math>x_6 = 15</math> <math>x_{14} = 105</math> <math>x_7 = 30</math> <math>x_{15} = 210</math>

4 KB (770 words) - 17:44, 11 October 2023
1967 AHSME Problems/Problem 15

1 KB (231 words) - 01:39, 16 August 2023
1962 AHSME Problems/Problem 15

1 KB (245 words) - 08:45, 30 January 2018
2013 AMC 8 Problems/Problem 15

2 KB (295 words) - 03:58, 29 December 2022
1952 AHSME Problems/Problem 15

1 KB (169 words) - 23:17, 2 January 2014
2014 AMC 10A Problems/Problem 15
...12A Problems|2014 AMC 12A #11]] and [[2014 AMC 10A Problems|2014 AMC 10A #15]]}} ...hour late if he continues at this speed. He increases his speed by <math>15</math> miles per hour for the rest of the way to the airport and arrives <m

3 KB (409 words) - 04:18, 20 June 2022
1994 AHSME Problems/Problem 15

1 KB (214 words) - 02:53, 28 May 2021
2014 AMC 12A Problems/Problem 15

4 KB (612 words) - 22:42, 2 August 2021
2014 AMC 10B Problems/Problem 15

3 KB (536 words) - 21:02, 31 October 2022
2014 AMC 12B Problems/Problem 15

1 KB (203 words) - 23:18, 3 January 2023
2014 AIME I Problems/Problem 15
== Problem 15 == ...theorem, we have <math>\frac{AF}{AB} = \frac{CF}{CB}</math>, so <math>AF = 15/7</math> and <math>CF = 20/7</math>.

10 KB (1,643 words) - 22:30, 28 January 2024
1993 AHSME Problems/Problem 15

897 bytes (140 words) - 13:39, 6 November 2020
1992 AHSME Problems/Problem 15

821 bytes (134 words) - 02:46, 20 February 2018
1968 AHSME Problems/Problem 15
<math>\text{(A) } 15\quad

1 KB (195 words) - 09:24, 1 January 2024
1990 AHSME Problems/Problem 15

643 bytes (92 words) - 05:37, 4 February 2016
1969 AHSME Problems/Problem 15

2 KB (250 words) - 18:29, 21 June 2018
1970 AHSME Problems/Problem 15

1 KB (226 words) - 22:26, 13 July 2019
1980 AHSME Problems/Problem 15

983 bytes (156 words) - 13:30, 9 March 2020
1958 AHSME Problems/Problem 15
{{AHSME 50p box|year=1958|num-b=13|num-a=15}}

444 bytes (62 words) - 06:13, 3 October 2014
Mock AIME 1 Pre 2005 Problems/Problem 15
{{Mock AIME box|year=Pre 2005|n=1|num-b=14|num-a=15|source=14769}}

2 KB (340 words) - 01:44, 3 March 2020
Mock AIME 5 2005-2006 Problems/Problem 15
{{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=14|num-a=15}}

955 bytes (157 words) - 21:20, 8 October 2014
2005 CEMC Gauss (Grade 7) Problems/Problem 15
...ext{(B)}\ 20 \qquad \text{(C)}\ 16 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 15</math>

2 KB (349 words) - 02:41, 23 October 2014
1987 AHSME Problems/Problem 15

699 bytes (107 words) - 12:36, 31 March 2018
1986 AHSME Problems/Problem 15

1 KB (185 words) - 18:43, 1 April 2018
2014 AMC 8 Problems/Problem 15

2 KB (280 words) - 11:10, 2 July 2023
2015 AMC 10A Problems/Problem 15

2 KB (372 words) - 14:45, 27 September 2020
2015 AMC 12A Problems/Problem 15

3 KB (411 words) - 15:53, 23 June 2022
2015 AMC 12B Problems/Problem 15

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2015 AMC 10B Problems/Problem 15

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2016 AMC 10A Problems/Problem 15

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2015 AIME I Problems/Problem 15
[[File:2015 AIME I 15.png|400px|right]] <cmath> A = \frac{15}{4} \int_0^4 \sqrt{ 64 - y^2 }dy </cmath>

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2015 AIME II Problems/Problem 15
...;O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=f ...label("$L$",(24/15,0.2));label("$n$",(-0.8,-0.12));label("$p$",((29/15,-48/15)));label("$\mathcal{P}$",(-1.6,1.1));label("$\mathcal{Q}$",(6,4));

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2015 AMC 8 Problems/Problem 15

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2016 AMC 12A Problems/Problem 15

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2016 AMC 10B Problems/Problem 15

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2016 AMC 12B Problems/Problem 15

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2016 AIME I Problems/Problem 15
pair d=intersectionpoints(circle(o1,R1),q--y+15*(y-q))[0]; pair c=intersectionpoints(circle(o2,R2),q--y+15*(y-q))[1];

14 KB (2,427 words) - 17:12, 8 January 2024
2016 AIME II Problems/Problem 15

2 KB (277 words) - 20:02, 19 April 2023
1975 AHSME Problems/Problem 15

887 bytes (148 words) - 23:42, 24 June 2019
1964 AHSME Problems/Problem 15
== Problem 15==

2 KB (260 words) - 21:01, 23 July 2019
1954 AHSME Problems/Problem 15
== Problem 15==

441 bytes (54 words) - 01:26, 28 February 2020
1965 AHSME Problems/Problem 15

298 bytes (49 words) - 11:12, 10 June 2016
1977 AHSME Problems/Problem 15

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2016 AMC 8 Problems/Problem 15

1 KB (156 words) - 22:58, 17 May 2024
1979 AHSME Problems/Problem 15
== Problem 15 ==

1 KB (240 words) - 13:21, 6 January 2017
2017 AMC 10A Problems/Problem 15
...al length of <math>2-0=2</math>. Therefore, the probability is <math>1.5/2=15/20=\boxed{\frac{3}{4} \space \text{(C)}}.</math>

4 KB (533 words) - 19:01, 15 March 2024
2017 AMC 12A Problems/Problem 15
Notice that <math>\frac{5\pi}{4}<\frac{3.15*5}{4}<4<\frac{3\pi}{2}, f(\frac{5\pi}{4})=3-\frac{3\sqrt{2}}{2}>0 .</math>

3 KB (564 words) - 14:12, 23 October 2021
2017 AMC 10B Problems/Problem 15
...xtbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{42}{25}\qquad\textbf{(C)}\ \frac{28}{15}\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{54}{25}</math>

4 KB (656 words) - 01:20, 4 December 2023
2017 AMC 12B Problems/Problem 15

9 KB (1,416 words) - 14:30, 23 November 2023
2017 AIME I Problems/Problem 15
==Problem 15== [[File:2017 AIME I 15.png|530px|right]]

22 KB (3,622 words) - 17:11, 6 January 2024
2017 AIME II Problems/Problem 15
[[File:2017 AIME II 15.png|300px|right]]

6 KB (971 words) - 02:08, 22 January 2024
1953 AHSME Problems/Problem 15
==Problem 15==

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2017 AMC 8 Problems/Problem 15

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2008 iTest Problems/Problem 15

1,002 bytes (137 words) - 13:25, 22 June 2018
2018 AMC 10A Problems/Problem 15
label("$A$", (-8.125,-10.15), S); label("$B$", (8.125,-10.15), S);

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2018 AMC 12A Problems/Problem 15

46 bytes (5 words) - 16:05, 8 February 2018
2018 AMC 12B Problems/Problem 15
...e a multiple of <math>3</math>, but does not contain 3. We have <math>(12, 15, 18, 21, 24, 27, 42, 45, 48, 51, 54, 57, 60, 66, 69, 72, 75, 78, 81, 84, 87 ...3) = 15</math> cases. For <math>\underline{3AB}</math>, we also have <math>15</math> cases, but when <math>B=3, 9</math>, <math>A</math> can equal <math>

8 KB (1,244 words) - 08:19, 30 May 2023
2018 AMC 10B Problems/Problem 15
...12B Problems|2018 AMC 12B #11]] and [[2018 AMC 10B Problems|2018 AMC 10B #15]]}}

6 KB (974 words) - 00:28, 30 May 2023
1960 AHSME Problems/Problem 15

2 KB (276 words) - 19:01, 17 May 2018
1961 AHSME Problems/Problem 15

3 KB (490 words) - 11:22, 16 September 2022
1963 AHSME Problems/Problem 15

1 KB (215 words) - 16:52, 7 June 2018
2018 AIME I Problems/Problem 15
==Problem 15== [[File:2018 AIME I 15.png|900px]]

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2018 AIME II Problems/Problem 15

9 KB (1,607 words) - 10:27, 4 February 2022
1973 AHSME Problems/Problem 15

2 KB (306 words) - 14:00, 20 February 2020
1983 AHSME Problems/Problem 15

1 KB (166 words) - 00:54, 20 February 2019
2018 AMC 8 Problems/Problem 15

3 KB (454 words) - 10:15, 10 April 2024
2006 iTest Problems/Problem 15

1 KB (197 words) - 23:31, 27 November 2018
Mock AIME I 2015 Problems/Problem 15

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2019 AIME II Problems/Problem 15
...ath>Y</math>. Suppose <math>XP=10</math>, <math>PQ=25</math>, and <math>QY=15</math>. The value of <math>AB\cdot AC</math> can be written in the form <ma label("$15$", Q--Y, SW);

7 KB (1,115 words) - 03:11, 7 January 2024
2019 AMC 12A Problems/Problem 15
...ndex.php/Perfect_square perfect squares] up to <math>14^2</math> (as <math>15^2</math> is larger than the largest possible sum of <math>x</math> and <mat

4 KB (686 words) - 04:22, 13 November 2022
2019 AMC 10A Problems/Problem 15
{{duplicate|[[2019 AMC 10A Problems|2019 AMC 10A #15]] and [[2019 AMC 12A Problems|2019 AMC 12A #9]]}} ...cursive formula, we find <math>a_3=\frac{3}{11}</math>, <math>a_4=\frac{3}{15}</math>, and so on. It appears that <math>a_n=\frac{3}{4n-1}</math>, for al

4 KB (687 words) - 08:11, 20 November 2023
2019 AMC 10B Problems/Problem 15

4 KB (704 words) - 23:59, 22 March 2023
2019 AMC 12B Problems/Problem 15

46 bytes (5 words) - 13:31, 14 February 2019
2019 AIME I Problems/Problem 15
By Power of a Point, <math>PX\cdot PY=PA\cdot PB=15</math>. Since <math>PX+PY=XY=11</math> and <math>XQ=11/2</math>, <cmath>XP= .../math>. Then we have <math>AP\cdot PB=XP\cdot PY</math>, that is, <math>xy=15</math>. Also, <math>XP+PY=x+y=XY=11</math>. Solve these above, we have <mat

13 KB (2,252 words) - 11:32, 1 February 2024
2006 SMT/Team Problems/Problem 15

3 KB (445 words) - 18:37, 14 January 2020
2020 AIME I Problems/Problem 15
A = (-15, 27); H = (-15, 13);

16 KB (2,678 words) - 22:45, 27 November 2023
2019 Mock AMC 10B Problems/Problem 15

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2019 AMC 8 Problems/Problem 15
==Problem 15==

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2020 AMC 10A Problems/Problem 15

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2020 AMC 12A Problems/Problem 15

4 KB (696 words) - 12:38, 13 September 2021
2020 AMC 10B Problems/Problem 15
...torname{lcm}(3,5)=15</math> before erasing. So, we first group <math>\frac{15}{5}=3</math> copies of the current cycle into one, then erase: As a quick confirmation, one cycle should have length <math>15\cdot\left(1-\frac{1}{3}\right)=10</math> at this point.</li>

10 KB (1,471 words) - 13:57, 30 October 2023
2020 AMC 12B Problems/Problem 15

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2023 AIME I Problems/Problem 15
...satisfy (7) and <math>a^2 + b^2 < 1000</math>, we have <math>1 \leq b \leq 15</math>.

8 KB (1,299 words) - 17:37, 3 June 2023
2021 JMPSC Accuracy Problems/Problem 15

2 KB (304 words) - 01:19, 12 July 2021
2021 AMC 10A Problems/Problem 15

4 KB (577 words) - 19:18, 28 October 2022
1959 AHSME Problems/Problem 15
...the legs. One of the acute angles of the triangle is: <math>\textbf{(A)}\ 15^{\circ} \qquad\textbf{(B)}\ 30^{\circ} \qquad\textbf{(C)}\ 45^{\circ} \qqua ...rect answer, one of the acute angles of the triangle will measure to <math>15</math> degrees. This implies that the other acute angle of the triangle wou

2 KB (263 words) - 11:33, 12 August 2020
2020 AIME II Problems/Problem 15
...aw segments <math>XM</math>, and <math>YM</math>. We have <math>MT=3\sqrt{15}</math>. ...larly, Ptolemy's theorem in <math>YTMC</math> gives<cmath>16TX=11TY+3\sqrt{15}CY</cmath> while Pythagoras' theorem in <math>\triangle CYT</math> gives <m

7 KB (1,221 words) - 16:46, 29 January 2023
2021 AIME I Problems/Problem 15

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2019 AMC 10C Problems/Problem 15

625 bytes (105 words) - 10:53, 6 August 2020
1955 AHSME Problems/Problem 15
== Problem 15==

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2022 AIME I Problems/Problem 15
so <math>(\alpha, \beta, \gamma) = (15/2^{\circ}, 45/2^{\circ}, 75/2^{\circ})</math>. Hence, <cmath> abc = (1-2\sin^2(\alpha))(1-2\sin^2(\beta))(1-2\sin^2(\gamma))=\cos(15^{\circ})\cos(45^{\circ})\cos(75^{\circ})=\frac{\sqrt{2}}{8}, </cmath>

15 KB (2,208 words) - 01:25, 1 February 2024
2020 CIME I Problems/Problem 15
==Problem 15==

523 bytes (80 words) - 11:58, 1 September 2020
2021 Fall AMC 10A Problems/Problem 15

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2021 Fall AMC 12A Problems/Problem 15

2 KB (311 words) - 19:01, 24 May 2023
2021 AMC 12A Problems/Problem 15
4 & 0 & \tbinom{6}{4}\tbinom{8}{0} & 15 \\ [1ex] 4 & 8 & \tbinom{6}{4}\tbinom{8}{8} & 15 \\ [1ex]

7 KB (1,152 words) - 14:13, 29 February 2024
2021 AMC 12B Problems/Problem 15
...20|2021 AMC 10B #20]] and [[2021 AMC 12B Problems#Problem 15|2021 AMC 12B #15]]}}

4 KB (537 words) - 13:49, 25 February 2024
2019 CIME I Problems/Problem 15

2 KB (264 words) - 17:57, 4 October 2020

Page text matches

Pythagorean Theorem
<cmath>8-15-17</cmath>

5 KB (886 words) - 13:51, 15 May 2024
2005 AMC 12A Problems/Problem 16
{{AMC12 box|year=2005|num-b=15|num-a=17|ab=A}}

2 KB (307 words) - 15:30, 30 March 2024
2016 AMC 10A Problems/Problem 2
...an rewrite <math>10^{x}\cdot 100^{2x}=1000^{5}</math> as <math>10^{5x}=10^{15}</math>: 10^{5x} & =10^{15}

1 KB (190 words) - 10:58, 16 June 2023
2016 AMC 10A Problems/Problem 4
&= \frac{3}{8}-\left(-\frac{2}{5}\right)\left\lfloor\frac{-15}{16}\right\rfloor\\

2 KB (257 words) - 10:57, 16 June 2023
2016 AMC 10A Problems/Problem 11
label("$1$",(15/2,0),dir(270)); label("$1$",(15/2,0),dir(270));

8 KB (1,016 words) - 00:17, 31 December 2023
2016 AMC 10A Problems/Problem 16
{{AMC10 box|year=2016|ab=A|num-b=15|num-a=17}}

1 KB (235 words) - 14:52, 25 June 2023
MathILy (serious mathematics infused with levity)
...age or mathematical development, had its first year in 2015, and grew from 15 attendees in 2015 to 31 attendees in 2023.

5 KB (706 words) - 23:49, 29 January 2024
ICTM Math Contest
The contest is given in-school May 1-15. Each school sets its own testing date and time within these guidelines.

8 KB (1,182 words) - 14:26, 3 April 2024
Math books
...roblems from the [[UK Junior Mathematical Olympiad]], for students ages 11-15.

24 KB (3,177 words) - 12:53, 20 February 2024
William Lowell Putnam Mathematical Competition
**School receives $15,000

4 KB (623 words) - 13:11, 20 February 2024
AMC historical results
*Achievement Roll: Awarded to students in 6th or younger who score 15 points or higher on the AMC 8. Awarded to students in 8th or younger who sc *AIME floor: 81 (top ~15%)

17 KB (1,921 words) - 20:53, 10 May 2024
Inequality
\\x+5-13+4x+20&\ge 3x+15

12 KB (1,798 words) - 16:20, 14 March 2023
American Invitational Mathematics Examination
...>: 3 Problem 6-10: 4 Problem 10-12: 5 Problem 12-15: 6}} ...hile correct answers receive one point of credit, making the maximum score 15. Problems generally increase in difficulty as the exam progresses - the fi

8 KB (1,057 words) - 12:02, 25 February 2024
Math competition books
...roblems from the [[UK Junior Mathematical Olympiad]], for students ages 11-15.

2 KB (302 words) - 16:45, 3 October 2019
American Regions Mathematics League
See the [[How to join an ARML team]] wiki page for more info. Teams have 15 students.

2 KB (267 words) - 17:06, 7 March 2020
Joining an ARML team
...The three top teams usually all place in the top 20, often even in the top 15 or 10. ...or is selected, although it does happen. Then the coaches select the best 15 young students (10th grade and below) to comprise a C team -- a team specif

21 KB (3,500 words) - 18:41, 23 April 2024
Stanford University Mathematics Camp
Students must be current 10-11th graders who will be 15-17 years old on the first day of the program. The cost is \$6500.

1 KB (166 words) - 17:54, 10 June 2016
Simon's Favorite Factoring Trick
...the equations, we have the solutions as <math>\boxed{(5, 25), (24, 6), (6, 15), (14, 7), (8, 10), (9, 9)}</math>.

7 KB (1,107 words) - 07:35, 26 March 2024
Pigeonhole Principle
# Prove that having 100 whole numbers, one can choose 15 of them so that the difference of any two is divisible by 7. ([[Pigeonhole

11 KB (1,985 words) - 21:03, 5 August 2023
Principle of Inclusion-Exclusion
15 take algebra, language arts, social studies, and history. 15 take algebra, language arts, biology, and history.

9 KB (1,703 words) - 07:25, 24 March 2024
Factorial
* <math>15! = 1307674368000</math>

10 KB (809 words) - 16:40, 17 March 2024
2006 AIME II
* [[2006 AIME II Problems/Problem 15]]

1 KB (133 words) - 12:32, 22 March 2011
Fermat's Little Theorem
...he units digit of <math>k^2 + 2^k</math>? ([[2008 AMC 12A Problems/Problem 15]])

16 KB (2,658 words) - 16:02, 8 May 2024
Alabama ARML
...9, Alabama did not send a team, but Grissom High in Alabama sent a team of 15 students to compete.

2 KB (258 words) - 00:50, 28 December 2021
LaTeX
<cmath>1+2+3+4+5+\sin \pi = \frac{5\cdot 6}{2}+0=15.</cmath>

1 KB (164 words) - 19:09, 14 February 2024
North Carolina MathCounts
* 2013 - Nikhil Reddy (15), Angela Deng (40), Lloyd Liu (42), Andrew Zhang, Coach: Michael Pillsbury

4 KB (582 words) - 21:40, 14 May 2024
Orthic triangle
...Y=15</math>. Compute the product <math>AB\cdot AC</math>. (AIME II, 2019, 15)

8 KB (1,408 words) - 11:54, 8 December 2021
Recursion
* [[2004_AIME_I_Problems#Problem_15| 2004 AIME I Problem 15]]

2 KB (316 words) - 16:03, 1 January 2024
Combinatorial identity
....com/wiki/index.php/2021_AMC_12A_Problems/Problem_15 2021 AMC 12A Problem 15]

12 KB (1,996 words) - 12:01, 18 May 2024
Composite number
4 6 8 9 10 12 14 15 16 18 20 21 22 24 25 26 27 28 30 32 33 34 35 36 38 39 40 42 44 45 46 48 49 ⇒ 13,15,17...... all are sum of 2 composites Hence, any odd positive ≥ 13 can be

6 KB (350 words) - 12:58, 26 September 2023
Circle
...cycle); dot("$A$",(8,0), 1.25*S); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*S); dot("$E$",(17,0), 1.25*S); dot("$F$",

9 KB (1,581 words) - 18:59, 9 May 2024
Ellipse
...prime integers, find <math> p+q. </math> ([[2005 AIME II Problems/Problem 15|Source]])

5 KB (892 words) - 21:52, 1 May 2021
Multiple
...nite]] number of multiples. As an example, some of the multiples of 15 are 15, 30, 45, 60, and 75.

860 bytes (142 words) - 22:51, 26 January 2021
Least common multiple
...multiply the results together. For example, to find the LCM of 8, 12 and 15, write: <math>15 = 3^1\cdot 5^1</math>

2 KB (383 words) - 10:49, 4 September 2022
Math Bee
...tes, while the second phase has 20 free-response questions to be solved in 15 minutes.

1 KB (197 words) - 10:59, 14 April 2024
Math Day at the Beach
This round lasts 45 minutes and consists of 15 multiple-choice questions. Scoring consists of:

4 KB (644 words) - 12:56, 29 March 2017
Writing scholarships
...uch as creative writing or journalism. Apply for this USD 500 award by May 15 of each year, including a 400-600 word short story in prose or in script.

7 KB (1,039 words) - 18:45, 18 January 2024
National Scholarships
* [[Best Buy @15 Scholarship]] of <dollar/>1,000 for students in grades 9-12. [https://www.a

4 KB (538 words) - 00:48, 28 January 2024
Law of Cosines
label("d",(15,0),(0,-1)); \qquad \mathrm{(B) \ } 8/\sqrt{15}

6 KB (1,003 words) - 00:02, 20 May 2024
Chicago ARML
The [[Chicago ARML]] team consists of four groups of approximately 15 high school students. The four groups are:

2 KB (227 words) - 11:47, 4 December 2023
Law of Sines
\qquad \mathrm{(B) \ } 8/\sqrt{15} ([[Mock AIME 4 2006-2007 Problems/Problem 15|Source]])

4 KB (658 words) - 16:19, 28 April 2024
Homothety
pair A=(15,15),B=(30,15),C=(30,30),D=(15,30),a=(60,60),b=(120,60),c=(120,120),d=(60,120); ...triangle <math>ABC</math> are <math>13,</math> <math>14,</math> and <math>15,</math> the radius of <math>\omega</math> can be represented in the form <m

3 KB (532 words) - 01:11, 11 January 2021
Pascal's triangle
...aring above it. For example, <math>{5 \choose 1}+{5 \choose 2} = 5 + 10 = 15 = {6 \choose 2}</math>. This property allows the easy creation of the firs

5 KB (838 words) - 17:20, 3 January 2023
2006 AMC 10B
* [[2006 AMC 10B Problems/Problem 15]]

2 KB (182 words) - 21:57, 23 January 2021
Mathematical Olympiad Summer Program
Top non-senior USAMO finishers: In addition to the winners, the next 15 or so non-senior non-Canadian finishers are invited to attend MOP. This gro ...ree instructional sessions: 8:30 AM - 10:00 AM, 10:15 AM - 11:45 AM, and 1:15 PM - 2:45 PM. Classes usually consist of a lecture followed by a problem se

6 KB (936 words) - 10:37, 27 November 2023
Power of a Point Theorem
* [[2005 AIME I Problems/Problem 15|2005 AIME I Problem 15]]

5 KB (827 words) - 17:30, 21 February 2024
2004 AIME I
* [[2004 AIME I Problems/Problem 15]]

1 KB (135 words) - 18:15, 19 April 2021
2004 AIME II
* [[2004 AIME II Problems/Problem 15]]

1 KB (135 words) - 12:24, 22 March 2011
2005 AIME I
* [[2005 AIME I Problems/Problem 15 | Problem 15]]

1 KB (154 words) - 12:30, 22 March 2011
2006 AIME I
* [[2006 AIME I Problems/Problem 15]]

1 KB (135 words) - 12:31, 22 March 2011
2005 AIME II
* [[2005 AIME II Problems/Problem 15]]

1 KB (135 words) - 12:30, 22 March 2011
2006 AMC 12B
* [[2006 AMC 12B Problems/Problem 15 | Problem 15]]

2 KB (210 words) - 00:06, 7 October 2014
Rhombus
* [[2006_AMC_10B_Problems/Problem_15 | 2006 AMC 10B Problem 15]]

3 KB (490 words) - 15:30, 22 February 2024
Mock AMC
[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9574 11-15]

51 KB (6,175 words) - 20:58, 6 December 2023
Modular arithmetic/Intermediate
<math>x^2 - 2x - 15 \equiv 0 \pmod{21}</math>. However, since <math>15 \equiv 99 \pmod{21}</math>, the original congruence is equivalent to

14 KB (2,317 words) - 19:01, 29 October 2021
2006 AMC 10A
* [[2006 AMC 10A Problems/Problem 15]]

2 KB (180 words) - 18:06, 6 October 2014
2006 AIME I Problems
The number <math> \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}</math> can be written as <math> a\sqrt{2}+b\sqrt{3}+c\sqrt{5}, </mat == Problem 15 ==

7 KB (1,173 words) - 03:31, 4 January 2023
2006 AIME I Problems/Problem 14
{{AIME box|year=2006|n=I|num-b=13|num-a=15}}

6 KB (980 words) - 21:45, 31 March 2020
2006 AIME I Problems/Problem 13
...>k=6</math>, then <math>n<1000</math> implies that <math>\frac{n+1}{64}\le 15</math>, so <math>n+1=64,64\cdot 3^2</math>.

10 KB (1,702 words) - 00:45, 16 November 2023
2006 AIME I Problems/Problem 5
The number <math> \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}</math> can be written as <math> a\sqrt{2}+b\sqrt{3}+c\sqrt{5}, </mat ...h> a\sqrt{2}+b\sqrt{3}+c\sqrt{5} = \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}</cmath>

3 KB (439 words) - 18:24, 10 March 2015
2006 AIME I Problems/Problem 4
...> have a factor of <math>10</math>. <math>86</math> have a factor of <math>15</math>. And so on. This gives us an initial count of <math>96 + 91 + 86 + \

2 KB (278 words) - 08:33, 4 November 2022
2006 AIME I Problems/Problem 3
...ber after digit removal could be 10,20,30) or <math>c=5.</math> Testing 5, 15, and 25 as the numbers after removal we find that our answer is clearly <ma

4 KB (622 words) - 03:53, 10 December 2022
2021 JMC 10
...f{(A) } 5 \qquad\textbf{(B) } 9 \qquad\textbf{(C) } 12 \qquad\textbf{(D) } 15 \qquad\textbf{(E) } 16</math> ...e written with digits in reverse order. A citizen in Malachar writes <math>15\cdot 73.</math> What does this Malacharian write as the answer?

12 KB (1,784 words) - 16:49, 1 April 2021
2006 AMC 12B Problems
\text {(A) } 5 \qquad \text {(B) } 10 \qquad \text {(C) } 15 \qquad \text {(D) } 20 \qquad \text {(E) } 25 ...three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle?

13 KB (2,058 words) - 12:36, 4 July 2023
2006 AMC 12A
* [[2006 AMC 12A Problems/Problem 15]]

1 KB (168 words) - 21:51, 6 October 2014
2004 AMC 12A
* [[2004 AMC 12A Problems/Problem 15]]

2 KB (186 words) - 17:35, 16 December 2019
2004 AMC 12B
* [[2004 AMC 12B Problems/Problem 15]]

2 KB (181 words) - 21:40, 6 October 2014
2005 AMC 12A
** [[2005 AMC 12A Problems/Problem 15|Problem 15]]

2 KB (202 words) - 21:30, 6 October 2014
2005 AMC 12B
* [[2005 AMC 12B Problems/Problem 15|Problem 15]]

2 KB (206 words) - 23:23, 21 June 2021
2000 AMC 12
* [[2000 AMC 12 Problems/Problem 15]]

1 KB (126 words) - 13:28, 20 February 2020
2001 AMC 12
* [[2001 AMC 12 Problems/Problem 15]]

1 KB (127 words) - 21:36, 6 October 2014
2002 AMC 12A
* [[2002 AMC 12A Problems/Problem 15]]

1 KB (158 words) - 21:33, 6 October 2014
2003 AMC 12A
* [[2003 AMC 12A Problems/Problem 15]]

1 KB (162 words) - 21:52, 6 October 2014
2002 AMC 12B
* [[2002 AMC 12B Problems/Problem 15]]

1 KB (154 words) - 00:32, 7 October 2014
2003 AMC 12B
* [[2003 AMC 12B Problems/Problem 15]]

1 KB (160 words) - 20:46, 1 February 2016
2006 AMC 12A Problems
<math> \mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 18\qquad \mathrm{(C) \ } 20\qquad \mathrm{(D) \ } 24\ label("$y$",(15,-4),N);

15 KB (2,223 words) - 13:43, 28 December 2020
2005 AMC 12A Problems
== Problem 15 == [[2005 AMC 12A Problems/Problem 15|Solution]]

13 KB (1,971 words) - 13:03, 19 February 2020
2004 AMC 12A Problems
...Players <math>A</math>, <math>B</math> and <math>C</math> start with <math>15</math>, <math>14</math> and <math>13</math> tokens, respectively. How many == Problem 15 ==

13 KB (1,953 words) - 00:31, 26 January 2023
2003 AMC 12A Problems
A solid box is <math>15</math> cm by <math>10</math> cm by <math>8</math> cm. A new solid is formed == Problem 15 ==

13 KB (1,955 words) - 21:06, 19 August 2023
2002 AMC 12A Problems
<math> \mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 34\qquad \mathrm{(C) \ } 43\qquad \mathrm{(D) \ } 51\ ...xt{(A)}\ 12 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 14 \qquad \text{(D)}\ 15 \qquad \text{(E)} 16</math>

12 KB (1,792 words) - 13:06, 19 February 2020
2000 AMC 12 Problems
== Problem 15 == [[2000 AMC 12 Problems/Problem 15|Solution]]

13 KB (1,948 words) - 12:26, 1 April 2022
2001 AMC 12 Problems
...ee numbers is <math>10</math> more than the least of the numbers and <math>15</math> == Problem 15 ==

13 KB (1,957 words) - 12:53, 24 January 2024
2002 AMC 12B Problems
== Problem 15 == [[2002 AMC 12B Problems/Problem 15|Solution]]

10 KB (1,547 words) - 04:20, 9 October 2022
2003 AMC 12B Problems
<cmath>\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}?</cmath> ...{(A) } 3 \qquad \text {(B) } 5 \qquad \text {(C) } 11 \qquad \text {(D) } 15 \qquad \text {(E) } 165

13 KB (1,987 words) - 18:53, 10 December 2022
2004 AMC 12B Problems
...{B}) 6 \qquad (\mathrm {C}) 9 \qquad (\mathrm {D}) 12 \qquad (\mathrm {E}) 15</math> <math>(\mathrm {A}) 13\qquad (\mathrm {B}) 14 \qquad (\mathrm {C}) 15 \qquad (\mathrm {D}) 16 \qquad (\mathrm {E}) 17</math>

13 KB (2,049 words) - 13:03, 19 February 2020
2005 AMC 12B Problems
...math>80</math> points, <math>20\%</math> got <math>85</math> points, <math>15\%</math> got <math>90</math> points, and the rest got <math>95</math> point == Problem 15 ==

12 KB (1,781 words) - 12:38, 14 July 2022
2006 AMC 12B Problems/Problem 4
\text {(A) } 5 \qquad \text {(B) } 10 \qquad \text {(C) } 15 \qquad \text {(D) } 20 \qquad \text {(E) } 25

1 KB (152 words) - 16:11, 8 December 2013
2006 AMC 12B Problems/Problem 10
...three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle? ...third side has length <math>15</math>, and so the perimeter is <math>21+7+15=43 \Rightarrow \boxed{\text {(A)}}</math>.

977 bytes (156 words) - 13:57, 19 January 2021
2006 AMC 12B Problems/Problem 14
...)=23</math>. The problem asks for the total cost of jam, or <math>N(5J)=11(15)=165</math> cents, or <math>1.65\implies\mathrm{(D)}</math> {{AMC12 box|year=2006|ab=B|num-b=13|num-a=15}}

1 KB (227 words) - 17:21, 8 December 2013
2006 AMC 12B Problems/Problem 16
{{AMC12 box|year=2006|ab=B|num-b=15|num-a=17}}

1 KB (203 words) - 16:36, 18 September 2023
2006 AMC 12B Problems/Problem 20
\mathrm{(D)}\ \frac 15 .../(1-r)</math> for the sum of an infinite geometric sequence, we get <math>(15/100)/(1-(1/10)) = \boxed{\frac 16}</math>.

3 KB (485 words) - 14:09, 21 May 2021
2006 AMC 12A Problems/Problem 14
{{AMC12 box|year=2006|ab=A|num-b=13|num-a=15}}

3 KB (442 words) - 03:13, 8 August 2022
2006 AMC 12A Problems/Problem 16
{{AMC12 box|year=2006|ab=A|num-b=15|num-a=17}}

2 KB (286 words) - 10:16, 19 December 2021
2006 AMC 12A Problems/Problem 22
...e OAP = OAB - 60^{\circ} = \frac{180^{\circ}-30^{\circ}}{2} - 60^{\circ} = 15^{\circ}</math>. Since <math>OA</math> is a radius and <math>OP</math> can b ...>, then we get a [[right triangle]]. Using simple trigonometry, <math>\cos 15^{\circ} = \frac{\frac{r}{2}}{2\sqrt{3}} = \frac{r}{4\sqrt{3}}</math>.

2 KB (343 words) - 15:39, 14 June 2023
2006 AMC 12A Problems/Problem 25
...-[[empty set | empty]] [[subset]]s <math>S</math> of <math>\{1,2,3,\ldots ,15\}</math> have the following two properties? <math>{15 \choose 1} + {14 \choose 2} + {13 \choose 3} + ... + {9 \choose 7} + {8 \ch

8 KB (1,405 words) - 11:52, 27 September 2022
2005 AMC 12B Problems/Problem 6
...(x)\left( \frac{-1}{7} \right) \rightarrow 64=49+x^2+2x \rightarrow x^2+2x-15=0</math>. The only positive value for <math>x</math> is <math>3</math>, whi \implies k^2+2k-15=0

2 KB (299 words) - 15:29, 5 July 2022
2005 AMC 12B Problems/Problem 9
...math>80</math> points, <math>20\%</math> got <math>85</math> points, <math>15\%</math> got <math>90</math> points, and the rest got <math>95</math> point The mean is equal to <math>10\%\cdot70 + 25\%\cdot80 + 20\%\cdot85 + 15\%\cdot90 + 30\%\cdot95 = 86</math>.

2 KB (280 words) - 15:35, 16 December 2021
2005 AMC 12B Problems/Problem 11
...12B Problems|2005 AMC 12B #11]] and [[2005 AMC 10B Problems|2005 AMC 10B #15]]}} ...th>6</math> bills left. <math>\dbinom{6}{2} = \dfrac{6\times5}{2\times1} = 15</math> ways. However, you counted the case when you have <math>2</math> ten

4 KB (607 words) - 21:01, 20 May 2023
2005 AMC 12B Problems/Problem 12
{{AMC10 box|year=2005|ab=B|num-b=15|num-a=17}}

2 KB (317 words) - 12:27, 16 December 2021
2005 AMC 12B Problems/Problem 14
{{AMC12 box|year=2005|ab=B|num-b=13|num-a=15}}

2 KB (278 words) - 21:12, 24 December 2020
2005 AMC 12B Problems/Problem 16
{{AMC12 box|year=2005|ab=B|num-b=15|num-a=17}}

2 KB (364 words) - 04:54, 16 January 2023
2005 AMC 12B Problems/Problem 18
xaxis(-1,15,Ticks(f, 2.0)); yaxis(-1,15,Ticks(f, 2.0));

2 KB (262 words) - 21:20, 21 December 2020
2005 AMC 12B Problems/Problem 19
...b=4</math> then <math>(a+b)+(a-b)=11+4</math>, <math>2a=15</math>, <math>a=15/2</math>, which is not a digit. Hence the only possible value for <math>a-b

2 KB (283 words) - 20:02, 24 December 2020
2005 AMC 12B Problems/Problem 23
\textbf{(A)}\ \frac {15}{2} \qquad \textbf{(C)}\ 15 \qquad

5 KB (786 words) - 11:36, 19 May 2024
2005 AMC 12B Problems/Problem 24
<math> \mathrm{(A)}\ {{{14}}}\qquad\mathrm{(B)}\ {{{15}}}\qquad\mathrm{(C)}\ {{{16}}}\qquad\mathrm{(D)}\ {{{17}}}\qquad\mathrm{(E)

4 KB (761 words) - 09:10, 1 August 2023
2006 AMC 10A Problems
<math>\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 18\qquad\mathrm{(C)}\ 20\qquad\mathrm{(D)}\ 24\qquad\mat label("$y$",(15,-4),N);

13 KB (2,028 words) - 16:32, 22 March 2022
2006 AMC 12A Problems/Problem 12
{{AMC10 box|year=2006|ab=A|num-b=13|num-a=15}}

2 KB (292 words) - 11:56, 17 December 2021
2006 AMC 10A Problems/Problem 16
<math>\textbf{(A) } \frac{35}{2}\qquad\textbf{(B) } 15\sqrt{2}\qquad\textbf{(C) } \frac{64}{3}\qquad\textbf{(D) } 16\sqrt{2}\qquad {{AMC10 box|year=2006|num-b=15|num-a=17|ab=A}}

5 KB (732 words) - 23:19, 19 September 2023
1991 AJHSME Problems
<math>\text{(A)}\ 5 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 20 \qquad \text{(E)}\ 25</math> fill((15,3)--(16,3)--(16,2)--(15,2)--cycle,black); fill((14,2)--(15,2)--(15,1)--(14,1)--cycle,black);

17 KB (2,246 words) - 13:37, 19 February 2020
2005 AIME II Problems/Problem 4
...[[positive integer]]s that are divisors of at least one of <math> 10^{10},15^7,18^{11}. </math> <math>15^7 = 3^7\cdot5^7</math> so <math>15^7</math> has <math>8\cdot8 = 64</math> divisors.

3 KB (377 words) - 18:36, 1 January 2024
2005 AIME II Problems
...r of positive integers that are divisors of at least one of <math> 10^{10},15^7,18^{11}. </math> In triangle <math> ABC, AB=13, BC=15, </math> and <math>CA = 14. </math> Point <math> D </math> is on <math> \ov

7 KB (1,119 words) - 21:12, 28 February 2020
2005 AIME II Problems/Problem 7
<cmath>(y^{14}+y^{12}+y^{10}+y^8+y^6+y^4+y^2+1)(y+1) = (y^{15}+y^{14}+y^{13}+y^{12}+y^{11}+y^{10}+y^9+y^8+y^7+y^6+y^5+y^4+y^3+y^2+y+1)=\f

2 KB (279 words) - 12:33, 27 October 2019
2005 AIME II Problems/Problem 10
import three; currentprojection = perspective(4,-15,4); defaultpen(linewidth(0.7));

3 KB (436 words) - 03:10, 23 September 2020
2005 AIME I Problems
== Problem 15 == [[2005 AIME I Problems/Problem 15|Solution]]

6 KB (983 words) - 05:06, 20 February 2019
2005 AIME I Problems/Problem 3
...9, 23, 29, 31, 37, 41, 43</math> and <math>47</math>) so there are <math> {15 \choose 2} =105</math> ways to choose a pair of primes from the list and th

2 KB (249 words) - 09:37, 23 January 2024
2005 AIME I Problems/Problem 4
15 & 330 & no\\ \hline

8 KB (1,248 words) - 11:43, 16 August 2022
2005 AIME I Problems/Problem 10
pair A=(15,32), B=(12,19), C=(23,20), M=B/2+C/2, P=(17,22); ...}p = \frac{107 \cdot 11 - 337}{11}</math>. Solving produces that <math>p = 15</math>. [[Substitution|Substituting]] backwards yields that <math>q = 32</m

5 KB (852 words) - 21:23, 4 October 2023
2005 AIME I Problems/Problem 12
...<math>S(4), S(5), \ldots, S(8)</math> are even, and <math>S(9), \ldots, S(15)</math> are odd, and so on.

4 KB (647 words) - 02:29, 4 May 2021
2005 AIME I Problems/Problem 13
label("$(13,15,18)$", (4,5), N); label("$(18,15,13)$", (5,4), N);

5 KB (897 words) - 00:21, 29 July 2022
2005 AIME I Problems/Problem 14
{{AIME box|year=2005|n=I|num-b=13|num-a=15}}

3 KB (561 words) - 14:11, 18 February 2018
2005 AIME II Problems/Problem 15
D((0,30)--(0,-10),Arrows(4));D((15,0)--(-25,0),Arrows(4));D((0,0)--MP("y=ax",(14,14 * (69/100)^.5),E),EndArrow [[Image:2005_AIME_II_-15.png||center|800px]]

12 KB (2,000 words) - 13:17, 28 December 2020
2005 AIME II Problems/Problem 14
In [[triangle]] <math> ABC, AB=13, BC=15, </math> and <math>CA = 14. </math> Point <math> D </math> is on <math> \ov pair C = rotate(15,A)*(A+dir(-50));

13 KB (2,129 words) - 18:56, 1 January 2024
2004 AIME I Problems/Problem 15
15&2^{13}-4\cdot 7\\ 17&2^{15}-16\cdot 9\\

9 KB (1,491 words) - 01:23, 26 December 2022
2004 AIME I Problems/Problem 14
currentprojection = perspective(-2,-50,15); size(200); {{AIME box|year=2004|n=I|num-b=13|num-a=15}}

4 KB (729 words) - 01:00, 27 November 2022
2004 AIME I Problems/Problem 11
...ath>\frac{72}{5x^2} - 1 = \frac{27}{x^3} - 1</math> and so <math>x = \frac{15}{8}</math>. Then <math>k = \frac{\frac{5}{3}\pi x^2}{24\pi - \frac{5}{3}\p ...knowledge, we obtain <math>\ell = 5</math> and lateral area <math>A_\ell = 15\pi</math>. The area of the base is <math>A_B = 3^2\pi = 9\pi</math>.

5 KB (839 words) - 22:12, 16 December 2015
2004 AIME I Problems/Problem 10
A [[circle]] of [[radius]] 1 is randomly placed in a 15-by-36 [[rectangle]] <math> ABCD </math> so that the circle lies completely ...h>[A'B'C'] = [ABC] \cdot \left(\frac{r_{A'B'C'}}{r_{ABC}}\right)^2 = \frac{15 \times 36}{2} \cdot \frac{25}{36} = \frac{375}{2}</math>.

5 KB (836 words) - 07:53, 15 October 2023
2004 AIME I Problems/Problem 5
\frac{a}{q} < \frac{160}{300} = \frac{8}{15} a + b < \frac{8}{15}q + (350 - \frac{7}{10}q)

3 KB (436 words) - 18:31, 9 January 2024
2004 AIME I Problems/Problem 7
...of <math>x</math> in <math>P(x)</math> is <math>-1 + 2 - 3 + \ldots + 14 - 15 = -8</math>, so <math>P(x) = 1 -8x + Cx^2 + Q(x)</math>, where <math>Q(x)</ ...roduct of each pair of distinct terms, or <math>C = \sum_{1 \le i \neq j}^{15} S_iS_j</math>. Also, we know that

5 KB (833 words) - 19:43, 1 October 2023
2004 AIME I Problems/Problem 8
...ath>. Thus <math>\phi(1000) = 1000\left(1 - \frac 12\right)\left(1 - \frac 15\right) = 400</math>, and the answer is <math>\frac{400}{2} - 1 = 199</math>

4 KB (620 words) - 21:26, 5 June 2021
2004 AIME I Problems
A circle of radius 1 is randomly placed in a 15-by-36 rectangle <math> ABCD </math> so that the circle lies completely with == Problem 15 ==

9 KB (1,434 words) - 13:34, 29 December 2021
2004 AIME II Problems/Problem 15
<cmath>s_{13, 4} = 2s_{15 - 13, 3} + 1 = 2s_{2, 3}+1</cmath> (46,15)&64&16\\

6 KB (899 words) - 20:58, 12 May 2022
2004 AIME II Problems/Problem 14
{{AIME box|year=2004|n=II|num-b=13|num-a=15}}

11 KB (1,857 words) - 21:55, 19 June 2023
2004 AIME II Problems/Problem 13
...AC \parallel DE, \angle ABC=120^\circ, AB=3, BC=5, </math> and <math>DE = 15. </math> Given that the [[ratio]] between the area of triangle <math> ABC < pair D=(0,0), E=(15,0), F=IP(CR(D, 75/7), CR(E, 45/7)), A=D+ (5+(75/7))/(75/7) * (F-D), C = E+

3 KB (486 words) - 22:15, 7 April 2023
2004 AIME II Problems/Problem 10
...d power, we will also satisfy the congruence. Thus, <math>2^{3}, 2^{9}, 2^{15},</math> etc. work; or, <cmath>y-x \equiv 3 \pmod 6</cmath> ...l and error soon gives us <math>2^9=512</math>. Notice a pattern? Trying 2^15 = 32768 also works. You could go on, but basically all powers of two 3 mod

8 KB (1,283 words) - 19:19, 8 May 2024
2004 AIME II Problems/Problem 7
pair A=(0,0),B=(0,25),C=(70/3,25),D=(70/3,0),E=(0,8),F=(70/3,22),G=(15,0); ...= B'E</math> (by SAS). By the [[Pythagorean Theorem]], we have <math>AB' = 15</math>. Similarly, from <math>BF = B'F</math>, we have

9 KB (1,501 words) - 05:34, 30 October 2023
2004 AIME II Problems/Problem 5
...math>\frac{250}{800}(60)=\frac{150}{8}</math>. The train then has <math>60-15-\frac{50}{3}-\frac{150}{8}=230/24</math> minutes left to travel 250 miles,

4 KB (592 words) - 19:02, 26 September 2020
2004 AIME II Problems
...|| AD, AC || DE, \angle ABC=120^\circ, AB=3, BC=5, </math> and <math>DE = 15. </math> Given that the ratio between the area of triangle <math> ABC </mat == Problem 15 ==

9 KB (1,410 words) - 05:05, 20 February 2019
2000 AIME II
* [[2000 AIME II Problems/Problem 15|Problem 15]]

1 KB (139 words) - 08:41, 7 September 2011
2001 AIME I
* [[2001 AIME I Problems/Problem 15|Problem 15]]

1 KB (139 words) - 08:41, 7 September 2011
2000 AIME I
* [[2000 AIME I Problems/Problem 15|Problem 15]]

1 KB (135 words) - 18:05, 30 May 2015
1999 AIME
* [[1999 AIME Problems/Problem 15|Problem 15]]

1 KB (118 words) - 08:41, 7 September 2011
1998 AIME
* [[1998 AIME Problems/Problem 15|Problem 15]]

1 KB (114 words) - 08:39, 7 September 2011
1997 AIME
* [[1997 AIME Problems/Problem 15|Problem 15]]

1 KB (114 words) - 08:39, 7 September 2011
1996 AIME
* [[1996 AIME Problems/Problem 15|Problem 15]]

1 KB (114 words) - 08:39, 7 September 2011
1995 AIME
* [[1995 AIME Problems/Problem 15|Problem 15]]

1 KB (114 words) - 08:38, 7 September 2011
1994 AIME
* [[1994 AIME Problems/Problem 15|Problem 15]]

1 KB (114 words) - 08:43, 7 September 2011
1983 AIME
* [[1983 AIME Problems/Problem 15|Problem 15]]

1 KB (114 words) - 20:35, 31 October 2020
1983 AIME Problems
Let <math>f(x)=|x-p|+|x-15|+|x-p-15|</math>, where <math>0 . Determine the [[minimum]] value taken by <math>f(x)</math> for <mat == Problem 15 ==

7 KB (1,104 words) - 12:53, 6 July 2022
1984 AIME Problems
...</math> is either <math>8</math> or <math>0</math>. Compute <math>\frac{n}{15}</math>. MP("t_1",(A+C+(C-A)*3/4+(A-C)*5/12)/2+(0,0.15),ESE);

6 KB (933 words) - 01:15, 19 June 2022
1986 AIME Problems
...ments. Let <math>S</math> be a set of positive integers, none greater than 15. Suppose no two disjoint subsets of <math>S</math> have the same sum. What ...ree HT, four TH, and five TT subsequences. How many different sequences of 15 coin tosses will contain exactly two HH, three HT, four TH, and five TT sub

5 KB (847 words) - 15:48, 21 August 2023
1987 AIME Problems
...or which there is a unique integer <math>k</math> such that <math>\frac{8}{15} < \frac{n}{n + k} < \frac{7}{13}</math>? == Problem 15 ==

6 KB (869 words) - 15:34, 22 August 2023
1988 AIME Problems
== Problem 15 == [[1988 AIME Problems/Problem 15|Solution]]

6 KB (902 words) - 08:57, 19 June 2021
1989 AIME Problems
== Problem 15 == [[1989 AIME Problems/Problem 15|Solution]]

7 KB (1,045 words) - 20:47, 14 December 2023
1990 AIME Problems
A triangle has vertices <math>P_{}^{}=(-8,5)</math>, <math>Q_{}^{}=(-15,-19)</math>, and <math>R_{}^{}=(1,-7)</math>. The equation of the bisector == Problem 15 ==

6 KB (870 words) - 10:14, 19 June 2021
1991 AIME Problems
...<math>\overline{DA}</math>, respectively. It is given that <math>PB^{}_{}=15</math>, <math>BQ^{}_{}=20</math>, <math>PR^{}_{}=30</math>, and <math>QS^{} == Problem 15 ==

7 KB (1,106 words) - 22:05, 7 June 2021
1992 AIME Problems
\text{Row 6: } & 1 & & 6 & &15& &20& &15 & & 6 & & 1 == Problem 15 ==

8 KB (1,117 words) - 05:32, 11 November 2023
1993 AIME Problems
...n{array}{|c|c|c|c|c|c|c|c|c|} \hline n & 0 & 1 & 2 & 3 & \dots & 13 & 14 & 15 \\ :(a) the winner caught <math>15</math> fish;

8 KB (1,275 words) - 06:55, 2 September 2021
1994 AIME Problems
The increasing sequence <math>3, 15, 24, 48, \ldots\,</math> consists of those positive multiples of 3 that are == Problem 15 ==

7 KB (1,141 words) - 07:37, 7 September 2018
1995 AIME Problems
== Problem 15 == [[1995 AIME Problems/Problem 15|Solution]]

6 KB (1,000 words) - 00:25, 27 March 2024
1996 AIME Problems
...>, <math>AB=\sqrt{30}</math>, <math>AC=\sqrt{6}</math>, and <math>BC=\sqrt{15}</math>. There is a point <math>D</math> for which <math>\overline{AD}</mat == Problem 15 ==

6 KB (931 words) - 17:49, 21 December 2018
1997 AIME Problems
== Problem 15 == [[1997 AIME Problems/Problem 15|Solution]]

7 KB (1,098 words) - 17:08, 25 June 2020
1998 AIME Problems
...th>\overline{CD},</math> respectively, so that <math>AP = 5, PB = 15, BQ = 15,</math> and <math>CR = 10.</math> What is the [[area]] of the [[polygon]] == Problem 15 ==

7 KB (1,084 words) - 02:01, 28 November 2023
1999 AIME Problems
...sides of the triangle have lengths <math>AB=13, BC=14,</math> and <math>CA=15,</math> and the tangent of angle <math>PAB</math> is <math>m/n,</math> wher == Problem 15 ==

7 KB (1,094 words) - 13:39, 16 August 2020
2000 AIME I Problems
== Problem 15 == [[2000 AIME I Problems/Problem 15|Solution]]

7 KB (1,204 words) - 03:40, 4 January 2023
2001 AIME I Problems
In triangle <math>ABC</math>, <math>AB=13</math>, <math>BC=15</math> and <math>CA=17</math>. Point <math>D</math> is on <math>\overline{A == Problem 15 ==

7 KB (1,212 words) - 22:16, 17 December 2023
2002 AIME I Problems
== Problem 15 == [[2002 AIME I Problems/Problem 15|Solution]]

8 KB (1,374 words) - 21:09, 27 July 2023
2003 AIME I Problems
== Problem 15 == [[2003 AIME I Problems/Problem 15|Solution]]

6 KB (965 words) - 16:36, 8 September 2019
2000 AIME II Problems
Given that <center><math>\frac 1{2!17!}+\frac 1{3!16!}+\frac 1{4!15!}+\frac 1{5!14!}+\frac 1{6!13!}+\frac 1{7!12!}+\frac 1{8!11!}+\frac 1{9!10! ...</math>. It is given that <math>AB=13</math>, <math>BC=14</math>, <math>CA=15</math>, and that the distance from <math>O</math> to triangle <math>ABC</ma

6 KB (947 words) - 21:11, 19 February 2019
2001 AIME II Problems
== Problem 15 == [[2001 AIME II Problems/Problem 15|Solution]]

8 KB (1,282 words) - 21:12, 19 February 2019
2002 AIME II Problems
== Problem 15 == [[2002 AIME II Problems/Problem 15|Solution]]

7 KB (1,177 words) - 15:42, 11 August 2023
2003 AIME II Problems
...e <math>ABC,</math> <math>AB = 13,</math> <math>BC = 14,</math> <math>AC = 15,</math> and point <math>G</math> is the intersection of the medians. Points ...ame side of line <math>AB</math> as <math>C</math> so that <math>AD = BD = 15.</math> Given that the area of triangle <math>CDM</math> may be expressed a

7 KB (1,127 words) - 09:02, 11 July 2023
1983 AIME Problems/Problem 2
Let <math>f(x)=|x-p|+|x-15|+|x-p-15|</math>, where <math>0 . Determine the [[minimum]] value taken by <math>f(x)</math> for <mat ...that <math>|x-p|=x-p</math>, <math>|x-15|=15-x</math>, and <math>|x-p-15|=15+p-x</math>.

1 KB (184 words) - 20:16, 14 January 2023
1983 AIME Problems/Problem 3
...for <math>x^2+18x+30</math>, so that the equation becomes <math>y=2\sqrt{y+15}</math>. ...> or <math>y=-6</math>. The second root is extraneous since <math>2\sqrt{y+15}</math> is always non-negative (and moreover, plugging in <math>y=-6</math>

3 KB (532 words) - 05:18, 21 July 2022
1983 AIME Problems/Problem 14
pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); ...pair a=(8,0),b=(20,0),t=(14,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--a--m--n--cycle); draw(

13 KB (2,149 words) - 18:44, 5 February 2024
1983 AIME Problems/Problem 15
 <cmath>0 \geq f(3) = 25 - 15\cos \alpha - 20 \sin \alpha</cmath>

19 KB (3,221 words) - 01:05, 7 February 2023
1984 AIME Problems/Problem 2
...</math> is either <math>8</math> or <math>0</math>. Compute <math>\frac{n}{15}</math>. Any multiple of 15 is a multiple of 5 and a multiple of 3.

1 KB (187 words) - 20:05, 29 May 2021
1984 AIME Problems/Problem 5
...ath>. This means that <math>\log ab^3 = 15\log 2 \Longrightarrow ab^3 = 2^{15}</math> and that <math>\log a^3 b = 21\log 2 \Longrightarrow a^3 b = 2^{21}

6 KB (863 words) - 16:10, 16 May 2024
1984 AIME Problems/Problem 9
...>AB</math> has length 3 cm. The area of [[face]] <math>ABC</math> is <math>15\mbox{cm}^2</math> and the area of face <math>ABD</math> is <math>12 \mbox { ...ghtanglemark(D,Da,Db));draw(rightanglemark(A,Db,D));draw(anglemark(Da,Db,D,15));

6 KB (947 words) - 20:44, 26 November 2021
1984 AIME Problems/Problem 10
...symbol{\cdots}&\boldsymbol{12}&\boldsymbol{13}&\boldsymbol{14}&\boldsymbol{15}&\boldsymbol{16}&\boldsymbol{17}&\boldsymbol{18}&\boldsymbol{19}&\boldsymbo

7 KB (1,163 words) - 23:53, 28 March 2022
1984 AIME Problems/Problem 12
...the double counted values. This means that <math>x = 0, \pm 5, \pm 10, \pm 15... \pm 1000</math> so the answer is 400 + 1 = <math>\boxed{401}</math>

3 KB (588 words) - 14:37, 22 July 2020
1984 AIME Problems/Problem 14
...or <math>4 \bmod{6}</math>. Notice that the numbers <math>9</math>, <math>15</math>, <math>21</math>, ... , and in general <math>9 + 6n</math> for nonne ...which yields <math>n=34=9+25</math> which does not work). Thus <math>n-9,n-15,n-21,n-27,</math> and <math>n-33</math> form a prime quintuplet. However, o

8 KB (1,346 words) - 01:16, 9 January 2024
1984 AIME Problems/Problem 15
\frac{x^2}{15}+\frac{y^2}{7}-\frac{z^2}{9}-\frac{w^2}{33}=1\\ \frac{x^2}{63}+\frac{y^2}{55}+\frac{z^2}{39}+\frac{w^2}{15}=1\\

6 KB (1,051 words) - 04:52, 8 May 2024
1985 AIME Problems/Problem 14
...t have <math>n > 10</math>, so <math>n = 15</math> and the answer is <math>15 + 10 = \boxed{25}</math>. ..., then the strongest <math>16</math> people get a total of <math>16*10-145=15</math> playing against the weakest <math>10</math> who gained <math>45</mat

5 KB (772 words) - 22:14, 18 June 2020
1985 AIME Problems/Problem 11
...th>d = \sqrt{(9 - 49)^2 + (20 + 55)^2} = \sqrt{40^2 + 75^2} = 5\sqrt{8^2 + 15^2} = 5\cdot 17 = 85</math>.

5 KB (932 words) - 17:00, 1 September 2020
1985 AIME Problems/Problem 10
<math>\frac{15}{24} \to 11</math> <math>\frac{20}{24}\to 15</math>

12 KB (1,859 words) - 18:16, 28 March 2022
1985 AIME Problems/Problem 9
real r = 8/15^0.5, a = 57.91, b = 93.135; real r = 8/15^0.5, a = 57.91, b = 93.135;

5 KB (763 words) - 16:20, 28 September 2019
1985 AIME Problems/Problem 6
<cmath>12+16\cdot \frac ab + 5\cdot \frac ba = \frac ba\cdot 15</cmath>

5 KB (789 words) - 03:09, 23 January 2023
1986 AIME Problems/Problem 15
<math>2a^2 + 5b^2 = - \frac {15}{2}ab \ \ \ \ (2)</math></div>

11 KB (1,722 words) - 09:49, 13 September 2023
1986 AIME Problems/Problem 14
...<math>2\sqrt{5}</math>, <math>\frac{30}{\sqrt{13}}</math>, and <math>\frac{15}{\sqrt{10}}</math>. Determine the [[volume]] of <math>P</math>. <cmath>\frac {hl}{\sqrt {h^2 + l^2}} = \frac {15}{\sqrt {10}}</cmath>

2 KB (346 words) - 13:13, 22 July 2020
1986 AIME Problems/Problem 13
...H</tt>, and five <tt>TT</tt> subsequences. How many different sequences of 15 coin tosses will contain exactly two <tt>HH</tt>, three <tt>HT</tt>, four <

4 KB (772 words) - 21:09, 7 May 2024
1986 AIME Problems/Problem 12
...ments. Let <math>S</math> be a set of positive integers, none greater than 15. Suppose no two disjoint subsets of <math>S</math> have the same sum. What ...> must have more than 4 elements, otherwise its sum would be at most <math>15+14+13+12=54</math>.

2 KB (364 words) - 19:41, 1 September 2020
1986 AIME Problems/Problem 11
...ath>1</math>). By the [[Binomial Theorem]], this is <math>(-1) \cdot (-1)^{15}{18 \choose 3} = \boxed{816}</math>. ...y^2</math> term is <math>{2\choose 0} + {3\choose 1} + \cdots + {17\choose 15}</math>. This is actually the sum of the first 16 triangular numbers, which

6 KB (872 words) - 16:51, 9 June 2023
1986 AIME Problems/Problem 10
...ath>a+b+c=\frac{3194+m}{222}>\frac{3194}{222}>14</math> so <math>a+b+c\geq 15</math>. Checking each of the multiples of <math>222</math> from <math>15\cdot222</math> to <math>18\cdot222</math> by subtracting <math>N</math> fro

3 KB (565 words) - 16:51, 1 October 2023
1986 AIME Problems/Problem 9
By similar triangles, <math>BE'=\frac{d}{510}\cdot450=\frac{15}{17}d</math> and <math>EC=\frac{d}{425}\cdot450=\frac{18}{17}d</math>. Sinc ...510 - \frac{6}{5}d + 510 - \frac{17}{15}d \Longrightarrow d\left(\frac{50}{15}\right) = 1020 \Longrightarrow d = \boxed{306}</math>.

11 KB (1,850 words) - 18:07, 11 October 2023
1986 AIME Problems/Problem 8
...\log 1 + \log 2 + \log 4 + \log 5 +\ldots + \log 1000000 = \log (2^05^0)(2^15^0)(2^25^0)\cdots (2^65^6).</cmath> Each power of <math>2</math> appears <ma

3 KB (487 words) - 20:52, 16 September 2020
1986 AIME Problems/Problem 2
...right)^2}{2\cdot 2\sqrt5\cdot2\sqrt6}=\frac{16}{8\sqrt{30}}=\sqrt{\frac{2}{15}},</cmath> so <math>\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{\frac{13}{15}}.</math>

3 KB (460 words) - 00:44, 5 February 2022
1987 AIME Problems/Problem 14
{{AIME box|year=1987|num-b=13|num-a=15}}

7 KB (965 words) - 10:42, 12 April 2024
1987 AIME Problems/Problem 10
...ance at rate <math>r</math> from the escalator, while Bob is getting <math>15</math> seconds of help at rate <math>r</math>. Solving for <math>r</math>,

7 KB (1,187 words) - 16:21, 27 January 2024
1987 AIME Problems/Problem 8
...or which there is a unique integer <math>k</math> such that <math>\frac{8}{15} < \frac{n}{n + k} < \frac{7}{13}</math>? <cmath>\begin{array}{ccccc}\frac{15}{8} &>& \frac{k + n}{n} &>& \frac{13}{7}\\

2 KB (393 words) - 16:59, 16 December 2020
1987 AIME Problems/Problem 4
...h>|x - 60| = 0</math>, which is when <math>x = 60</math> and <math>y = \pm 15</math>. Since the graph is [[symmetry|symmetric]] about the y-axis, we just ...using the [[Shoelace Theorem]], we get <math>2 \cdot \frac{1}{2}(80 - 48)(15) = \boxed{480}</math>.

2 KB (371 words) - 17:25, 13 February 2024
1987 AIME Problems/Problem 3
...6,\ 2^3 = 8,\ 2 \cdot 5 = 10,</math> <math>\ 2 \cdot 7 = 14,\ 3 \cdot 5 = 15,\ 3 \cdot 7 = 21,</math> <math>\ 2 \cdot 11 = 22,\ 2 \cdot 13 = 26,</math>

3 KB (511 words) - 09:29, 9 January 2023
1988 AIME Problems/Problem 14
{{AIME box|year=1988|num-b=13|num-a=15}}

4 KB (700 words) - 17:21, 3 May 2021
1988 AIME Problems/Problem 13
...+ F_{15}) + 1 &\Longrightarrow (aF_{17} + bF_{16})\cdot x + (aF_{16} + bF_{15} + 1) = 0,\ x\not\in Q \\ &\Longrightarrow aF_{17} + bF_{16} = 0 \text{ and } aF_{16} + bF_{15} + 1 = 0 \\

10 KB (1,585 words) - 03:58, 1 May 2023
1988 AIME Problems/Problem 11
<math>504 = 3m + 15</math>

2 KB (422 words) - 00:22, 6 September 2020
1988 AIME Problems/Problem 6
...use the arithmetic progression from left to right has difference <math>x - 15</math>. Therefore, we have <math>x = 50</math>, and because the desired ast

5 KB (878 words) - 23:06, 20 November 2023
1989 AIME Problems/Problem 15
...ac{[APC] + [BPC]}{[APB]} = 3</math>, so <math>CP = \dfrac{3}{4} \cdot CF = 15</math>. ...ed out, so <math>w_C = 1</math> and <math>w_F = 3</math>. Thus, <math>CP = 15</math> and <math>PF = 5</math>.

13 KB (2,091 words) - 00:20, 26 October 2023
1989 AIME Problems/Problem 14
{{AIME box|year=1989|num-b=13|num-a=15}}

2 KB (408 words) - 17:28, 16 September 2023
1989 AIME Problems/Problem 11
<math>n = 4: 5000+15*116 = 6740</math>

5 KB (851 words) - 18:01, 28 December 2022
1989 AIME Problems/Problem 6
label("$P$",(6,15),N); label("$X$",(12.5,15),N);

6 KB (980 words) - 15:08, 14 May 2024
1989 AIME Problems/Problem 4
<math>m = 15</math> gives a solution for k. <math>10 + 5a = 15^3</math>

3 KB (552 words) - 12:41, 3 March 2024
1990 AIME Problems/Problem 14
{{AIME box|year=1990|num-b=13|num-a=15}}

7 KB (1,086 words) - 08:16, 29 July 2023
1990 AIME Problems/Problem 12
...th of each of the 12 sides is <math>2 \cdot 12\sin 15</math>. <math>24\sin 15 = 24\sin (45 - 30) = 24\frac{\sqrt{6} - \sqrt{2}}{4} = 6(\sqrt{6} - \sqrt{2

6 KB (906 words) - 13:23, 5 September 2021
1990 AIME Problems/Problem 7
...gle]] has [[vertex|vertices]] <math>P_{}^{}=(-8,5)</math>, <math>Q_{}^{}=(-15,-19)</math>, and <math>R_{}^{}=(1,-7)</math>. The [[equation]] of the [[ang pair P=(-8,5),Q=(-15,-19),R=(1,-7),S=(7,-15),T=(-4,-17);

8 KB (1,319 words) - 11:34, 22 November 2023
1990 AIME Problems/Problem 5
The [[prime factorization]] of <math>75 = 3^15^2 = (2+1)(4+1)(4+1)</math>. For <math>n</math> to have exactly <math>75</ma

1 KB (175 words) - 03:45, 21 January 2023
2006 AMC 10B Problems/Problem 10
...times as long as a second side, and the length of the third side is <math>15</math>. What is the greatest possible perimeter of the triangle? The lengths of the sides are: <math>x</math>, <math>3x</math>, and <math>15</math>.

900 bytes (132 words) - 13:57, 26 January 2022
2006 AMC 10B Problems/Problem 14
{{AMC10 box|year=2006|ab=B|num-b=13|num-a=15}}

2 KB (264 words) - 21:10, 19 September 2023
1991 AIME Problems/Problem 14
{{AIME box|year=1991|num-b=13|num-a=15}}

2 KB (284 words) - 03:56, 23 January 2023
1991 AIME Problems/Problem 12
...<math>\overline{DA}</math>, respectively. It is given that <math>PB^{}_{}=15</math>, <math>BQ^{}_{}=20</math>, <math>PR^{}_{}=30</math>, and <math>QS^{} ...",P,N);label("$Q$",Q,E);label("$R$",R,SW);label("$S$",S,W); label("$15$",B/2+P/2,N);label("$20$",B/2+Q/2,E);label("$O$",O,SW); </asy></center

8 KB (1,270 words) - 23:36, 27 August 2023
2006 AMC 10B Problems/Problem 16
{{AMC10 box|year=2006|ab=B|num-b=15|num-a=17}}

2 KB (336 words) - 10:51, 11 May 2024
1991 AIME Problems/Problem 11
...</math> and <math>m \angle MOA = 15^\circ</math>. Thus <math>AM = (1) \tan{15^\circ} = 2 - \sqrt {3}</math>, which is the radius of one of the circles. T Note that it is very useful to understand the side lengths of a 15-75-90 triangle, as these triangles often appear on higher level math contes

4 KB (740 words) - 19:33, 28 December 2022
1991 AIME Problems/Problem 10
bab & 2 & 4 & 15 \\

5 KB (813 words) - 06:10, 25 February 2024
1991 AIME Problems/Problem 9
...only the [[positive]] root will work, so the value of <math>y = \frac{29}{15}</math> and <math>m + n = \boxed{044}</math>. <cmath>\csc x + \cot x = \frac {841}{435} = \frac {29}{15},</cmath>

10 KB (1,590 words) - 14:04, 20 January 2023
1992 AIME Problems/Problem 14
{{AIME box|year=1992|num-b=13|num-a=15}}

4 KB (667 words) - 01:26, 16 August 2023
1992 AIME Problems/Problem 4
\text{Row 6: } & 1 & & 6 & &15& &20& &15 & & 6 & & 1

3 KB (476 words) - 14:13, 20 April 2024
2006 AMC 10B Problems/Problem 23
<math> \mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 17\qquad \mathrm{(C) \ } \frac{35}{2}\qquad \mathrm{( label("$7$",(1.45,0.15));

5 KB (861 words) - 00:53, 25 November 2023
2006 AMC 10B Problems/Problem 24
pair top=X+15*dir(X--A), bottom=X+15*dir(X--B);

4 KB (558 words) - 14:38, 6 April 2024
1993 AIME Problems/Problem 14
{{AIME box|year=1993|num-b=13|num-a=15}}

3 KB (601 words) - 09:25, 19 November 2023
1993 AIME Problems/Problem 13
...ac{2\cdot \left(\frac{1}{4}\right)}{1-\left(\frac{1}{4}\right)^2}=\frac{8}{15}</math>. Therefore, <math>t = CD = CX\cdot\tan(\angle BXA) = 100 \cdot \frac{8}{15} = \frac{160}{3}</math>, and the answer is <math>\boxed{163}</math>.

8 KB (1,231 words) - 20:06, 26 November 2023
1993 AIME Problems/Problem 3
...n{array}{|c|c|c|c|c|c|c|c|c|} \hline n & 0 & 1 & 2 & 3 & \dots & 13 & 14 & 15 \\ :(a) the winner caught <math>15</math> fish;

2 KB (364 words) - 00:05, 9 July 2022
1994 AIME Problems/Problem 15
.../2,Fb,12,-35); D(La,bluedots);D(Lb,bluedots);D(endptproject((C+P)/2,Fc,18,-15),bluedots);D(IP(La,Lb),blue);

4 KB (717 words) - 22:20, 3 June 2021
1994 AIME Problems/Problem 14
{{AIME box|year=1994|num-b=13|num-a=15}}

2 KB (303 words) - 00:03, 28 December 2017
1994 AIME Problems/Problem 11
..., and <math>15</math>'s. Because <math>0</math>, <math>6</math>, and <math>15</math> are all multiples of <math>3</math>, the change will always be a mul

4 KB (645 words) - 15:12, 15 July 2019
1994 AIME Problems/Problem 6
draw((-15,-10)--(15,-10)); draw((-15,10)--(15,10));

4 KB (721 words) - 16:14, 8 March 2021
1994 AIME Problems/Problem 1
The increasing [[sequence]] <math>3, 15, 24, 48, \ldots\,</math> consists of those [[positive]] multiples of 3 that

946 bytes (139 words) - 21:05, 1 September 2023
1995 AIME Problems/Problem 15
...ve. This is an infinite geometric series whose sum is <math>\frac{3/64}{1-(15/32)}=\frac{3}{34}</math>, so the answer is <math>\boxed{037}</math>. ...e roll <math>4</math> consecutive <tt>H</tt>'s, and there is a <math>\frac{15}{16}</math> probability we roll a <tt>T</tt>. Thus,

6 KB (979 words) - 13:20, 11 April 2022
1995 AIME Problems/Problem 14
{{AIME box|year=1995|num-b=13|num-a=15}}

3 KB (484 words) - 13:11, 14 January 2023
1995 AIME Problems/Problem 8
For <math>y=3</math>, we have <math>15,27,\cdots ,99</math>, or <math>8</math> cases.

4 KB (646 words) - 17:37, 1 January 2024
1996 AIME Problems/Problem 15
...\Longrightarrow \cos 2\theta = \frac{\sqrt{3}}{2} \Longrightarrow \theta = 15^{\circ}</math>. The answer is <math>\lfloor 1000r \rfloor = \left\lfloor 10 ...cdot 4}}{8} = \frac {\sqrt {6} \pm \sqrt {2}}{4}</math>, so <math>\theta = 15^{\circ}</math> (as the other roots are too large to make sense in context).

5 KB (710 words) - 21:04, 14 September 2020
1996 AIME Problems/Problem 14
{{AIME box|year=1996|num-b=13|num-a=15}}

5 KB (923 words) - 21:21, 22 September 2023
1996 AIME Problems/Problem 13
...>, <math>AB=\sqrt{30}</math>, <math>AC=\sqrt{6}</math>, and <math>BC=\sqrt{15}</math>. There is a point <math>D</math> for which <math>\overline{AD}</mat pair B=(0,0), C=(15^.5, 0), A=IP(CR(B,30^.5),CR(C,6^.5)), E=(B+C)/2, D=foot(B,A,E);

3 KB (521 words) - 01:18, 25 February 2016
1996 AIME Problems/Problem 12
<math>\sum\limits_{k = 1}^{9}\sum\limits_{j = 1}^{k}j = 45+36+28+21+15+10+6+3+1 = 165</math>.

5 KB (879 words) - 11:23, 5 September 2021
1996 AIME Problems/Problem 11
Hence, the answer is <math>\frac{18\pi}{15}+\frac{5\pi}{15}=\frac{23\pi}{15}\cdot\frac{180}{\pi}=\boxed{276}</math>

6 KB (1,022 words) - 20:23, 17 April 2021
1996 AIME Problems/Problem 6
<math>\frac{5}{16}+\frac{5}{16}-\frac{5}{32}=\frac{15}{32}</math>

3 KB (461 words) - 00:33, 16 May 2024
1997 AIME Problems/Problem 14
{{AIME box|year=1997|num-b=13|num-a=15}}

5 KB (874 words) - 22:30, 1 April 2022
1997 AIME Problems/Problem 9
...+ 6\cdot 3^5\sqrt{5} + 15\cdot 3^4 \cdot 5 + 20\cdot 3^3 \cdot 5\sqrt{5} + 15 \cdot 3^2 \cdot 25 + 6 \cdot 3 \cdot 25\sqrt{5} + 5^3\right)</math> <math>=

4 KB (586 words) - 21:53, 30 December 2023
1998 AIME Problems/Problem 14
{{AIME box|year=1998|num-b=13|num-a=15}}

2 KB (390 words) - 21:05, 29 May 2023
1998 AIME Problems/Problem 11
...th>\overline{CD},</math> respectively, so that <math>AP = 5, PB = 15, BQ = 15,</math> and <math>CR = 10.</math> What is the area of the polygon that is triple P=(5,0,0),Q=(20,0,15),R=(20,10,20),Pa=(15,20,20),Qa=(0,20,5),Ra=(0,10,0);

7 KB (1,084 words) - 11:48, 13 August 2023
1998 AIME Problems/Problem 9
real m=60-12*sqrt(15); <math>60 - m = 12\sqrt{15}</math> 

4 KB (624 words) - 18:34, 18 February 2018
1998 AIME Problems/Problem 2
<math>\sum_{i=1}^{15} i=\frac{(15)(16)}{2}</math> ordered pairs. For <math>x > 15</math>, <math>y</math> must follow <math>x < y\le 30</math>. Hence, there a

6 KB (913 words) - 16:34, 6 August 2020
1999 AIME Problems/Problem 15
...has pairwise parallel planar and oppositely equal length (<math>4\sqrt{13},15,17</math>) edges and can be inscribed in a parallelepiped (rectangular box) <math>q^2+r^2=15^2</math>

7 KB (1,169 words) - 15:28, 13 May 2024

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