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• 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) - 19:06, 17 June 2024
• 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
  4 KB (658 words) - 22:12, 21 November 2023
• 2015 AMC 10B Problems/Problem 15
  2 KB (275 words) - 01:44, 10 June 2024
• 2016 AMC 10A Problems/Problem 15
  1 KB (211 words) - 14:53, 25 June 2023
• 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>
  9 KB (1,407 words) - 19:37, 17 February 2024
• 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));
  31 KB (5,086 words) - 19:15, 20 December 2023
• 2015 AMC 8 Problems/Problem 15
  2 KB (333 words) - 01:41, 15 January 2024
• 2016 AMC 12A Problems/Problem 15
  8 KB (1,255 words) - 09:05, 5 September 2022
• 2016 AMC 10B Problems/Problem 15
  2 KB (378 words) - 21:13, 18 June 2022
• 2016 AMC 12B Problems/Problem 15
  3 KB (554 words) - 01:25, 4 August 2023
• 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

Page text matches

• 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 < p < 15</math>. 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 (2,151 words) - 17:48, 27 May 2024
• 1983 AIME Problems/Problem 15
  <!-- [[Image:1983_AIME-15.png|200px]] --> <cmath>0 \geq f(3) = 25 - 15\cos \alpha - 20 \sin \alpha</cmath>
  20 KB (3,497 words) - 15:37, 27 May 2024
• 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.
  2 KB (257 words) - 19:23, 6 June 2024
• 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

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