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• 2021 Fall AMC 12B Problems/Problem 15
  ...shapes like <math>\triangle ABC</math>. Then <math>\angle BAC=360^\circ/24=15^\circ</math>, and <math>\angle EAC = 45^\circ</math>, so <math>\angle{EAB}
  5 KB (805 words) - 01:45, 22 May 2024
• 2021 Fall AMC 10B Problems/Problem 15
  11 KB (1,780 words) - 17:33, 6 June 2024
• 2022 AMC 8 Problems/Problem 15
  7 KB (914 words) - 14:49, 23 November 2023
• 2022 AMC 12A Problems/Problem 15
  46 bytes (5 words) - 01:36, 12 November 2022
• 2022 AMC 10A Problems/Problem 15
  ...rilateral <math>ABCD</math> with side lengths <math>AB=7, BC=24, CD=20, DA=15</math> is inscribed in a circle. The area interior to the circle but exteri D = intersectionpoints(Circle(A,15),Circle(C,20))[1];
  5 KB (768 words) - 21:00, 21 February 2024
• 2022 AMC 10B Problems/Problem 15
  ...6}{a}=\frac{6a+30}{2a+2}</math>, from which <cmath>\frac{3a+6}{a}=\frac{3a+15}{a+1}.</cmath> <cmath>3a^2+9a+6=3a^2+15a</cmath> <cmath>6a=6</cmath>
  2 KB (402 words) - 16:52, 6 June 2024
• 2022 AMC 12B Problems/Problem 15
  46 bytes (5 words) - 00:14, 5 January 2023
• 2023 AMC 8 Problems/Problem 15
  6 KB (1,028 words) - 17:51, 11 May 2024
• 2023 AIME II Problems/Problem 15
  If \({\rm Rem} \left( n , 11 \right) = 8\), then \(k_n = 15\) and \(b_n = 167\). The first few values of <math>b_n</math> are <math>11, 17, 20, 10, 5, 14, 7, 15, 19, 21,</math> and <math>22</math>. We notice that <math>b_{12} = b_1 = 11
  7 KB (1,144 words) - 08:13, 1 February 2024
• Mock AIME 3 2006-2007 Problems/Problem 15
  210 bytes (41 words) - 09:51, 18 June 2023
• 2023 AMC 10A Problems/Problem 15
  4th area = <math>64\pi-49\pi = 15\pi</math>
  6 KB (979 words) - 11:54, 19 May 2024
• 2023 AMC 12A Problems/Problem 15
  draw((-50,15)--(50,15)); draw((50,15)--(50,-15));
  7 KB (1,093 words) - 17:23, 18 June 2024
• 2023 AMC 10B Problems/Problem 15
  <math>(2^{15}) \times (3^{14}) \times ((2^2)^{13}) \times (5^{12}) \times ((2 \times 3)^ ...immediately eliminate B, D, and E since 13 only appears in <math>13!, 14!, 15, 16!</math>, so <math>13\cdot 13\cdot 13\cdot 13</math> is a perfect square
  10 KB (1,510 words) - 14:27, 29 March 2024
• 2023 AMC 12B Problems/Problem 15
  ...18|2023 AMC 10B #18]] and [[2023 AMC 12B Problems/Problem 15|2023 AMC 12B #15]]}} ...<math>c</math> are positive integers such that<cmath>\frac{a}{14}+\frac{b}{15}=\frac{c}{210}.</cmath>Which of the following statements are necessarily tr
  7 KB (1,115 words) - 13:34, 21 April 2024
• 2025 AMC 8 Problems/Problem 15
  5 bytes (1 word) - 09:56, 17 June 2024
• Mock AIME 6 2006-2007 Problems/Problem 15
  751 bytes (134 words) - 10:27, 23 November 2023
• 2024 AIME I Problems/Problem 15
  12 KB (2,049 words) - 16:13, 7 June 2024
• 2021 WSMO Team Round/Problem 15
  335 bytes (52 words) - 13:01, 14 December 2023
• 2022 SSMO Team Round Problems/Problem 15
  859 bytes (143 words) - 13:07, 14 December 2023
• 2021 WSMO Team Round Problems/Problem 15
  361 bytes (55 words) - 22:16, 15 December 2023

Page text matches

• 2003 AIME II Problems/Problem 11
  ...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 <math>DM=\sqrt{15^2-\left(\frac{25} {2}\right)^2}=\frac{5} {2} \sqrt{11}.</math>
  5 KB (772 words) - 19:47, 1 August 2023
• 2003 AIME II Problems/Problem 8
  <math>21*1848-35*1716+15*1440=1848+20*132+(20*1716-35*1716+15*1716)+15*(-276)</math> <math>=1848+5*132+15(132-276)</math>
  5 KB (793 words) - 15:18, 14 July 2023
• 2003 AIME II Problems/Problem 6
  ...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 ...-14-15</math> triangle is a <math>5-12-13</math> triangle and a <math>9-12-15</math> triangle "glued" together on the <math>12</math> side, <math>[ABC]=\
  5 KB (787 words) - 17:38, 30 July 2022
• 2002 AIME II Problems/Problem 14
  {{AIME box|year=2002|n=II|num-b=13|num-a=15}}
  4 KB (658 words) - 19:15, 19 December 2021
• 2002 AIME II Problems/Problem 13
  [[WLOG]], let <math>W_C=15</math>. <math>W_P=W_C+W_X=15+11=26</math>.
  6 KB (935 words) - 13:23, 3 September 2021
• 2002 AIME II Problems/Problem 12
  ...7}>.4</math>. All the rest work. Therefore there are <math>3\cdot5=\textbf{15}</math> possibilities here. Taking all these cases into account, we find that there are <math>4+15+4=23</math> ways to have <math>a_{10} = .4</math> and <math>a_n\leq .4</mat
  7 KB (1,127 words) - 13:34, 19 June 2022
• 2002 AIME II Problems/Problem 7
  Thus, <math>k \equiv 0, 15 \pmod{16}</math>. <math>k \equiv 15 \pmod{16}</math>
  3 KB (403 words) - 12:10, 9 September 2023
• 2001 AIME II Problems/Problem 14
  ...>, we see that the solutions common to both equations have arguments <math>15^\circ , 105^\circ, 195^\circ, </math> and <math>\ 285^\circ</math>. We can {{AIME box|year=2001|n=II|num-b=13|num-a=15}}
  2 KB (380 words) - 15:03, 22 July 2018
• 2001 AIME II Problems/Problem 13
  ...> use Law of Cosines on <math>\triangle ABD</math> to find <math>AD=2\sqrt{15}</math>
  4 KB (743 words) - 03:32, 23 January 2023
• 2001 AIME II Problems/Problem 10
  ...next 90 numbers (6 each), so our total is <math>4\cdot 16 + 6 \cdot \frac{15 \cdot 16}{2} = \boxed{784}</math>.
  4 KB (549 words) - 23:16, 19 January 2024
• 2001 AIME II Problems/Problem 7
  ..., 120), S=(0, 60), T=(45, 60), U = (60,0), V=(60, 40), O1 = (30,30), O2 = (15, 75), O3 = (70, 10); D(CR(D(O1), 30)); D(CR(D(O2), 15)); D(CR(D(O3), 10));
  7 KB (1,112 words) - 02:15, 26 December 2022
• 2001 AIME II Problems/Problem 6
  ...r corresponding side lengths, so <math>\frac{[EFGH]}{[ABCD]} = \left(\frac 15\right)^2 = \frac{1}{25}</math>, and the answer is <math>10n + m = \boxed{25
  4 KB (772 words) - 19:31, 6 December 2023
• 2001 AIME II Problems/Problem 4
  D((0,0)--(10/(15/8),10),EndArrow(6)); D((0,0)--(13,13 * 3/10),EndArrow(6));
  2 KB (240 words) - 20:34, 4 July 2013
• 2000 AIME II Problems/Problem 14
  ...}\right)-\left(1+\sum_{k=1}^{32m-1} {k\cdot k!}\right) = \sum_{k=32m}^{32m+15}k\cdot k!.</math> &=16! +\sum_{m=1}^{62}\sum_{k=32m}^{32m+15}k\cdot k!
  7 KB (1,131 words) - 14:49, 6 April 2023
• 2000 AIME II Problems/Problem 12
  ...</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 <math>\triangle ABC</m ...ternatively, a <math>13-14-15</math> triangle may be split into <math>9-12-15</math> and <math>5-12-13</math> [[right triangle]]s):
  3 KB (532 words) - 13:14, 22 August 2020
• 2000 AIME II Problems/Problem 7
  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! <cmath>\frac {19!}{2!17!}+\frac {19!}{3!16!}+\frac {19!}{4!15!}+\frac {19!}{5!14!}+\frac {19!}{6!13!}+\frac {19!}{7!12!}+\frac {19!}{8!11
  2 KB (281 words) - 12:09, 5 April 2024
• 2005 AMC 10B Problems/Problem 14
  {{AMC10 box|year=2005|ab=B|num-b=13|num-a=15}}
  5 KB (882 words) - 22:12, 30 April 2024
• 2005 AMC 10B Problems/Problem 24
  ...>. Testing <math>a-b=4</math> gives us <math>2a=15 \Longrightarrow a=\frac{15}{2}, b=\frac{7}{2}</math>, which is impossible, as <math>a</math> and <math
  5 KB (845 words) - 19:23, 17 September 2023
• 2006 AMC 12A Problems/Problem 1
  ...cost <math>8\cdot 2=16</math> dollars. In total, the purchase costs <math>15+16=\boxed{\textbf{(A) }31}</math> dollars.
  846 bytes (115 words) - 17:20, 16 December 2021
• 2006 AMC 12A Problems/Problem 3
  <math>\textbf{(A)}\ 15\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 24\qquad\tex
  1 KB (155 words) - 17:30, 16 December 2021

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