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• 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]]
  7 KB (1,148 words) - 23:33, 6 January 2024
• 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
  2 bytes (1 word) - 02:02, 7 December 2019
• 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
  283 bytes (49 words) - 00:23, 14 December 2019

Page text matches

• 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) - 17:46, 24 May 2024
• 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

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