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• 1950 AHSME Problems/Problem 45
  ...lso counts the 100 sides of the polygon, so the actual answer is <math>4950-100=\boxed{\textbf{(A)}\ 4850 }</math>. ...the number of diagonals of a polygon with <math>n</math> sides is <math>n(n-3)/2</math>. Taking <math>n=100</math>, we see that the number of diagonals
  1 KB (217 words) - 21:42, 3 May 2024
• 1953 AHSME Problems/Problem 45
  Since both lengths are positive, the [[AM-GM Inequality]] is satisfied. The correct relationship between <math>a</math {{AHSME 50p box|year=1953|num-b=44|num-a=46}}
  689 bytes (111 words) - 23:02, 14 February 2020
• 1951 AHSME Problems/Problem 45
  == Problem 45 == {{AHSME 50p box|year=1951|num-b=44|num-a=46}}
  1 KB (194 words) - 12:27, 5 July 2013
• 1952 AHSME Problems/Problem 45
  Using the RMS-AM-GM-HM inequality, we can see that the answer is <math>\fbox{E}</math>. {{AHSME 50p box|year=1952|num-b=44|num-a=46}}
  624 bytes (104 words) - 17:36, 9 July 2015
• 1958 AHSME Problems/Problem 45
  ...ars and <math> y</math> cents, <math> x</math> and <math> y</math> both two-digit numbers. In error it is cashed for <math> y</math> dollars and <math> {{AHSME 50p box|year=1958|num-b=44|num-a=46}}
  795 bytes (124 words) - 06:29, 3 October 2014
• 2007 iTest Problems/Problem 45
  {{iTest box|year=2007|num-b=44|num-a=46}}
  3 KB (446 words) - 05:07, 16 June 2018
• 1955 AHSME Problems/Problem 45
  == Problem 45== ...h>, we get <math>r^2+2-2r = 2</math>, we can simplify this to get <math>r(r-2)=0</math>, so <math>r=0</math> or <math>2</math>. But since <math>r\neq 0<
  1 KB (232 words) - 17:23, 2 July 2020
• 2008 iTest Problems/Problem 45
  {{2008 iTest box|num-b=44|num-a=46}}
  2 KB (287 words) - 18:54, 12 July 2018
• 1959 AHSME Problems/Problem 45
  34 bytes (6 words) - 10:15, 14 June 2019
• 1954 AHSME Problems/Problem 45
  {{AHSME 50p box|year=1954|num-b=44|num-a=46}}
  1 KB (193 words) - 10:42, 6 July 2020
• 1956 AHSME Problems/Problem 45
  542 bytes (86 words) - 21:00, 27 March 2021
• 45-45-90 triangle
  a 45-45-90 triangle has 2 angles with the value of 45 and one that is 90 degrees. it is special because it has a ratio for it's s
  232 bytes (45 words) - 14:29, 7 January 2022

Page text matches

• 2005 AMC 12A Problems/Problem 16
  pair P0=O0+9*dir(-45), P3=O3+dir(70); [[Image:2005_12A_AMC-16b.png]]
  2 KB (307 words) - 15:30, 30 March 2024
• 2016 AMC 10A Problems/Problem 8
  ...bf{(B)}\ 30\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 45</math> .../math> as the amount of money Foolish Fox started with we have <math>2(2(2x-40)-40)-40=0.</math> Solving this we get <math>\boxed{\textbf{(C) }35}</math
  1 KB (169 words) - 14:59, 8 August 2021
• MATHCOUNTS
  ...ter), 2-2.5 (State/National)<br><u>Target:</u> 1.5 (School), 2 (Chapter), 2-2.5 (State/National)}} ...dents have 40 minutes to complete the Sprint Round. This round is very fast-paced and requires speed and accuracy as well. The first 20 problems are usu
  10 KB (1,506 words) - 21:31, 14 May 2024
• Chemistry competitions
  ...National Chemistry Olympiad national exam (USNCO) is a 3-part, 4 hour and 45 minute exam administered in mid or late April by ACS Local Sections. Approx ...www.amazon.com/gp/product/0618221565/ref=pd_lpo_k2_dp_k2a_1_img/102-5655201-2084940?%5Fencoding=UTF8 ''Chemistry''] by Steven S. Zumdahl, Susan A. Zumda
  2 KB (258 words) - 19:31, 8 March 2023
• United States of America Mathematical Talent Search
  ...f|difficulty=3-6|breakdown=<u>Problem 1-2</u>: 3-4<br><u>Problem 3-5</u>: 5-6}} ...ed to high scorers at the end of the year. These typically include a free t-shirt, along with other prizes like books or software of the participant's c
  4 KB (613 words) - 13:08, 18 July 2023
• 1955 AHSME Problems/Problem 38
  ...ogether, we get: <math>2(a+b+c+d)=90</math>. This means that <math>a+b+c+d=45</math>.
  1 KB (200 words) - 23:35, 28 August 2020
• Joining an ARML team
  ...as well as practices of previous years' team rounds. Please email Xinke Guo-Xue at xinkeguoxue@gmail.com, or message Xinke's AoPS account "hurdler", if ...room 2112, on Thursdays at 7pm. For more information, e-mail Eric Brattain-Morrin at [mailto:eric.brattain@gmail.com eric.brattain@gmail.com] and visit
  21 KB (3,500 words) - 18:41, 23 April 2024
• Factorial
  ...[[recursion|recursive definition]] for the factorial is <math>n!=n \cdot (n-1)!</math>. * <math>45! = 119622220865480194561963161495657715064383733760000000000</math>
  10 KB (809 words) - 16:40, 17 March 2024
• North Carolina MathCounts
  ...Eli Park (20), Brian Zhang, Sukrith Velmineti, Eric Wu, Coach: Ann Chapoton-Genna ...rry Zhao (13), Benjamin Wu (51), Jeffrey Liu, David Li, Coach: Ann Chapoton-Genna
  4 KB (582 words) - 21:40, 14 May 2024
• Constructive counting
  ''How many four-digit numbers are there?'' '''Solution''': We can construct a four-digit by picking the first digit, then the second, and so on until the fourt
  12 KB (1,896 words) - 23:55, 27 December 2023
• Complementary counting
  ...see that there are <math>19</math> of them. Thus, our answer is is <math>99-19 = 80</math>. <math>\square</math> ...2006 AMC 10A Problems/Problem 21 | 2006 AMC 10A Problem 21]]: How many four-digit positive integers have at least one digit that is a 2 or a 3?''
  8 KB (1,192 words) - 17:20, 16 June 2023
• Composite number
  ...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 50 51 52 54 55 56 57 58 60 62 63 64 65 66 68 69 70 72 74 75 76 77
  6 KB (350 words) - 12:58, 26 September 2023
• Multiple
  ...umber 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
• Math Day at the Beach
  This round lasts 45 minutes and consists of 15 multiple-choice questions. Scoring consists of: This round lasts for 25 minutes and consists of 5 short-answer questions. Your score is 5 times the number of correct answers, for a
  4 KB (644 words) - 12:56, 29 March 2017
• 2011 AMC 10B Problems/Problem 22
  ...f the triangle is the diagonal of the pyramid's base. This is a <math>45-45-90</math> triangle. Also, we can let the dimensions of the rectangle be <mat ...because the rectangle is perpendicular to the base, and they share a <math>45^\circ</math> angle with the larger triangle. Therefore, the legs of the rig
  4 KB (691 words) - 18:38, 19 September 2021
• 2011 AMC 10B Problems/Problem 4
  LeRoy and Bernardo went on a week-long trip together and agreed to share the costs equally. Over the week, eac ...nardo half of the difference, which is <math>\boxed{\textbf{(C) } \;\frac{B-A}{2}}</math>
  1 KB (249 words) - 13:05, 24 January 2024
• Mathematical Olympiad Summer Program
  The '''Mathematical Olympiad Program''' (abbreviated '''MOP''') is a 3-week intensive problem solving camp held at the Carnegie Mellon University t The other important purpose of MOP is to train younger students in Olympiad-level problem solving and broaden their mathematical horizons.
  6 KB (936 words) - 10:37, 27 November 2023
• Right triangle
  ...ies. One of these is the [[isosceles triangle|isosceles]] <math>45^{\circ}-45^{\circ}-90^{\circ}</math> triangle, where the hypotenuse is equal to <math> label("$45^{\circ}$", A, 6*dir(290));
  3 KB (499 words) - 23:41, 11 June 2022
• 2006 AIME I Problems/Problem 15
  Given that a sequence satisfies <math> x_0=0 </math> and <math> |x_k|=|x_{k-1}+3| </math> for all integers <math> k\ge 1, </math> find the minimum possi ...2006\implies (k + 1)^2 = 2007</math> and <math>44^2 = 1936 < 2007 < 2025 = 45^2</math>. Plugging in <math>k = 44</math> yields <math>(3/2)(2025 - 2007) =
  6 KB (910 words) - 19:31, 24 October 2023
• 2006 AIME I Problems/Problem 8
  ...B=(4.2,0), C=(5.85,-1.6), D=(4.2,-3.2), EE=(0,-3.2), F=(-1.65,-1.6), G=(0.45,-1.6), H=(3.75,-1.6), I=(2.1,0), J=(2.1,-3.2), K=(2.1,-1.6); ...ath>44.5^2</math> is also less than <math>2006</math>, so we have numbers 1-44, times two because 0.5 can be added to each of time, plus 1, because 0.5
  5 KB (730 words) - 15:05, 15 January 2024
• 2006 AIME I Problems/Problem 6
  ...e digits, 0 through 9, is 45. So the sum of all the numbers is <math>\frac{45\times72\times111}{999}= \boxed{360} </math>. {{AIME box|year=2006|n=I|num-b=5|num-a=7}}
  2 KB (237 words) - 19:14, 20 November 2023
• 2006 AIME I Problems/Problem 2
  ...e possible values of S, so the number of possible values of S is <math>4995-4095+1=901</math>. ...math>, so the number of possible values of T, and therefore S, is <math>955-55+1=\boxed{901}</math>.
  1 KB (189 words) - 20:05, 4 July 2013
• 2021 JMC 10
  <math>\textbf{(A) }\pi-e \qquad\textbf{(B) }2\pi-2e\qquad\textbf{(C) }2e\qquad\textbf{(D) }2\pi \qquad\textbf{(E) }2\pi +e</m ...h>74</math> and <math>83</math> are pretentious. How many pretentious three-digit numbers are odd?
  12 KB (1,784 words) - 16:49, 1 April 2021
• 2006 AMC 12B Problems
  ...1.99</math>, and <math>\textdollar0.99</math>. Mary will pay with a twenty-dollar bill. Which of the following is closest to the percentage of the <ma How many even three-digit integers have the property that their digits, read left to right, are
  13 KB (2,058 words) - 12:36, 4 July 2023
• 2005 AMC 12A Problems
  ...they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike's rate. How many miles had Mike ridden when they met? ...on all six faces and then cut into <math>n^3</math> unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is <math
  13 KB (1,971 words) - 13:03, 19 February 2020
• 2002 AMC 12A Problems
  <math>(2x+3)(x-4)+(2x+3)(x-6)=0 </math> ..., and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.
  12 KB (1,792 words) - 13:06, 19 February 2020
• 2001 AMC 12 Problems
  two-digit number such that <math>N = P(N)+S(N)</math>. What is the units digit o A telephone number has the form <math>\text{ABC-DEF-GHIJ}</math>, where each letter represents
  13 KB (1,957 words) - 12:53, 24 January 2024
• 2002 AMC 12B Problems
  The positive integers <math>A, B, A-B, </math> and <math>A+B</math> are all prime numbers. The sum of these four For how many integers <math>n</math> is <math>\dfrac n{20-n}</math> the square of an integer?
  10 KB (1,547 words) - 04:20, 9 October 2022
• 2004 AMC 12B Problems
  In the expression <math>c\cdot a^b-d</math>, the values of <math>a</math>, <math>b</math>, <math>c</math>, and Minneapolis-St. Paul International Airport is 8 miles southwest of downtown St. Paul and
  13 KB (2,049 words) - 13:03, 19 February 2020
• 2006 AMC 12B Problems/Problem 10
  \text {(A) } 43 \qquad \text {(B) } 44 \qquad \text {(C) } 45 \qquad \text {(D) } 46 \qquad \text {(E) } 47 {{AMC12 box|year=2006|ab=B|num-b=9|num-a=11}}
  977 bytes (156 words) - 13:57, 19 January 2021
• 2006 AMC 12B Problems/Problem 23
  ...arrow 49 = 36 + s^2 - 12s\cos(\alpha) \Rightarrow \cos(\alpha) = \dfrac{s^2-13}{12s}. ...rrow 121 = 36 + s^2 - 12s\sin(\alpha) \Rightarrow \sin(\alpha) = \dfrac{s^2-85}{12s}.
  7 KB (1,169 words) - 14:04, 10 June 2022
• 2006 AMC 12A Problems/Problem 17
  But it can also be seen that <math>\angle BDA = 45^\circ</math>. Therefore, since <math>D</math> lies on <math>\overline{BE}</ ...\circ.</math> Also, <math>ED = EG,</math> which implies <math>\angle EGD = 45^\circ</math>, so <math>\triangle EDG</math> is an isosceles right triangle.
  6 KB (958 words) - 23:29, 28 September 2023
• 2006 AMC 12A Problems/Problem 22
  Project any two non-adjacent and non-opposite sides of the [[hexagon]] to the [[circle]]; the [[arc]] between the [[Image:2006_12A_AMC-22.png]]
  2 KB (343 words) - 15:39, 14 June 2023
• 2005 AMC 12B Problems/Problem 15
  The sum of four two-digit numbers is <math>221</math>. None of the eight digits is <math>0</math <math>221</math> can be written as the sum of four two-digit numbers, let's say <math>\overline{ae}</math>, <math>\overline{bf}</ma
  2 KB (411 words) - 21:02, 21 December 2020
• 2005 AMC 12B Problems/Problem 18
  ..., which is <math>\frac{14^2}{2}-\frac{4^2}{2}-(\frac{5\sqrt{2}}{2})^2\pi=98-8-\frac{25\pi}{2}</math>, which is approximately <math>51</math>. The answer {{AMC12 box|year=2005|ab=B|num-b=17|num-a=19}}
  2 KB (262 words) - 21:20, 21 December 2020
• 2006 AMC 10A Problems
  Define <math>x\otimes y=x^3-y</math>. What is <math>h\otimes (h\otimes h)</math>? Doug and Dave shared a pizza with 8 equally-sized slices. Doug wanted a plain pizza, but Dave wanted anchovies on half
  13 KB (2,028 words) - 16:32, 22 March 2022
• 2006 AMC 10A Problems/Problem 15
  <math>\textbf{(A) } 29\qquad\textbf{(B) } 42\qquad\textbf{(C) } 45\qquad\textbf{(D) } 47\qquad\textbf{(E) } 50\qquad</math> ...et exactly twice per lap (once at the starting point, the other at the half-way point). Thus, there are <math>\frac{30}{\frac{2\pi}{5}} \approx 23.8</ma
  3 KB (532 words) - 17:49, 13 August 2023
• 2006 AMC 10A Problems/Problem 17
  ...>BZ = \frac 12AH = 1</math>, so <math>\triangle BWZ</math> is a <math>45-45-90 \triangle</math>. Hence <math>WZ = \frac{1}{\sqrt{2}}</math>, and <math>[ ...ways to come to the same conclusion using different [[right triangle|45-45-90 triangles]].
  6 KB (1,066 words) - 00:21, 2 February 2023
• Newman's Tauberian Theorem
  (which is well-defined by this formula for <math>\Re s>0</math>) admits an to some [[open set | open]] domain <math>E</math> containing the closed half-plane
  6 KB (1,034 words) - 07:55, 12 August 2019
• 2005 AIME II Problems/Problem 5
  ...e case <math> b=a^2 </math>, note that <math> 44^2=1936 </math> and <math> 45^2=2025 </math>. Thus, all values of <math>a</math> from <math>2</math> to < ...re <math> 44-2+1=43 </math> possibilities for the square case and <math> 12-2+1=11 </math> possibilities for the cube case. Thus, the answer is <math> 4
  3 KB (547 words) - 19:15, 4 April 2024
• 2005 AIME II Problems
  ...math> a_{k+1} = a_{k-1} - \frac 3{a_k} </math> for <math> k = 1,2,\ldots, m-1. </math> Find <math> m. </math> ...> and <math> E </math> between <math> A </math> and <math> F, m\angle EOF =45^\circ, </math> and <math> EF=400. </math> Given that <math> BF=p+q\sqrt{r},
  7 KB (1,119 words) - 21:12, 28 February 2020
• 2005 AIME II Problems/Problem 12
  ...> and <math> E </math> between <math> A </math> and <math> F, m\angle EOF =45^\circ, </math> and <math> EF=400. </math> Given that <math> BF=p+q\sqrt{r}, ...0)); pair A=(0,9), B=(9,9), C=(9,0), D=(0,0), E=(2.5-0.5*sqrt(7),9), F=(6.5-0.5*sqrt(7),9), G=(4.5,9), O=(4.5,4.5); draw(A--B--C--D--A);draw(E--O--F);dr
  13 KB (2,080 words) - 21:20, 11 December 2022
• 2005 AIME I Problems/Problem 6
  ...math> P </math> be the product of the nonreal roots of <math> x^4-4x^3+6x^2-4x=2005. </math> Find <math> \lfloor P\rfloor. </math> The left-hand side of that [[equation]] is nearly equal to <math>(x - 1)^4</math>. T
  4 KB (686 words) - 01:55, 5 December 2022
• 2005 AIME I Problems/Problem 11
  ...semicircle is tangent to only one side of the square, we will have "wiggle-room" to increase its size. Once it is tangent to two adjacent sides of the We can just look at a quarter circle inscribed in a <math>45-45-90</math> right triangle. We can then extend a radius, <math>r</math> to one
  4 KB (707 words) - 11:11, 16 September 2021
• 2005 AIME I Problems/Problem 12
  ...05 </math> with <math> S(n) </math> [[even integer | even]]. Find <math> |a-b|. </math> It is well-known that <math>\tau(n)</math> is odd if and only if <math>n</math> is a [[
  4 KB (647 words) - 02:29, 4 May 2021
• 2004 AIME II Problems/Problem 15
  <cmath>s_{82, 9} = 2s_{82, 8} = 4s_{82, 7} = 8s_{127 - 82, 6} = 8s_{45, 6}</cmath> <cmath>s_{45, 6} = 2s_{63 - 45, 5} + 1 = 2s_{18, 5} + 1 = 4s_{31 - 18, 4} + 1 = 4s_{13, 4} + 1</cmath>
  6 KB (899 words) - 20:58, 12 May 2022
• 2004 AIME II Problems/Problem 13
  ...CR(E, 45/7)), A=D+ (5+(75/7))/(75/7) * (F-D), C = E+ (3+(45/7))/(45/7) * (F-E), B=IP(CR(A,3), CR(C,5)); == Additional Trigonometry-Free Alternative ==
  3 KB (486 words) - 22:15, 7 April 2023
• 1983 AIME Problems
  Let <math>f(x)=|x-p|+|x-15|+|x-p-15|</math>, where <math>0 < p < 15</math>. Determine the [[minimum]] value t ...the real roots of the equation <math>x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}</math>?
  7 KB (1,104 words) - 12:53, 6 July 2022
• 1990 AIME Problems
  Find the value of <math>(52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2}</math>. <center><math>\frac 1{x^2-10x-29}+\frac1{x^2-10x-45}-\frac 2{x^2-10x-69}=0</math></center>
  6 KB (870 words) - 10:14, 19 June 2021
• 1995 AIME Problems
  ...h> and <math>n</math> are relatively prime positive integers. Find <math>m-n.</math> ...h>d_{},</math> the equation <math>x^4+ax^3+bx^2+cx+d=0</math> has four non-real roots. The product of two of these roots is <math>13+i</math> and the
  6 KB (1,000 words) - 00:25, 27 March 2024
• 1999 AIME Problems
  Consider the parallelogram with vertices <math>(10,45),</math> <math>(10,114),</math> <math>(28,153),</math> and <math>(28,84).</ Find the sum of all positive integers <math>n</math> for which <math>n^2-19n+99</math> is a perfect square.
  7 KB (1,094 words) - 13:39, 16 August 2020
• 2000 AIME I Problems
  ...is, and let <math>E</math> be the reflection of <math>D</math> across the y-axis. The area of pentagon <math>ABCDE</math> is <math>451</math>. Find <mat The diagram shows a rectangle that has been dissected into nine non-overlapping squares. Given that the width and the height of the rectangle ar
  7 KB (1,204 words) - 03:40, 4 January 2023
• 2001 AIME I Problems
  Find the sum of all positive two-digit integers that are divisible by each of their digits. ...the roots, real and non-real, of the equation <math>x^{2001}+\left(\frac 12-x\right)^{2001}=0</math>, given that there are no multiple roots.
  7 KB (1,212 words) - 22:16, 17 December 2023
• 2000 AIME II Problems
  ...ht distinguishable rings, let <math>n</math> be the number of possible five-ring arrangements on the four fingers (not the thumb) of one hand. The order The equation <math>2000x^6+100x^5+10x^3+x-2=0</math> has exactly two real roots, one of which is <math>\frac{m+\sqrt{n
  6 KB (947 words) - 21:11, 19 February 2019
• 2003 AIME II Problems
  ...th>C</math> is never immediately followed by <math>A</math>. How many seven-letter good words are there? ...to the axis of the cylinder, and the plane of the second cut forms a <math>45^\circ</math> angle with the plane of the first cut. The intersection of the
  7 KB (1,127 words) - 09:02, 11 July 2023
• 1983 AIME Problems/Problem 3
  ...] [[root]]s of the [[equation]] <math>x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}</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>, we get <math>-6=6</ma
  3 KB (532 words) - 05:18, 21 July 2022
• 1983 AIME Problems/Problem 4
  A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of A=r*dir(45),B=(A.x,A.y-r);
  11 KB (1,741 words) - 22:40, 23 November 2023
• 1984 AIME Problems/Problem 14
  .../math> and <math>29</math>, yielding a maximal answer of 38. Since <math>38-25=13</math>, which is prime, the answer is <math>\boxed{038}</math>. ...could possibly work by Chicken McNugget is <math>9 \cdot 25 - 9 - 25 = 225-34 = 191</math>. We then bash from top to bottom:
  8 KB (1,346 words) - 01:16, 9 January 2024
• 1984 AIME Problems/Problem 15
  ...h> \frac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}+\frac{z^2}{8^2-5^2}+\frac{w^2}{8^2-7^2}=1 </math></div> ...frac{x^{2}}{t-1}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1.</cmath>
  6 KB (1,051 words) - 04:52, 8 May 2024
• 1985 AIME Problems/Problem 14
  ...> games and so earned 45 points playing each other. Then they also earned 45 points playing against the stronger <math>n</math> players. Since every po ...145=15</math> playing against the weakest <math>10</math> who gained <math>45</math> points vs them, which is a contradiction since it must be larger. Th
  5 KB (772 words) - 22:14, 18 June 2020
• 1986 AIME Problems/Problem 14
  ...{h^2l^2}{h^2 + l^2} = \frac {1}{\frac {1}{h^2} + \frac {1}{l^2}} = \frac {45}{2}</cmath> <cmath>\frac {1}{l^2} + \frac {1}{w^2} = \frac {45}{900}</cmath>
  2 KB (346 words) - 13:13, 22 July 2020
• 1987 AIME Problems/Problem 10
  Similarly, Al will take <math>\frac{x}{3b-e}=\frac{150}{e}</math> time to get to the bottom. ...{ex}{3b-e}=2\cdot75=\frac{2ex}{b+e}=\frac{6ex}{3b+3e}=\frac{(ex)-(6ex)}{(3b-e)-(3b+3e)}=\frac{5x}{4}</math>
  7 KB (1,187 words) - 16:21, 27 January 2024
• 1987 AIME Problems/Problem 7
  <math>1000 = 2^35^3</math> and <math>2000 = 2^45^3</math>. By [[LCM#Using prime factorization|looking at the prime factoriza {{AIME box|year=1987|num-b=6|num-a=8}}
  3 KB (547 words) - 22:54, 4 April 2016
• 1989 AIME Problems/Problem 15
  real x = 0.4, y = 0.2, z = 1-x-y; <math>PDR</math> is a <math>3-4-5</math> [[right triangle]], so <math>\angle PDR</math> (<math>\angle ADQ</m
  13 KB (2,091 words) - 00:20, 26 October 2023
• 1989 AIME Problems/Problem 12
  ...",C,left,p); dot("$D$",D,up,p); dot("$M$",P,dir(-45),p); dot("$N$",Q,0.2*(Q-P),p); ...h>. The formula for the length of a median is <math>m=\sqrt{\frac{2a^2+2b^2-c^2}{4}}</math>, where <math>a</math>, <math>b</math>, and <math>c</math> ar
  2 KB (376 words) - 13:49, 1 August 2022
• 1989 AIME Problems/Problem 2
  ...e 2} = 45</math> have 2 members. Thus the answer is <math>1024 - 1 - 10 - 45 = \boxed{968}</math>. {{AIME box|year=1989|num-b=1|num-a=3}}
  911 bytes (135 words) - 08:30, 27 October 2018
• 1990 AIME Problems/Problem 12
  ...]] 12. The [[sum]] of the lengths of all sides and [[diagonal]]s of the 12-gon can be written in the form <math>a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6 [[Image:1990 AIME-12.png]]
  6 KB (906 words) - 13:23, 5 September 2021
• 1990 AIME Problems/Problem 6
  First, we notice that there are 45 tags left, after 25% of the original fish have went away/died. Then, some < ...math>\frac{3}{70}</math> of the fish in September were tagged, <math>\frac{45}{5n/4} = \frac{3}{70}</math>, where <math>n</math> is the number of fish in
  2 KB (325 words) - 13:16, 26 June 2022
• 1990 AIME Problems/Problem 5
  ...Therefore, <math>n = 2^43^45^2</math> and <math>\frac{n}{75} = \frac{2^43^45^2}{3 \cdot 5^2} = 16 \cdot 27 = \boxed{432}</math>. {{AIME box|year=1990|num-b=4|num-a=6}}
  1 KB (175 words) - 03:45, 21 January 2023
• 1990 AIME Problems/Problem 4
  <center><math>\frac 1{x^2-10x-29}+\frac1{x^2-10x-45}-\frac 2{x^2-10x-69}=0</math></center> Simplifying, <math>-64a + 40 \times 16 = 0</math>, so <math>a = 10</math>. Re-substituting, <math>10 = x^2 - 10x - 29 \Longleftrightarrow 0 = (x - 13)(x +
  1 KB (156 words) - 07:35, 4 November 2022
• 2006 AMC 10B Problems/Problem 10
  <math> \textbf{(A) } 43\qquad \textbf{(B) } 44\qquad \textbf{(C) } 45\qquad \textbf{(D) } 46\qquad \textbf{(E) } 47 </math> {{AMC10 box|year=2006|ab=B|num-b=9|num-a=11}}
  900 bytes (132 words) - 13:57, 26 January 2022
• 1992 AIME Problems/Problem 14
  ...ath>y=z=1</math> we get <math>\frac{2}{x}+2(x+1)=92 \implies x+\frac{1}{x}=45</math>, and so <math>\frac{2}{x}(x+1)^2=2(x+\frac{1}{x}+2)=2 \cdot 47 = 94< {{AIME box|year=1992|num-b=13|num-a=15}}
  4 KB (667 words) - 01:26, 16 August 2023
• 1993 AIME Problems/Problem 6
  ...ast possible value of <math>b = 45</math>, so the answer is <math>10(45) + 45 = 495</math>. <math>n = 9a + 36 = 10b + 45 = 11c + 55</math>
  3 KB (524 words) - 18:06, 9 December 2023
• 1993 AIME Problems/Problem 2
  ...two distances evaluate to <math>8(45) + 10\cdot 4 = 400</math> and <math>8(45) + 10\cdot 6 = 420</math>. By the [[Pythagorean Theorem]], the answer is <m {{AIME box|year=1993|num-b=1|num-a=3}}
  2 KB (241 words) - 11:56, 13 March 2015
• 1994 AIME Problems/Problem 5
  ...integer <math>n\,</math>, let <math>p(n)\,</math> be the product of the non-zero digits of <math>n\,</math>. (If <math>n\,</math> has only one digits, ...\equiv 005</math>), and since our <math>p(n)</math> ignores all of the zero-digits, replace all of the <math>0</math>s with <math>1</math>s. Now note th
  2 KB (275 words) - 19:27, 4 July 2013
• 1995 AIME Problems/Problem 12
  ...\overline{OC},</math> and <math>\overline{OD},</math> and <math>\angle AOB=45^\circ.</math> Let <math>\theta</math> be the measure of the dihedral angle ...>AP = 1.</math> It follows that <math>\triangle OPA</math> is a <math>45-45-90</math> [[right triangle]], so <math>OP = AP = 1,</math> <math>OB = OA = \
  8 KB (1,172 words) - 21:57, 22 September 2022
• 1995 AIME Problems/Problem 10
  3&45&&&& \\ ...ath> 0 \leq b < 42 </math>. Then note that <math> b, b + 42, ... , b + 42(a-1) </math> are all primes.
  3 KB (436 words) - 19:26, 2 September 2023
• 1996 AIME Problems/Problem 12
  <cmath>|a_1-a_2|+|a_3-a_4|+|a_5-a_6|+|a_7-a_8|+|a_9-a_{10}|.</cmath> ...obtained from these paired sequences are also obtained in another <math>2^5-1</math> ways by permuting the adjacent terms <math>\{a_1,a_2\},\{a_3,a_4\},
  5 KB (879 words) - 11:23, 5 September 2021
• 1996 AIME Problems/Problem 10
  ...h> <math>\dfrac{2\sin{141^{\circ}}\cos{45^{\circ}}}{2\cos{141^{\circ}}\sin{45^{\circ}}} = \tan{141^{\circ}}</math>. ...]] function is <math>180^\circ</math>, and the tangent function is [[one-to-one]] over each period of its domain.
  4 KB (503 words) - 15:46, 3 August 2022
• 1996 AIME Problems/Problem 3
  ...itive [[integer]] <math>n</math> for which the expansion of <math>(xy-3x+7y-21)^n</math>, after like terms have been collected, has at least 1996 terms. ...fter <math>1996</math> is <math>2025 = 45^2</math>, so our answer is <math>45 - 1 = \boxed{044}</math>.
  3 KB (515 words) - 04:29, 27 November 2023
• 1997 AIME Problems/Problem 11
  ...la <math>\cos x + \cos y = 2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)</math> ...+\cos(\frac{41}{2})+\dots+\cos(\frac{1}{2})]} \Rightarrow \frac{\cos(\frac{45}{2})}{\cos(\frac{135}{2})}</cmath>
  10 KB (1,514 words) - 14:35, 29 March 2024
• 1997 AIME Problems/Problem 4
  [[Image:1997_AIME-4.png]] \sqrt{10r+r^2}&=& 4-r\\
  2 KB (354 words) - 22:33, 2 February 2021
• 1998 AIME Problems/Problem 11
  ...font-size:100%">"For non-asymptote version of image, see [[:Image:1998_AIME-11.png]]"</span> ...[hypotenuse]]s are <math>5\sqrt{5}</math>. The other two are of <math>45-45-90 \triangle</math>s with legs of length 15, so their hypotenuses are <math>
  7 KB (1,084 words) - 11:48, 13 August 2023
• 1998 AIME Problems/Problem 10
  [[Image:1998_AIME-10a.png|450px]] [[Image:1998_AIME-10b.png|450px]]
  3 KB (496 words) - 13:02, 5 August 2019
• 1998 AIME Problems/Problem 9
  ...inutes past 9 a.m.). The two mathematicians meet each other when <math>|M_1-M_2| \leq m</math>. Also because the mathematicians arrive between 9 and 10, real m=60-12*sqrt(15);
  4 KB (624 words) - 18:34, 18 February 2018
• 1999 AIME Problems/Problem 10
  ...lso be picked. Since the triangle accounts for 3 segments, there are <math>45 - 3 = 42</math> segments remaining. ...se3} \cdot 42}{{45\choose4}} = \frac{10 \cdot 9 \cdot 8 \cdot 42 \cdot 4!}{45 \cdot 44 \cdot 43 \cdot 42 \cdot 3!} = \frac{16}{473}</math>. The solution
  3 KB (524 words) - 17:25, 17 July 2023
• 1999 AIME Problems/Problem 4
  pair W=dir(225), X=dir(315), Y=dir(45), Z=dir(135), O=origin; {{AIME box|year=1999|num-b=3|num-a=5}}
  3 KB (398 words) - 13:27, 12 December 2020
• 1999 AIME Problems/Problem 2
  Consider the [[parallelogram]] with [[vertex|vertices]] <math>(10,45)</math>, <math>(10,114)</math>, <math>(28,153)</math>, and <math>(28,84)</m ...The slope of the line (since it passes through the origin) is <math>\frac{45 + \frac{135}{19}}{10} = \frac{99}{19}</math>, and the solution is <math>m +
  3 KB (423 words) - 11:06, 27 April 2023
• 1999 AIME Problems/Problem 8
  [[Image:1999_AIME-8.png]] [[Image:1999_AIME-8a.png]]
  3 KB (445 words) - 19:40, 4 July 2013
• 2000 AIME I Problems/Problem 4
  The diagram shows a [[rectangle]] that has been dissected into nine non-overlapping [[square]]s. Given that the width and the height of the rectangl draw((34,36)--(34,45)--(25,45));
  3 KB (485 words) - 00:31, 19 January 2024
• 2001 AIME I Problems/Problem 9
  ...=(0,0),B=(13,0),C=IP(CR(A,17),CR(B,15)), D=A+p*(B-A), E=B+q*(C-B), F=C+r*(A-C); ...ot \sin \angle CAB}{\frac 12 \cdot AB \cdot AC \cdot \sin \angle CAB} = p(1-r)
  4 KB (673 words) - 20:15, 21 February 2024
• 2001 AIME I Problems/Problem 6
  Recast the problem entirely as a block-walking problem. Call the respective dice <math>a, b, c, d</math>. In the ...combination]] of four numbers, there is only one way to order them in a non-decreasing order. It suffices now to find the number of combinations for fou
  11 KB (1,729 words) - 20:50, 28 November 2023
• 2001 AIME I Problems/Problem 4
  ...math>A</math> and <math>B</math> measure <math>60</math> degrees and <math>45</math> degrees, respectively. The bisector of angle <math>A</math> intersec <math>\angle ABC=45^{\circ}</math> and <math>\angle BAC=60^{\circ}</math>, so <math>BC = 12\sqr
  3 KB (534 words) - 03:22, 23 January 2023
• 2001 AIME I Problems/Problem 1
  Find the sum of all positive two-digit integers that are divisible by each of their digits. ...r expand on solution 2, it would be tedious to test all <math>90</math> two-digit numbers. We can reduce the amount to look at by focusing on the tens d
  4 KB (687 words) - 18:37, 27 November 2022
• 2002 AIME I Problems/Problem 14
  ...>. After any element <math>x</math> is removed, we are given that <math>n|N-x</math>, so <math>x\equiv N\pmod{n}</math>. Since <math>1\in\mathcal{S}</ma ...02</math>, so <math>n \leq 44</math>. The largest factor of 2001 less than 45 is 29, so <math>n=29</math> and <math>n+1</math> <math>\Rightarrow{\fbox{30
  2 KB (267 words) - 19:18, 21 June 2021
• 2002 AIME I Problems/Problem 1
  ...ree-digit arrangement that reads the same left-to-right as it does right-to-left) is <math>\dfrac{m}{n}</math>, where <math>m</math> and <math>n</math> ...larly, there is a <math>\frac 1{26}</math> probability of picking the three-letter palindrome.
  3 KB (369 words) - 23:36, 6 January 2024
• 2003 AIME I Problems/Problem 11
  ...ility is symmetric around <math>45^\circ</math>. Thus, take <math>0 < x < 45</math> so that <math>\sin x < \cos x</math>. Then <math>\cos^2 x</math> is ...math>\cos 2x > \frac 12 \sin 2x</math>. Since we've chosen <math>x \in (0, 45)</math>, <math>\cos 2x > 0</math> so
  2 KB (284 words) - 13:42, 10 October 2020
• 2003 AIME II Problems/Problem 15
  ...that <math> \left( x^{23} + x^{22} + \cdots + x^2 + x + 1 \right) \cdot (x-1) = x^{24} - 1 </math>. The five-element sum is just <math>\sin 30^\circ + \sin 60^\circ + \sin 90^\circ + \s
  4 KB (675 words) - 17:23, 30 July 2022
• 2003 AIME II Problems/Problem 13
  ...ing the starting vertex in the next move. Thus <math>P_n=\frac{1}{2}(1-P_{n-1})</math>. ...tex is <math>{10 \choose 5} + {10 \choose 8} + {10 \choose 2} = 252 + 45 + 45 = 342</math>. Since the ant has two possible steps at each point, there ar
  15 KB (2,406 words) - 23:56, 23 November 2023
• 2003 AIME II Problems/Problem 5
  ...to the axis of the cylinder, and the plane of the second cut forms a <math>45^\circ</math> angle with the plane of the first cut. The intersection of the {{AIME box|year=2003|n=II|num-b=4|num-a=6}}
  1 KB (204 words) - 17:41, 30 July 2022
• 2002 AIME II Problems/Problem 8
  <cmath>\left\lfloor\frac{2002}{45}\right\rfloor=44 </cmath> <cmath>\left\lfloor\frac{2002}{44}\right\rfloor=45 </cmath>
  6 KB (908 words) - 14:22, 14 July 2023
• 2001 AIME II Problems/Problem 7
  pair P = (0,0), Q = (90, 0), R = (0, 120), S=(0, 60), T=(45, 60), U = (60,0), V=(60, 40), O1 = (30,30), O2 = (15, 75), O3 = (70, 10); ...h>. Or, the inradius could be directly by using the formula <math>\frac{a+b-c}{2}</math>, where <math>a</math> and <math>b</math> are the legs of the ri
  7 KB (1,112 words) - 02:15, 26 December 2022
• 2000 AIME II Problems/Problem 15
  ...least positive integer <math>n</math> such that <center><math>\frac 1{\sin 45^\circ\sin 46^\circ}+\frac 1{\sin 47^\circ\sin 48^\circ}+\cdots+\frac 1{\sin ...um_{k=23}^{67} \frac{1}{\sin (2k-1) \sin 2k} = \frac{1}{\sin 1} \left(\cot 45 - \cot 46 + \cot 47 - \cdots + \cot 133 - \cot 134 \right).</cmath>
  3 KB (469 words) - 21:14, 7 July 2022
• 2000 AIME II Problems/Problem 10
  ...cmath>\frac{\tfrac{45}{r}}{1-19\cdot\tfrac{26}{r^2}}+\frac{\tfrac{60}{r}}{1-37\cdot\tfrac{23}{r^2}}=0.</cmath> {{AIME box|year=2000|n=II|num-b=9|num-a=11}}
  2 KB (399 words) - 17:37, 2 January 2024
• 2000 AIME II Problems/Problem 4
  ...43</math>, while the smallest example of the latter is <math>3^2 \cdot 5 = 45</math>. ...math>2</math> with <math>\frac{18}{6} = 3</math> factors, namely <math>2^{3-1} = 4</math>. Thus, our answer is <math>2^2 \cdot 3^2 \cdot 5 = \boxed{180}
  2 KB (397 words) - 15:55, 11 May 2022
• 2005 AMC 10B Problems/Problem 21
  ...imes 3 \times 4 \times 3</math> ways to pick the slips, so <math>q = \frac{45 \cdot 6 \cdot 4^2 \cdot 3^2}{40 \cdot 39 \cdot 38 \cdot 37}</math>. Hence, the answer is <math>\frac{q}{p} = \frac{45 \cdot 6 \cdot 4^2 \cdot 3^2}{10\cdot 4!} = \boxed{\textbf{(A) }162}</math>.
  3 KB (398 words) - 19:17, 17 September 2023
• 2005 AMC 10B Problems/Problem 24
  Let <math>x</math> and <math>y</math> be two-digit integers such that <math>y</math> is obtained by reversing the digits Then, <math>x^2 - y^2 = (x-y)(x+y) = (9a - 9b)(11a + 11b) = 99(a-b)(a+b) = m^2</math>.
  5 KB (845 words) - 19:23, 17 September 2023
• 2007 iTest
  **[[2007 iTest Problems/Problem 45|Problem 45]]
  3 KB (305 words) - 15:10, 5 November 2023
• 2006 AMC 10B Problems
  ...</math> and <math>y</math>, define <math> x \mathop{\spadesuit} y = (x+y)(x-y) </math>. What is <math> 3 \mathop{\spadesuit} (4 \mathop{\spadesuit} 5) < ...rcle is painted blue. What is the ratio of the blue-painted area to the red-painted area?
  14 KB (2,059 words) - 01:17, 30 January 2024
• 2000 AMC 12 Problems/Problem 8
  <cmath>S_{100}=\frac{100[2\cdot 4+(100-1)4]}{2}</cmath> ...se Figure 0 exists) <math>\dbinom{101-1}{0}+4\dbinom{101-1}{1}+4\dbinom{101-1}{2}=20201</math> or <math>\textbf{(C)}</math>
  8 KB (1,202 words) - 16:17, 10 May 2024
• Mock AIME 1 2006-2007 Problems/Problem 6
  ...<math>P_{1}: y=x^{2}+\frac{101}{100}</math> and <math>P_{2}: x=y^{2}+\frac{45}{4}</math> be two [[parabola]]s in the [[Cartesian plane]]. Let <math>\math ...+ by = c</math> has a unique [[root]] so <math>b^2 - 4\cdot a \cdot (\frac{45}4a - c) = 0</math> or equivalently <math>b^2 - 45a^2 + 4ac = 0</math>. We
  3 KB (460 words) - 15:52, 3 April 2012
• Mock AIME 1 2006-2007 Problems/Problem 8
  ...th> be a point in <math>ABCDE</math> such that <math>\angle ABP=\angle BAP=45^\circ</math>. We see that <math>\angle CBP=115^\circ</math> and thus <math> ...e PBE=\frac{45}{2}</math>. <math>\angle ABE=\angle ABP+\angle PBE=45+\frac{45}{2}^\circ=\frac{135}{2}^\circ\implies m+n=\boxed{137}</math>.
  1 KB (244 words) - 14:54, 21 August 2020
• Mock AIME 1 2006-2007 Problems/Problem 9
  ...e sum <math>S = a_{10} + a_{11} + \ldots = 1 + r + r^2 +\ldots = \frac{1}{1-r}</math> has value <math>\frac 1{1 - \omega / 2}</math>. Different choices <math>\sum\frac1{1-z/2}</math> where <math>z^{10}=1</math>.
  5 KB (744 words) - 19:46, 20 October 2020
• Mock AIME 1 2006-2007/Problems
  [[Mock AIME 1 2006-2007 Problems/Problem 1|Solution]] [[Mock AIME 1 2006-2007 Problems/Problem 2|Solution]]
  8 KB (1,355 words) - 14:54, 21 August 2020
• 1959 IMO Problems/Problem 1
  ...simplifies to proving <math> \frac{7n+1}{14n+3} </math> irreducible. We re-write this fraction as: Trying 3, we end up with <math> \frac{67}{45} </math>. Again, both end up 1 away from being multiples of 22. This is whe
  5 KB (767 words) - 10:59, 23 July 2023
• 1959 IMO Problems/Problem 4
  ...he triangle as <math>a</math> and <math>b</math>. We also observe the well-known fact that in a right triangle, the median to the hypotenuse is of half ...desired length. Next draw the circular locus of points X such that MXO is 45 degrees. To accomplish this, simply find the point on the perpendicular bis
  6 KB (939 words) - 17:31, 15 July 2023
• 2006 Romanian NMO Problems
  In the acute-angle triangle <math>ABC</math> we have <math>\angle ACB = 45^\circ</math>. The points <math>A_1</math> and <math>B_1</math> are the feet We define a ''pseudo-inverse'' <math>B\in \mathcal M_n(\mathbb C)</math> of a matrix <math>A\in\
  11 KB (1,779 words) - 14:57, 7 May 2012
• 2006 Romanian NMO Problems/Grade 7/Problem 3
  In the acute-angle triangle <math>ABC</math> we have <math>\angle ACB = 45^\circ</math>. The points <math>A_1</math> and <math>B_1</math> are the feet ...ac{B_1C}{BC}</math>. However <math>\triangle{BB_1C}</math> is a <math>45-45-90</math> triangle, so <math>\dfrac{B_1C}{BC}=\dfrac{1}{\sqrt{2}}</math> and
  2 KB (294 words) - 22:24, 26 November 2023
• 1969 Canadian MO Problems/Problem 10
  Let <math>ABC</math> be the right-angled isosceles triangle whose equal sides have length 1. <math>P</math> is ...les]] [[right triangle]]s. Hence, <math>PQ=RC=x</math> and <math>QC=PR=BR=1-x.</math>
  1 KB (211 words) - 13:13, 30 December 2015
• 2005 AMC 10A Problems/Problem 14
  How many three-digit numbers satisfy the property that the middle digit is the average of t ...(B) } 42\qquad \textbf{(C) } 43\qquad \textbf{(D) } 44\qquad \textbf{(E) } 45 </math>
  4 KB (694 words) - 19:25, 13 December 2021
• 2005 AMC 10A Problems/Problem 16
  ...m the number. The units digit of the result is <math>6</math>. How many two-digit numbers have this property? So the numbers that have this property are <math>40, 41, 42, 43, 44, 45, 46, 47, 48, 49</math>.
  2 KB (279 words) - 11:57, 17 July 2023
• Orthocenter
  ...0), C=(10,0), A1 = (B+C)/2, O = circumcenter(A,B,C), G = (A+B+C)/3, H = 3*G-2*O; ...triangles <math>AGH</math>, <math>A'GO</math> are [[similar]] by side-angle-side similarity. It follows that <math>AH</math> is parallel to <math>OA'</
  5 KB (829 words) - 13:11, 20 February 2024
• MathPath
  During the years 2020 and 2021, MathPath was held virtual due to COVID-19. | 7:00 - 7:05 || Wake-up calls
  2 KB (266 words) - 21:38, 13 December 2023
• 2007 iTest Problems
  Compute the sum of all twenty-one terms of the geometric series <cmath>|x-y|=c</cmath>
  30 KB (4,794 words) - 23:00, 8 May 2024
• 2005 AMC 10A Problems
  <math> (a \star b) = \frac{a+b}{a-b} </math>. ...they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike's rate. How many miles had Mike ridden when they met?
  14 KB (2,026 words) - 11:45, 12 July 2021
• 2005 AMC 10A Problems/Problem 20
  ...s and the octagon, all inside a square. The right triangles are <math>45-45-90</math> triangles with hypotenuse <math>\frac{\sqrt{2}}{2}</math>, so the ...math>\sqrt{2}</math> and realise that it is the hypotenuse of a <math>45-45-90</math> triangle with side length ratios <math>1:1:\sqrt{2}</math>.).
  3 KB (414 words) - 10:25, 15 May 2023
• 1999 AMC 8 Problems/Problem 16
  ...her test overall, she needed to get <math>60\% \cdot 75 = 0.60 \cdot 75 = 45</math> questions right. Therefore, she needed to answer <math>45 - 40 = 5</math> more questions to pass, so the correct answer is <math>\box
  1 KB (167 words) - 20:30, 11 January 2024
• Pythagorean triple
  ...+ b^2 = c^2</math>. Pythagorean triples arise in [[geometry]] as the side-lengths of [[right triangle]]s. <math>(28, 45, 53)</math><nowiki>*</nowiki>
  4 KB (684 words) - 16:45, 1 August 2020
• 2020 AMC 10A Problems
  ...{(A) } 0 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 30 \qquad\textbf{(D) } 45 \qquad\textbf{(E) } 60</math> <cmath>\frac{a-3}{5-c} \cdot \frac{b-4}{3-a} \cdot \frac{c-5}{4-b}</cmath>
  13 KB (1,968 words) - 18:32, 29 February 2024
• Pigeonhole Principle/Solutions
  ...riends, then the remaining friends must have from <math>1</math> to <math>n-2</math> friends for the remaining friends not to also have no friends. By p ...same remainder when divided by 5. Since there are 5 possible remainders (0-4), by the pigeonhole principle, at least two of the integers must share the
  10 KB (1,617 words) - 01:34, 26 October 2021
• 2002 USAMO Problems/Problem 2
  ...\frac{s-a}{r} \right)^2 + 4\left( \frac{s-b}{r} \right)^2 + 9\left( \frac{s-c}{r} \right)^2 = \left( \frac{6s}{7r} \right)^2 \frac{(s-a)^2}{36} + \frac{(s-b)^2}{9} + \frac{(s-c)^2}{4} = \frac{s^2}{36 + 9 + 4}
  2 KB (298 words) - 22:32, 6 April 2016
• Mock AIME 3 Pre 2005 Problems/Problem 11
  ...anging variables gives <math>45x=8\cdot 15+bx=15^2\implies x=5</math>. Back-substitution yields <math>AC=21</math> and consequently <math>AB=24</math>. {{Mock AIME box|year=Pre 2005|n=3|num-b=10|num-a=12}}
  2 KB (325 words) - 15:32, 22 March 2015
• 2007 AIME I Problems/Problem 9
  [[Image:AIME I 2007-9.png]] ...{180 - \theta}{2}</math>. Apply the [[trigonometric identities|tangent half-angle formula]] (<math>\tan \frac{\theta}{2} = \sqrt{\frac{1 - \cos \theta}{
  11 KB (1,851 words) - 12:31, 21 December 2021
• 2007 AIME I Problems/Problem 11
  ...p)</math> over this range is <math>(2k)k=2k^2</math>. <math>44<\sqrt{2007}<45</math>, so all numbers <math>1</math> to <math>44</math> have their full ra ...re <math>2007 - 1981 + 1 = 27</math> (1 to be inclusive) of them. <math>27*45=1215</math>, and <math>215+740=
  3 KB (562 words) - 20:02, 30 December 2023
• 2007 AIME I Problems/Problem 12
  ...s</math>, where <math>p,q,r,s</math> are integers. Find <math>\frac{p-q+r-s}2</math>. ..."75^\circ",C,B,A,2); MA("30^\circ",A,C,B,4); MA("30^\circ",A,Cp,Bp,4); MA("45^\circ",A,extension(A,C,Bp,Cp),Bp,3); MA("15^\circ",C,Y,Cp,8); MA("15^\circ
  10 KB (1,458 words) - 20:50, 3 November 2023
• 2007 AIME II Problems/Problem 3
  ...ince <math>\triangle ABE \cong \triangle CDF</math> and are both <math>5-12-13</math> [[right triangle]]s, we can come to the conclusion that the two ne A slightly more analytic/brute-force approach:
  6 KB (933 words) - 00:05, 8 July 2023
• 2007 AIME II Problems/Problem 15
  R=rotate(90,C)*(C+r*(A-C)/length(A-C)); Z=extension(C,P,R,R+A-C); label("$A$",A,N); label("$B$",B,0.15*(B-P)); label("$C$",C,0.1*(C-P));
  11 KB (2,099 words) - 17:51, 4 January 2024
• 2007 AMC 12A Problems
  ...thrm{(A)}\ \frac 23\qquad \mathrm{(B)}\ \frac 34\qquad \mathrm{(C)}\ \frac 45\qquad \mathrm{(D)}\ \frac 56\qquad \mathrm{(E)}\ \frac 78</math> A finite [[sequence]] of three-digit integers has the property that the tens and units digits of each term
  11 KB (1,750 words) - 13:35, 15 April 2022
• 2007 AMC 12A Problems/Problem 20
  ...}{3}\qquad \mathrm{(D)}\ \frac{8\sqrt{2}-11}{3}\qquad \mathrm{(E)}\ \frac{6-4\sqrt{2}}{3}</math> ...es of an octagon and one straight edge. The diagonal edges form <math>45-45-90 \triangle</math> [[right triangle]]s, making the distance on the edge of
  2 KB (317 words) - 10:10, 16 September 2022
• 2007 USAMO Problems/Problem 2
  ...to cover all grid points by an infinite family of [[circle|discs]] with non-overlapping interiors if each disc in the family has [[radius]] at least 5? ...in the sense that any further enlargement would cause it to violate the non-overlap condition. Then <math>D(O,r)</math> is tangent to at least three dis
  5 KB (754 words) - 03:41, 7 August 2014
• Pascal Triangle Related Problems
  10: 1 10 45 120 210 252 210 120 45 10 1 1 10 45 120 210 252 210 120 45 10 1
  4 KB (513 words) - 20:18, 3 January 2023
• 2007 Cyprus MO/Lyceum/Problem 22
  [[Image:2007 CyMO-22.PNG|200px]] Since <math>AMN</math> is a <math>45-45-90</math> triangle, the length of <math>AM</math> is <math>\frac{3a}{\sqrt{2
  979 bytes (166 words) - 02:33, 19 January 2024
• 2007 AMC 12B Problems/Problem 3
  [[Image:2007_12B_AMC-3.png]] <math>\mathrm {(A)} 35 \qquad \mathrm {(B)} 40 \qquad \mathrm {(C)} 45 \qquad \mathrm {(D)} 50 \qquad \mathrm {(E)} 60</math>
  905 bytes (130 words) - 10:39, 27 February 2022
• 1951 AHSME
  ** [[1951 AHSME Problems/Problem 45|Problem 45]]
  3 KB (258 words) - 14:25, 20 February 2020
• LaTeX:Pictures
  ...mputer. (Alternatively, depending on your browser, you may be able to right-click on the link to the image and choose "Save link as...") Save the image ...enter set up to produce PDF documents (LaTeX => PDF in the appropriate drop-menu). If you don't, you'll get a bunch of errors because LaTeX will expect
  9 KB (1,454 words) - 16:52, 19 February 2024
• LaTeX:Layout
  ...each of the examples as you go. It takes almost no time at all to just copy-paste, compile, and view the results. ...kage[margin=2.5cm]{geometry}</code>. Check out the [http://tug.ctan.org/tex-archive/macros/latex/contrib/geometry/geometry.pdf geometry package user man
  30 KB (5,171 words) - 10:16, 4 April 2021
• Chicken McNugget Theorem
  ...am + bn</math> for [[nonnegative]] integers <math>a, b</math> is <math>mn-m-n</math>. ...e proof is based on the fact that in each pair of the form <math>(k, mn-m-n-k)</math>, exactly one element is expressible.
  17 KB (2,748 words) - 19:22, 24 February 2024
• 2007 AMC 12A Problems/Problem 9
  ...thrm{(A)}\ \frac 23\qquad \mathrm{(B)}\ \frac 34\qquad \mathrm{(C)}\ \frac 45\qquad \mathrm{(D)}\ \frac 56\qquad \mathrm{(E)}\ \frac 78</math> ...the following equation: <math>x-r = r + \frac{x}{7}</math>. We get <math>x-2r=\frac{x}{7}</math>, so <math>\frac{6}{7}x = 2r</math>, and <math>\frac{x}
  5 KB (804 words) - 14:55, 21 August 2022
• 2007 AMC 12A Problems/Problem 14
  <math>(6-a)(6-b)(6-c)(6-d)(6-e)=45</math> ...3, \pm 1, \pm 5</math>. The product of all six of these is <math>-225=(-5)(45)</math>, so the factors are <math>-3, -1, 1, 3,</math> and <math>5.</math>
  2 KB (278 words) - 02:10, 16 February 2021
• 2007 AMC 12A Problems/Problem 19
  [[Image:2007_12A_AMC-19.png]] ...f the parallel lines is either <math>1556</math> units above or below the y-intercept of line <math>\overline{DE}</math>; hence the equation of the para
  4 KB (565 words) - 17:01, 2 April 2023
• 2007 AMC 12A Problems/Problem 25
  <div style="text-align:center;"><math>S_{n + 1} = S_{n} + S_{n - 2}</math></div> ...the number of size <math>k</math> subsets of the set of the first <math>14-2k</math> positive integers. For instance, the arrangment o | | o | | o | |
  9 KB (1,461 words) - 23:07, 27 January 2024
• 2005 AMC 12A Problems/Problem 11
  How many three-digit numbers satisfy the property that the middle digit is the [[mean|avera ...2 \qquad (\mathrm {C})\ 43 \qquad (\mathrm {D}) \ 44 \qquad (\mathrm {E})\ 45</math>
  2 KB (266 words) - 00:59, 19 October 2020
• 2006 AIME II Problems/Problem 3
  Therefore, we have a total of <math>97-48=\boxed{049}</math> threes. ...duct of the first 100 odd integers, so our new sequence is actually 9, 27, 45...189. Divide every term by 9 to get a new sequence; 1, 3, 5...21, which cl
  4 KB (562 words) - 18:37, 30 October 2020
• 2006 AIME II Problems/Problem 6
  ...E = \sqrt{6} - \sqrt{2}</math>. <math>\triangle CEF</math> is a <math>45-45-90 \triangle</math>, so <math>CE = \frac{AE}{\sqrt{2}} = \sqrt{3} - 1</math> ...[altitude]] of the equilateral triangle and the altitude of the <math>45-45-90 \triangle</math>. The former is <math>\frac{AE\sqrt{3}}{2}</math>, and th
  7 KB (1,067 words) - 12:23, 8 April 2024
• 1997 PMWC Problems/Problem I11
  <math>l=45</math> {{PMWC box|year=1997|num-b=I10|num-a=I12}}
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• 2005 PMWC Problems/Problem T8
  ...[[rectangle]] of unequal sides remains. If the sum of the areas of the cut-off pieces is <math>200</math> and the lengths of the legs of the triangles Using subtraction of areas or <math>45-45-90</math> triangles, we find that the area of the rectangle is <math>(x + y)
  1 KB (212 words) - 22:58, 1 January 2020
• 2006 Cyprus MO/Lyceum/Problem 17
  [[Image:2006 CyMO-17.PNG|250px|right]] ...qquad\mathrm{(B)}\ 50^\circ\qquad\mathrm{(C)}\ 40^\circ\qquad\mathrm{(D)}\ 45^\circ\qquad\mathrm{(E)}\ 70^\circ</math>
  789 bytes (123 words) - 22:00, 30 November 2015
• 2006 Cyprus MO/Lyceum/Problem 19
  [[Image:2006 CyMO-19.PNG|250px|right]] ...sosceles triangle]] with<math> AB=A\Gamma=\sqrt2</math> and <math>\angle A=45^\circ</math>. If <math>B\Delta</math> is an [[altitude]] of the [[triangle]
  1 KB (214 words) - 23:44, 22 December 2016
• 2007 AMC 12B Problems
  <center>[[Image:2007_12B_AMC-3.png]]</center> <math>\mathrm {(A)} 35\qquad \mathrm {(B)} 40\qquad \mathrm {(C)} 45\qquad \mathrm {(D)} 50\qquad \mathrm {(E)} 60</math>
  12 KB (1,814 words) - 12:58, 19 February 2020
• 1960 IMO Problems/Problem 2
  ...gives <math>a<\frac{7}{2}</math> and hence <math>-\frac{1}{2} \le x<\frac{45}{8}</math>. So, answer: <math>-\frac{1}{2} \le x<\frac{45}{8}</math>, except <math>x=0</math>.
  2 KB (239 words) - 05:51, 7 April 2024
• Asymptote: Useful functions
  markangle("$45^o$", B, O, A, p=orange); markangle("$45^\circ$", B, O, A, p=orange);
  4 KB (646 words) - 21:18, 26 March 2024
• 2004 AMC 12A Problems/Problem 12
  [[Image:2004_AMC_12A-12.png]] ...math> and <math>(4,4)</math>. Using the [[distance formula]] or <math>45-45-90</math> [[right triangle]]s, the answer is <math>2\sqrt{2}\ \mathrm{(B)}</
  1 KB (181 words) - 20:17, 3 July 2013
• 2004 AMC 12A Problems/Problem 21
  .... Using the formula <math>\cos 2\theta = 2\cos^2 \theta - 1 = 2\left(\frac 45\right) - 1 = \frac 35 \Rightarrow \mathrm{(D)}</math>. <cmath>\cos^{0}\theta=5-5*\cos^{2}\theta</cmath>
  1 KB (176 words) - 13:19, 15 July 2022
• 2002 AMC 12B Problems/Problem 24
  ...h>P</math> in its interior such that <math>PA = 24, PB = 32, PC = 28, PD = 45</math>. Find the perimeter of <math>ABCD</math>. ...A = (0,0), B = (40,0), C = (25.6 * 52 / 24, 19.2 * 52 / 24), D = (40 - (40-25.6)*77/32,19.2*77/32), P = (25.6,19.2), Q = (25.6, 18.5);
  2 KB (313 words) - 10:23, 4 July 2013
• 2005 iTest
  *[[2005 iTest Problems/Problem 45|Problem 45]]
  3 KB (283 words) - 02:37, 24 January 2024
• 2000 AMC 12 Problems/Problem 17
  ...is a good value for <math>\theta</math>, because then we have a <math>30-60-90 \triangle</math> -- <math>\angle BAC=90</math> because <math>AB</math> is Let <math>OC=x</math>. Then <math>CA=1-x</math>. since <math>BC</math> bisects <math>\angle ABO</math>, we can use
  6 KB (979 words) - 12:50, 17 July 2022
• 2007 AMC 10A Problems/Problem 17
  ...>3</math>. Thus the minimum value of <math>m</math> is <math>3^2 \cdot 5 = 45</math>, which makes <math>n = \sqrt[3]{3^3 \cdot 5^3} = 15</math>. The mini ...th> = 45, and since <math>5^3 \cdot 3^3 = 15^3</math>, our answer is <math>45 + 15</math> = <math>\boxed {(D)60}</math>
  1 KB (201 words) - 08:04, 11 February 2023
• 2007 AMC 10A Problems
  ...thrm{(A)}\ \frac 23\qquad \mathrm{(B)}\ \frac 34\qquad \mathrm{(C)}\ \frac 45\qquad \mathrm{(D)}\ \frac 56\qquad \mathrm{(E)}\ \frac 78</math> pair X=O0+2*dir(30), Y=O2+dir(45);
  13 KB (2,058 words) - 17:54, 29 March 2024
• 1995 AHSME Problems
  ...product for three easy payments of <math>\textdollar 29.98</math> and a one-time shipping and handling charge of <math>\textdollar 9.98</math>. How many How many base 10 four-digit numbers, <math>N = \underline{a} \underline{b} \underline{c} \underlin
  17 KB (2,387 words) - 22:44, 26 May 2021
• 1995 AHSME Problems/Problem 29
  For how many three-element sets of distinct positive integers <math>\{a,b,c\}</math> is it true ...} \qquad \mathrm{(C) \ 40 } \qquad \mathrm{(D) \ 43 } \qquad \mathrm{(E) \ 45 } </math>
  2 KB (371 words) - 14:33, 15 January 2023
• 1951 AHSME Problems
  The largest number by which the expression <math>n^3-n</math> is divisible for all possible integer values of <math>n</math>, is: ...t{be tangent to the x-axis}\\ \qquad\textbf{(D)}\ \text{be tangent to the y-axis}\qquad\textbf{(E)}\ \text{lie in one quadrant only} </math>
  23 KB (3,641 words) - 22:23, 3 November 2023
• 1959 AHSME
  * [[1959 AHSME Problems/Problem 45|Problem 45]]
  3 KB (257 words) - 14:19, 20 February 2020
• 1959 AHSME Problems
  The value of <math>x^2-6x+13</math> can never be less than: ...is herd of <math>n</math> cows among his four sons so that one son gets one-half the herd,
  22 KB (3,345 words) - 20:12, 15 February 2023
• 2002 AMC 12B Problems/Problem 18
  \qquad\mathrm{(D)}\ \frac 45 [[Image:2002_12B_AMC-18.png]]
  3 KB (376 words) - 19:16, 20 August 2019
• Mock AIME 1 Pre 2005 Problems/Problem 1
  ...the sum of the hundreds places is <math>(1+2+3+\cdots+9)(72) \times 100 = 45 \cdot 72 \cdot 100 = 324000</math>. ...y also be <math>0</math>). Thus, the the sum of the tens digit gives <math>45 \cdot 64 \cdot 10 = 28800</math>.
  1 KB (194 words) - 13:44, 5 September 2012
• 2007 AMC 10A Problems/Problem 15
  pair X=O0+2*dir(30), Y=O2+dir(45); draw((-h-1,-h-1)--(-h-1,h+1)--(h+1,h+1)--(h+1,-h-1)--cycle);
  2 KB (326 words) - 10:29, 4 June 2021
• Construction
  23. Construct <math>15^\circ, 30^\circ, 45^\circ, 60^\circ, 75^\circ</math> angles. Hence or otherwise, construct a ri
  3 KB (443 words) - 20:52, 28 August 2014
• 2007 AMC 8 Problems
  The average cost of a long-distance call in the USA in <math>1985</math> was <math>41</math> cents per minute, and the average cost of a long-distance
  12 KB (1,800 words) - 20:01, 8 May 2023
• 2007 AMC 10A Problems/Problem 24
  pair A=(-2*sqrt(2),0), B = (2*sqrt(2),0), C = A+2*dir(45), D = B+2*dir(135), E = A+2*dir(90), F = B+2*dir(90); ...frac{1}{8}</math> the area of its circle since <math>\angle{OAC}</math> is 45 degrees and <math>\angle{OAE}</math> forms a right [[triangle]]. Thus <mat
  3 KB (439 words) - 15:39, 3 June 2021
• 2004 AMC 12B Problems/Problem 24
  ...a \neq 0</math>, we have <math>\tan^4 DBE = 1 \Longrightarrow \angle DBE = 45^{\circ}</math>. The second condition tells us that <math>2\cot (45 - \alpha) = 1 + \cot \alpha</math>. Expanding, we have <math>1 + \cot \alph
  2 KB (302 words) - 19:59, 3 July 2013
• 2008 AMC 12A Problems/Problem 20
  pair C=(0,0),A=(0,3),B=(4,0),D=(4-2.28571,1.71429); label("\(45^{\circ}\)",(.2,.1),E);
  6 KB (951 words) - 16:31, 2 August 2019
• 2007 AMC 10B Problems/Problem 18
  ...t(2)), C=(-1-sqrt(2),1+sqrt(2)), D=(-1-sqrt(2),-1-sqrt(2)), E=(1+sqrt(2),-1-sqrt(2)); ...2sqrt(2),1+sqrt(2)), C1=(0,1+sqrt(2)), D1=(0,-1-sqrt(2)), E1=(2+2sqrt(2),-1-sqrt(2));
  2 KB (385 words) - 14:17, 4 June 2021
• 2008 AMC 12B Problems
  ...ine{AB}</math>, <math>\angle AOC = 30^\circ</math>, and <math>\angle DOB = 45^\circ</math>. What is the ratio of the area of the smaller sector <math>COD label ("\(45^\circ\)", (-0.3,0.1), WNW);
  14 KB (2,199 words) - 13:43, 28 August 2020
• 2008 AMC 12B Problems/Problem 4
  ...ine{AB}</math>, <math>\angle AOC = 30^\circ</math>, and <math>\angle DOB = 45^\circ</math>. What is the ratio of the area of the smaller sector <math>COD label ("\(45^\circ\)", (-0.3,0.1), WNW);
  1 KB (183 words) - 22:35, 10 June 2017
• 2008 AMC 12B Problems/Problem 25
  ...=(7-5)(7+5)=DX^2-CY^2</math> and <math>DX+CY=19-11=8</math>. Thus, <math>DX-CY=3</math>. Over to the other side: <math>\triangle BCY</math> is <math>30-60-90</math>, and is therefore congruent to <math>\triangle BCQ</math>. So <mat
  12 KB (2,015 words) - 20:54, 9 October 2022
• 1985 AJHSME Problems
  A "stair-step" figure is made of alternating black and white squares in each row. Row If you walk for <math>45</math> minutes at a rate of <math>4 \text{ mph}</math> and then run for <ma
  12 KB (1,670 words) - 17:42, 24 November 2021
• 2008 AIME I Problems
  On a long straight stretch of one-way single-lane highway, cars all travel at the same speed and all obey the safety rule D((0,s)--r*dir(45)--(s,0));
  9 KB (1,536 words) - 00:46, 26 August 2023
• 2008 AIME I Problems/Problem 11
  ...A</math>, or by appending two <math>B</math>s to a string of length <math>n-2</math> ending in a <math>B</math>. Thus, we have the [[recursion]]s a_n &= a_{n-2} + b_{n-2}\\
  7 KB (1,173 words) - 22:39, 28 November 2023
• 2008 AIME I Problems/Problem 15
  D((0,s)--r*dir(45)--(s,0)); D((0,0)--r*dir(45));
  6 KB (1,041 words) - 00:54, 1 February 2024
• Mock AIME 5 Pre 2005 Problems
  Two 5-digit numbers are called "responsible" if they are: &\text {iii.} a + b + c + d + e + f + g + h + i + j = 45\end{align*}</cmath>
  6 KB (909 words) - 07:27, 12 October 2022
• Mock AIME 1 2007-2008 Problems/Problem 15
  <cmath>\sum_{x=2}^{44} 2\sin{x}\sin{1}[1 + \sec (x-1) \sec (x+1)]</cmath> ...ta_1,\, \theta_2, \, \theta_3, \, \theta_4</math> are degrees <math>\in [0,45]</math>. Find <math>\theta_1 + \theta_2 + \theta_3 + \theta_4</math>.
  2 KB (276 words) - 16:27, 26 December 2015
• Mock AIME 1 2007-2008 Problems
  [[Mock AIME 1 2007-2008 Problems/Problem 1|Solution]] [[Mock AIME 1 2007-2008 Problems/Problem 2|Solution]]
  6 KB (992 words) - 14:15, 13 February 2018
• 2008 AIME II Problems/Problem 5
  label("\(504-x\)",(G+D)/2,S); ...ac{BF}{AF} = \frac {DG}{CG} \Longleftrightarrow \frac{h}{504+x} = \frac{504-x}{h} \Longrightarrow x^2 + h^2 = 504^2.</cmath>
  8 KB (1,338 words) - 23:15, 28 November 2023
• 2008 AIME II Problems/Problem 15
  where <math>a</math> and <math>b</math> are non-negative integers. Now this is a [[Pell equation]], with solutions in the fo ...h> to be integer, the squares must be odd. So we generate <math>N = 1, 21, 45, 73, 105, 141, 181, 381, 441, 721, 801</math>. <math>N</math> cannot exceed
  4 KB (628 words) - 16:23, 2 January 2024
• 1952 AHSME
  ** [[1952 AHSME Problems/Problem 45|Problem 45]]
  3 KB (257 words) - 14:25, 20 February 2020
• 1953 AHSME
  ** [[1953 AHSME Problems/Problem 45|Problem 45]]
  3 KB (257 words) - 14:24, 20 February 2020
• 1954 AHSME
  ** [[1954 AHSME Problems/Problem 45|Problem 45]]
  3 KB (257 words) - 14:23, 20 February 2020
• 1955 AHSME
  ** [[1955 AHSME Problems/Problem 45|Problem 45]]
  3 KB (257 words) - 14:23, 20 February 2020
• 1956 AHSME
  * [[1956 AHSME Problems/Problem 45|Problem 45]]
  3 KB (257 words) - 14:22, 20 February 2020
• 1957 AHSME
  * [[1957 AHSME Problems/Problem 45|Problem 45]]
  3 KB (257 words) - 14:21, 20 February 2020
• 1958 AHSME
  * [[1958 AHSME Problems/Problem 45|Problem 45]]
  3 KB (257 words) - 14:20, 20 February 2020
• 2001 IMO Shortlist Problems/G1
  ...2AB+\angle BAC=45^{\circ}+\alpha</math>, and likewise <math>\angle C_2BC = 45^{\circ}+\beta</math>. It then follows from the Law of Sines on triangles <m <cmath>\frac{d}{\sin{(\alpha+45^{\circ})}}=\frac{s}{\sin{\theta}}</cmath>
  3 KB (584 words) - 15:07, 12 December 2011
• 1989 AHSME Problems/Problem 29
  Using [[De Moivre's Theorem]], <math>(1+i)^{99}=[\sqrt{2}cis(45^\circ)]^{99}=\sqrt{2^{99}}\cdot cis(99\cdot45^\circ)=2^{49}\sqrt{2}\cdot ci {{AHSME box|year=1989|num-b=28|num-a=30}}
  1 KB (173 words) - 19:52, 30 December 2020
• 1990 AHSME Problems/Problem 29
  <cmath>\begin{array}{ccccc}1&3&9&27&81\\2&6&18&54\\4&12&36\\5&15&45\\7&21&63\\8&24&72\\10&30&90\\11&33&99\\13&39\\\vdots&\vdots\end{array}</cma ...h> multiples of three in that range, so there are <math>88-29=59</math> non-multiples, and <math>3+14+59=76</math>, which is <math>\fbox{D}</math>
  2 KB (285 words) - 19:25, 25 September 2020
• 2008 iTest
  *[[2008 iTest Problems/Problem 45|Problem 45]]
  5 KB (424 words) - 15:18, 24 January 2020
• 2008 iTest Problems
  Single-day passes cost <math>\textdollar{33}</math> for adults (Jerry and Hannah), ...tting average is <math>.417</math> after six games, and the team is <math>6-0</math> (six wins and no losses). They are off to their best start in years
  71 KB (11,749 words) - 01:31, 2 November 2023
• Intermediate Math Open
  ...Power Round, where teams work together on a series of related problems for 45 minutes. Following that is the Team Round, consisting of 10 group problems
  1 KB (243 words) - 17:53, 1 November 2014
• 2002 AMC 10A Problems
  Given that <math>a, b,</math> and <math>c</math> are non-zero real numbers, define <math>(a, b, c) = \frac{a}{b} + \frac{b}{c} + \fra ..., and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.
  11 KB (1,733 words) - 11:04, 12 October 2021
• 2002 AMC 10B Problems
  <math>(3x-2)(4x+1)-(3x-2)4x+1</math> For how many positive integers <math>n</math> is <math>n^2-3n+2</math> a prime number?
  10 KB (1,540 words) - 22:53, 19 December 2023
• 2002 AMC 12A Problems/Problem 7
  A <math>45^\circ</math> arc of circle A is equal in length to a <math>30^\circ</math> Using that here, the arc of circle A has length <math>\frac{45}{360}\cdot2\pi{r_1}=\frac{r_1\pi}{4}</math>. The arc of circle B has length
  3 KB (492 words) - 14:46, 31 January 2024
• 2002 AMC 12A Problems/Problem 11
  <math>\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 55 \qq ...frac{1}{20}\right)60</math>. Setting the two equal, we have <math>40t+2=60t-3</math> and we find <math>t=\frac{1}{4}</math> of an hour. Substituting t b
  2 KB (396 words) - 00:19, 1 February 2024
• 2004 AMC 10B Problems
  How many two-digit positive integers have at least one <math>7</math> as a digit? A standard six-sided die is rolled, and <math>P</math> is the product of the five numbers t
  13 KB (1,988 words) - 20:19, 15 May 2024
• 2001 AMC 10 Problems
  two-digit number such that <math>N = P(N)+S(N)</math>. What is the units digit o ...t), and the rest sells for half price. How much money is raised by the full-price tickets?
  14 KB (1,983 words) - 16:25, 2 June 2022
• 1998 AHSME Problems/Problem 19
  label("\((5\cos\theta,5\sin\theta)\)", 5*dir(-45), SE, red ); {{AHSME box|year=1998|num-b=18|num-a=20}}
  2 KB (368 words) - 11:08, 4 February 2017
• 2000 AMC 10 Problems/Problem 16
  ..., hence we can compute the slope, <math>m_1</math> to be <math>\frac{0-3}{6-0}=-\frac{1}{2}</math>, so <math>l_1</math> is the line <math>y=\frac{-1}{2} ...th> gives us <math>b_2=-2</math>, so <math>l_2</math> is the line <math>y=x-2</math>.
  7 KB (966 words) - 23:22, 31 July 2023
• 1985 AJHSME Problems/Problem 2
  ...7)+(4+6)+5[/mathjax]. [mathjax]4\cdot10+5 = 45[/mathjax], and [mathjax]900+45=\boxed{\text{(B)}~945}[/mathjax]. ...th> and then add the negatives using <math>\frac{n(n+1)}{2}=1+2+3+\cdots+(n-1)+n</math> to get the answer.
  2 KB (282 words) - 13:43, 4 April 2024
• 1985 AJHSME Problems/Problem 23
  ...xt{(A)}\ 30 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 40 \qquad \text{(D)}\ 45 \qquad \text{(E)}\ 50</math> {{AJHSME box|year=1985|num-b=22|num-a=24}}
  1 KB (181 words) - 09:06, 22 January 2023
• 1985 AJHSME Problems/Problem 21
  ...ext{(C)}\ 44\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ \text{more than }45\% </math> ...ould have <math>146.41</math>. As the total increase is greater than <math>45\%</math>, the answer is <math>\boxed{\text{E}}</math>.
  905 bytes (133 words) - 08:39, 7 January 2023
• 1985 AJHSME Problems/Problem 13
  If you walk for <math>45</math> minutes at a [[rate]] of <math>4 \text{ mph}</math> and then run for <math>45</math> minutes is <math>\frac{3}{4}</math> of an hour, so the walking contr
  1 KB (170 words) - 08:13, 13 January 2023
• 1999 AHSME Problems/Problem 22
  The graphs of <math>y = -|x-a| + b</math> and <math>y = |x-c| + d</math> intersect at points <math>(2,5)</math> and <math>(8,3)</math>. ...orthogonal half-lines. In the first graph both point downwards at a <math>45^\circ</math> angle, in the second graph they point upwards. One can easily
  3 KB (538 words) - 12:02, 17 October 2020
• 2004 AMC 12B Problems/Problem 15
  ...\qquad \mathrm{(C) \ } 27 \qquad \mathrm{(D) \ } 36\qquad \mathrm{(E) \ } 45 </math> ...olution is <math>(a,b)=(3,1)</math>, and the difference in ages is <math>31-13=\boxed{\mathrm{(B)\ }18}</math>.
  2 KB (326 words) - 09:47, 17 October 2020
• 2002 AMC 10B Problems/Problem 17
  for (int i=1; i<8; ++i) A[i] = A[i-1] + 2*dir(45*(i-1)); {{AMC10 box|year=2002|ab=B|num-b=16|num-a=18}}
  2 KB (273 words) - 13:27, 21 May 2021
• 2004 AMC 10B Problems/Problem 25
  ...th>\sqrt 3</math> and not a <math>\sqrt 2</math> in them (There's no <math>45^{\circ}</math> angle either). So, our answer is <math>\boxed{\mathrm{(B)\ } {{AMC10 box|year=2004|ab=B|num-b=24|after=Last Question}}
  4 KB (703 words) - 22:37, 2 November 2022
• 2009 AMC 12A Problems
  ...}\ 15 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 45 \qquad \textbf{(E)}\ 55</math> ...rom <math>A</math> to <math>B</math>, turns through an angle between <math>45^{\circ}</math> and <math>60^{\circ}</math>, and then sails another <math>20
  13 KB (2,105 words) - 13:13, 12 August 2020
• 2009 AMC 10A Problems
  \mathrm{(D)}\ 45 \mathrm{(D)}\ \frac{45}{4}
  14 KB (2,130 words) - 11:32, 7 November 2021
• 2009 AMC 12A Problems/Problem 25
  ...quiv \theta_{n + 1} + \theta_{n} \pmod{180}</math>. Since <math>\theta_1 = 45, \theta_2 = 30</math>, all terms in the sequence <math>\{\theta_1, \theta_2 <cmath>\tan{\theta_n}=\tan{(\theta_{n-1}+\theta_{n-2})}</cmath>
  7 KB (990 words) - 07:23, 24 October 2022
• 2009 AMC 12A Problems/Problem 13
  ...rom <math>A</math> to <math>B</math>, turns through an angle between <math>45^{\circ}</math> and <math>60^{\circ}</math>, and then sails another <math>20 ...>AC^2</math> for two different angles, preferably for both extremes (<math>45</math> and <math>60</math> degrees). You can use the law of cosines to do s
  5 KB (739 words) - 10:24, 9 February 2015
• 2009 AMC 12A Problems/Problem 4
  ...}\ 15 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 45 \qquad \textbf{(E)}\ 55</math> * <math>45 = 25+10+5+5</math>
  1 KB (194 words) - 20:40, 29 January 2020
• 2009 AMC 10A Problems/Problem 20
  ...t <math>1\mathrm km/min = 60\mathrm km/h</math>.) This solves to <math>v_A=45</math> and <math>v_L=15</math>. ...te, after <math>5</math> minutes the distance between them will be <math>20-5=15</math> kilometers.
  3 KB (472 words) - 15:40, 4 June 2023
• 2009 AMC 10A Problems/Problem 17
  pair EF=rotate(90)*(D-B); pair E=intersectionpoint( (0,-100)--(0,100), (B-100*EF)--(B+100*EF) );
  4 KB (684 words) - 21:14, 23 October 2023
• 2001 AMC 12 Problems/Problem 12
  ...<math>15</math>, <math>20</math>, <math>30</math>, <math>40</math>, <math>45</math> and <math>60</math> are divisible by <math>5</math>. Therefore in the set <math>\{1,\dots,60\}</math> there are precisely <math>30-6=24</math> numbers that satisfy all criteria from the problem statement.
  7 KB (1,110 words) - 15:20, 30 May 2022
• 2001 AMC 12 Problems/Problem 24
  In <math>\triangle ABC</math>, <math>\angle ABC=45^\circ</math>. Point <math>D</math> is on <math>\overline{BC}</math> so that ...<math>DH=\frac{\sqrt{3}}{3}kt</math>. The comparable ratio is that <math>BH-DH=t</math>. If we involve our <math>k</math>, we get:
  8 KB (1,316 words) - 22:48, 7 March 2024
• 2009 AMC 10A Problems/Problem 5
  <math>\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81</math> <math>1+2+3+4+5\cdots n+(n-1)+(n-2)\cdots+1</math>
  2 KB (231 words) - 15:59, 9 February 2023
• 2009 AMC 10A Problems/Problem 4
  \mathrm{(D)}\ \frac{45}{4} {{AMC10 box|year=2009|ab=A|num-b=3|num-a=5}}
  1 KB (159 words) - 14:30, 11 September 2022
• 2002 AMC 12A Problems/Problem 18
  Let <math>C_1</math> and <math>C_2</math> be circles defined by <math>(x-10)^2 + y^2 = 36</math> and <math>(x+15)^2 + y^2 = 81</math> First examine the formula <math>(x-10)^2+y^2=36</math>, for the circle <math>C_1</math>. Its center, <math>D_1<
  3 KB (485 words) - 03:13, 1 September 2023
• 2009 AMC 12B Problems/Problem 23
  ...enlarged by a factor of <math>\frac{3\sqrt{2}}{4}</math> and rotated <math>45</math> degrees that lies within the original square. This skips all the abs ...equations gives us the four lines <cmath>x-y=\frac{4}{3},</cmath> <cmath>x-y=-\frac{4}{3},</cmath> <cmath>x+y=\frac{4}{3},</cmath> <cmath>x+y=-\frac{4}
  2 KB (422 words) - 13:25, 20 January 2020
• 2009 AMC 10B Problems/Problem 5
  Twenty percent less than 60 is one-third more than what number? Twenty percent less than 60 is <math>\frac 45 \cdot 60 = 48</math>. One-third more than a number ''n'' is <math>\frac 43n</math>. Therefore <math>\f
  693 bytes (97 words) - 15:03, 22 April 2021
• 2009 AMC 12B Problems/Problem 16
  \mathrm{(B)}\ \frac 45\qquad so <math>CD = \boxed{\frac 45}.</math>
  1 KB (166 words) - 21:07, 16 February 2024
• 2009 AMC 12B Problems/Problem 24
  ...As <math>\frac{180}{7}</math> is not on the interval <math>30 \leq x \leq 45</math>, this yields no solution. ...in(6x)) = 6x - 360</math>. As <math>72</math> is not on the interval <math>45 \leq x \leq 60</math>, this yields no solution.
  7 KB (1,287 words) - 07:09, 22 December 2022
• 2009 AMC 10B Problems
  Each morning of her five-day workweek, Jane bought either a <math>50</math>-cent muffin or a <math>75 Twenty percent less than 60 is one-third more than what number?
  15 KB (2,262 words) - 00:53, 18 June 2021
• 2009 AMC 10B Problems/Problem 22
  ...math>M=(2,1)</math>, and the line <math>QM</math> has the equation <math>2y-x=0</math>. ...th>y+2x-2=0</math>, we get <math>y=\frac 25</math>, and then <math>x=\frac 45</math>.
  12 KB (1,868 words) - 03:36, 30 September 2023
• 1988 AJHSME Problems
  filldraw((b,14-b)--(b+1,14-b)--(b+1,15-b)--(b,15-b)--cycle,black,gray); filldraw((b-1,14-b)--(b,14-b)--(b,15-b)--(b-1,15-b)--cycle,white,black);
  14 KB (1,872 words) - 15:23, 17 January 2023
• 2005 USAMO Problems/Problem 6
  ...always 9, the sum of the digits will be <math>9p</math>. Since <math>10^{p-1} < n(n+1)/2</math>, this example shows that <cmath>f(n) < 9\log_{10}n^5 = 45\log_{10}n.</cmath>
  4 KB (725 words) - 23:59, 29 March 2016
• 1989 AJHSME Problems
  ...xt{(B)}\ 20 \qquad \text{(C)}\ 30 \qquad \text{(D)}\ 35 \qquad \text{(E)}\ 45</math> Which of the five "T-like shapes" would be symmetric to the one shown with respect to the dashed
  13 KB (1,765 words) - 11:55, 22 November 2023
• 2009 AIME II Problems/Problem 4
  ...math> grapes, and the child in <math>k</math>-th place had eaten <math>n+2-2k</math> grapes. The total number of grapes eaten in the contest was <math> ...l <math>k</math>, the child in <math>k</math>-th place had eaten <math>n+2-2k</math> grapes".
  4 KB (762 words) - 18:40, 12 January 2024
• 2009 AIME II Problems/Problem 14
  b_2 & = \frac 45\cdot \frac 35 + \frac 35 \sqrt{1 - \left(\frac 35\right)^2} = \frac{24}{25} b_3 & = \frac 45\cdot \frac {24}{25} + \frac 35 \sqrt{1 - \left(\frac {24}{25}\right)^2} = \
  4 KB (680 words) - 13:49, 23 December 2023
• 1988 AJHSME Problems/Problem 20
  ...auge on a [[cylinder|cylindrical]] coffee maker shows that there are <math>45</math> cups left when the coffee maker is <math>36\% </math> full. How man ...coffee the maker will hold when full be <math>x</math>. Then, <cmath>.36x=45 \Rightarrow x=125 \rightarrow \boxed{\text{C}}</cmath>
  963 bytes (130 words) - 23:56, 4 July 2013
• 1990 AJHSME Problems
  <math>\text{(A)}\ \text{45 dollars} \qquad \text{(B)}\ \text{52 dollars} \qquad \text{(C)}\ \text{54 d There are twenty-four <math>4</math>-digit numbers that use each of the four digits <math>2</
  15 KB (2,059 words) - 15:03, 6 October 2021
• 1989 AJHSME Problems/Problem 7
  ...xt{(B)}\ 20 \qquad \text{(C)}\ 30 \qquad \text{(D)}\ 35 \qquad \text{(E)}\ 45</math> {{AJHSME box|year=1989|num-b=6|num-a=8}}
  830 bytes (116 words) - 17:14, 29 April 2021
• 1990 AJHSME Problems/Problem 8
  <math>\text{(A)}\ \text{45 dollars} \qquad \text{(B)}\ \text{52 dollars} \qquad \text{(C)}\ \text{54 d {{AJHSME box|year=1990|num-b=7|num-a=9}}
  748 bytes (107 words) - 00:05, 5 July 2013
• 2009 IMO Problems/Problem 4
  ...e the incentre of triangle <math>ADC</math>. Suppose that <math>\angle BEK=45^\circ</math>. Find all possible values of <math>\angle CAB</math>. {{IMO box|year=2009|num-b=3|num-a=5}}
  584 bytes (94 words) - 01:17, 19 November 2023
• 2009 IMO Problems
  ...1,\ldots,k-1</math>. Prove that <math>n</math> doesn't divide <math>a_k(a_1-1)</math>. ...e the incentre of triangle <math>ADC</math>. Suppose that <math>\angle BEK=45^\circ</math>. Find all possible values of <math>\angle CAB</math>.
  3 KB (509 words) - 09:23, 10 September 2020
• 1991 AJHSME Problems/Problem 19
  ...umbers, we let them be <math>1,2,3,4,5,6,7,8,9</math>. Their sum is <math>45</math>. ...he sum of all ten is <math>100</math> so the last number must be <math>100-45=55\rightarrow \boxed{\text{C}}</math>.
  895 bytes (135 words) - 12:16, 8 October 2015
• 1992 AJHSME Problems
  <math>\dfrac{10-9+8-7+6-5+4-3+2-1}{1-2+3-4+5-6+7-8+9}=</math> label("$c$",(2-sqrt(3)/10,0.1),WNW);
  17 KB (2,346 words) - 13:36, 19 February 2020
• 1993 AJHSME Problems
  label("$45$",(4,-.25),S); ...y of 0's, 1's, 3's, 4's, 5's, 6's, 7's, 8's and 9's, but he has only twenty-two 2's. How far can he number the pages of his scrapbook with these digits
  14 KB (1,797 words) - 11:13, 28 December 2022
• 2005 AMC 10A Problems/Problem 19
  ...ed with their bases on a line. The center square is lifted out and rotated 45 degrees, as shown. Then it is centered and lowered into its original locati ...>DE</math> is 1 inch. Then, because <math>DEC</math> is a <math>45^{\circ}-45^{\circ}-90^{\circ}</math> triangle, <math>EC=\frac{\sqrt{2}}{2}</math>, and
  3 KB (315 words) - 12:46, 14 December 2021
• 2010 AMC 12A Problems
  What is <math>\left(20-\left(2010-201\right)\right)+\left(2010-\left(201-20\right)\right)</math>? Halfway through a 100-shot archery tournament, Chelsea leads by 50 points. For each shot a bullsey
  12 KB (1,817 words) - 15:00, 12 August 2020
• 2010 AMC 12A Problems/Problem 19
  <math>\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 63 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 201 \q ...math> marbles, we must draw a white marble from boxes <math>1, 2, \ldots, n-1,</math> and draw a red marble from box <math>n.</math> Thus, <cmath>P(n) =
  2 KB (379 words) - 18:30, 10 July 2022
• 2010 AMC 12A Problems/Problem 25
  ...ormed, so the total number of quadrilaterals for this case is <math>3\cdot{45} = 135</math>. ...th> square case, and <math>7</math> rectangle cases, there are <math>2255-1-2\cdot7=2240</math> quadrilaterals counted 4 times. Thus there are <math>1+7
  8 KB (1,315 words) - 03:33, 23 January 2021
• 2010 AMC 10A Problems
  .... The numbers <math>x</math> and <math>x+32</math> are three-digit and four-digit palindromes, respectively. What is the sum of the digits of <math>x</m ...3,4,5,6,7,8\}</math> and also arranges them in descending order to form a 3-digit number. What is the probability that Bernardo's number is larger than
  13 KB (1,902 words) - 11:20, 5 March 2023
• 2010 AMC 10B Problems
  What is <math>100(100-3)-(100\cdot100-3)</math>? ...ings during her <math>9</math>-hour work day. The first meeting took <math>45</math> minutes and the second meeting took twice as long. What percent of h
  12 KB (1,817 words) - 22:44, 22 December 2020
• 2010 AMC 12B Problems
  ...ings during her <math>9</math>-hour work day. The first meeting took <math>45</math> minutes and the second meeting took twice as long. What percent of h In <math>\triangle ABC</math>, <math>\cos(2A-B)+\sin(A+B)=2</math> and <math>AB=4</math>. What is <math>BC</math>?
  12 KB (1,845 words) - 13:00, 19 February 2020
• Mock AIME 1 2010 Problems
  Find the last three digits of the number of 7-tuples of positive integers <math>(a_1, a_2, a_3, a_4, a_5, a_6, a_7)</math> ...be the set of all points <math>P</math> such that <math>m \angle APB \geq 45^{\circ}</math>. Find the last three digits of the largest integer less than
  7 KB (1,297 words) - 01:29, 25 November 2016
• Proofs without words
  http://mathoverflow.net/questions/8846/proofs-without-words http://gurmeet.net/computer-science/mathematical-recreations-proofs-without-words/
  55 KB (7,986 words) - 17:04, 20 December 2018
• 2010 AIME I Problems/Problem 9
  ...c {15}{7}</math> and so the sum is <math>28 - \frac {45}{7} = \frac {196 - 45}{7}</math> giving <math>151 + 7 = \fbox{158}</math>. .... Move 28 over, divide both sides by 3, then cube to get <math>k^3-6k^2+12k-8 = k^3+22k^2+18k</math>. The <math>k^3</math> terms cancel out, so solve th
  3 KB (483 words) - 07:35, 4 November 2022
• 2010 AIME I Problems/Problem 15
  ...p(ABM)}{p(CBM)} = \frac {12 + d + x}{28 + d - x}</math>. Equating and cross-multiplying yields <math>25x + 2dx = 15d + 180</math> or <math>d = \frac {25 ...the polynomial by <math>2</math>. The only such <math>x</math> in the above-stated range is <math>\frac {22}3</math>.''
  14 KB (2,210 words) - 13:14, 11 January 2024
• 2010 AIME II Problems
  Let <math>K</math> be the product of all factors <math>(b-a)</math> (not necessarily distinct) where <math>a</math> and <math>b</math> ...of <math>P(z)</math> are <math>w+3i</math>, <math>w+9i</math>, and <math>2w-4</math>, where <math>i^2=-1</math>. Find <math>|a+b+c|</math>.
  8 KB (1,246 words) - 21:58, 10 August 2020
• 2010 AIME II Problems/Problem 14
  ...math>ABC</math> with [[right angle]] at <math>C</math>, <math>\angle BAC < 45^\circ</math> and <math>AB = 4</math>. Point <math>P</math> on <math>\overli ...inewidth(0.7)+fontsize(10)); size(250); real lsf = 0.5; /* changes label-to-point distance */
  10 KB (1,507 words) - 00:31, 19 November 2023
• 1994 AJHSME Problems
  ...day Maria must work <math>8</math> hours. This does not include the <math>45</math> minutes she takes for lunch. If she begins working at <math>\text{7 draw(circle((45.5,-1.5),1));
  16 KB (2,292 words) - 13:36, 19 February 2020
• 2010 AIME II Problems/Problem 11
  Define a <i>T-grid</i> to be a <math>3\times3</math> matrix which satisfies the following Find the number of distinct <i>T-grids</i>.
  6 KB (1,057 words) - 01:58, 8 January 2023
• 2010 USAMO Problems/Problem 4
  ...</math>, as the other two angles in triangle <math>BIC</math> add to <math>45^{\circ}</math>. Assume that only <math>AB, AC, BI</math>, and <math>CI</mat pair D = extension(B, bisectorpoint(C, B, A), A, C); ldot(D, "$D$", D-B);
  7 KB (1,230 words) - 19:47, 31 January 2024
• 2010 USAJMO Problems/Problem 1
  ...closed under multiplication and a non-square times a square is always a non-square. ...lement <math>g_m</math> of the equivalence class <math>C_m</math> is square-free, since if it were divisible by the square of a prime, the quotient woul
  12 KB (2,338 words) - 20:30, 13 February 2024
• 1995 AJHSME Problems
  <math>\text{(A)}\ 45\% \qquad \text{(B)}\ 47\dfrac{1}{2}\% \qquad \text{(C)}\ 50\% \qquad \text{ ...distance. They set off in opposite directions around the outside of the 18-block area as shown. When they meet for the first time, they will be closes
  14 KB (2,096 words) - 18:29, 2 January 2023
• 2010 AMC 10B Problems/Problem 6
  <math>\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 65</math> {{AMC10 box|year=2010|ab=B|num-b=5|num-a=7}}
  2 KB (260 words) - 17:00, 1 August 2022
• 2010 AMC 12B Problems/Problem 1
  ...ings during her <math>9</math>-hour work day. The first meeting took <math>45</math> minutes and the second meeting took twice as long. What percent of h The total amount of time spend in meetings in minutes is <math>45 + 45 \times 2 = 135.</math>
  1 KB (151 words) - 16:59, 1 August 2022
• 2010 AMC 10A Problems/Problem 7
  ...ion is <math>\sqrt{2}</math> miles, because it is the hypotenuse of a 45-45-90 triangle with side length one (mile). Therefore, Crystal's distance from label("$45^{\circ}$", (0,1.25), NE);
  3 KB (370 words) - 00:16, 29 June 2022
• 1996 AJHSME Problems
  <math>\text{(A)}\ P-Q \qquad \text{(B)}\ P\cdot Q \qquad \text{(C)}\ \dfrac{S}{Q}\cdot P \qquad <math>\text{(A)}\ 3+x \qquad \text{(B)}\ 3-x \qquad \text{(C)}\ 3\cdot x \qquad \text{(D)}\ 3/x \qquad \text{(E)}\ x/3<
  13 KB (1,880 words) - 13:35, 19 February 2020
• 2007 AMC 8 Problems/Problem 20
  ...t page--><onlyinclude>Before the district play, the Unicorns had won <math>45</math>% of their basketball games. During district play, they won six more ...the information given in the problem, we can say that <math>\frac{x}{y}=0.45.</math> Next, the Unicorns win 6 more games and lose 2 more, for a total of
  6 KB (958 words) - 18:32, 20 January 2024
• 1975 IMO Problems
  ...CAQ</math> are constructed externally with <math>\angle CBP = \angle CAQ = 45^\circ, \angle BCP = \angle ACQ = 30^\circ, \angle ABR = \angle BAR = 15^\ci
  2 KB (416 words) - 16:06, 29 January 2021
• 2004 AMC 8 Problems
  How many different four-digit numbers can be formed by rearranging the four digits in <math>2004</ma ...lton’s eighth-grade class wants to participate in the annual three-person-team basketball tournament. Lance, Sally, Joy, and Fred are chosen for the t
  13 KB (1,860 words) - 19:58, 8 May 2023
• 2006 AMC 8 Problems
  How many two-digit numbers have digits whose sum is a perfect square? ...% </math> on a 30-problem test. If the three tests are combined into one 60-problem test, which percent is closest to her overall score?
  16 KB (2,215 words) - 19:18, 10 April 2024
• 1997 AJHSME Problems
  Ahn chooses a two-digit integer, subtracts it from 200, and doubles the result. What is the l ...speech for her class. Her speech must last between one-half hour and three-quarters of an hour. The ideal rate of speech is 150 words per minute. If
  12 KB (1,702 words) - 12:35, 6 November 2022
• 1998 AJHSME Problems
  ...frac{6}{x} \qquad \text{(B)}\ \dfrac{6}{x+1} \qquad \text{(C)}\ \dfrac{6}{x-1} \qquad \text{(D)}\ \dfrac{x}{6} \qquad \text{(E)}\ \dfrac{x+1}{6}</math> ...by <math>20\% </math>. Later the price is lowered again, this time by one-half the reduced price. The price is now
  14 KB (1,920 words) - 19:31, 31 January 2024
• 2007 AMC 8 Problems/Problem 17
  tint and <math>45\%</math> water. Five liters of yellow tint are added to ...}\ 25 \qquad \mathrm{(B)}\ 35 \qquad \mathrm{(C)}\ 40 \qquad \mathrm{(D)}\ 45 \qquad \mathrm{(E)}\ 50</math>
  1 KB (166 words) - 00:28, 2 July 2023
• 2013 IMO Problems/Problem 3
  ...of triangle <math>ABC</math>. Prove that triangle <math>ABC</math> is right-angled. ...I_c}</math> be <math>M_a</math>, which lies on <math>(ABC)</math>, the nine-point circle of <math>\triangle I_aI_bI_c</math>; analogously define <math>M
  3 KB (553 words) - 09:41, 17 January 2016
• 2003 AMC 12A Problems/Problem 14
  Using Law of Cosines, we find that side <math>(KL)^2=16+16-2(16)cos{150}\implies (KL)^2=32+\frac{32\sqrt{3}}{2}\implies (KL)^2=32+16\sq ...s are equal, so the triangles are congruent. Notice that <math>\angle{KNL}=45^\circ</math>, etc, so <math>\angle{KNM}=90^\circ</math>. So we have a squar
  3 KB (483 words) - 17:26, 22 June 2023
• 2011 AMC 12A Problems/Problem 19
  ...er of players given elite status is equal to <math>2^{1+\lfloor \log_{2} (N-1) \rfloor}-N</math>. Suppose that <math>19</math> players are given elite s ...1)\rfloor}-N = 19</math>. After rearranging, we get <math>\lfloor\log_{2}(N-1)\rfloor = \log_{2} \left(\frac{N+19}{2}\right)</math>.
  4 KB (563 words) - 11:12, 3 December 2023
• 2011 AMC 12A Problems/Problem 22
  ...ger. A point <math>X</math> in the interior of <math>R</math> is called ''n-ray'' partitional if there are <math>n</math> rays emanating from <math>X</m ...that is <math>100</math>-ray partitional (let this point be the bottom-left-most point).
  11 KB (1,818 words) - 17:38, 6 September 2021
• 1958 AHSME Problems
  ...of a circle with center <math> O</math>. Arc <math> AB</math> is a quarter-circle. Then the ratio of the area of triangle <math> CED</math> to the area \textbf{(C)}\ 45\qquad
  25 KB (3,872 words) - 14:21, 20 February 2020
• 1993 AHSME Problems
  ...th> and <math>c</math>, define <math>\boxed{a,b,c}</math> to mean <math>a^b-b^c+c^a</math>. Then <math>\boxed{1,-1,2}</math> equals <math>\frac{15^{30}}{45^{15}} =</math>
  20 KB (2,814 words) - 08:15, 27 June 2021
• 2011 AMC 10A Problems/Problem 21
  ...r you can switch sides of the scale. But since all numbers are increased 8-fold, it will cancel out when calculating the probability. ...lowed by the six left-over coins. We can do that in <math>\binom{10}{2} = 45</math> different ways.
  4 KB (646 words) - 21:36, 17 March 2024
• 2011 AMC 10B Problems/Problem 18
  <math> \textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 75</math> ...{MC} = 6</math>. We notice that <math>\triangle BMC</math> is a <math>30-60-90</math> triangle, and <math>\angle BMC = 30^{\circ}</math>. If we let <mat
  5 KB (782 words) - 14:29, 1 April 2024
• 2011 AMC 12B Problems
  LeRoy and Bernardo went on a week-long trip together and agreed to share the costs equally. Over the week, ea ...)}\ \frac{A-B}{2} \qquad \textbf{(C)}\ \frac{B-A}{2} \qquad \textbf{(D)}\ B-A \qquad \textbf{(E)}\ A+B</math>
  13 KB (1,978 words) - 16:28, 12 July 2020
• 2011 AMC 12B Problems/Problem 23
  ...+ |y - 2|</math>. Since the minimum value for the difference between the y-coordinates is at <math>y = 0</math>, we get <math>2x + 4 = 20</math> or <ma ...ossible detours from any point would look like a rhombus or square rotated 45 degrees (with centre at a point on or inside the rectangle). Drawing this o
  4 KB (756 words) - 21:17, 31 October 2023
• 2011 AMC 12B Problems/Problem 20
  <math>y_0 = 6-\frac{45}{24} = \frac{33}{8}</math> ...the circumradius quite easily with the formula <math>\sqrt{(s)(s-a)(s-b)(s-c)} = \frac{abc}{4R}</math>, such that <math>s=\frac{a+b+c}{2}</math> and <m
  8 KB (1,200 words) - 19:31, 7 August 2023
• 2001 AMC 10 Problems/Problem 22
  <math> \textbf{(A)}\ 43 \qquad \textbf{(B)}\ 44 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 46 \qquad \textbf{(E)}\ 47 </math> Thus <math> 66-18-25=66-43=v=23 </math>.
  5 KB (721 words) - 16:44, 9 August 2022
• 2011 AIME I Problems
  Jar <math>A</math> contains four liters of a solution that is <math>45\%</math> acid. Jar <math>B</math> contains five liters of a solution that i The vertices of a regular nonagon (9-sided polygon) are to be labeled with the digits 1 through 9 in such a way t
  10 KB (1,634 words) - 22:21, 28 December 2023
• 2011 AIME I Problems/Problem 1
  Jar A contains four liters of a solution that is 45% acid. Jar B contains five liters of a solution that is 48% acid. Jar C c There are <math>\frac{45}{100}(4)=\frac{9}{5}</math> L of acid in Jar A. There are <math>\frac{48}{1
  4 KB (786 words) - 16:43, 5 February 2022
• 2011 AIME I Problems/Problem 2
  <math>HB=\sqrt{17^2-10^2}=3\sqrt{21}</math> ...ath>d^2</math> and moving everything to the LHS, we see that <math>d^2+24d-45=0</math>. Applying the quadratic formula, <math>d=\frac{-24\pm \sqrt{756}}{
  5 KB (873 words) - 04:33, 17 December 2023
• 2011 AIME I Problems/Problem 15
  ...h>. All three of <math>a</math>, <math>b</math>, and <math>c</math> are non-zero: say, if <math>a=0</math>, then <math>b=-c=\pm\sqrt{2011}</math> (which ...w that <math>b</math>, <math>c</math> have the same sign. So <math>|a| \ge 45 = \lceil \sqrt{2011} \rceil</math>.
  8 KB (1,302 words) - 04:07, 24 July 2023
• 2011 AIME I Problems/Problem 14
  ...s 2\left(90-\theta\right)=2\cos^{2}\left(90-\theta\right)-1=2\sin^{2}\theta-1.</cmath> ...2\sqrt{2}\right)}{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}=\frac{5-4\sqrt{2}}{1}=5-\sqrt{32}.</cmath> The answer is <math>\boxed{037}</math>.
  8 KB (1,344 words) - 18:39, 9 February 2023
• 2011 AIME II Problems/Problem 10
  <cmath>K = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{28\cdot 16\cdot 8\cdot 4} = 32\sqrt{14}.</cmath> ...of triangle <math>\triangle EOF</math> is <math>R = \frac{abc}{4K} = \frac{45}{\sqrt{14}}</math>. Looking at <math>EPFO</math>, we see that <math>\angle
  11 KB (1,720 words) - 03:12, 18 December 2023
• 2011 AIME II Problems/Problem 13
  ...d <math>AC</math> be <math>Q</math>. Then <math>BQP</math> is a <math>30-90-60</math> triangle, hence <math>QP=\frac{6\sqrt{2}}{\sqrt{3}}</math> (We kno ...D=45^{\circ}</math>, <math>m\stackrel{\frown}{PD}=m\stackrel{\frown}{PB}=2(45^{\circ})=90^{\circ}</math>. Thus, <math>O_{1}PB</math> and <math>O_{2}PD</m
  13 KB (2,055 words) - 05:25, 9 September 2022
• 2011 AIME II Problems/Problem 15
  P(9) &= 45 \\ ...</cmath>Since <math>a-2b<a+2b</math>, it must be that <math>(a-2b,a+2b)=(1,45),(3,15),(5,9)</math>, which gives solutions <math>(23,11),(9,3),(7,1)</math
  8 KB (1,273 words) - 14:03, 7 January 2023
• 1999 AMC 8 Problems
  <math>\text{(A)}\ 30 \qquad \text{(B)}\ 45 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 75 \qquad \text{(E)}\ 90</math> label("$45$",(0,9),W);
  17 KB (2,394 words) - 19:51, 8 May 2023
• 2011 USAMO Problems
  In hexagon <math>ABCDEF</math>, which is nonconvex but not self-intersecting, no pair of opposite sides are parallel. The internal angles s ...ots</math>, <math>A_{11}</math> of <math>A</math> such that <math>|A_i | = 45</math> for <math>1 \le i \le 11</math> and <math>|A_i \cap A_j| = 9</math>
  3 KB (484 words) - 12:26, 17 April 2016
• 2011 USAJMO Problems/Problem 1
  <math>2^{n-2} + 3^n \cdot 4^{n-1} = a^2 + a + \dfrac {1}{4} (1 - 2011^n)</math>. Since <math>n \ge 2</math>, <math>n-2 \ge 0</math>, so <math>2^{n-2}</math> similarly, the entire LHS is an integer, and so are <math>a^2</mat
  6 KB (990 words) - 21:49, 1 October 2021
• 2011 USAMO Problems/Problem 6
  ...ots</math>, <math>A_{11}</math> of <math>A</math> such that <math>|A_i | = 45</math> for <math>1 \le i \le 11</math> and <math>|A_i \cap A_j| = 9</math> ...tyle \binom{10}2 = 45,</math> each set is in 45 triples and thus will have 45 elements. We can now throw in 60 more elements outside the union of the <ma
  7 KB (1,209 words) - 12:50, 25 August 2023
• 2003 AMC 10B Problems/Problem 23
  ...ac{\sqrt2}{2}\qquad\textbf{(B)}\ \frac{\sqrt2}{4}\qquad\textbf{(C)}\ \sqrt2-1\qquad\textbf{(D)}\ \frac{1}2\qquad\textbf{(E)}\ \frac{1+\sqrt2}{4} </math> https://www.youtube.com/watch?v=LREcUjK-56U&feature=youtu.be
  5 KB (817 words) - 20:02, 28 April 2024
• 2011 AMC 10B Problems
  LeRoy and Bernardo went on a week-long trip together and agreed to share the costs equally. Over the week, eac ...egers <math>a</math> and <math>b</math>, Ron reversed the digits of the two-digit number <math>a</math>. His erroneous product was <math>161</math>. Wha
  13 KB (2,090 words) - 18:05, 7 January 2021
• 2007 AMC 10B Problems
  <math>\textbf{(A) } 35 \qquad\textbf{(B) } 40 \qquad\textbf{(C) } 45 \qquad\textbf{(D) } 50 \qquad\textbf{(E) } 60</math> ...s the average of <math>a</math> and <math>c.</math> How many different five-digit numbers satisfy all these properties?
  15 KB (2,297 words) - 12:57, 19 February 2020
• 2007 AMC 10B Problems/Problem 4
  <math>\textbf{(A) } 35 \qquad\textbf{(B) } 40 \qquad\textbf{(C) } 45 \qquad\textbf{(D) } 50 \qquad\textbf{(E) } 60</math> {{AMC10 box|year=2007|ab=B|num-b=3|num-a=5}}
  928 bytes (140 words) - 12:19, 4 July 2013
• 2008 AMC 12B Problems/Problem 15
  ...qrt{3}</math>. <math>S=4+2\sqrt{3}-(2-\sqrt{3})=2+3\sqrt{3}</math>, <math>S-R=1</math>. {{AMC12 box|year=2008|ab=B|num-b=14|num-a=16}}
  3 KB (505 words) - 22:58, 7 October 2021
• 1998 AJHSME Problems/Problem 24
  ...ular numbers up to <math>120</math> are <math>1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120</math>. When you divide each of those numbers by {{AJHSME box|year=1998|num-b=23|num-a=25}}
  5 KB (686 words) - 15:35, 29 May 2021
• 2002 AMC 10B Problems/Problem 23
  <math> \mathrm{(A) \ } 45\qquad \mathrm{(B) \ } 56\qquad \mathrm{(C) \ } 67\qquad \mathrm{(D) \ } 78\ ...ore, <math>a_m = a_{m-1} + m, a_{m-1}=a_{m-2}+(m-1), a_{m-2} = a_{m-3} + (m-2)</math>, and so on until <math>a_2 = a_1 + 2</math>.
  5 KB (781 words) - 06:38, 23 October 2023
• 2011 AMC 12B Problems/Problem 12
  ...e center square is just <math>x^2</math>. The triangles are all <math>45-45-90</math> triangles, with a side length ratio of <math>1:1:\sqrt{2}</math>. {{AMC12 box|year=2011|ab=B|num-b=11|num-a=13}}
  2 KB (322 words) - 13:08, 10 September 2018
• 2011 AMC 12B Problems/Problem 10
  <math>\textrm{(A)}\ 15 \qquad \textrm{(B)}\ 30 \qquad \textrm{(C)}\ 45 \qquad \textrm{(D)}\ 60 \qquad \textrm{(E)}\ 75</math> {{AMC12 box|year=2011|ab=B|num-b=9|num-a=11}}
  781 bytes (119 words) - 10:11, 5 August 2019
• 1998 AHSME Problems/Problem 6
  ...= 2025</math>, the factors should be as close to <math>44</math> or <math>45</math> as possible. {{AHSME box|year=1998|num-b=5|num-a=7}}
  1 KB (157 words) - 14:28, 5 July 2013
• 1998 AJHSME Problems/Problem 12
  <math>\text{(A)}\ 45 \qquad \text{(B)}\ 49 \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 54 \qquad \t <math>1-\frac{1}{n}=\frac{n-1}{n}</math>
  1 KB (157 words) - 11:48, 20 December 2018
• 1999 AMC 8 Problems/Problem 2
  <math>\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 45 \qquad \textbf{(C)}\ 60 \qquad \textbf{(D)}\ 75 \qquad \textbf{(E)}\ 90</ma {{AMC8 box|year=1999|num-b=1|num-a=3}}
  1 KB (209 words) - 21:45, 15 January 2024
• 1999 AMC 8 Problems/Problem 4
  label("$45$",(0,9),W); ...t Bjorn biked 45 miles, and Alberto biked 60. Thus the answer is <math>60-45=15</math> <math>\boxed{\text{(A)}}</math>.
  1 KB (196 words) - 21:02, 14 April 2017
• 1984 AHSME Problems
  <math> \frac{1000^2}{252^2-248^2} </math> equals [[File:AHSME-1984-Q4.jpg]]
  13 KB (1,879 words) - 14:00, 19 February 2020
• 1984 AHSME Problems/Problem 8
  ...math>, <math> AB=5 </math>, <math> BC=3\sqrt{2} </math>, <math> \angle BCD=45^\circ </math>, and <math> \angle CDA=60^\circ </math>. The length of <math> label("$45^\circ$",(8,-3),WNW);
  2 KB (266 words) - 12:49, 5 July 2013
• 1999 AMC 8 Problems/Problem 21
  draw(arc((19/3,0),(19/3-8/17,-15/17),(22/3,0),CCW)); ...xt{(B)}\ 30 \qquad \text{(C)}\ 35 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 45</math>
  2 KB (332 words) - 22:27, 24 January 2024
• 1984 AHSME Problems/Problem 13
  <math> =\frac{4\sqrt{3}-6\sqrt{2}-2\sqrt{30}}{-6-2\sqrt{15}}</math> ...=\frac{-6\sqrt{3}+9\sqrt{2}+3\sqrt{30}+2\sqrt{45}-3\sqrt{30}-\sqrt{450}}{9-15} </math>
  2 KB (258 words) - 16:43, 1 March 2015
• 1984 AHSME Problems/Problem 29
  ...s of [[real numbers]] <math> (x, y) </math> which satisfy <math> (x-3)^2+(y-3)^2=6 </math>. ...g this out and forming it into a [[quadratic]] yields <math> (1+k^2)x^2+(-6-6k)x+12=0 </math>.
  4 KB (614 words) - 20:09, 12 September 2022
• 1967 AHSME Problems
  ...<math>2a3</math> is added to the number <math>326</math> to give the three-digit number <math>5b9</math>. If <math>5b9</math> is divisible by 9, then ...ht)\left(\frac{y^2+1}{y}\right)+\left(\frac{x^2-1}{y}\right)\left(\frac{y^2-1}{x}\right)</math>, <math>xy \not= 0</math>,
  20 KB (3,108 words) - 14:14, 20 February 2020
• 1998 AJHSME Problems/Problem 23
  draw((45,0)--(57,0)--(51,6sqrt(3))--cycle); ...C)}\ \frac{7}{16}\qquad\text{(D)}\ \frac{9}{16}\qquad\text{(E)}\ \frac{11}{45} </math>
  2 KB (279 words) - 15:23, 29 May 2021
• 1984 AHSME Problems/Problem 15
  ...thrm{(B) \ }30^\circ \qquad \mathrm{(C) \ } 36^\circ \qquad \mathrm{(D) \ }45^\circ \qquad \mathrm{(E) \ } 60^\circ </math> ...can be defined by the slope of the line that makes that angle with the [[x-axis]].
  3 KB (493 words) - 18:16, 4 June 2021
• 2005 CEMC Gauss (Grade 7) Problems
  MarkPoint("9.4",dir(45),dir(45)); ...it vector - use NE to move it slightly NorthEast of this point (or use dir(45))
  16 KB (2,317 words) - 03:54, 24 October 2014
• 2009 AMC 8 Problems
  pair ac=C+2.828*dir(45), ba=(-6-2.828, 0);
  18 KB (2,551 words) - 18:46, 27 February 2024
• 1953 AHSME Problems/Problem 46
  {{AHSME 50p box|year=1953|num-b=45|num-a=47}}
  1 KB (185 words) - 01:11, 4 January 2019
• 1999 AMC 8 Problems/Problem 12
  ...}\ 24\% \qquad \text{(B)}\ 27\% \qquad \text{(C)}\ 36\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ 73\%</math> {{AMC8 box|year=1999|num-b=11|num-a=13}}
  1 KB (189 words) - 19:56, 9 June 2022
• 2000 AMC 8 Problems/Problem 25
  Adding these up, we get <math>45</math>, and subtracting that from <math>72</math>, we get <math>27</math>, {{AMC8 box|year=2000|num-b=24|after=Last<br />Question}}
  3 KB (484 words) - 13:59, 22 October 2023
• 1993 AJHSME Problems/Problem 11
  label("$45$",(4,-.25),S); {{AJHSME box|year=1993|num-b=10|num-a=12}}
  2 KB (238 words) - 00:11, 5 July 2013
• 1999 AMC 8 Problems/Problem 19
  ...e <math>3</math> tablespoons of butter per pan, meaning <math>3 \cdot 15 = 45</math> tablespoons of butter are required for <math>15</math> pans. ...partial sticks of butter are forbidden! Thus, we need <math>\lceil \frac{45}{8} \rceil = 6</math> sticks of butter, and the answer is <math>\boxed{B}</
  2 KB (266 words) - 22:26, 24 January 2024
• 1997 AJHSME Problems/Problem 18
  ...}\ 30\% \qquad \text{(B)}\ 35\% \qquad \text{(C)}\ 40\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ 65\%</math> ..., the percent decrease is <math>\frac{1.25 - 0.8}{1.25}\cdot 100\% = \frac{45}{125} \cdot 100\% = 36\%</math>, which is closest to <math>\boxed{B}</math>
  918 bytes (130 words) - 19:49, 31 October 2016
• 1996 AJHSME Problems/Problem 4
  ...arly, the denominator has <math>17</math> terms of <math>3 + 6 + 9 + ... + 45 + 48 + 51</math>. There are <math>8</math> pairs of numbers that add up to {{AJHSME box|year=1996|num-b=3|num-a=5}}
  2 KB (348 words) - 00:24, 5 July 2013
• 1996 AJHSME Problems/Problem 23
  ...ss than what was needed. Instead the manager gave each employee a <math>\$45</math> bonus and kept the remaining <math>\$95</math> in the company fund. ...th>, the only number that is <math>95</math> more than a multiple of <math>45</math> out of options <math>A</math> and <math>E</math> is option <math>\bo
  2 KB (303 words) - 17:00, 29 January 2019
• 1995 AJHSME Problems/Problem 10
  <math>\text{(A)}\ 45\% \qquad \text{(B)}\ 47\dfrac{1}{2}\% \qquad \text{(C)}\ 50\% \qquad \text{ He saved <math>\frac{54}{120}\cdot 100\% = 0.45 \cdot 100\% = 45\%</math> off the total of the original prices, and the answer is <math>\box
  1,012 bytes (136 words) - 17:02, 12 November 2019
• 1997 AHSME Problems/Problem 8
  ...e <math>11n > 490</math>, leading to <math>n > 44.5</math>. Thus, <math>n=45, 46, 47, 48</math> will be more expensive than <math>n=49</math>. {{AHSME box|year=1997|num-b=7|num-a=9}}
  2 KB (293 words) - 22:58, 15 February 2018
• 1997 AHSME Problems/Problem 14
  <math>(x - 84)(x + 45) = 0</math> {{AHSME box|year=1997|num-b=13|num-a=15}}
  1 KB (217 words) - 14:13, 5 July 2013
• 2003 AMC 10B Problems/Problem 24
  The first four terms in an arithmetic sequence are <math>x+y</math>, <math>x-y</math>, <math>xy</math>, and <math>\frac{x}{y}</math>, in that order. What ...an also express the third and fourth terms as <math>x-3y</math> and <math>x-5y.</math> Then we can set them equal to <math>xy</math> and <math>\frac{x}{
  4 KB (779 words) - 16:16, 12 March 2024
• 2008 AMC 8 Problems/Problem 7
  If <math>\frac{3}{5}=\frac{M}{45}=\frac{60}{N}</math>, what is <math>M+N</math>? \textbf{(C)}\ 45 \qquad
  572 bytes (80 words) - 11:21, 27 April 2024
• 2008 AMC 8 Problems/Problem 15
  ...tenth game, she scored fewer than <math>10</math> points and her points-per-game average for the <math>10</math> games was also an integer. What is the ...points. The closest multiple of <math>9</math> is <math>45</math>. <math>45-37=8</math>. Now we have to add a number to get a multiple of 10. The next m
  1 KB (190 words) - 19:24, 8 August 2021
• 2006 AMC 8 Problems/Problem 14
  ...in the class. Alice reads a page in 20 seconds, Bob reads a page in <math>45</math> seconds and Chandra reads a page in <math>30</math> seconds. ...ll we need for this problem is that there's 760 pages, Bob reads a page in 45 seconds and Chandra reads a page in 30 seconds. A lot of people will find h
  1 KB (196 words) - 01:22, 5 July 2013
• Primitive Pythagorean Triple
  27 36 45 [3 - 4 - 5] 24 45 51 [8 - 15 - 17]
  55 KB (3,565 words) - 11:05, 21 May 2020
• 2006 AMC 8 Problems/Problem 16
  ...a, are in the class. Alice reads a page in 20 seconds, Bob reads a page in 45 seconds and Chandra reads a page in 30 seconds. ...erefore, the amount of seconds each person reads is simply <math>160 \cdot 45 = \boxed{\textbf{(E)}\ 7200}</math>.
  2 KB (361 words) - 19:12, 11 January 2023
• 2009 AMC 8 Problems/Problem 7
  pair ac=C+2.828*dir(45), ba=(-6-2.828, 0);
  2 KB (315 words) - 08:23, 30 May 2023
• 1994 AJHSME Problems/Problem 2
  <math> 1+ 2+ 3 + 4 + 5 + 6 + 7 + 8 + 9 = \dfrac{(9)(10)}{2} = 45 </math> <math>\frac{45+55}{10} = \dfrac{100}{10} = \boxed{\text{(D)}\ 10}</math>
  527 bytes (50 words) - 00:12, 5 July 2013
• 1994 AJHSME Problems/Problem 3
  ...day Maria must work <math>8</math> hours. This does not include the <math>45</math> minutes she takes for lunch. If she begins working at <math>\text{7 8 hours from 7:25 AM is 15:25 or 3:25 PM. 45 minutes from 25 minutes is 10 minutes after the hour, so her working day en
  703 bytes (114 words) - 13:17, 30 October 2016
• 1992 AJHSME Problems/Problem 18
  <math>\text{(A)}\ 45 \qquad \text{(B)}\ 50 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 75 \qquad \t <cmath>\frac{80+0+100}{4} = \boxed{\text{(\textbf{A})}\ 45}</cmath>
  666 bytes (101 words) - 05:42, 31 August 2015
• 2010 PUMaC Problems
  Find the sum of the coefficients of the polynomial <math>(63x-61)^4</math>. Calculate <math>{\sum_{n=1}^\infty\left(\lfloor\sqrt[n]{2010}\rfloor-1\right)}</math> where <math>\lfloor x\rfloor</math> is the largest integer
  22 KB (3,694 words) - 23:58, 3 June 2022
• 2016 AMC 10A Problems/Problem 19
  label("$B$", B, dir(45)); label("$C$", C, dir(-45));
  8 KB (1,386 words) - 15:10, 8 October 2023
• 1995 AJHSME Problems/Problem 13
  ...BED=180</cmath> <cmath>2\angle BED=90</cmath> <cmath>\angle BED=\angle BDE=45^\circ</cmath> So <math>\angle AED=\angle AEB +\angle BED=40 +45=85^\circ</math> Since ACDE is a quadrilateral, the sum of its angles is 360
  1 KB (210 words) - 21:08, 27 October 2016
• 1995 AJHSME Problems/Problem 16
  Altogether, the summer project totaled <math>(7)(3)+(4)(5)+(5)(9)=21+20+45=86</math> days of work for a single student. This equals <math>744/86=9</ma {{AJHSME box|year=1995|num-b=15|num-a=17}}
  1 KB (141 words) - 10:15, 19 August 2019
• 1995 AJHSME Problems/Problem 17
  ...graders and <math>300</math> total students, so the percent is <math>\frac{45}{300}=\frac{15}{100}= \boxed{\text{(D)}\ 15\%}</math> {{AJHSME box|year=1995|num-b=16|num-a=18}}
  1 KB (125 words) - 03:15, 23 December 2012
• 1995 AJHSME Problems/Problem 25
  ...uses travel on the same highway, how many Dallas-bound buses does a Houston-bound bus pass in the highway (not in the station)? ...many buses you pass. At 12:45 pm, the Dallas bus that left at 8:00 am is 4:45 away (Note - <math>a:b</math> - <math>a</math> is for hrs. and <math>b</mat
  2 KB (283 words) - 22:37, 5 June 2020
• 2007 Alabama ARML TST Problems/Problem 15
  ...>DE=EC+CD=2CP=12\sqrt{2}</math>. This shows that <math>DEF</math> is a 5-12-13 right triangle, so it has area <math>\frac{EF\cdot DE}{2}=60</math>, so ...\circ</math>. Note that <math>\angle APC = \angle APP' + PP'C = 90^\circ + 45^\circ = 135^\circ</math>, so by Law of Cosines on triangle <math>APC</math>
  2 KB (389 words) - 18:12, 21 March 2018
• 1950 AHSME
  ** [[1950 AHSME Problems/Problem 45|Problem 45]]
  3 KB (254 words) - 14:26, 20 February 2020
• 1950 AHSME Problems
  Let <math> R=gS-4 </math>. When <math>S=8</math>, <math>R=16</math>. When <math>S=10</math>, The sum of the roots of the equation <math> 4x^{2}+5-8x=0 </math> is equal to:
  22 KB (3,306 words) - 19:50, 3 May 2023
• 2006 AMC 8 Problems/Problem 11
  How many two-digit numbers have digits whose sum is a perfect square? ...9 </math> integers whose digits sum to <math> 9 </math>: <math>18, 27, 36, 45, 54, 63, 72, 81, \text{and } 90</math>.
  971 bytes (130 words) - 13:44, 25 April 2021
• 1985 AHSME Problems
  ...ame as <math>a(b-c)</math> in ordinary algebraic notation. If <math>a\div b-c+d</math> is evaluated in such a language, the result in ordinary algebraic ...\qquad \mathrm{(D) \ } \frac{a}{b-c+d} \qquad \mathrm{(E) \ }\frac{a}{b-c-d} </math>
  17 KB (2,488 words) - 03:26, 20 March 2024
• 2005 CEMC Gauss (Grade 7) Problems/Problem 3
  MarkPoint("9.4",dir(45),dir(45)); ...it vector - use NE to move it slightly NorthEast of this point (or use dir(45))
  2 KB (272 words) - 01:43, 23 October 2014
• 1985 AHSME Problems/Problem 16
  ...n A+\tan B}{1-\tan A\tan B},</cmath> giving <cmath>\begin{align*}1 &= \tan 45^{\circ} \\ &= \tan(A+B) \\ &= \frac{\tan A+\tan B}{1-\tan A\tan B},\end{ali ...cos B+\sin B\cos A,\end{align*}</cmath> this reduces to <cmath>\frac{\cos(A-B)+\sin(A+B)}{\cos A\cos B}.</cmath>
  5 KB (904 words) - 22:25, 19 March 2024
• 1985 AHSME Problems/Problem 17
  pair W=A+a*dir(135), X=B+a*dir(45), Y=C+a*dir(-45), Z=D+a*dir(-135); {{AHSME box|year=1985|num-b=16|num-a=18}}
  2 KB (340 words) - 22:39, 19 March 2024
• 1985 AHSME Problems/Problem 26
  ...ve integer <math>n</math> for which <math>\frac{n-13}{5n+6}</math> is a non-zero reducible fraction. <math> \mathrm{(A)\ } 45 \qquad \mathrm{(B) \ }68 \qquad \mathrm{(C) \ } 155 \qquad \mathrm{(D) \
  954 bytes (144 words) - 01:44, 20 March 2024
• 2010 AMC 8 Problems
  ...); label("$ 20$",(0.2,-1.85),SE*lsf,fp); label("$\mathrm{Price}$",(0.16,-3.45),SE*lsf,fp); label("$1$",(1.54,-5.97),SE*lsf,fp); label("$2$",(2.53,-5.95), <math>\textbf{(B)}\ \text{non-square rhombus} </math>
  18 KB (2,768 words) - 21:05, 9 January 2024
• 2006 AMC 8 Problems/Problem 15
  ...a, are in the class. Alice reads a page in 20 seconds, Bob reads a page in 45 seconds and Chandra reads a page in 30 seconds. <math>30x = 45(760-x)</math> Distribute the <math>45</math>
  2 KB (323 words) - 22:54, 6 January 2023
• 2011 AMC 8 Problems
  ...h> apples at a cost of <math> 50 </math> cents per apple. She paid with a 5-dollar bill. How much change did Margie receive? Karl's rectangular vegetable garden is <math> 20 </math> feet by <math> 45 </math> feet, and Makenna's is <math> 25 </math> feet by <math> 40 </math>
  16 KB (2,371 words) - 17:34, 9 January 2024
• 2011 AMC 8 Problems/Problem 2
  Karl's rectangular vegetable garden is <math> 20 </math> feet by <math> 45 </math> feet, and Makenna's is <math> 25 </math> feet by <math> 40 </math> ...et. Since <math>1000 > 900,</math> Makenna's garden is larger by <math>1000-900=100</math> square feet. <math>\Rightarrow \boxed{ \textbf{(E)}\ \text{Ma
  1 KB (155 words) - 12:35, 13 November 2016
• 2011 AMC 8 Problems/Problem 6
  ...a car, motorcycle, or both. If <math>331</math> adults own cars and <math>45</math> adults own motorcycles, how many of the car owners do not own a moto <math>\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 306 \qquad \textbf{(E)}\ 351</math>
  1 KB (194 words) - 00:34, 27 December 2022
• 2011 AMC 8 Problems/Problem 14
  ...\dfrac{22}{45} \qquad\textbf{(D) } \dfrac12 \qquad\textbf{(E) } \dfrac{23}{45} </math> ...\frac{120+100}{270+180} = \frac{220}{450} = \boxed{\textbf{(C)}\ \frac{22}{45}}</math>
  2 KB (250 words) - 15:10, 17 December 2023
• 2011 AMC 8 Problems/Problem 16
  '''25-25-30''' <cmath>x^2 = (25 + 15)(25-15)</cmath>
  2 KB (371 words) - 16:51, 21 January 2024
• 1950 AHSME Problems/Problem 46
  {{AHSME 50p box|year=1950|num-b=45|num-a=47}}
  1 KB (172 words) - 16:49, 15 July 2023
• 2011 AMC 8 Problems/Problem 11
  label("T",(-25,45),W); {{AMC8 box|year=2011|num-b=10|num-a=12}}
  2 KB (346 words) - 15:00, 17 December 2023
• 1950 AHSME Problems/Problem 30
  ...girls leave. There are then left two boys for each girl. After this <math>45</math> boys leave. There are then <math>5</math> girls for each boy. The nu From the first sentence, we get that <math>2(g-15)=b</math>.
  1,015 bytes (161 words) - 01:01, 12 October 2020
• 1950 AHSME Problems/Problem 44
  {{AHSME 50p box|year=1950|num-b=43|num-a=45}}
  1 KB (188 words) - 16:16, 9 May 2015
• 1950 AHSME Problems/Problem 50
  A privateer discovers a merchantman <math>10</math> miles to leeward at 11:45 a.m. and with a good breeze bears down upon her at <math>11</math> mph, whi <math>\textbf{(A)}\ 3\text{:}45\text{ p.m.} \qquad
  2 KB (309 words) - 16:53, 15 July 2023
• 1989 AHSME Problems
  ...qquad\textrm{(C)}\ 30^\circ\qquad\textrm{(D)}\ 40^\circ\qquad\textrm{(E)}\ 45^\circ </math> ...th>n</math> between <math>1</math> and <math>100</math> does <math> x^{2}+x-n </math> factor into the product of two linear factors with integer coeffic
  15 KB (2,247 words) - 13:44, 19 February 2020
• 2012 AMC 12A Problems/Problem 10
  pair A=(-5,0), B=(5,0), C=(sqrt(45),6), D=(0,0), E=(sqrt(45),0); {{AMC12 box|year=2012|ab=A|num-b=9|num-a=11}}
  2 KB (379 words) - 01:59, 16 February 2021
• 2012 AMC 12A Problems/Problem 18
  ..., and <math>z+y=26</math>. Subtracting the last 2 equations we have <math>x-z=-1</math> and adding this to the first equation we have <math>x=13</math>. ...} + 26\overline{BB'}^2</math>. Solving we get <math>\overline{BB'} = \frac{45}{2}</math>. Plugging it in we get <math>\overline{BI} = 15</math>. Therefor
  4 KB (717 words) - 19:07, 28 July 2021
• 1992 AHSME Problems/Problem 17
  The 2-digit integers from 19 to 92 are written consecutively to form the integer < ...ath>9</math>. For each one of these cycles, we add <math>0 + 1 + ... + 9 = 45</math>. This is divisible by <math>9</math>, thus we can ignore the sum. Ho
  2 KB (342 words) - 13:19, 4 January 2021
• 1989 AHSME Problems/Problem 19
  ...owledge, the angle that cuts off the arc of length <math>3</math> is <math>45</math> degrees (an inscribed angle is <math>\frac{1}{2}</math> the measure ...ude and also part of the 45-45-90 triangle, it makes both legs of the 45-45-90 triangle to be x√3. We sum together the areas of both triangles and get
  4 KB (649 words) - 10:04, 20 May 2021
• 1989 AHSME Problems/Problem 13
  transform t = rotate(-45,(3,.5)); transform t = rotate(-45,(3,.5));
  2 KB (242 words) - 07:51, 22 October 2014
• 2012 AMC 10B Problems
  ...bbits. How many more students than rabbits are there in all 4 of the third-grade classrooms? The point in the xy-plane with coordinates <math>(1000, 2012)</math> is reflected across the lin
  18 KB (2,350 words) - 18:48, 9 July 2023
• 2012 AMC 12B Problems
  ...abbits. How many more students than rabbits are there in all 4 of the third-grade classrooms? In order to estimate the value of <math>x-y</math> where <math>x</math> and <math>y</math> are real numbers with <math
  20 KB (2,681 words) - 09:47, 29 June 2023
• 2012 AMC 12B Problems/Problem 7
  ...Since the question asks for the answer in feet, the answer is <math>\frac{45*6}{12} \rightarrow \boxed{\textbf{(E)}\ 22.5}</math>. {{AMC12 box|year=2012|ab=B|num-b=6|num-a=8}}
  1 KB (180 words) - 07:37, 29 June 2023
• 2012 AMC 10B Problems/Problem 12
  ...math> and <math>C</math> is <math>10\sqrt 2</math>, and <math>\angle BAC = 45^\circ</math>. Point <math>D</math> is <math>20</math> meters due north of p ...f <math>\angle BAC = 45^\circ</math>, <math>\triangle CBA</math> is a 45-45-90 triangle. Thus, the lengths of sides <math>CB</math>, <math>BA</math>, an
  2 KB (293 words) - 20:22, 3 September 2021
• 2012 AMC 12B Problems/Problem 19
  label("$P_3$", P[3], dir(-45)); [[File:2012_AMC-12B-19‎.jpg]]
  5 KB (815 words) - 21:59, 19 September 2023
• 1989 AHSME Problems/Problem 16
  ...h>, and the difference in the <math>x</math>-coordinates is <math>48 - 3 = 45</math>. The gcd of 264 and 45 is 3, so the line segment joining <math>(3,17)</math> and <math>(48,281)</m
  1 KB (201 words) - 03:22, 11 July 2020
• 2010 AMC 8 Problems/Problem 19
  unitsize(45); ...the radius of the larger circle. The area of the annulus is <math>\pi(AC^2-BC^2)=\pi AB^2=64\pi</math>.
  2 KB (279 words) - 09:04, 10 March 2023
• 2012 AMC 10B Problems/Problem 19
  ...<math>AG=\frac{AD}{2}=15,</math> <math>\triangle AGB=\frac{15\times 6}{2}=45</math>. Now, <math>\triangle AED\sim \triangle BEF</math>, so <math>\frac{A ...fore, <cmath>\begin{align*}BFDG&=ABCD-\triangle AGB-\triangle DCF \\ &=180-45-\frac{135}{2} \\ &=\boxed{\frac{135}{2}}\end{align*}</cmath>
  3 KB (509 words) - 14:23, 23 August 2022
• 2015 AIME I Problems
  ...ates a sequence according to the rule <math>a_n=a_{n-1}\cdot | a_{n-2}-a_{n-3} |</math> for all <math>n\ge 4</math>. Find the number of such sequences f Let <math>f(x)</math> be a third-degree polynomial with real coefficients satisfying
  10 KB (1,615 words) - 21:48, 13 January 2024
• 2012 AIME I Problems/Problem 12
  ...CDE = 120-B</math>, <math>\angle CDA = 60+B</math>, and <math>\angle A = 90-B</math>. ...CE}{\sin (120-B)}</math>. This simplifies to <math>16 = \frac{CE}{\sin (120-B)}</math>.
  9 KB (1,523 words) - 12:23, 7 September 2022
• 2012 AIME I Problems/Problem 13
  ...ath>BX = 4,</math> and <math>CX = 3.</math> A <math>60^\circ</math> counter-clockwise rotation about vertex <math>A</math> maps <math>X</math> to <math> <cmath>BC^2 = BX^2+CX^2 - 2 \cdot BX \cdot CX \cdot \cos150^\circ = 4^2+3^2-24 \cdot \frac{-\sqrt{3}}{2} = 25+12\sqrt{3}</cmath>
  11 KB (1,889 words) - 20:42, 25 January 2023
• 1975 IMO Problems/Problem 4
  ...math>. Note that out of all of the positive integers less than or equal to 45, the maximal sum of the digits is 12. It's not hard to prove that any base-10 number is equivalent to the sum of its digits modulo 9. Therefore <math>4
  2 KB (273 words) - 09:45, 24 April 2024
• 2012 AIME II Problems/Problem 7
  ...spaces, there are <math>\binom{11}{8}=165</math>, which includes the first 45. You're getting the handle. For 12 spaces, there are <math>\binom{12}{8}= ...But here <math>\binom{7}{4}=35</math>, so N must be the last or largest 7-digit number with 4 1's. Thus the last 8 digits of <math>N</math> must be <
  2 KB (345 words) - 14:55, 17 October 2020
• 2017 AMC 8 Problems
  label("45\%", (0.7, 0)); Let <math>Z</math> be a 6-digit positive integer, such as 247247, whose first three digits are the sam
  12 KB (1,771 words) - 21:13, 20 January 2024
• Mock AIME II 2012 Problems/Problem 13
  .../math> in order to bisect the area of the octahedron. We see that the cross-section will be a hexagon as it passes through six of the eight faces. By sy ...math>\sin 45 = \frac{1}{\sqrt 2}</math> and so have values less than <math>45</math>.
  4 KB (628 words) - 19:32, 27 December 2012
• Mock AIME II 2012 Problems/Problem 14
  ...michael number}</math>, <math>n</math>, <math>(p-1)</math> divides <math>(n-1)</math> and <math>n</math> is not prime. Find the sum of all two digit <ma ...math>(a+1)^n\pmod{a}\equiv 1^n\pmod{a}</math>, and therefore we get <math>1-1\equiv 0\pmod{a}</math>. Therefore, some two digit <math>\textit{near Carm
  4 KB (789 words) - 03:23, 5 April 2012
• Mock AIME I 2012 Problems/Problem 8
  ...cmath> Taking the fourth power, the desired answer is <math>\dfrac{7+\sqrt{45}}{2} \implies r+s+t=\boxed{054}</math>. ..., connect <math>z^2</math> to the origin and construct an altitude to the x-axis. Extend this line one unit above <math>z^2</math> and connect that poin
  3 KB (543 words) - 18:51, 7 May 2020
• Mock AIME I 2012 Problems/Problem 15
  First, note that <math>x^{2k}+x^k+1=(x^{3k}-1)/(x^k-1)</math>. Let <math>z_1,z_2,\dots,z_n</math> be the distinct complex number ...^{20}(x^{2k}+x^k+1)=\prod_{k=1}^{20}\frac{x^{3k}-1}{x^k-1}=\prod_{i=1}^n (x-z_i)^{t(z_i)}.\tag{1}
  3 KB (488 words) - 20:05, 10 March 2015
• 1951 AHSME Problems/Problem 12
  ...that the angle is <math>\frac{|60(2)-11(15)|}{2}=\frac{|120-165|}{2}=\frac{45}{2}=\boxed{\textbf{(C)}\ 22\frac {1}{2}^{\circ} }</math> {{AHSME 50p box|year=1951|num-b=11|num-a=13}}
  698 bytes (97 words) - 12:20, 5 July 2013
• 1951 AHSME Problems/Problem 44
  ...uad\textbf{(B)}\ \frac{2abc}{ab+bc+ac}\qquad\textbf{(C)}\ \frac{2abc}{ab+ac-bc} </math> ...h> \textbf{(D)}\ \frac{2abc}{ab+bc-ac}\qquad\textbf{(E)}\ \frac{2abc}{ac+bc-ab} </math>
  1 KB (168 words) - 12:26, 5 July 2013
• 2010 AMC 8 Problems/Problem 18
  <cmath>AD=45</cmath> ...he area is <math>225\pi</math>. The area of the rectangle is <math>30\cdot 45=1350</math>. We calculate the ratio:
  2 KB (301 words) - 09:04, 10 March 2023
• 1974 AHSME Problems
  <math> \mathrm{(A)\ } 4x+3y=xy \qquad \mathrm{(B) \ }y=\frac{4x}{6-y} \qquad \mathrm{(C) \ } \frac{x}{2}+\frac{y}{3}=2 \qquad </math> <math> \mathrm{(D) \ } \frac{4y}{y-6}=x \qquad \mathrm{(E) \ }\text{none of these} </math>
  15 KB (2,151 words) - 14:04, 19 February 2020
• 1974 AHSME Problems/Problem 23
  unitsize(45); unitsize(45);
  2 KB (337 words) - 12:44, 5 July 2013
• 1950 AHSME Answer Key
  #A<!--45--> 6: <math>\textbf{(C)}\ y^{2}+10y-7=0</math>
  3 KB (443 words) - 06:25, 20 January 2023
• 1952 AHSME Problems
  The expression <math> a^3-a^{-3} </math> equals: ...ac{1}{a^2}\right) \qquad \textbf{(C) \ }\left(a-\frac{1}{a}\right)\left(a^2-2+\frac{1}{a^2}\right) \qquad </math>
  23 KB (3,556 words) - 15:35, 30 December 2023
• 1952 AHSME Problems/Problem 44
  ...d \textbf{(B) \ } 10-k \qquad \textbf{(C) \ } 11-k \qquad \textbf{(D) \ } k-1 \qquad \textbf{(E) \ } k+1 </math> ...h>, we have <math>11 = k + x</math>, or <math>x = \boxed{\textbf{(C) \ } 11-k}</math>.
  743 bytes (129 words) - 04:38, 15 January 2023
• 1953 AHSME Problems
  <math>\textbf{(A)}\ (x+y)(x-y) \qquad \textbf{(D)}\ (x+iy)(x-iy)\qquad
  21 KB (3,123 words) - 14:24, 20 February 2020
• 1954 AHSME Problems
  The square of <math>5-\sqrt{y^2-25}</math> is: ...extbf{(C)}\ y^2 \\ \textbf{(D)}\ (5-y)^2\qquad\textbf{(E)}\ y^2-10\sqrt{y^2-25} </math>
  23 KB (3,535 words) - 16:29, 24 April 2020
• 1955 AHSME Problems
  The equality <math>\frac{1}{x-1}=\frac{2}{x-2}</math> is satisfied by: The graph of <math>x^2-4y^2=0</math>:
  22 KB (3,509 words) - 21:29, 31 December 2023
• 1980 AHSME Problems
  If the ratio of <math>2x-y</math> to <math>x+y</math> is <math>\frac{2}{3}</math>, what is the ratio pair D=origin, A=(13,0), C=D+12*dir(r), B=A+3*dir(180-(90-r+s));
  15 KB (2,302 words) - 10:47, 30 April 2021
• 1989 AHSME Problems/Problem 23
  ...36,45)\qquad\text{(C)}\ (37,45)\qquad\text{(D)}\ (44,35)\qquad\text{(E)}\ (45,36)</math> <cmath>\sum_{k=1}^n1+2\sum_{k=1}^nk=n+n(n+1)=n(n+2)=(n+1)^2-1</cmath>
  3 KB (387 words) - 14:24, 21 June 2023
• 2020 Mock Combo AMC 10
  ...ths from the origin <math>(0, 0)</math> to a point on the line <math>y=2020-2x</math> such that each step is from <math>(x, y)</math> to either <math>(x <cmath>r_1 = 1</cmath><cmath>r_n = nr_{n-1} + n</cmath>Let <math>S</math> be the sum of all <math>n</math> such that
  15 KB (2,452 words) - 03:03, 4 July 2020
• 2008 AMC 8 Problems/Problem 25
  <math> \textbf{(A)}\ 42\qquad \textbf{(B)}\ 44\qquad \textbf{(C)}\ 45\qquad \textbf{(D)}\ 46\qquad \textbf{(E)}\ 48\qquad</math> ...ath>144\pi</math>. The area of the black regions is <math>(100-64)\pi + (36-16)\pi + 4\pi = 60\pi</math>. The percentage of the design that is black is
  1 KB (227 words) - 20:38, 2 January 2023
• 2008 AMC 8 Problems
  ...F LUCK}</math> represents the ten digits <math>0-9</math>, in order. What 4-digit number is represented by the code word <math>\text{CLUE}</math>? draw((1.88,3.25)--(9.45,3.25));
  14 KB (2,035 words) - 15:23, 26 January 2024
• 2015 AMC 10B Problems/Problem 24
  If we substitute <math>k = 22</math> into the equation: <math>44(45) = 1980 < 2015</math>. So he has <math>35</math> moves to go. This makes So <math>p_{2015} = (0+1-2+3-4+5...-44+35, 0+1-2+3-4+5...-44)</math>.
  5 KB (916 words) - 17:42, 13 August 2023
• 2004 AMC 8 Problems/Problem 10
  ...th> minutes on Tuesday, from 8:20 to 10:45 on Wednesday morning, and a half-hour on Friday. He is paid <math>\textdollar 3</math> per hour. How much did {{AMC8 box|year=2004|num-b=9|num-a=11}}
  1 KB (142 words) - 02:09, 29 January 2023
• 2004 AMC 8 Problems/Problem 25
  ...\qquad \text{(C)}\ 28-4\pi \qquad \text{(D)}\ 28-2\pi \qquad \text{(E)}\ 32-2\pi</math> ...the circle. This value can be found with Pythagorean or a <math>45^\circ - 45^\circ - 90^\circ</math> circle to be <math>2\sqrt{2}</math>. The radius is
  2 KB (346 words) - 21:43, 2 January 2023
• 2005 AMC 8 Problems/Problem 3
  Rotating square <math>ABCD</math> counterclockwise <math>45^\circ</math> so that the line of symmetry <math>BD</math> is a vertical lin {{AMC8 box|year=2005|num-b=2|num-a=4}}
  1 KB (152 words) - 13:33, 19 October 2020
• 2013 AMC 12A Problems
  <math> \textbf {(A) } 35 \qquad \textbf {(B) } 40 \qquad \textbf {(C) } 45 \qquad \textbf {(D) } 50 \qquad \textbf {(E) } 55 </math> ...th> dollars, and Dorothy gave Sammy <math>d</math> dollars. What is <math>t-d</math>?
  14 KB (2,206 words) - 19:31, 15 May 2024
• 2013 AMC 12A Problems/Problem 2
  <math> \textbf {(A) } 35 \qquad \textbf {(B) } 40 \qquad \textbf {(C) } 45 \qquad \textbf {(D) } 50 \qquad \textbf {(E) } 55 </math> Therefore, the total runs by the opponent is <math>(2+4+6+8+10)+(1+2+3+4+5) = 45</math>, which is <math>C</math>
  1,004 bytes (158 words) - 18:24, 23 August 2021
• 2013 AMC 12A Problems/Problem 16
  ...teger, and the only fraction that satisfies both conditions is <math>\frac{45}{46}</math> <math>44 + \frac{46}{3} * \frac{45}{46} = 44 + 15 = 59</math>, which is choice E
  7 KB (1,207 words) - 01:43, 29 October 2023
• 2013 AMC 10A Problems/Problem 25
  A[i] = dir(45*i); ...ed to subtract <math>\dbinom{4}{2} -1 = 5</math> from this count, <math>70-5 = 65</math>. Note that diagonals like <math>\overline{AD}</math>, <math>\
  6 KB (1,054 words) - 09:21, 28 December 2021
• 2013 AMC 10A Problems/Problem 4
  <math> \textbf{(A)}\ 35 \qquad\textbf{(B)}\ 40 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 50 \qquad\textbf{(E)}\ 55 </math> ...ores is <math>2 + 1 + 4 + 2 + 6 + 3 + 8 + 4 + 10 + 5 = \boxed{\textbf{(C) }45}</math>
  2 KB (303 words) - 21:40, 4 July 2023
• 2013 AMC 12A Problems/Problem 22
  ...e forwards and backwards when written in base 10 with no leading zeros. A 6-digit palindrome <math>n</math> is chosen uniformly at random. What is the p ...tiply 5-digit palindromes <math>ABCBA</math> by <math>11</math>, giving a 6-digit palindrome:
  3 KB (423 words) - 00:52, 16 April 2023
• 2013 AMC 10A Problems/Problem 20
  A unit square is rotated <math>45^\circ</math> about its center. What is the area of the region swept out by path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;
  4 KB (701 words) - 17:55, 23 July 2021
• 2013 AMC 10B Problems/Problem 22
  .../math>, or <math>9</math> to preserve symmetry, since the sum of 1 to 9 is 45, and we need the remaining 8 to be divisible by 4 (otherwise we will have u <math>4S=45+3J</math>
  4 KB (746 words) - 17:29, 30 September 2023
• 2013 AMC 10A Problems
  A taxi ride costs \$1.50 plus \$0.25 per mile traveled. How much does a 5-mile taxi ride cost? <math> \textbf{(A)}\ 35 \qquad\textbf{(B)}\ 40 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 50 \qquad\textbf{(E)}\ 55 </math>
  12 KB (1,894 words) - 15:59, 3 January 2024
• 2013 AMC 12A Problems/Problem 20
  ...are just chosen from the two end-points of the arc, and there are <math>19-k</math> possible places for the third person. Once the three places of <mat <cmath>\sum_{k=11}^{18} 3\cdot 19 \cdot (19-k) = 3\cdot 19 \cdot (1+\cdots+8) = 3\cdot 19\cdot 36</cmath>
  5 KB (885 words) - 10:14, 29 October 2023
• 2013 AMC 10B Problems/Problem 12
  ...in which you end up with two segments of the same length. <math>\frac{20}{45}</math> is equivalent to <math>\boxed{\textbf{(B) }\frac{4}{9}}</math>. ...inom{5}{2} + \binom{5}{2}}{\binom{10}{2}} = \dfrac{10+10}{45} = \dfrac{20}{45}=\boxed{\textbf{(B) }\dfrac49}</cmath>.
  2 KB (316 words) - 11:53, 27 April 2024
• 2013 AMC 10B Problems/Problem 18
  <math> \textbf{(A)}\ 33\qquad\textbf{(B)}\ 34\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ 58 </math> ...each pair uniquely determines the value of <math>d</math>, so we get <math>45</math> numbers with the given property.
  3 KB (480 words) - 22:23, 26 March 2023
• 2013 AMC 12B Problems
  ...<math>33</math>. What is the average age of all of these parents and fifth-graders? guide squiggly(path g, real stepsize, real slope=45)
  16 KB (2,459 words) - 02:46, 30 January 2021
• 2013 AMC 12B Problems/Problem 13
  ..., <math>\alpha, \beta, \gamma</math> also form an arithmetic sequence with 45, 60, and 75, and the largest two angles of the quadrilateral add up to 240 {{AMC12 box|year=2013|ab=B|num-b=12|num-a=14}}
  3 KB (443 words) - 12:32, 8 January 2021
• 2013 AMC 12B Problems/Problem 7
  ...nth turn, the eighth number will be the 53rd number said, because <math>53-45=8</math>. Since we are starting from 1 every turn, the 53rd number said wi ...s have been said so far, <math>\frac{9(9+1)}{2} = 45</math> and <math>53 - 45 = \boxed{\textbf{(E) }8}</math>
  2 KB (361 words) - 16:18, 16 October 2022
• 2013 AMC 12B Problems/Problem 12
  guide squiggly(path g, real stepsize, real slope=45) {{AMC12 box|year=2013|ab=B|num-b=11|num-a=13}}
  3 KB (533 words) - 22:11, 3 October 2022
• 2013 AMC 10B Problems
  ...ir parents is 33. What is the average age of all of these parents and fifth-graders? ...y attempted 50% more two-point shots than three-point shots. How many three-point shots did they attempt?
  12 KB (1,926 words) - 21:54, 6 October 2022
• 1951 AHSME Problems/Problem 45
  == Problem 45 == {{AHSME 50p box|year=1951|num-b=44|num-a=46}}
  1 KB (194 words) - 12:27, 5 July 2013
• 2013 AIME I Problems/Problem 15
  (c) <math>p</math> divides <math>A-a</math>, <math>B-b</math>, and <math>C-c</math>, and ...</math> still gives <math>16</math> solutions because <math>C_\text{max}=2B-1=101>100</math>. Likewise, <math>B=54</math> gives <math>15</math> solution
  4 KB (661 words) - 23:14, 26 May 2023
• 2013 AIME I Problems/Problem 9
  pair M = (a*B+(12-a)*K) / 12; pair N = (b*C+(12-b)*K) / 12;
  9 KB (1,530 words) - 17:12, 18 April 2024
• 2013 AIME I Problems/Problem 12
  First, find that <math>\angle R = 45^\circ</math>. ...coordinates. Call <math>Q</math> the origin and <math>QP</math> be on the x-axis. It is easy to see that <math>F</math> is the vertex on <math>RP</math>
  9 KB (1,490 words) - 02:25, 2 May 2024
• 2013 AIME I Problems/Problem 13
  ...e AB_{n-1}C_{n-1}</math>. The area of the union of all triangles <math>B_{n-1}C_nB_n</math> for <math>n\geq1</math> can be expressed as <math>\tfrac pq< ...I say some area has ratio <math>\frac{1}{2}</math>, that means its area is 45.
  7 KB (1,085 words) - 22:48, 17 July 2023
• 2013 AIME II Problems/Problem 10
  ...e <math>k</math>, then the equation for line<math>AKL</math> is <math>y=k(x-4-\sqrt{13})</math>. Then we get <math>(k^2+1)x^2-2k^2(4+\sqrt{13})x+k^2\cdot (4+\sqrt{13})^2-13=0</math>. According to [[Vieta's Formulas]], we get
  5 KB (823 words) - 17:57, 29 December 2023
• 2013 USAMO Problems/Problem 1
  pair M = 2*foot(P,relpoint(O_B--O_C,0.5-10/lisf),relpoint(O_B--O_C,0.5+10/lisf))-P; draw((abs(dot(unit(M-P),unit(B-P))) < 1/2011) ? rightanglemark(M,P,B) : anglemark(M,P,B), rgb(0.0,0.8,0.8))
  14 KB (1,830 words) - 18:22, 10 May 2023
• 1989 AHSME Problems/Problem 7
  ...qquad\textrm{(C)}\ 30^\circ\qquad\textrm{(D)}\ 40^\circ\qquad\textrm{(E)}\ 45^\circ </math> {{AHSME box|year=1989|num-b=6|num-a=8}}
  2 KB (221 words) - 18:04, 21 October 2018
• 1989 AHSME Problems/Problem 12
  ...e following is closest to the number of westbound vehicles present in a 100-mile section of highway? At the beginning of the five-minute interval, say the eastbound driver is at the point <math>x=0</math>,
  2 KB (240 words) - 07:50, 22 October 2014
• 2014 AIME II Problems/Problem 2
  draw("10",(45,15)); draw("$y$",(10,45));
  2 KB (371 words) - 15:19, 28 February 2022
• 2014 AIME II Problems/Problem 11
  ...>, <math>\measuredangle DRE=75^{\circ}</math> and <math>\measuredangle RED=45^{\circ}</math>. <math>RD=1</math>. Let <math>M</math> be the midpoint of se ...(d--r); cm = foot(r, e, m); c=extension(r,cm,d,e); p=midpoint(r--c); a=p+(e-m);
  6 KB (1,059 words) - 18:24, 20 January 2024
• 2014 AIME II Problems/Problem 9
  ...we get <math>10 + 10 + 50 + 10 + 40 + 100 + 10 + 30 + 60 + 15 + 70 + 120 + 45 + 10 + 1 = \boxed{581}.</math> ...o arrange <math>3</math> adjacent chairs, but then we subtract <math>n 2^{n-4}</math> ways to arrange <math>4.</math> Finally, we add <math>1</math> to
  4 KB (693 words) - 16:55, 25 December 2023
• 2014 AIME II Problems/Problem 14
  ...math>\triangle{ABC}, AB=10, \angle{A}=30^\circ</math> , and <math>\angle{C=45^\circ}</math>. Let <math>H, D,</math> and <math>M</math> be points on the l <math>AHC</math> is a <math>45-45-90</math> triangle, so <math>\angle{HAB}=15^\circ</math>.
  11 KB (1,442 words) - 19:28, 21 October 2023
• 1967 AHSME Problems/Problem 16
  \textbf{(C)}\ 45\qquad {{AHSME 40p box|year=1967|num-b=15|num-a=17}}
  1 KB (176 words) - 01:39, 16 August 2023
• 1967 AHSME Problems/Problem 19
  \textbf{(D)}\ \frac{45}{2}\qquad {{AHSME 40p box|year=1967|num-b=18|num-a=20}}
  1 KB (211 words) - 01:40, 16 August 2023
• Geometry Solutions
  tri=(y,-1)--(y-1,0)--(y,1); label("C", (y-1, -0.1), S);
  23 KB (3,182 words) - 12:30, 5 April 2014
• Northeastern WOOTers Mock AIME I Problems
  ...<math>y</math> be digits, not necessarily distinct and not necessarily non-zero. For how many quadruples <math>(u,v,x,y)</math> is it true that <cmath> ...loor\frac{3n+4}{13}\right\rfloor-\left\lfloor\frac{n-28+\left\lfloor\frac{n-7}{13}\right\rfloor}{4}\right\rfloor.</cmath>
  8 KB (1,336 words) - 09:10, 30 May 2020
• 2016 AMC 10A Problems
  ...and <math>y</math> with <math>y \neq 0</math> by <cmath>\text{rem} (x ,y)=x-y\left \lfloor \frac{x}{y} \right \rfloor</cmath>where <math>\left \lfloor \ ...bf{(B)}\ 30\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 45</math>
  14 KB (2,104 words) - 22:26, 16 September 2022
• 1975 IMO Problems/Problem 3
  ...CAQ</math> are constructed externally with <math>\angle CBP = \angle CAQ = 45^\circ, \angle BCP = \angle ACQ = 30^\circ, \angle ABR = \angle BAR = 15^\ci ...<math> \triangle BXC</math> are equilateral. Now, <math> \angle PBX = 60 - 45 = 15</math>. Similarly, <math> \angle PXB = 15</math>, so <math> \triangle
  4 KB (750 words) - 16:14, 29 January 2021
• 1962 AHSME Problems/Problem 6
  ...<math>\dfrac{18}{4}=4.5</math>. If we draw the diagonal, we create a 45-45-90 triangle, whose hypotenuse (also the diagonal of the square) is <math>\bo {{AHSME 40p box|year=1962|before=Problem 5|num-a=7}}
  1 KB (174 words) - 15:55, 18 April 2018
• 1962 AHSME Problems/Problem 10
  ...150}{3\frac{1}{3}}</math>. Simplifying, you get <math>r=45</math>. <math>45-40=5 \rightarrow \boxed{\text{A}}</math> {{AHSME 40p box|year=1962|before=Problem 9|num-a=11}}
  2 KB (277 words) - 22:14, 3 October 2014
• 1962 AHSME Problems/Problem 16
  <cmath>a^2=\frac{45}2</cmath> {{AHSME 40p box|year=1962|before=Problem 15|num-a=17}}
  1 KB (170 words) - 22:16, 3 October 2014
• 1962 AHSME Problems/Problem 35
  ...{(B)}\ 40\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 42.4\qquad\textbf{(E)}\ 45 </math>
  1 KB (189 words) - 17:22, 17 April 2014
• 1962 AHSME Problems
  The expression <math>\frac{1^{4y-1}}{5^{-1}+3^{-1}}</math> is equal to: <math>\textbf{(A)}\ \frac{4y-1}{8} \qquad
  17 KB (2,459 words) - 22:40, 10 April 2023
• 1951 AHSME Problems/Problem 46
  {{AHSME 50p box|year=1951|num-b=45|num-a=47}}
  2 KB (283 words) - 17:26, 1 January 2014
• 2013 AMC 8 Problems
  A sign at the fish market says, "50% off, today only: half-pound packages for just \$3 per package." What is the regular price for a fu What is the value of <math>4 \cdot (-1+2-3+4-5+6-7+\cdots+1000)</math>?
  15 KB (2,162 words) - 20:05, 8 May 2023
• 2013 AMC 8 Problems/Problem 7
  ...ey counted 6 cars in the first 10 seconds. It took the train 2 minutes and 45 seconds to clear the crossing at a constant speed. Which of the following w ...conds: <math>2</math> minutes and <math>45</math> seconds<math>=2\cdot60 + 45 = 165</math> seconds. We then set up a ratio: <cmath>\frac{3}{5}=\frac{x}{1
  2 KB (238 words) - 19:05, 15 April 2023
• 2013 AMC 8 Problems/Problem 12
  ...}\ 30 \qquad \textbf{(C)}\ 33 \qquad \textbf{(D)}\ 40 \qquad \textbf{(E)}\ 45</math> {{AMC8 box|year=2013|num-b=11|num-a=13}}
  1 KB (160 words) - 21:39, 3 January 2024
• 2013 AMC 8 Problems/Problem 13
  <math>\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 46 \qquad \textbf{(C)}\ 47 \qquad \textbf{(D)}\ 48 \qq ...only answer choice that is a multiple of 9 is <math>\boxed{\textbf{(A)}\ 45}</math>.
  1 KB (203 words) - 21:40, 3 January 2024
• 2013 AMC 8 Problems/Problem 22
  for (int i=0; i<3; ++i){for (int j=0; j>-2; --j){if ((i-j)<3){add(corner,(50i,50j));}}} draw((5,-100)--(45,-100));
  2 KB (256 words) - 19:06, 22 January 2024
• 2013 AMC 8 Problems/Problem 25
  ==Video Solution for Problems 21-25== ...it loses <math>2\pi</math> inches each, because <math>\dfrac{1}{2} 2\pi (x-2) - \dfrac{1}{2} 2\pi (x)= -2 \pi</math>
  3 KB (519 words) - 19:28, 21 December 2023
• 1981 AHSME Problems
  If <math> \dfrac{x}{x-1}=\dfrac{y^2+2y-1}{y^2+2y-2} </math>, then <math>x</math> equals ...tbf{(C)}\ y^2+2y+2 \qquad \\ \textbf{(D)}\ y^2+2y+1\qquad\textbf{(E)}\ -y^2-2y+1 </math>
  17 KB (2,633 words) - 15:44, 16 September 2023
• 1968 AHSME Problems
  ...lue of <math>x</math> such that <math>64^{x-1}</math> divided by <math>4^{x-1}</math> equals <math>256^{2x}</math> is: ...hrough the point <math>(0,4)</math> is perpendicular to the line <math>x-3y-7=0</math>. Its equation is:
  16 KB (2,571 words) - 14:13, 20 February 2020
• 2014 AMC 10A Problems
  \textbf{(II)}\ x-y < a-b\qquad</math> <math> \textbf{(A)}\ 45\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 54\qquad\textbf{(D)}\ 60\qquad\tex
  15 KB (2,190 words) - 15:21, 22 December 2020
• 2014 AMC 10A Problems/Problem 5
  Thus, the solution is <cmath>90-87=3\implies\boxed{\textbf{(C)} \ 3}</cmath> ...the upper <math>25\%</math> is <math>100</math> points and the lower <math>45\%</math> is <math>70</math> or <math>80</math>).The mean is <math>10\%\cdot
  3 KB (390 words) - 00:10, 27 June 2023
• 2014 AMC 10A Problems/Problem 14
  <math> \textbf{(A)}\ 45\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 54\qquad\textbf{(D)}\ 60\qquad\tex draw((-3,0)--(9,0),linewidth(1),Arrows); //x-axis
  6 KB (1,001 words) - 13:07, 25 July 2022
• 2014 AMC 10A Problems/Problem 17
  Three fair six-sided dice are rolled. What is the probability that the values shown on two ...ide this values by <math>6^3</math> because there are 3 dice. <math>\dfrac{45}{216}=\boxed{\dfrac{5}{24}}</math>.
  3 KB (496 words) - 22:43, 21 November 2022
• 2014 AMC 10A Problems/Problem 22
  ...we have <math>DE=10\sqrt 3</math>. Since <math>ADE</math> is a <math>30-60-90</math> triangle, <math>AE=2 \cdot AD=2 \cdot 10=\boxed{\textbf{(E)} \: 20 ...DE = 10\sqrt{3}</math>. Because <math>\triangle ADE</math> is a <math>30-60-90</math> triangle, <math>AE = \boxed{\textbf{(E)}~20}</math>.
  12 KB (1,821 words) - 18:16, 29 October 2023
• 1994 AHSME Problems/Problem 17
  ...uad\textbf{(C)}\ 4\pi-4 \qquad\textbf{(D)}\ 2\pi+4 \qquad\textbf{(E)}\ 4\pi-2 </math> ...o radii of circle <math>O</math> and the segment of the rectangle are 45-45-90 triangles.
  2 KB (287 words) - 03:07, 28 May 2021
• 1994 AHSME Problems/Problem 19
  ...hinking of the worst case scenario, essentially an adaptation of the Pigeon-hole principle. ...if we have all those disks we won't have 10 of any one disk. This gives us 45 disks.
  1 KB (209 words) - 03:10, 28 May 2021
• 1958 AHSME Problems/Problem 44
  {{AHSME box|year=1958|num-b=43|num-a=45}}
  2 KB (320 words) - 19:10, 26 April 2014
• 1960 AHSME Problems
  The fraction <math>\frac{a^2+b^2-c^2+2ab}{a^2+c^2-b^2+2ac}</math> is (with suitable restrictions of the values of a, b, and c) <math>\text{(D) reducible to} \frac{a-b+c}{a+b-c}\qquad</math>
  21 KB (3,242 words) - 21:27, 30 December 2020
• 2014 AMC 12A Problems
  The difference between a two-digit number and the number obtained by reversing its digits is <math>5</mat ...d and breakfast inn has <math>5</math> rooms, each with a distinctive color-coded decor. One day <math>5</math> friends arrive to spend the night. The
  12 KB (1,863 words) - 19:04, 11 April 2024
• 2014 AMC 12A Problems/Problem 12
  <cmath> \dfrac{x^2}{y^2} = \dfrac{1}{2 - \sqrt{3}} = \dfrac{2 + \sqrt{3}}{4-3} = 2 + \sqrt{3}=\boxed{\textbf{(D)}}</cmath> ...r\sqrt{6}}{3}</math>. We can also see that <math>\angle YBO_1 = 75^{\circ}-45^{\circ} = 30^{\circ}=\angle YO_1B</math>, so <math>\triangle YO_1B</math> i
  7 KB (1,191 words) - 23:37, 23 June 2022
• 2014 AMC 12A Problems/Problem 15
  ...here <math>a</math> is non-zero. Let <math>S</math> be the sum of all five-digit palindromes. What is the sum of the digits of <math>S</math>? \textbf{(E) }45\qquad</math>
  4 KB (612 words) - 22:42, 2 August 2021
• 2014 AMC 12A Problems/Problem 25
  ...ight\rvert^2}{3^2+4^2} </cmath> which rearranges to <math>(4x+3y)^2 = 25(6x-8y+25)</math>. <cmath>25k^2 = 6x-8y+25</cmath><cmath>25k = 4x+3y.</cmath>
  4 KB (661 words) - 16:18, 2 September 2022
• 2014 AMC 12A Problems/Problem 23
  <cmath>\dfrac1{99^2}=0.\overline{b_{n-1}b_{n-2}\ldots b_2b_1b_0},</cmath> ...the repeating decimal expansion. What is the sum <math>b_0+b_1+\cdots+b_{n-1}</math>?
  3 KB (453 words) - 09:35, 26 May 2023
• 2014 AMC 10B Problems/Problem 13
  draw((30,0)--(25,8.66025404)--(30, 17.3205081)--(40, 17.3205081)--(45, 8.66025404)--(40, 0)--(30, 0)); draw((30,0)--(25,-8.66025404)--(30, -17.3205081)--(40, -17.3205081)--(45, -8.66025404)--(40, 0)--(30, 0));
  3 KB (488 words) - 19:08, 19 December 2023
• Mock AIME 3 2006-2007 Problems/Problem 11
  ...2 22.5 + 2r^2\cos22.5\sin22.5 - 6 = 0</cmath> <cmath>r^2 \cos 45 + r^2\sin 45 = 6</cmath> <cmath>r^2 = 3 \sqrt 2</cmath>. Since <math>(x^2+y^2)^2 = r^4</
  4 KB (699 words) - 01:53, 30 April 2022
• 2014 AIME II Problems
  ...> are roots of <math>p(x)=x^3+ax+b</math>, and <math>r+4</math> and <math>s-3</math> are roots of <math>q(x)=x^3+ax+b+240</math>. Find the sum of all po Charles has two six-sided dice. One of the die is fair, and the other die is biased so that it c
  8 KB (1,410 words) - 00:04, 29 December 2021
• 2014 AIME I Problems
  Jon and Steve ride their bicycles along a path that parallels two side-by-side train tracks running the east/west direction. Jon rides east at <math>2 ...<math>2014</math>, respectively, and each graph has two positive integer x-intercepts. Find <math>h</math>.
  9 KB (1,472 words) - 13:59, 30 November 2021
• 2014 AIME I Problems/Problem 7
  ...1</math> and <math>|z| = 10</math>. Let <math>\theta = \arg \left(\tfrac{w-z}{z}\right) </math>. The maximum possible value of <math>\tan^2 \theta</mat ...{\sin x}{\cos x-10}\right)=\frac{\cos x(\cos x-10)-(-\sin x)\sin x}{(\cos x-10)^2}=\dfrac{1 - 10\cos{x}}{(\cos{x} - 10)^2}</cmath>
  5 KB (782 words) - 20:25, 10 October 2023
• 2014 AIME I Problems/Problem 9
  ...x_2<x_3</math> be the three real roots of the equation <math>\sqrt{2014}x^3-4029x^2+2=0</math>. Find <math>x_2(x_1+x_3)</math>. ...for the middle term <math>-4029</math> is equal to <cmath>-2{\sqrt{2014}}^2-1</cmath>.
  10 KB (1,653 words) - 00:30, 27 January 2024
• 2014 AIME I Problems/Problem 13
  dot("$D$",D,dir(45)); ...A,dir(135)); dot("$B$",B,dir(215)); dot("$C$",C,dir(305)); dot("$D$",D,dir(45)); dot("$H$",H,dir(90)); dot("$F$",F,dir(270)); dot("$E$",E,dir(180)); dot
  9 KB (1,404 words) - 21:07, 13 October 2023
• 2014 AIME I Problems/Problem 15
  ...rc</math>. But <math>DF=FE</math>, so <math>\triangle DEF</math> is a 45-45-90 triangle. Letting <math>DG=3x</math>, we have that <math>EG=4x</math>, <m ...while the segment <math>CJ</math> is <math>2x\sqrt{2}</math> since its 3-4-5 again. Now adding all those segments together we can find that <math>AC=5=
  10 KB (1,643 words) - 22:30, 28 January 2024
• 1993 AHSME Problems/Problem 3
  <math>\frac{15^{30}}{45^{15}} =</math> First we must convert these to the same bases. We can rewrite <math>45^{15}</math> as <math>15^{15} \cdot 3^{15}</math> Now
  658 bytes (74 words) - 00:34, 30 November 2016
• 1993 AHSME Problems/Problem 22
  ...--(45,5)--(40,5)--(35,5)--(30,5)--(31,7)--(36,7)--(41,7)--(46,7)--(46,2)--(45,0), dashed); draw((45,0)--(45,5)--(46,7), dashed);
  7 KB (780 words) - 08:19, 27 June 2021
• 1973 AHSME Problems
  Two congruent 30-60-90 are placed so that they overlap partly and their hypotenuses coincide. If <math> \textbf{(A)}\ 90\qquad\textbf{(B)}\ 72\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 15 </math>
  18 KB (2,788 words) - 13:55, 20 February 2020
• 1992 AHSME Problems/Problem 26
  fill((0,-1)--arc((0,-1),2-sqrt(2),225,315)--cycle,grey); draw(arc((0,-1),2-sqrt(2),225,315),black+linewidth(1));
  3 KB (414 words) - 14:29, 20 February 2018
• 1991 AHSME Problems/Problem 5
  In the arrow-shaped polygon [see figure], the angles at vertices <math>A,C,D,E</math> and ...ll us that AB=AG and that angle A is a right angle, triangle ABG is a 45-45-90 triangle. Thus we can find the legs of the triangle by dividing <math>20<
  1 KB (209 words) - 18:52, 21 November 2014

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