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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) Target: 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=Problem 1-2: 3-4 Problem 3-5: 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 (3,435 words) - 00:56, 23 May 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 . 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) - 03:13, 27 May 2024
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]]" ...[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

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