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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}}

944 bytes (154 words) - 12:44, 13 August 2014
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>

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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

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