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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) - 10:00, 18 June 2024
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 digit, t ...\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) - 10:27, 14 June 2024
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) - 12:34, 14 June 2024
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

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