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1990 AIME Problems/Problem 10
...</math>. We can write the numbers of set <math>A</math> as <math>\{n^8, n^{16}, \ldots n^{144}\}</math> and of set <math>B</math> as <math>\{n^3, n^6, \l ...right)</math> respectively, where <math>\text{cis}\,\theta = \cos \theta + i \sin \theta</math> and <math>k_1</math> and <math>k_2</math> are integers f

3 KB (564 words) - 04:47, 4 August 2023
1994 AIME Problems/Problem 8
...left(\mathrm{cis}\,60^{\circ}\right) = (a+11i)\left(\frac 12+\frac{\sqrt{3}i}2\right)=b+37i.</cmath> ...Equating the two vectors, we get <math>a+b=26\sqrt{3}</math> and <math>a-b=16\sqrt{3}</math>. Therefore, <math>a=21\sqrt{3}</math> and <math>b=5\sqrt{3}<

5 KB (787 words) - 02:09, 28 June 2024
1995 AIME Problems/Problem 15
...s the sum of all numbers of the form <math>\frac 32\left( \left(\frac {15}{16}\right)^n \cdot \left(\frac 12\right)^n \right) \cdot \left(\frac 1{32}\rig ...ll <math>4</math> consecutive <tt>H</tt>'s, and there is a <math>\frac{15}{16}</math> probability we roll a <tt>T</tt>. Thus,

6 KB (990 words) - 03:08, 9 July 2024
1995 AIME Problems/Problem 9
...ath> such that <math>\text{Arg}((11+xi)^3)=\text{Arg}(1331-33x^2+(363x-x^3)i)=\text{Arg}(1+xi)</math>. This will happen when <math>\frac{363x-x^3}{1331- <cmath>20\cos ^3 y = 16 \cos y</cmath>

7 KB (1,181 words) - 13:47, 3 February 2023
1995 AIME Problems/Problem 5
...math> and the sum of the other two roots is <math>3+4i,</math> where <math>i=\sqrt{-1}.</math> Find <math>b.</math> <cmath>m\cdot n = 13 + i,m' + n' = 3 + 4i\Longrightarrow m'\cdot n' = 13 - i,m + n = 3 - 4i.</cmath>

3 KB (457 words) - 14:08, 4 July 2024
1995 AIME Problems/Problem 1
...> are the perpendicular bisectors of two adjacent sides of square <math>S_{i+2}.</math> The total area enclosed by at least one of <math>S_{1}, S_{2}, ...+ \left(\frac{1}{4}\right)^2 + \left(\frac{1}{8}\right)^2 + \left(\frac{1}{16}\right)^2</math>

2 KB (302 words) - 19:29, 4 July 2013
1996 AIME Problems/Problem 7
for(int i=0;i<7;++i) D(shift(i,j)*unitsquare);

4 KB (551 words) - 11:44, 26 June 2020
1998 AIME Problems/Problem 13
...ed to be <math>a_1i + a_2i^2+ a_3i^3 + \cdots + a_ni^n,</math> where <math>i^2 = - 1.</math> Let <math>S_n</math> be the sum of the complex power sums ...2(-176 -64i) + 144i = -352 + 16i</math>. Thus, <math>|p| + |q| = |-352| + |16| = 368</math>.

2 KB (384 words) - 14:47, 14 June 2024
1998 AIME Problems/Problem 2
<div style="text-align:center;"><math>A = I + \frac B2 - 1</math></div> ...yle="text-align:center;"><math>450 = I + \frac {60}2 - 1</math><br /><math>I = 421</math></div>

6 KB (913 words) - 16:34, 6 August 2020
1999 AIME Problems/Problem 12
pair A=(0,0),B=(50,0),C=IP(circle(A,23+245/2),circle(B,27+245/2)), I=incenter(A,B,C); Let the incenter be denoted <math>I</math>. It is commonly known that the incenter is the intersection of the

3 KB (472 words) - 19:03, 21 June 2024
1999 AIME Problems/Problem 9
...finition of <math>f(z)</math>, <cmath>f(1+i) = (a+bi)(1+i) = (a-b) + (a+b)i,</cmath> this image must be equidistant to <math>(1,1)</math> and <math>(0, ...55}}{16}</math>. Finally, <math>b = |a+bi|\sin\theta = 8\frac{\sqrt{255}}{16} = \frac{\sqrt{255}}{2}</math>, and <math>b^2 = \frac{255}{4}</math>, so th

6 KB (1,010 words) - 19:01, 24 May 2023
2000 AIME I Problems/Problem 15
Round 7: <math>b</math> to <math>b</math>, <math>15</math> to right, <math>16</math> left in deck, <math>n = -2 + 8k</math>, because <math>n = 2 + 4k</ma ...index, categorizing them further as <math>I\equiv 2\pmod 4</math> or <math>I\equiv 4\pmod 4</math> in that order (The card with the smaller remainder wi

15 KB (2,673 words) - 19:16, 6 January 2024
2000 AIME I Problems/Problem 8
...the container minus the volume of the liquid, which is <math>\frac{925\pi}{16}</math>, which is <math>\frac{37}{64}</math> of the volume of the container The volume of the cone with height 9 is <math>\frac{675}{16}\pi</math>.

4 KB (677 words) - 16:33, 30 December 2023
2000 AIME I Problems/Problem 6
<cmath>((x^2 + 2xy + y^2) + 16 - 2(4)(x+y)) = 4xy</cmath> <cmath>x^2 - 2xy + y^2 + 16 - 8x - 8y = 0</cmath>

6 KB (966 words) - 11:31, 24 June 2024
2000 AIME I Problems/Problem 5
...18 = 7</math> white marbles total, the denominator becomes <math>(9 + w_1)(16 - w_1) = 150</math>. Testing <math>w_1 = 1</math> gives a solution, and th {{AIME box|year=2000|n=I|num-b=4|num-a=6}}

7 KB (1,011 words) - 20:09, 4 January 2024
2001 AIME I Problems/Problem 15
...H</math>, and thus to an octagonal circuit of these vertices. The cube has 16 diagonal segments that join nonadjacent vertices. In effect, the problem as ...for any vertex, it can be linked to at most one different-colored vertex, i.e. its opposite vertex. Thus, there are only two possible types of circuits

11 KB (1,837 words) - 18:53, 22 January 2024
2001 AIME I Problems/Problem 14
n&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17&18&19\\\hline a_n&1&1&1&2&2&3&4&5&7&9&12&16&21&28&37&49&65&86\\\hline

13 KB (2,298 words) - 19:46, 9 July 2020
2001 AIME I Problems/Problem 12
triple I = (2/3,2/3,2/3); draw(L--F--N--E--M--G--L--I--M--I--N--I--J);

6 KB (1,050 words) - 18:44, 27 September 2023
2001 AIME I Problems/Problem 9
/* -- arbitrary values, I couldn't find nice values for pqr please replace if possible -- */ /* -- arbitrary values, I couldn't find nice values for pqr please replace if possible -- */

4 KB (673 words) - 20:52, 28 June 2024
2001 AIME I Problems/Problem 5
...e side length of the equilateral triangle is twice of this, so <math>\frac{16\sqrt{3}}{13}</math>. ...n=\dfrac{4\sqrt{3}}{13}</math> which means the side length is <math>\dfrac{16\sqrt{3}}{13}=\sqrt{\dfrac{768}{169}}.</math>

6 KB (1,043 words) - 10:09, 15 January 2024

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