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2005 AIME II Problems/Problem 12
...BG</math>). Then <math>\tan \angle EOG = \frac{x}{450}</math>, and <math>\tan \angle FOG = \frac{y}{450}</math>. ...frac{y}{450}}{1 - \frac{x}{450} \cdot \frac{y}{450}}.</cmath> Since <math>\tan 45 = 1</math>, this simplifies to <math>1 - \frac{xy}{450^2} = \frac{x + y}

13 KB (2,080 words) - 21:20, 11 December 2022
1991 AIME Problems/Problem 11
for (int i=0; i<12; ++i) dot((cos(i*pi/6), sin(i*pi/6)));

4 KB (740 words) - 19:33, 28 December 2022
1999 AIME Problems/Problem 11
...</math> are relatively prime positive integers that satisfy <math>\frac mn<90,</math> find <math>m+n.</math> ...manipulate this sum by wrapping the terms around (since the first half is equal to the second half), but it quickly becomes apparent that this way is diffi

4 KB (614 words) - 04:38, 8 December 2023
Mock AIME 1 2006-2007/Problems
...angle BAC</math> hits <math>BC</math> at <math>D</math>. If <math>\angle C=90^\circ</math>, and the maximum value of <math>\frac{[ABD]}{[ACD]}=\frac{m}{n .../math>. <math>\lfloor x\rfloor</math> is the greatest integer less than or equal to <math>x</math>. If for fixed <math>n</math>, there exists an integer <ma

8 KB (1,355 words) - 14:54, 21 August 2020
Euler line
Acute triangle ...vec BD}{\vec DA} = n = \frac {\tan \alpha – \tan \gamma}{\tan \beta – \tan \gamma} > 0.</math>

59 KB (10,203 words) - 04:47, 30 August 2023
1985 IMO Problems/Problem 1
...</math> is clearly a cyclic quadrilateral, the angle <math>COE </math> is equal to half the angle <math>GAO </math>. Then \begin{matrix} {CE} & = & r \tan(COE) \\

4 KB (684 words) - 07:28, 3 October 2021
2007 AIME I Problems/Problem 9
...n the exterior of the triangle and are the centers of two [[circle]]s with equal [[radius|radii]]. The circle with center <math>O_1</math> is tangent to th [[Image:AIME I 2007-9.png]]

11 KB (1,851 words) - 12:31, 21 December 2021
2007 AIME I Problems/Problem 12
...their [[complement]]s are <math>15^{\circ}</math>, and <math>\angle DAB' = 90 - 2(15) = 60^{\circ}</math>. We know that <math>\angle DB'A = 75^{\circ}</m ...h>30-60-90</math> triangle is <math>10</math>, the base of the <math>45-45-90</math> is <math>10\sqrt{3}</math>, and their common height is <math>10\sqrt

10 KB (1,458 words) - 20:50, 3 November 2023
2007 AIME II Problems/Problem 15
Q=r*(rotate(-90)*A/length(A)); X=extension(A,P,Q,Q+A); Y=extension(B,P,Q,Q+A); R=rotate(90,C)*(C+r*(A-C)/length(A-C)); Z=extension(C,P,R,R+A-C);

11 KB (2,099 words) - 17:51, 4 January 2024
LaTeX:Commands
| <math>\textstyle 2^{23}</math>||2^{23}||<math>\textstyle n_{i-1}</math>||n_{i-1} | <math>a^{i+1}_3</math>||a^{i+1}_3||<math>x^{3^2}</math>||x^{3^2}

12 KB (1,898 words) - 15:31, 22 February 2024
1999 AHSME Problems
I. The digit is 1. <math> \mathrm{(A) \ I\ is\ true} \qquad \mathrm{(B) \ I\ is\ false} \qquad \mathrm{(C) \ II\ is\ true} \qquad \mathrm{(D) \ III\ is

13 KB (1,945 words) - 18:28, 19 June 2023
2008 AMC 12A Problems
Which of the following is equal to the [[product]] ...rs <math>15\%</math> off the sticker price followed by a <math>\textdollar 90</math> rebate, and store <math>B</math> offers <math>25\%</math> off the sa

13 KB (2,025 words) - 13:56, 2 February 2021
2008 AMC 12B Problems/Problem 24
...he resulting heights <math>h_n</math> and <math>h_{n+1}</math>. From 30-60-90 rules, the distance between the points where these altitudes meet the x-axi ...th> coordinates are the squares of the heights). Setting these expressions equal and dividing throughout by <math>h_{n+1}+h_n</math> leaves <math>h_{n+1}-h_

9 KB (1,482 words) - 13:52, 4 April 2024
2008 AMC 10B Problems/Problem 24
...rm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95</math> ...</math> equals <math>55</math> degrees, thus making angle <math>AYB</math> equal to <math>60</math> degrees. We can also find out that angle <math>CYB</math

12 KB (1,944 words) - 17:15, 20 January 2024
2008 iTest Problems
for(int i=0;i<=5;i=i+1) draw(circle(7/2+d*i,3/2));

71 KB (11,749 words) - 01:31, 2 November 2023
2009 AMC 10A Problems/Problem 10
Assume the length of <math>BD</math> is equal to <math>h</math>. Then, by Pythagoras, we have, <math>A): 4 \sqrt{3}</math>. I am too lazy to go over this, but we immediately see that this is very impro

5 KB (879 words) - 18:57, 30 April 2024
2009 AIME I Problems/Problem 12
...> be the circle having <math>\overline{CD}</math> as a diameter. Let <math>I</math> be a point outside <math>\triangle ABC</math> such that <math>\overl ...length <math>x</math>. Then the perimeter of <math>\triangle ABI</math> is equal to <cmath>2(x+AD+DB)=2(x+37).</cmath> It remains to compute <math>\dfrac{2(

12 KB (1,970 words) - 22:53, 22 January 2024
2009 AIME I Problems/Problem 15
...t{3})=14-7\sqrt3</math>. Then, the area of [<math>\triangle BPC</math>] is equal to <math>7 \cdot OP=98-49\sqrt{3}</math>, and so the answer is <math>98+49+ {{AIME box|year=2009|n=I|num-b=14|after=Last Question}}

6 KB (1,048 words) - 19:35, 2 January 2023
Mock AIME 2 2010 Answer Key
Notice that removing as many lemmings as possible at each step (i.e., using the greedy algorithm) is optimal. This can be easily proved throu ...{10})</math>, with <math>0 \le b_i \le 2, 3|a_i-b_i</math> for <math>1 \le i \le 10</math>. Given that <math>f</math> can take on <math>K</math> distinc

36 KB (6,214 words) - 20:22, 13 July 2023
2011 AIME I Problems/Problem 14
...A_6}</math>, and <math>\overline{A_7 A_8}</math>, respectively. For <math>i = 1, 3, 5, 7</math>, ray <math>R_i</math> is constructed from <math>M_i</ma label("$M_1$",M,dir(90));

8 KB (1,344 words) - 18:39, 9 February 2023
2011 USAJMO Problems/Problem 3
...th> is on the circumcircle of <math>\Delta XYZ</math>, and similarly <math>I</math> is on the circumcircle of <math>\Delta XYZ</math>. Therefore <math>\ ...th>HM=HB/2=HZ/2</math>. Therefore <math>\Delta HMZ</math> is a <math>30-60-90</math> right triangle and so <math>\angle ZHB=60^{\circ}</math>. So <math>\

15 KB (2,593 words) - 13:37, 29 January 2021
1985 AHSME Problems
if the sum of the remaining terms is to equal <math>1</math>? int i,j;

17 KB (2,488 words) - 03:26, 20 March 2024
2012 AMC 12B Problems
'''Note:''' 1 foot is equal to 12 inches. ...the first <math>10</math> positive integers such that for each <math>2 \le i \le 10</math> either <math>a_i+1</math> or <math>a_i-1</math> or both appea

20 KB (2,681 words) - 09:47, 29 June 2023
2012 AIME I Problems
{{AIME Problems|year=2012|n=I}} [[2012 AIME I Problems/Problem 1|Solution]]

10 KB (1,617 words) - 14:49, 2 June 2023
1991 AHSME Problems
...<math>t=0</math> to time <math>t=1</math> a population increased by <math>i\%</math>, and from time <math>t=1</math> to time <math>t=2</math> the popul <math>\textbf{(A) } (i+j)\% \qquad

16 KB (2,451 words) - 04:27, 6 September 2021
2014 AIME I Problems
{{AIME Problems|year=2014|n=I}} path laceL=(-20,-30)..tension 0.75 ..(-90,-135)..(-102,-147)..(-152,-150)..tension 2 ..(-155,-140)..(-135,-40)..(-50,

9 KB (1,472 words) - 13:59, 30 November 2021
1973 AHSME Problems
Two congruent 30-60-90 are placed so that they overlap partly and their hypotenuses coincide. If t Of the following five statements, I to V, about the binary operation of averaging (arithmetic mean),

18 KB (2,788 words) - 13:55, 20 February 2020
1987 AHSME Problems
have a solution <math>(x, y)</math> inside Quadrant I if and only if \textbf{(E)}\ 90\pi </math>

16 KB (2,291 words) - 13:45, 19 February 2020
1983 AHSME Problems
...angle at <math>C</math>. If <math>\sin A = \frac{2}{3}</math>, then <math>\tan B</math> is If <math>\tan{\alpha}</math> and <math>\tan{\beta}</math> are the roots of <math>x^2 - px + q = 0</math>, and <math>\co

15 KB (2,309 words) - 23:43, 2 December 2021
1978 AHSME Problems
...then <math>(a+ b+ c-d) + (a + b- c+ d) +(a-b+ c+d)+ (-a+ b+c+d)</math> is equal to ...etween <math>\mathit{A}</math> and <math>\mathit{B}</math> is less than or equal to the distance between <math>\mathit{A}</math>

15 KB (2,432 words) - 01:06, 22 February 2024
2007 IMO Problems/Problem 4
pair I=incenter(A,B,C); ...90+\dfrac{C}{2}</math>, and similarly <math>\angle{RPK} = 90+\angle{PCK} = 90+\dfrac{C}{2}</math>. Therefore, <math>\angle{RQL} = \angle{RPK}</math>. Usi

8 KB (1,480 words) - 14:52, 5 August 2022
2015 AIME II Problems/Problem 15
...the areas of <math>\triangle DBA</math> and <math>\triangle ACE</math> are equal. This common area is <math>\frac{m}{n}</math>, where <math>m</math> and <ma ...midpoint of <math>\overline{BC},</math> implying that <math>\angle{BAC} = 90.</math> Now draw <math>\overline{PA}, \overline{PB}, \overline{PM},</math>

31 KB (5,086 words) - 19:15, 20 December 2023
2016 AIME I Problems/Problem 9
...os(m)\cos(n)=\frac{1}{2}(\cos(m+n)+\cos(m-n))</math>. <math>\cos(m+n)=\sin(90-m-n)=\sin(BAC)=\frac{1}{5}</math>. The maximum possible value of <math>\cos ...at <math>\sin \alpha = \frac{1}{5}</math>. Since <math>\sin \alpha = \cos (90- \alpha)</math>,

9 KB (1,526 words) - 02:31, 29 December 2021
2016 AIME II Problems/Problem 14
...and <math>KQ=y</math>, assuming WLOG <math>x>y</math>, we must have <math>\tan(120^{\circ})=-\sqrt{3}=\dfrac{\dfrac{x+y}{100 \sqrt{3}}}{\dfrac{30000-xy}{3 ...qrt{33}}{2}+\frac{15\pm\sqrt{33}}{7\pm\sqrt{33}} &= 2d\end{align*}</cmath> I'll use <math>a=\frac{9+\sqrt{33}}{2}</math> because both values should give

15 KB (2,560 words) - 01:44, 1 July 2023
2020 AMC 12A Problems
for (int i = 0; i < 3; ++i) { pair A = (j,i);

14 KB (2,073 words) - 15:15, 21 October 2021
2018 AIME I Problems/Problem 13
...rac A2 \sin\frac B2\sin\frac C2,</cmath> with equality when <math>\alpha = 90^\circ</math>, that is, when <math>X</math> is the foot of the perpendicular ..., and that occurs when <math>\alpha=45^{\circ}</math>, or <math>\angle AXB=90^{\circ}</math>. Ergo, <math>X</math> is the foot of the altitude from <math

13 KB (2,200 words) - 21:36, 6 January 2024
2006 iTest Problems/Problem U6
The smallest possible value of <math>\frac{y}{x}</math> is equal to <math>\frac{m}{n}</math> where <math>m</math> and <math>n</math> are rel ...r(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs);

4 KB (722 words) - 20:53, 27 March 2019
2019 AIME II Problems
Call a positive integer <math>n</math> <math>k</math>-pretty if <math>n</math> has exactly <math>k</math> positive divisors and <math>n ...th> such that for nonnegative integers <math>n,</math> the value of <math>\tan(2^n\theta)</math> is positive when <math>n</math> is a multiple of <math>3<

7 KB (1,254 words) - 14:45, 21 August 2023
2019 AIME II Problems/Problem 1
label("$9$",(A+B)/2,dir(-90)); label("$6$",(A+E)/2,dir(-90));

11 KB (1,794 words) - 15:32, 14 January 2024
2020 AIME I Problems/Problem 13
label("$I$", (-3.7912668488110084,8.257242529248508), NE * labelscalefactor); pair A, B, C, D, E, F, I, P, MA, MB, MC;

35 KB (5,215 words) - 23:08, 29 October 2023
2023 AIME I Problems/Problem 12
...erty that <cmath>\angle AEP = \angle BFP = \angle CDP.</cmath> Find <math>\tan^2(\angle AEP).</math> A = 55*sqrt(3)/3 * dir(90);

16 KB (2,592 words) - 15:40, 13 April 2024
2020 AMC 12A Problems/Problem 12
...es of the triangle must be <math>180^{\circ}</math> and one angle is <math>90^{\circ}</math>. This means the last leg angle must also be <math>45^{\circ} ...-5)</math>, the vector <math><3,-5></math>. This is <math>\perp</math> and equal to the original line segment. The difference between the two vectors is <ma

7 KB (1,145 words) - 20:27, 5 November 2023
2020 AIME I Problems/Problem 1
A = (0, tan(3 * pi / 7)); F = rotate(90/7, A) * (A - (0, 2));

6 KB (968 words) - 15:01, 24 January 2024
2020 AIME II Problems/Problem 4
...etween <math>A</math> and <math>A'</math>, <math>x+y=18</math>. Thus <math>90+18=\boxed{108}</math>. ...has the solution <math>(x,y) = (21,-3)</math>. The requested sum is <math>90+21-3=108</math>.

10 KB (1,542 words) - 13:29, 19 January 2024
2021 AIME I Problems/Problem 2
...math>AFCE</math> are rectangles, we know that <math>\angle AFG= \angle CDG=90 ^{\circ}</math>. ...the area of the parallelogram is <math>63-98x</math>. These equations are equal, so we can solve for <math>x</math> and obtain <math>x = \frac{3}{8}</math>

8 KB (1,294 words) - 00:59, 23 August 2022
2021 AIME I Problems/Problem 11
~MRENTHUSIASM (credit given to Math Jams's 2021 AIME I Discussion) ...<math>\sin\theta=\frac{4\sqrt{210}}{59}.</math> Since <math>0^\circ<\theta<90^\circ,</math> we have <math>\cos\theta>0.</math> It follows that <cmath>\co

16 KB (2,635 words) - 19:56, 24 December 2023
2023 AIME II Problems
...<math>\triangle ABC</math> be an isosceles triangle with <math>\angle A = 90^\circ.</math> There exists a point <math>P</math> inside <math>\triangle AB ...mega = \cos\frac{2\pi}{7} + i \cdot \sin\frac{2\pi}{7},</math> where <math>i = \sqrt{-1}.</math> Find the value of the product <cmath>\prod_{k=0}^6 \lef

8 KB (1,370 words) - 21:34, 28 January 2024
2023 AIME I Problems
{{AIME Problems|year=2023|n=I}} [[2023 AIME I Problems/Problem 1|Solution]]

7 KB (1,154 words) - 12:54, 20 February 2024
2022 AMC 12B Problems/Problem 25
draw(shift((sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw(shift((1-sqrt(3)/2,1/2))*rotate(90)*polygon(6));

20 KB (2,980 words) - 18:17, 2 January 2024
2023 AIME I Problems/Problem 8
Rhombus <math>ABCD</math> has <math>\angle BAD < 90^\circ.</math> There is a point <math>P</math> on the incircle of the rhombu dot("$S$",S,1.5*dir(-90),linewidth(4.5));

17 KB (2,612 words) - 14:54, 3 July 2023
2023 AIME I Problems/Problem 13
Denote <math>\alpha = \tan^{-1} \frac{\sqrt{21}}{\sqrt{31}}</math>. Case 1: <math>\angle AOC = \angle BOC = 2 \alpha</math> or <math>2 \left( 90^\circ - \alpha \right)</math>.

13 KB (2,130 words) - 01:52, 31 January 2024
DVI exam
Solution ...BC</math> equal to <math>\frac {3}{2},</math> distance from incenter <math>I</math> to vertex <math>C</math> is <math>IC = 4 - \frac {3}{2} = \frac {5}{

31 KB (5,191 words) - 08:10, 30 April 2024

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