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- ...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
- 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
- '''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 appea20 KB (2,681 words) - 09:47, 29 June 2023
- {{AIME Problems|year=2012|n=I}} [[2012 AIME I Problems/Problem 1|Solution]]10 KB (1,617 words) - 14:49, 2 June 2023
- ...<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)\% \qquad16 KB (2,451 words) - 04:27, 6 September 2021
- {{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
- 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
- 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
- ...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>\co15 KB (2,309 words) - 23:43, 2 December 2021
- ...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
- 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>. Usi8 KB (1,480 words) - 14:52, 5 August 2022
- ...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
- ...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
- ...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 give15 KB (2,560 words) - 01:44, 1 July 2023
- for (int i = 0; i < 3; ++i) { pair A = (j,i);14 KB (2,073 words) - 15:15, 21 October 2021
- ...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 <math13 KB (2,200 words) - 21:36, 6 January 2024
- 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
- Call a positive integer <math>n</math> <math>k</math>-<i>pretty</i> 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
- label("$9$",(A+B)/2,dir(-90)); label("$6$",(A+E)/2,dir(-90));11 KB (1,794 words) - 15:32, 14 January 2024
- 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
- ...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
- ...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 <ma7 KB (1,145 words) - 20:27, 5 November 2023
- A = (0, tan(3 * pi / 7)); F = rotate(90/7, A) * (A - (0, 2));6 KB (968 words) - 15:01, 24 January 2024
- ...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
- ...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
- ~MRENTHUSIASM (credit given to Math Jams's <b>2021 AIME I Discussion</b>) ...<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>\co16 KB (2,635 words) - 19:56, 24 December 2023
- ...<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 \lef8 KB (1,370 words) - 21:34, 28 January 2024
- {{AIME Problems|year=2023|n=I}} [[2023 AIME I Problems/Problem 1|Solution]]7 KB (1,154 words) - 12:54, 20 February 2024
- 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
- 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
- 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
- <i><b>Standard Solution</b></i> <i><b>Short Solution</b></i>32 KB (5,375 words) - 11:43, 5 May 2024
