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- ...{2,1,3\}, \{2,3,1\},\{3,1,2\}</math> and <math>\{3,2,1\}</math>. We often drop the brackets and commas, so the permutation <math>\{2,1,3\}</math> would ju3 KB (422 words) - 11:01, 25 December 2020
- ...1</math> boxes (which divide the nonconsecutive numbers) into which we can drop the <math>n - k</math> remaining elements, with the caveat that each of the8 KB (1,405 words) - 11:52, 27 September 2022
- <math>\dagger</math> Alternatively, drop an altitude from <math>O_1</math> to <math>O_3T</math> at <math>O_3'</math>4 KB (693 words) - 13:03, 28 December 2021
- ...angle OF_1F_2'</math> is isosceles, and as <math>F_1F_2'=20</math>, we can drop an altitude to <math>F_1F_2'</math> to easily find that <math>\tan \angle O12 KB (2,001 words) - 20:26, 23 July 2024
- Drop the altitude from A and call the base of the altitude Q. Also, drop the altitudes from E and D to AB and AC respectively. Call the feet of the14 KB (2,340 words) - 16:38, 21 August 2024
- ...AB</math> so we just need to find <math>CB</math>, call it <math>x</math>. Drop an altitude from <math>F</math> to <math>AB</math> and call it <math>H</mat9 KB (1,501 words) - 05:34, 30 October 2023
- Drop perpendiculars from <math>O</math> to <math>AB</math> (with foot <math>T_1<11 KB (1,741 words) - 22:40, 23 November 2023
- ...already given in the later section [Proof that R,A, and B are collinear]). Drop a perpendicular from <math>P</math> to the lines that the centers are on. Y13 KB (2,151 words) - 17:48, 27 May 2024
- ...h>BC</math> and call its intersection with <math>BC</math> <math>K</math>. Drop the perpendicular from <math>F</math> to <math>KO</math> and call its inter ...efore, <math>DC = \sqrt{10} \Longleftrightarrow BA = \sqrt{2}</math>. Now, drop the altitude from <math>O</math> to <math>BA</math> and call its intersecti20 KB (3,497 words) - 15:37, 27 May 2024
- ...2</math> in succession, into <math>(\dag)</math>. In each case, most terms drop, and we end up with6 KB (1,051 words) - 19:35, 1 August 2024
- ...th>, <math>CZ=b</math>, <math>AX=c</math>, and <math>WD=d</math>. First we drop a perpendicular from <math>Q</math> to a point <math>R</math> on <math>BC</3 KB (530 words) - 07:46, 1 June 2018
- As in Solution 1, drop an altitude <math>h</math> to <math>c</math>. Let <math>h</math> meet <math8 KB (1,401 words) - 21:41, 20 January 2024
- Alternatively, we can drop an altitude from <math>C</math> and arrive at the same answer. Now, if we drop an altitude from point<math>M</math>, we get :6 KB (980 words) - 15:08, 14 May 2024
- ...bserve that <math>[PQRS]=\frac{1}{2}\cdot30\cdot40=600</math>. Thus, if we drop an altitude from <math>P</math> to <math>\overline{SR}</math> to point <mat8 KB (1,270 words) - 23:36, 27 August 2023
- First, drop a perpendicular from <math>O</math> to <math>AB</math>. Call this midpoint5 KB (787 words) - 02:09, 28 June 2024
- Now looking at triangle <math>\triangle PAC,</math> we drop the perpendicular from <math>P</math> to <math>AC</math>, and call the foot8 KB (1,172 words) - 21:57, 22 September 2022
- ...2} + 1}{2}}}{\sqrt{\frac{\frac{-\sqrt{2}}{2} + 1}{2}}}</cmath> in which we drop the negative roots (as it is clear cosine of <math>22.5</math> and <math>6710 KB (1,512 words) - 17:16, 18 June 2024
- .../math>, then the number of possible values of <math>y</math> will begin to drop again, equaling the amount when <math>x = 14</math>. Then when we finally s6 KB (913 words) - 16:34, 6 August 2020
- Drop [[perpendicular]]s from <math>P</math> to the three sides of <math>\triangl7 KB (1,184 words) - 13:25, 22 December 2022
- ...(a+bi)z</math> forms an angle of <math>\theta</math> with <math>z</math>. Drop the altitude/median from <math>Q</math> to base <math>OP</math>, and you en6 KB (1,010 words) - 19:01, 24 May 2023