Search results

Create the page "Ac 2007" on this wiki! See also the search results found.

2007 BMO Problems/Problem 1
...>ABCD </math> be a convex quadrilateral with <math>AB=BC=CD </math>, <math>AC \neq BD </math>, and let <math>E </math> be the intersection point of its d ...eta </math>, then triangles <math>ABC, BCD </math> are congruent and <math>AC = BD </math>, a contradiction. Thus we conclude that <math>AE = DE </math>

3 KB (566 words) - 23:59, 14 September 2014
2007 Cyprus MO/Lyceum/Problem 21
[[Image:2007 CyMO-21.PNG|250px]] ...ween any two pulley center points are <math>AB=3\,\mathrm{m}</math>, <math>AC=4\,\mathrm{m}</math> and <math>BC=5\,\mathrm{m}</math>, then the length of

1 KB (175 words) - 22:21, 23 July 2020
2007 AMC 12B Problems/Problem 6
...math> has side lengths <math>AB = 5</math>, <math>BC = 6</math>, and <math>AC = 7</math>. Two bugs start simultaneously from <math>A</math> and crawl alo {{AMC12 box|year=2007|ab=B|num-b=5|num-a=7}}

792 bytes (121 words) - 04:21, 15 December 2020
2006 AIME II Problems
...<math> D </math> so that <math> AD=13, </math> and extend <math> \overline{AC} </math> through <math> C </math> to point <math> E </math> so that <math> {{AIME box|year = 2006|n=II|before=[[2006 AIME I Problems]]|after=[[2007 AIME I Problems]]}}

8 KB (1,350 words) - 12:00, 4 December 2022
2007 AMC 12B Problems
{{AMC12 Problems|year=2007|ab=B}} [[2007 AMC 12B Problems/Problem 1 | Solution]]

12 KB (1,814 words) - 12:58, 19 February 2020
Excircle
...</math> and <math>E_2</math> the points on sides <math>BC</math> and <math>AC</math>, respectively, such that <math>CD_2 = BD_1</math> and <math>CE_2 = A ...gle <math>ABC</math> opposite <math>A</math>. (Source: Problem 13.2 - MOSP 2007)

5 KB (843 words) - 03:02, 1 July 2020
2007 AMC 10A Problems/Problem 24
...}</math> is a right triangle. We know <math>AO=2\sqrt{2}</math> and <math>AC=2</math>, so <math>\triangle{ACO}</math> is an isosceles right triangle, an {{AMC10 box|year=2007|ab=A|num-b=23|num-a=25}}

3 KB (439 words) - 15:39, 3 June 2021
2007 AMC 10B Problems/Problem 11
2R = \frac {AC}{\sin \angle ABC} = \frac {3}{\frac {2\sqrt {2}}{3}} \Longrightarrow R = \f {{AMC10 box|year=2007|ab=B|num-b=10|num-a=12}}

5 KB (851 words) - 22:02, 26 July 2021
Isogonal conjugate
...th> and <math>AE</math> are constructed on sides <math>AB</math> and <math>AC</math> to the outer (inner) side of <math>\triangle ABC.</math> Let <math>\ <math>AB</math> and <math>AC</math> are isogonals with respect to the pair <math>(\ell,A).</math>

54 KB (9,416 words) - 08:40, 18 April 2024
2008 AMC 12A Problems
Triangle <math>ABC</math> has <math>AC=3</math>, <math>BC=4</math>, and <math>AB=5</math>. Point <math>D</math> is {{AMC12 box|year=2008|ab=A|before=[[2007 AMC 12B Problems|2007 AMC 12B]]|after=[[2008 AMC 12B Problems|2008 AMC 12B]]}}

13 KB (2,025 words) - 13:56, 2 February 2021
2007 AMC 12B Problems/Problem 23
...ments such that <math>AB = x + y</math>, <math>BC = x + z</math> and <math>AC = y + z</math>. Since ABC is a right triangle, one of <math>x</math>, <math ...ths then become <math>AB = x + y</math>, <math>BC = x + 6</math> and <math>AC = y + 6</math>. Plugging into Pythagorean's theorem:

4 KB (725 words) - 19:59, 4 January 2024
Mock AIME 1 2007-2008 Problems
[[Mock AIME 1 2007-2008 Problems/Problem 1|Solution]] [[Mock AIME 1 2007-2008 Problems/Problem 2|Solution]]

6 KB (992 words) - 14:15, 13 February 2018
Mock AIME 1 2007-2008 Problems/Problem 11
...ters <math>O_1,O_2,O_3</math>, respectively. Also, <math>AB = 23, BC = 25, AC=24</math>, and <math>\stackrel{\frown}{BF} = \stackrel{\frown}{EC},\ \stack {{Mock AIME box|year=2007-2008|n=1|num-b=10|num-a=12|source=196980}}

2 KB (265 words) - 20:02, 29 January 2020
2008 USAMO Problems
...ndicular]] [[bisect]]ors of <math>\overline{AB}</math> and <math>\overline{AC}</math> intersect ray <math>AM</math> in points <math>D</math> and <math>E< {{USAMO newbox|year=2008|before=[[2007 USAMO Problems|2007 USAMO]]|after=[[2009 USAMO Problems|2009 USAMO]]}}

4 KB (674 words) - 21:48, 12 August 2014
2008 iTest Problems
...math>(x-1)^{2007}+2(x-2)^{2006}+3(x-3)^{2005}+\cdots+2006(x-2006)^2+2007(x-2007)</math>. ...hen the sum of the reciprocals of the <math>2007</math> terms on the <math>2007^\text{th}</math> row gets divided by <math>2008</math>.

71 KB (11,749 words) - 01:31, 2 November 2023
2007 iTest Problems/Problem TB2
* <math>b+ac+d=2</math> * <math>ac+d=1</math>

3 KB (550 words) - 20:05, 2 January 2020
AoPS Wiki:Competition ratings
...line{AB}</math>, and let <math>E</math> be the midpoint of <math>\overline{AC}</math>. The angle bisector of <math>\angle BAC</math> intersects <math>\ov ...th>, <math>\angle CBA \leq 90^{\circ}</math>, <math>BC=1</math>, and <math>AC \geq AB</math>. Let <math>H</math>, <math>I</math>, and <math>O</math> be t

41 KB (6,786 words) - 04:09, 7 July 2024
2007 IMO Problems
[[2007 IMO Problems/Problem 1 | Solution]] [[2007 IMO Problems/Problem 2 | Solution]]

3 KB (505 words) - 09:24, 10 September 2020
2007 AMC 10B Problems
{{AMC10 Problems|year=2007|ab=B}} [[2007 AMC 10B Problems/Problem 1|Solution]]

15 KB (2,297 words) - 12:57, 19 February 2020
2007 AMC 10B Problems/Problem 4
Therefore, the measure of <math>\text{arc}AC</math> is also <math>100^\circ.</math> Since the measure of an [[inscribed {{AMC10 box|year=2007|ab=B|num-b=3|num-a=5}}

928 bytes (140 words) - 12:19, 4 July 2013

Page

Toolbox

Search

Search results