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• 2010 AIME I Problems/Problem 13
  ...d <math>N'</math>. Notice that <math>UNN'</math> and <math>TUT'</math> are similar. Thus: ...<math>ZT = a-84</math>. Since <math>NYU</math> and <math>UZT</math> are [[similar]],
  10 KB (1,418 words) - 23:05, 20 October 2021
• 2010 AIME I Problems/Problem 15
  ...<math>M</math> to <math>I_1 I_2</math> is of length <math>r</math>, so by similar triangles we find that <math>r^2 = MP \cdot MQ = (x + y - 6)( - x + y + 1)< === Solution 6 (Similar to Solution 1 with easier computation) ===
  14 KB (2,210 words) - 13:14, 11 January 2024
• 2010 AIME II Problems/Problem 14
  ...>, so <math>AD=a\cdot{AP}</math>. Moreover, since <math>CD=DP=a</math>, by similar triangle ratios, <math>\frac{AP}{a+a\cdot{AP}}=a</math>. Therefore, <math>A [[Category:Intermediate Geometry Problems]]
  10 KB (1,507 words) - 00:31, 19 November 2023
• 2010 AIME II Problems/Problem 9
  Using a similar argument, <math>NI=MH</math>, and ...ngruent. You can use the AA similarity to see that the small triangles are similar to the large triangles. Now you can proceed as in Solution 1.
  5 KB (857 words) - 22:22, 27 August 2023
• 1993 USAMO Problems/Problem 2
  ...t reflection of <math>E</math> over all the points of <math>XYZW</math> is similar to <math>XYZW</math> with a scale of <math>2</math> with center <math>E</ma [[Category:Olympiad Geometry Problems]]
  2 KB (446 words) - 08:09, 10 April 2023
• 2010 USAMO Problems/Problem 1
  angle <math>\gamma</math>, therefore they are similar, and so the ratio <math>PY : Of course, as with any geometry problem, DRAW A HUGE DIAGRAM spanning at least one page. And label all your
  13 KB (2,271 words) - 14:08, 23 May 2024
• 2010 AIME II Problems/Problem 15
  ...P = \angle MDP</math>, triangles <math>NEP</math> and <math>MDP</math> are similar. Hence ...math>. The information needed to use the Ratio Lemma can be found from the similar triangle section above.
  9 KB (1,523 words) - 15:24, 21 November 2023
• 2010 IMO Problems/Problem 2
  Key Lemma. Triangles <math>DKI</math> and <math>DIJ</math> are similar. ...ce, by SAS Similarity, triangles <math>DKI</math> and <math>DIJ</math> are similar, as desired.
  3 KB (525 words) - 14:52, 16 July 2023
• 2011 AMC 12A Problems/Problem 15
  ...> is also <math> 90^{\circ} </math>. Hence <math> \triangle EOM </math> is similar to <math>\triangle EPO </math>. [[Category:Introductory Geometry Problems]]
  2 KB (376 words) - 23:14, 5 January 2024
• 2011 AMC 12A Problems/Problem 22
  Using a method similar to before, if a point is <math>60</math>-ray partitional, then we must be a [[Category:Intermediate Geometry Problems]]
  11 KB (1,818 words) - 17:38, 6 September 2021
• 2011 AMC 12B Problems/Problem 20
  We can calculate the distances with coordinate geometry. (Note that <math>XA = XB = XC</math> because <math>X</math> is the circumc ...<math>\frac{XB}{XC} = \frac{BE}{CE} = 1</math>, <math>XB = XC</math>. In a similar fashion <math>XA = XB = XC = R</math>, where <math>R</math> is the circumci
  8 KB (1,200 words) - 19:31, 7 August 2023
• 2001 AMC 10 Problems/Problem 21
  ...</math>. Examining the cross section of the cone and cylinder, we find two similar triangles. Hence, <math> \frac{12-2r}{12}=\frac{2r}{10} </math> which we so ==Solution 3 (Very similar to solution 2 but explained more)==
  5 KB (697 words) - 23:14, 10 February 2024
• 2011 AIME I Problems/Problem 4
  Note: This is similar to Solution 2 after the first four lines [[Category:Intermediate Geometry Problems]]
  6 KB (1,068 words) - 18:52, 2 August 2023
• 2011 AIME I Problems/Problem 14
  ...t(1+\sqrt{2}\right)=0.\end{align*}</cmath> By the [https://www.cuemath.com/geometry/distance-between-two-lines/ distance between parallel lines formula], a cor ...til they intersect. Denote their intersection as <math>I_1</math>. Through similar triangles & the <math>45-45-90</math> triangles formed, we find that <math>
  8 KB (1,344 words) - 18:39, 9 February 2023
• 2011 AIME I Problems/Problem 8
  ...et <math>\{u, v, w, x, y, z\} = \{AU, AV, CW, CX, BY, BZ\}</math>. Then by similar triangles ...h_c = \frac{40\sqrt{221}}{30}, h_a = \frac{40\sqrt{221}}{23}</math>. From similar triangles we can see that <math>\frac{27h}{h_a}+\frac{27h}{h_c} \le 27 \rig
  6 KB (1,077 words) - 21:47, 12 April 2022
• 2011 AIME II Problems/Problem 13
  ==Solution 8 (similar to solution 4)== [[Category:Intermediate Geometry Problems]]
  13 KB (2,055 words) - 05:25, 9 September 2022
• 2011 USAMO Problems/Problem 3
  If <math>r-t=s-t=0</math>, then <math>P</math> must be similar to <math>P'</math> and the conclusion is obvious. [[Category:Olympiad Geometry Problems]]
  11 KB (1,925 words) - 12:07, 31 August 2023
• 2011 AMC 10B Problems/Problem 9
  [[Category: Introductory Geometry Problems]] ...of sides (in similar figures) because area is a second-degree property of similar figures. So like solution 3, the ratio of sides is <math>\sqrt{\frac{1}{3}}
  4 KB (587 words) - 22:08, 31 August 2023
• 2011 USAMO Problems/Problem 5
  ...through <math>R</math>, contradicting <math>Q_1 Q_2 \parallel AB</math>. A similar contradiction occurs if <math>Q_1 Q_2 \parallel CD</math> but <math>Q_1 Q_2 [[Category:Olympiad Geometry Problems]]
  4 KB (660 words) - 01:04, 15 February 2024
• 1996 AHSME Problems/Problem 30
  import olympiad; import geometry; size(150); defaultpen(linewidth(0.8)); ...ngle ADE</math>. Hence triangles <math>ABQ</math> and <math>EDQ</math> are similar. Therefore, <cmath>\frac{AQ}{EQ}=\frac{BQ}{DQ}=\frac{AB}{ED}=\frac{3}{5}.</
  10 KB (1,515 words) - 13:09, 20 December 2023

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