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2019 AIME II Problems/Problem 11
...math> <cmath>\implies O_{1}B = O_{1}A=\frac{147}{16\sqrt{5}}</cmath> Using similar logic we obtain <math>O_{2}A =\frac{189}{16\sqrt{5}}.</math> ==Solution 5 (Olympiad Geometry)==

12 KB (1,985 words) - 19:52, 28 January 2024
2019 AIME I Problems/Problem 6
==Solution 3 (Similar triangles)== ==Solution 4 (Similar triangles, orthocenters)==

11 KB (1,717 words) - 20:11, 19 January 2024
2019 AIME I Problems/Problem 3
...gle PQR= \frac{25}{2}\cdot \frac{3}{5}=\frac{15}{2}</math>. Preceding in a similar fashion for <math>\triangle PAF</math>, the area of <math>\triangle PAF</ma ==Solution 3 (Easiest, uses only basic geometry too)==

6 KB (901 words) - 09:38, 10 May 2024
2019 AIME II Problems/Problem 1
...the new triangle formed has length <math>4.5</math>. Notice we have formed similar triangles, and we can solve for <math>h</math>. ...pythagorean theorem, we have the height is <math>8,</math> and we can use similar triangles to finish.

11 KB (1,794 words) - 15:32, 14 January 2024
2019 AIME I Problems/Problem 11
...d <math>EN=r_N</math>. Triangles <math>INB</math> and <math>BNE</math> are similar to <math>IBE</math>, so <math>\frac{IN}{BN}=\frac{BN}{EN}</math>, or <math> ...= 90^{\circ}</math>, triangle <math>AGI</math> and <math>AHI_A</math> are similar by AA. Then <math>\frac{r_{I_A}}{r} = \frac{h+r_{I_A}}{h-r} \implies r_{I_A

21 KB (3,915 words) - 19:55, 10 October 2023
2019 AIME I Problems/Problem 13
...GF = \frac{BG + 4}{3}</math>. Also note that <math>\triangle GBC</math> is similar to <math>\triangle GFE</math>, which gives us <math>GF = \frac{7 \cdot BG}{ ...th>. So <math>BP=\frac{5}{4}</math> and <math>FP=\frac{7}{4}</math>. These similar triangles also gives us <math>DP=\frac{5}{3}x</math> so <math>DE=\frac{16}{

10 KB (1,620 words) - 20:44, 20 December 2023
What is the definition of Pure Mathematics?
That may be true for some areas of math. However, what about geometry, trigonometry, and calculus? And what is the definition of numbers? Now you ...math but not the whole thing. For example, the definition of geometry is ''Geometry is concerned with the properties and relations of points, lines, surfaces,

35 KB (5,882 words) - 18:08, 28 June 2021
2005 Indonesia MO Problems/Problem 7
By using similar reasoning, the coordinates of <math>L</math> are <math>(b_1-b_2-c,b_2+b_1+c [[Category:Olympiad Geometry Problems]]

5 KB (959 words) - 20:57, 12 October 2019
2023 AIME I Problems/Problem 12
...ath>P</math> passes through the circumcircle of triangle <math>BFD.</math> Similar reasoning leads us to the fact that <math>P</math> also passes through the ...hem <math>Q,R,</math> and <math>S</math> respectively. This gives us three similar right triangles <math>FQP</math>, <math>ERP</math>, and <math>DSP.</math>

16 KB (2,592 words) - 15:40, 13 April 2024
2019 AMC 8 Problems/Problem 24
Triangles <math>BEF</math> and <math>EXD</math> are similar, and since <math>BE = ED</math>, they are also congruent, and so <math>XE=E ==Solution 5 (Similar Triangles)==

27 KB (4,256 words) - 19:30, 17 January 2024
SAS Similarity
SAS stands for Side-Angle-Side, for two [[triangle]]s to be [[similar|similar triangles]] by SAS similarity, they must have a pair of congruent angles an ...orresponding sides with identical ratios as shown below, the triangles are similar.

1 KB (145 words) - 21:12, 28 January 2021
2020 AMC 10B Problems/Problem 8
==Solution 1 (Geometry)== ...ath>R=(x,-3),</math> we have <math>R=\left(\pm\sqrt7,-3\right)</math> by a similar process. <p>

5 KB (822 words) - 14:53, 6 June 2023
2020 AMC 12B Problems/Problem 10
==Solution 1 (Similar Triangles)== ...=\frac{\sqrt{5}}{2}.</math> Now we just need to find <math>AP</math> using similar triangles <math>\triangle APN\sim\triangle ANM:</math>

9 KB (1,328 words) - 16:14, 11 September 2023
2020 AIME II Problems/Problem 15
Corresponding sides of similar <math>\triangle XTM \sim \triangle YMT</math> is <math>MT,</math> so ...ath> onto line <math>AB</math>. Let M be midpoint BC. Then triangle ABC is similar to triangle XTM.

7 KB (1,221 words) - 16:46, 29 January 2023
2020 AIME II Problems/Problem 7
...ht triangles <math>\triangle ABD</math> and <math>\triangle CBE</math> are similar, implying that the radius of the sphere is<cmath>CE = AD \cdot\frac{BC}{AB} == Solution (Clean analytic geometry) ==

8 KB (1,255 words) - 13:43, 24 June 2024
2021 AIME I Problems/Problem 2
==Solution 1 (Similar Triangles)== Let <math>AG=x</math>. Then, we have <math>DG=11-x</math>. By similar triangles, we know that <math>FG=\frac{7}{3}(11-x)</math> and <math>CG=\fra

8 KB (1,294 words) - 00:59, 23 August 2022
2021 AIME I Problems/Problem 9
==Solution 1 (Similar Triangles and Pythagorean Theorem)== ==Solution 2 (Similar Triangles and Pythagorean Theorem)==

23 KB (3,640 words) - 18:16, 25 January 2024
2021 AIME I Problems/Problem 13
...>\overline{OF}\parallel\overline{O_{1}O_{2}}</math> giving us the pairs of similar triangles <cmath>\triangle O_{1}GD\sim\triangle OFD~\text{and}~\triangle O_ ...th>O_1O_2 = O_1Y - O_2Y = 2 \cdot 961 - 2\cdot 625 = \boxed{672}</cmath>by similar triangles.

17 KB (2,852 words) - 03:59, 7 February 2024
2022 AIME I Problems/Problem 3
Additionally, we can see that <math>\triangle XZW</math> is similar to <math>\triangle PQZ</math> and <math>\triangle AZB</math>. We know that ...BQC</math> yields <math>QC = 333\sqrt{1-k^2}</math>. We then proceed using similar triangles: <math>\angle BYQ = \angle BQC = 90^{\circ}</math> and <math>\ang

24 KB (3,832 words) - 20:59, 2 March 2024
2022 AIME I Problems/Problem 11
...ath>MN</math> at <math>I</math>. Then, <math>\bigtriangleup{IMC}</math> is similar to <math>\bigtriangleup{AIN}</math>. Therefore, <math>\frac{CI}{MC}=\frac{A [[Category:Intermediate Geometry Problems]]

16 KB (2,517 words) - 20:22, 31 January 2024

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