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• Pythagorean Theorem
  ...one of the many tools in a good geometer's arsenal. A very large number of geometry problems can be solved by building right triangles and applying the Pythago ...e <math>ABC, CBH, ACH</math> are similar right triangles, and the areas of similar triangles are proportional to the squares of corresponding side lengths,
  5 KB (886 words) - 21:12, 22 January 2024
• MATHCOUNTS
  MATHCOUNTS curriculum includes [[arithmetic]], [[algebra]], [[counting]], [[geometry]], [[number theory]], [[probability]], and [[statistics]]. The focus of MA ...states (most notably Florida), there is an optional ciphering round. Very similar to countdown (in both difficulty and layout), a team sends up a representat
  10 KB (1,497 words) - 11:42, 10 March 2024
• Art of Problem Solving
  * [https://artofproblemsolving.com/store/item/intro-geometry Introduction to Geometry] ...n] is an online interactive game for students to participate in activities similar to the [[MATHCOUNTS]] Countdown Round.
  5 KB (667 words) - 17:09, 3 July 2023
• Resources for mathematics competitions
  *[https://www.clevermath.org/ Clevermath] Is similar to above ...olving.com/resources/articles/bary.pdf Barycentric Coordinates in Olympiad Geometry] by Max Schindler and Evan Chen
  16 KB (2,152 words) - 21:46, 6 May 2024
• Math books
  ...eek9], currently a junior in high school. It covers the basics of algebra, geometry, combinatorics, and number theory, along with sets of accompanying practice ...919597 Undergraduate Algebra] by [[Serge Lang]]. Some compare it to being similar to Dummit and Foote with regards to rigor, although this text is slightly m
  24 KB (3,177 words) - 12:53, 20 February 2024
• Menelaus' theorem
  ===Proof with Similar Triangles=== [[Category:Geometry]]
  5 KB (804 words) - 03:01, 12 June 2023
• Fermat's Little Theorem
  A similar version can be used to prove [[Euler's Totient Theorem]], if we let <math>S === Proof 4 (Geometry) ===
  16 KB (2,658 words) - 16:02, 8 May 2024
• Brahmagupta's Formula
  == Similar formulas == A similar formula which Brahmagupta derived for the area of a general quadrilateral i
  3 KB (465 words) - 18:31, 3 July 2023
• Contest strategies
  ..., a visual aid can help solvers visualize and look for clues. Even in non-geometry problems, drawing pictures can help formulate ideas. ...lked about “wishful thinking” — trying to make a problem look like a similar one solved before.
  3 KB (538 words) - 13:13, 16 January 2021
• Incenter/excenter lemma
  In [[geometry]], the '''incenter/excenter lemma''', sometimes called the '''Trillium theo The incenter/excenter lemma makes frequent appearances in olympiad geometry. Along with the larger lemma, two smaller results follow: first, <math>A</m
  2 KB (291 words) - 16:31, 18 May 2021
• Triangle
  ...ctions of the sides summing to 180 degrees. Triangles exist in Euclidean [[geometry]], and are the simplest possible polygon. In [[physics]], triangles are not ...des and is also [[equiangular]]. Note that all equilateral triangles are [[similar]]. All the angles of equilateral triangles are <math>60^{\circ}</math>
  4 KB (628 words) - 17:17, 17 May 2018
• American Scholastic Mathematics Association
  It features questions with probability, counting, arithmetic, algebra, and geometry. It is 35 minutes long and contains 7 questions. The difficulty of this math competition is similar to MOEMS (easier problems) and AMC 8 (harder problems)
  1 KB (153 words) - 13:11, 14 May 2019
• Ptolemy's Inequality
  ...math>P </math> such that the [[triangles]] <math>APB, \; DCB </math> are [[similar]] and have the same [[orientation]]. In particular, this means that ...rity]], we also know that the triangles <math>ABD, \; PBC </math> are also similar, which implies that
  3 KB (602 words) - 09:01, 7 June 2023
• 2011 AMC 10B Problems/Problem 22
  [[Category: Introductory Geometry Problems]] Notice that the smaller pyramid on top of the cube is similar to the larger pyramid. Thus, letting <math>x</math> be the edge length of t
  4 KB (691 words) - 18:38, 19 September 2021
• Homothety
  * The resulting image of a polygon from a homothety is [[similar]] to the original polygon. * https://brilliant.org/wiki/euclidean-geometry-homothety/ (contains sample problems and related proofs)
  3 KB (532 words) - 01:11, 11 January 2021
• Power of a Point Theorem
  Draw extra lines to create similar triangles! (Hint: Draw <math>AD</math> on all three figures. Draw another l * [[Geometry]]
  5 KB (827 words) - 17:30, 21 February 2024
• 2006 AIME I Problems/Problem 7
  ...by attaching this triangle with the next trapezoid. Parallel lines give us similar triangles, so we know the proportion of this triangle to the previous trian ...ratio we seed is <math>\frac{33(ay+yx)}{11xy}.</math> Finally note that by similar triangles <math>\frac{x}{x+a} =\frac{y}{y+b} \implies bx = ya.</math> There
  4 KB (709 words) - 01:50, 10 January 2022
• 2006 AMC 12A Problems/Problem 17
  (Similar to Solution 1) [[Category:Introductory Geometry Problems]]
  6 KB (958 words) - 23:29, 28 September 2023
• 2006 AMC 10A Problems/Problem 16
  ...\sim \triangle AEO_2 \sim \triangle AFC</math>. Using the first pair of [[similar triangles]], we write the proportion: [[Category:Introductory Geometry Problems]]
  5 KB (732 words) - 23:19, 19 September 2023
• 2006 AMC 10A Problems/Problem 17
  There are many different similar ways to come to the same conclusion using different [[right triangle|45-45- [[Category:Introductory Geometry Problems]]
  6 KB (1,066 words) - 00:21, 2 February 2023
• 2006 AMC 10A Problems/Problem 19
  How many non-[[similar]] triangles have angles whose degree measures are distinct positive integer [[Category:Introductory Geometry Problems]]
  2 KB (259 words) - 03:10, 22 June 2023
• 2005 AIME II Problems/Problem 8
  ...<math>O_3</math> to <math>\overline{AB}</math> be <math>T</math>. From the similar [[right triangle]]s <math>\triangle HO_1T_1 \sim \triangle HO_2T_2 \sim \tr ...=10-4=6</math>, and <math>O_1O_2=14</math>. But <math>O_1O_3O_3'</math> is similar to <math>O_1O_2O_2'</math> so <math>O_3O_3'=\frac{6}{14} \cdot 10=\frac{30}
  4 KB (693 words) - 13:03, 28 December 2021
• 2005 AIME II Problems/Problem 12
  === Solution 3 (similar triangles)=== ...>, and <math>KA = EB</math> (90 degree rotation), and now we can bash on 2 similar triangles <math>\triangle GAK \sim \triangle GHO</math>.
  13 KB (2,080 words) - 21:20, 11 December 2022
• 2005 AIME II Problems/Problem 15
  ...y=ax </math> contains the center of a circle that is externally [[tangent (geometry)|tangent]] to <math> w_2 </math> and internally tangent to <math> w_1. </ma ...stance from this tangent point to the origin is <math>\sqrt{69}.</math> By similar triangles, the slope of this line is then <math>\frac{\sqrt{69}}{5\sqrt{3}}
  12 KB (2,000 words) - 13:17, 28 December 2020
• 2005 AIME II Problems/Problem 14
  import olympiad; import cse5; import geometry; size(150); == Solution 2 (Similar Triangles)==
  13 KB (2,129 words) - 18:56, 1 January 2024
• 2004 AIME I Problems/Problem 14
  ...riangle]]s <math>\triangle CDA</math> and <math>\triangle CEB</math> are [[similar]] [[right triangle]]s. By the Pythagorean Theorem <math>CD=8\cdot\sqrt{6}< [[Category:Intermediate Geometry Problems]]
  4 KB (729 words) - 01:00, 27 November 2022
• 2004 AIME I Problems/Problem 11
  ...se the plane cut is parallel to the base of our solid, <math>C</math> is [[similar]] to the uncut solid and so the height and slant height of cone <math>C</ma <math>V</math> and <math>C</math> are similar cones, because the plane that cut out <math>C</math> was parallel to the ba
  5 KB (839 words) - 22:12, 16 December 2015
• 2004 AIME I Problems/Problem 10
  ...{A'B'C'} = r_{ABC} - 1 = 5</math>, and since the ratio of the areas of two similar figures is equal to the square of the ratio of their corresponding lengths, ...dn't know the formula for the distance from a point to a line, you can use similar triangles to get the ratio:
  5 KB (836 words) - 07:53, 15 October 2023
• 2004 AIME II Problems/Problem 12
  We use a similar argument with the line <math>DO</math>, and find the height from the top of [[Category:Intermediate Geometry Problems]]
  3 KB (431 words) - 23:21, 4 July 2013
• 2004 AIME II Problems/Problem 7
  ...theta)=\frac{BC'}{17}=\frac53</math>, so <math>BC'=\frac{85}{3}</math>. By similar triangles, <math>CC'=\frac{3}{17}BC'=\frac{15}{3}</math>, so <math>BC=\frac ...le, and so are B'DG and C'FG. Since C'F is 3, then using the properties of similar triangles GF is 51/8. DF is 22, so DG is 125/8. Finally, DB can to calculat
  9 KB (1,501 words) - 05:34, 30 October 2023
• 1983 AIME Problems/Problem 14
  ...we can find the third side of the triangle using [[Stewart's Theorem]] or similar approaches. We get <math>AC = \sqrt{56}</math>. Now we have a kite <math>AQ ...=\frac{BR}{NR},</math> triangles <math>BNR</math> and <math>AMR</math> are similar. If we let <math>y=BN</math>, we have <math>AM=3BN=3y</math>.
  13 KB (2,149 words) - 18:44, 5 February 2024
• 1983 AIME Problems/Problem 15
  This solution, while similar to Solution 2, is arguably more motivated and less contrived. === Solution 4 (coordinate geometry) ===
  19 KB (3,221 words) - 01:05, 7 February 2023
• 1984 AIME Problems/Problem 3
  ...that go through <math>P</math>, all four triangles are [[similar triangles|similar]] to each other by the <math>AA</math> postulate. Also, note that the lengt Alternatively, since the triangles are similar by <math>AA</math>, then the ratios between the bases and the heights of ea
  4 KB (726 words) - 13:39, 13 August 2023
• 1985 AIME Problems/Problem 4
  A small [[square (geometry) | square]] is constructed inside a square of [[area]] 1 by dividing each s ...c{n-1}{\sqrt{1985}}</math>. Notice that <math>\triangle CEL</math> is also similar to <math>\triangle CDF</math> by <math>AA</math> similarity. Thus, <math>\f
  3 KB (484 words) - 21:40, 2 March 2020
• 1986 AIME Problems/Problem 9
  ...ll three smaller triangles and the larger triangle are [[similar triangles|similar]] (<math>\triangle ABC \sim \triangle DPD' \sim \triangle PEE' \sim \triang By similar triangles, <math>BE'=\frac{d}{510}\cdot450=\frac{15}{17}d</math> and <math>
  11 KB (1,850 words) - 18:07, 11 October 2023
• 1987 AIME Problems/Problem 15
  Because all the [[triangle]]s in the figure are [[similar]] to triangle <math>ABC</math>, it's a good idea to use [[area ratios]]. In [[Category:Intermediate Geometry Problems]]
  5 KB (838 words) - 18:05, 19 February 2022
• 1988 AIME Problems/Problem 12
  This problem is quite similar to the 1992 AIME problem 14, which also featured concurrent cevians and is [[Category:Intermediate Geometry Problems]]
  4 KB (727 words) - 23:37, 7 March 2024
• 1989 AIME Problems/Problem 15
  ...onally we now see that triangles <math>FPE</math> and <math>CPB</math> are similar, so <math>FE \parallel BC</math> and <math>\frac{FE}{BC} = \frac{1}{3}</mat [[Category:Intermediate Geometry Problems]]
  13 KB (2,091 words) - 00:20, 26 October 2023
• 1989 AIME Problems/Problem 10
  By identical logic, we can find similar expressions for the sums of the other two cotangents: [[Category:Intermediate Geometry Problems]]
  8 KB (1,401 words) - 21:41, 20 January 2024
• 2006 AMC 10B Problems/Problem 15
  Rhombus <math>ABCD</math> is similar to rhombus <math>BFDE</math>. The area of rhombus <math>ABCD</math> is <mat [[Category:Introductory Geometry Problems]]
  3 KB (445 words) - 22:01, 20 August 2022
• 1991 AIME Problems/Problem 12
  By similar logic, we have <math>APOS</math> is a cyclic quadrilateral. Let <math>AP = ...iangle ABD \sim \triangle OQP</math> by <math>AA</math> [[similar triangle|similar]]ity. From here, it's clear that
  8 KB (1,270 words) - 23:36, 27 August 2023
• 1991 AIME Problems/Problem 2
  ...r of the rectangle would have length <math>\frac{1}{168}\cdot{5}</math> by similar triangles. If you add the two lengths together, it is <math>\frac{167}{168} [[Category:Intermediate Geometry Problems]]
  4 KB (595 words) - 12:51, 17 June 2021
• 1992 AIME Problems/Problem 12
  ...square]]s. To take a bite, a player chooses one of the remaining [[square (geometry) | squares]], then removes ("eats") all squares in the quadrant defined by This game is similar to an AoPS book.
  2 KB (443 words) - 22:41, 22 December 2021
• 1992 AIME Problems/Problem 9
  ...the equations in <math>(1)</math> without directly resorting to trig. From similar triangles, [[Category:Intermediate Geometry Problems]]
  5 KB (874 words) - 10:27, 22 August 2021
• 2006 AMC 10B Problems/Problem 20
  ...has legs <math>2</math> and <math>20</math>. Aha! The two triangles are similar by SAS, with one triangle having side lengths <math>100</math> times the ot ...and 2 is one of the lengths of the adjacent sides. Those two triangles are similar because <math>AD</math> and <math>AB</math> are perpendicular. <math> \frac
  4 KB (594 words) - 15:45, 30 July 2023
• 2006 AMC 10B Problems/Problem 23
  We use mass points (similar to above). Let the triangle be <math>ABC</math> with cevians (lines to oppo [[Category:Introductory Geometry Problems]]
  5 KB (861 words) - 00:53, 25 November 2023
• 2006 AMC 10B Problems/Problem 24
  == Solution 1 (Similar Triangles) == ...point <math>E</math>. Triangles <math>EBO</math> and <math>ECP</math> are similar, and by symmetry, so are triangles <math>EAO</math> and <math>EDP</math>. T
  4 KB (558 words) - 14:38, 6 April 2024
• 1994 AIME Problems/Problem 10
  Using similar right triangles, we identify that <math>CD = \sqrt{AD \cdot BD}</math>. Let [[Category:Intermediate Geometry Problems]]
  3 KB (534 words) - 16:23, 26 August 2018
• 1994 AIME Problems/Problem 6
  ...agon to the area of a regular triangle. Since the ratio of the area of two similar figures is the square of the ratio of their side lengths, we see that the r Picturing the diagram in your head should give you an illustration similar to the one above. The distance from parallel sides of the center hexagon is
  4 KB (721 words) - 16:14, 8 March 2021
• 1995 AIME Problems/Problem 12
  Similar to Solution 1, <math>\angle APC</math> is the dihedral angle we want. WLOG, [[Category:Intermediate Geometry Problems]]
  8 KB (1,172 words) - 21:57, 22 September 2022
• 1995 AIME Problems/Problem 9
  In a similar fashion, we encode the angles as complex numbers, so if <math>BM=x</math>, [[Category:Intermediate Geometry Problems]]
  7 KB (1,181 words) - 13:47, 3 February 2023
• 1995 AIME Problems/Problem 4
  ...th>O_9A_9 = \frac{2 \cdot O_6A_6 + 1 \cdot O_3A_3}{3} = 5</math> (consider similar triangles). Applying the [[Pythagorean Theorem]] to <math>\triangle O_9A_9P == Solution 2 (Analytic Geometry) ==
  3 KB (605 words) - 11:30, 5 May 2024
• 1996 AIME Problems/Problem 14
  ...gh the interiors of how many of the <math>1\times 1\times 1</math> [[cube (geometry) | cube]]s? ...(plane of <math>x=0</math> is not considered since <math>m \ne 0</math>). Similar arguments for slices along <math>y</math>-planes and <math>z</math>-planes
  5 KB (923 words) - 21:21, 22 September 2023
• 1996 AIME Problems/Problem 4
  ...= 49</math>, and so the sides of the shadow are <math>7</math>. Using the similar triangles in blue, [[Category:Intermediate Geometry Problems]]
  2 KB (257 words) - 17:50, 4 January 2016
• 1998 AIME Problems/Problem 12
  In a similar vein, using LoC on <math>\Delta PEQ</math> and <math>\Delta CEQ,</math> res [[Category:Intermediate Geometry Problems]]
  5 KB (876 words) - 20:27, 9 June 2022
• 1998 AIME Problems/Problem 6
  There are several [[similar triangles]]. <math>\triangle PAQ\sim \triangle PDC</math>, so we can write [[Category:Intermediate Geometry Problems]]
  2 KB (254 words) - 19:38, 4 July 2013
• 1999 AIME Problems/Problem 8
  ...responding angles we see that all of the triangles are [[similar triangles|similar]], so they are all equilateral triangles. We can solve for their side lengt [[Category:Intermediate Geometry Problems]]
  3 KB (445 words) - 19:40, 4 July 2013
• 2000 AIME I Problems/Problem 8
  By [[similar triangle]]s, we find that the dimensions of the liquid in the first cone to ...of the container is <math>100\pi</math>. The cone formed by the liquid is similar to the original, but scaled down by <math>\frac{3}{4}</math> in all directi
  4 KB (677 words) - 16:33, 30 December 2023
• 2001 AIME I Problems/Problem 7
  ...> is <math>\frac{43}{63}</math>, which is the scale factor between the two similar triangles, and thus <math>DE = \frac{43}{63} \times 20 = \frac{860}{63}</ma ...ratio of <math>FP:PA</math> is <math>20:43</math>. Therefore, the smaller similar triangle <math>ADE</math> is <math>43/63</math> the height of the original
  9 KB (1,540 words) - 08:31, 1 December 2022
• 2002 AIME I Problems/Problem 11
  We can use a similar trick as with reflections in 2D: Imagine that the entire space is divided i [[Category:Intermediate Geometry Problems]]
  3 KB (591 words) - 15:11, 21 August 2019
• 2002 AIME I Problems/Problem 2
  ...e triangle is therefore <math>2\sqrt{3}+2</math> and we use simplification similar to as showed above, and we reach the result <math>\frac{1}{2} \cdot (\sqrt{ [[Category:Intermediate Geometry Problems]]
  2 KB (287 words) - 19:54, 4 July 2013
• 2003 AIME II Problems/Problem 14
  ...<math>c-b=8\sqrt{3}/3,</math> and thus <math>c=18\sqrt{3}/3</math>. Using similar methods (or symmetry), we determine that <math>D=(10\sqrt{3}/3,10)</math>, This is similar to solution 2 but faster and easier.
  9 KB (1,461 words) - 15:09, 18 August 2023
• 2003 AIME II Problems/Problem 11
  By AA similarity, <math>\Delta EMB</math> is similar to <math>\Delta ACB</math>, <math>\frac{EM}{AC}=\frac{MB}{CB}</math>. Solvi [[Category: Intermediate Geometry Problems]]
  5 KB (772 words) - 19:47, 1 August 2023
• 2003 AIME II Problems/Problem 6
  ...hus <math>[A'DE]=\frac{1}{9}[A'B'C']=\frac{1}{9}[ABC].</math> We can apply similar reasoning to the other small triangles in <math>A'B'C'</math> not contained [[Category: Intermediate Geometry Problems]]
  5 KB (787 words) - 17:38, 30 July 2022
• 2003 AIME II Problems/Problem 4
  ...th>Z</math> on <math>ABD</math>. Clearly, the two regular tetrahedrons are similar, so if we can find the ratio of the sides, we can find the ratio of the vol [[Category: Intermediate Geometry Problems]]
  3 KB (563 words) - 17:36, 30 July 2022
• 2002 AIME II Problems/Problem 14
  ...h>O</math> on <math>\overline{AP}</math> is drawn so that it is [[Tangent (geometry)|tangent]] to <math>\overline{AM}</math> and <math>\overline{PM}</math>. Gi ...ower of a point]]. Using the fact that the ratio of corresponding sides in similar triangles is equal to the ratio of their perimeters, we have
  4 KB (658 words) - 19:15, 19 December 2021
• 2002 AIME II Problems/Problem 13
  Notice now that <math>\triangle{PBQ}</math> is similar to <math>\triangle{EBA}</math>. Therefore, Also, <math>\triangle{PRA}</math> is similar to <math>\triangle{DBA}</math>. Therefore,
  6 KB (935 words) - 13:23, 3 September 2021
• 2001 AIME II Problems/Problem 15
  ...is at <math>(8,8,8)</math>. Using a little visualization (involving some [[similar triangles]], because we have parallel lines) shows that the tunnel meets th [[Category:Intermediate Geometry Problems]]
  4 KB (518 words) - 15:01, 31 December 2021
• 2001 AIME II Problems/Problem 13
  ...ine{CD}</math> at <math>M</math>. Note that <math>\triangle{DAB}</math> is similar to <math>\triangle{BMC}</math> by AA similarity, since <math>\angle{ABD}=\a ...th>x=12</math>.So <math>AM=DM=20</math> similarly, we use the same pair of similar triangle we get <math>\frac{CM}{BM}=\frac{BM}{DM}</math>, we get that <math
  4 KB (743 words) - 03:32, 23 January 2023
• 2001 AIME II Problems/Problem 12
  ...math>\frac 12</math> of the original one. Since the ratio of the volume of similar polygons is the cube of the ratio of their corresponding lengths, it follow [[Category:Intermediate Geometry Problems]]
  2 KB (380 words) - 00:28, 5 June 2020
• 2001 AIME II Problems/Problem 7
  ...STR \sim \triangle UQV</math>. Since the ratio of corresponding lengths of similar figures are the same, we have ...we find that <math>A_{1}Q = 60</math>, <math>A_{1}R = 90</math>. Using the similar triangles, <math>RA_{2} = 45</math>, <math>QA_{3} = 20</math>, so <math>A_{
  7 KB (1,112 words) - 02:15, 26 December 2022
• 2001 AIME II Problems/Problem 6
  ...hs are positive, we discard the other root). The ratio of the areas of two similar figures is the square of the ratio of their corresponding side lengths, so [[Category:Intermediate Geometry Problems]]
  4 KB (772 words) - 19:31, 6 December 2023
• 2000 AIME II Problems/Problem 6
  ...e segment into three partitions with lengths <math>x_1, 75, x_2</math>. By similar triangles, we easily find that <math>\frac{x - 75}{100} = \frac{x_1+x_2}{10 [[Category:Intermediate Geometry Problems]]
  3 KB (433 words) - 19:42, 20 December 2021
• 2005 AMC 10B Problems/Problem 14
  ...ve that <math>\triangle AYC</math> and <math>\bigtriangleup MXC</math> are similar. By the [[Pythagorean theorem]], <math>AY</math> is <math>\sqrt3</math>. [[Category:Introductory Geometry Problems]]
  5 KB (882 words) - 22:12, 30 April 2024
• 2005 IMO Shortlist Problems/G3
  ...ath>. It follows that triangles <math> \displaystyle ABK, LDA </math> are similar. Then since <math> \displaystyle ABCD </math> is a parallelogram, ...th>, this implies that triangles <math> \displaystyle KBC, CDL </math> are similar. This means that
  3 KB (453 words) - 10:53, 24 June 2007
• 1959 IMO Problems/Problem 5
  ...{MB} = \frac{AN}{NB} </math> since the triangles <math>ABN, BCM</math> are similar. Then <math>NM </math> bisects <math>ANB </math>. [[Category:Olympiad Geometry Problems]]
  2 KB (408 words) - 01:40, 2 January 2023
• Geometry/Introduction
  == Introductory Topics in Geometry == The following topics make a good introduction to [[geometry]].
  1 KB (122 words) - 16:25, 18 May 2021
• 1995 AHSME Problems/Problem 8
  <math>\triangle BDE</math> is similar to <math>\triangle BAC</math> by angle-angle similarity since <math>E=C = 9 [[Category:Introductory Geometry Problems]]
  1 KB (199 words) - 13:58, 5 July 2013
• Geometry/Olympiad
  An olympiad-level study of [[geometry]] involves familiarity with intermediate topics to a high level, a multitud === Synthetic geometry ===
  2 KB (242 words) - 10:16, 18 June 2023
• Equilateral triangle
  ...me angle measurements, so, by AA similarity, all equilateral triangles are similar). [[Category:Geometry]]
  1 KB (186 words) - 19:57, 15 September 2022
• Midpoint
  In [[Euclidean geometry]], the '''midpoint''' of a [[line segment]] is the [[point]] on the segment ...these segment lengths, <math>\Delta ABC \sim \Delta EFD (SSS)</math> with similar ratio 2:1. The area ratio is then 4:1; this tells us
  4 KB (596 words) - 17:09, 9 May 2024
• Volume
  ...</math>, where <math>(a_z,b_z)</math> are the bounds of <math>z</math> and similar bounds are defined for <math>x</math> and <math>y</math>. *[[Center]]s of adjacent [[face]]s of a unit [[cube (geometry) | cube]] are joined to form a regular [[octahedron]]. What is the [[volume
  3 KB (523 words) - 20:24, 17 August 2023
• Orthocenter
  ...2 GO</math>. Then the triangles <math>AGH</math>, <math>A'GO</math> are [[similar]] by side-angle-side similarity. It follows that <math>AH</math> is parall Euclidean Geometry in Mathematical Olympiads by Evan Chen - Section 1.3
  5 KB (829 words) - 13:11, 20 February 2024
• Similarity (geometry)
  ...tation. For example, a globe and the surface of the earth are, in theory, similar. ...tion, rotation and reflection ([[rigid motion]]s). We say two objects are similar if they are congruent up to a [[dilation]].
  2 KB (261 words) - 20:42, 25 November 2023
• 1970 IMO Problems/Problem 1
  By similar triangles and the fact that both centers lie on the angle bisector of <math [[Category:Olympiad Geometry Problems]]
  2 KB (380 words) - 22:12, 19 May 2015
• Euler line
  ...er [[Leonhard Euler]]. Its existence is a non-trivial fact of Euclidean [[geometry]]. Certain fixed orders and distance [[ratio]]s hold among these points. I .... Consider the [[medial triangle]] <math>\triangle O_AO_BO_C</math>. It is similar to <math>\triangle ABC</math>. Specifically, a rotation of <math>180^\circ<
  59 KB (10,203 words) - 04:47, 30 August 2023
• Nine-point circle
  That such a circle exists is a non-trivial theorem of Euclidean [[geometry]]. ...h>O_bE_b</math> and <math>O_cE_c</math> are diagonals of the circumcircle. Similar logic to the above gives us that <math>O_aO_cE_aE_c</math> is a rectangle w
  6 KB (994 words) - 16:02, 12 March 2024
• Pick's Theorem
  It is similar to the [[Shoelace Theorem]], and although it is less powerful, it is a good [[Category:Geometry]]
  2 KB (301 words) - 13:08, 20 February 2024
• 2003 AMC 10A Problems/Problem 22
  Therefore <math>\triangle GFA</math> and <math>\triangle ABH</math> are similar. <math>\triangle GCH</math> and <math>\triangle GEA</math> are also similar.
  9 KB (1,446 words) - 22:48, 8 May 2024
• 1987 IMO Problems/Problem 2
  ...ular. Furthermore, we have triangles <math>ABN</math> and <math>ALC</math> similar because two corresponding angles are equal. [[Category:Olympiad Geometry Problems]]
  4 KB (736 words) - 15:39, 21 September 2014
• 2006 Canadian MO Problems/Problem 2
  Now, we use [[vector]] geometry: intersection <math>I</math> of the diagonals of <math>DEFG</math> is also ..., and <math>MY // AH</math>, so <math>MYX</math> and <math>A'AX</math> are similar, and so <math>X</math> lies on <math>A'M</math>, as desired. Reversing the
  2 KB (416 words) - 20:00, 21 September 2014
• 2002 USAMO Problems/Problem 2
  ...and [[inradius]], respectively. Prove that triangle <math>ABC </math> is similar to a triangle <math>T </math> whose side lengths are all positive integers [[Category:Olympiad Geometry Problems]]
  2 KB (298 words) - 22:32, 6 April 2016
• 2007 AIME I Problems/Problem 9
  Use a similar solution to the aforementioned solution. Instead, call <math>\angle CAB = 2 ...s that triangles <math>I_{C}AB</math> and <math>I_{C}O_{1}O_{2}</math> are similar.
  11 KB (1,851 words) - 12:31, 21 December 2021
• 2002 IMO Shortlist Problems/G4
  ...hen we must also have all triangles <math> \displaystyle QA_1X </math> are similar. Since <math> \displaystyle Q </math> is fixed, this means that there exis [[Category:Olympiad Geometry Problems]]
  3 KB (470 words) - 07:32, 28 March 2007
• 2007 AIME II Problems/Problem 15
  Now, triangles <math>ABC</math> and <math>A'B'C'</math> are similar by parallel sides, so we can find ratios of two quantities in each triangle [[Category:Intermediate Geometry Problems]]
  11 KB (2,099 words) - 17:51, 4 January 2024
• 2005 IMO Shortlist Problems/G1
  ...les <math> \displaystyle MCP </math>, <math> \displaystyle DBI </math> are similar. ...th> are similar. Thus triangles <math> \displaystyle MCP, NPI </math> are similar. But we note that by measures of intercepted arcs, <math> \angle ICP = \fr
  7 KB (1,088 words) - 16:57, 30 May 2007
• 2004 IMO Shortlist Problems/G2
  ...suffices to show that triangles <math> \displaystyle DFC, DAF </math> are similar. Since these triangles share a common angle, it then suffices to show <mat ...AOE </math> to <math> \displaystyle FDE </math>, i.e., these triangles are similar. Since <math> \displaystyle OA = OE </math>, it follows that <math> \displ
  10 KB (1,539 words) - 23:37, 6 June 2007
• Circumradius
  ...gle and <math>s = (a+b+c)/2</math> is the semiperimeter. Note that this is similar to the previously mentioned formula; the reason being that <math>A = rs</ma [[Category:Geometry]]
  4 KB (729 words) - 16:52, 19 February 2024
• 2007 AMC 12A Problems/Problem 10
  ...circumcircle, so it has length <math>2 \cdot 3 = 6</math>. The triangle is similar to a 3-4-5 triangle with the ratio of their side lengths equal to <math>\fr [[Category:Introductory Geometry Problems]]
  2 KB (231 words) - 14:02, 3 June 2021
• 2006 AIME II Problems/Problem 6
  ...d AA~ we know that triangles <math>AA'D'</math> and <math>D'C'E</math> are similar. Thus, the sides are proportional: <math>\frac{AA'}{A'D'} = \frac{D'C'}{C'E ...olve in terms of the side <math>x</math> only (single-variable beauty)? By similar triangles we obtain that <math>BE=\frac{x}{1-x}</math>, therefore <math>CE=
  7 KB (1,067 words) - 12:23, 8 April 2024

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