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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
2006 AIME II Problems/Problem 12
Because the ratio of the area of two similar figures is the square of the ratio of the corresponding sides, <math>[GBC] ==Solution 2: Analytic Geometry/Coord Bash==

6 KB (1,033 words) - 02:36, 19 March 2022
Primary Mathematics World Contest
...similar to [[AMC 10]]/[[AMC 12]] level, with an emphasis on number theory, geometry and logic.

1 KB (183 words) - 13:57, 15 October 2018
2004 AMC 12A Problems/Problem 18
...>\overline{AB}</math> is constructed inside the square, and the [[tangent (geometry)|tangent]] to the semicircle from <math>C</math> intersects side <math>\ove ...h>CDE</math> are in arithmetic progression. Thus it is [[similar triangles|similar]] to the triangle <math>3 - 4 - 5</math> and since <math>DC = 2</math>, <ma

5 KB (738 words) - 13:11, 27 March 2023
Lexington Math Tournament
...nd is replaced with an Accuracy Round, but the rest of the event follows a similar format. ...ach competition, but some examples of past mini-events are the estimathon, geometry bee, science bowl (or ball), and Tetris.

3 KB (484 words) - 20:04, 12 March 2024
Concurrence/Problems
Hallie is teaching geometry to Warren. She tells him that the three medians, the three angle bisectors, To find the similar formula for <math>CE</math>, we just switch the signs of <math>BC^2</math>

3 KB (458 words) - 15:44, 1 December 2015
2000 AMC 12 Problems/Problem 21
...e angle chasing shows that the two right triangles are [[similar triangles|similar]]. Thus the ratio of the sides of the triangles are the same. Since <math>A Since triangle <math>A</math> is similar to the large triangle, it has <math>h_A = a(\frac{c}{b}) = \frac{ac}{b}</ma

4 KB (666 words) - 09:22, 28 October 2022
2007 AMC 10A Problems/Problem 21
...th> be the surface area of the inner square. The ratio of the areas of two similar figures is equal to the square of the ratio of their sides. As the diagonal [[Category:Introductory Geometry Problems]]

4 KB (609 words) - 14:41, 3 December 2023
1951 AHSME Problems
In connection with proof in geometry, indicate which one of the following statements is ''incorrect'': <math> \textbf{(A)}\ \text{the triangles are similar in opposite pairs}\qquad\textbf{(B)}\ \text{the triangles are congruent in

23 KB (3,641 words) - 22:23, 3 November 2023
2002 AMC 12B Problems/Problem 20
...ht triangles <math>\triangle MON</math> and <math>\triangle XOY</math> are similar by Leg-Leg with a ratio of <math>\frac{1}{2}</math>, so <math>XY=2(MN)=\box [[Category:Introductory Geometry Problems]]

3 KB (447 words) - 15:02, 17 August 2023
2002 AMC 12B Problems/Problem 25
Similar to Solution 1, we proceed to get the area of the circle satisfying <math>f( [[Category:Introductory Geometry Problems]]

2 KB (365 words) - 14:48, 7 March 2022
2007 AMC 10A Problems/Problem 18
...at triangles <math>\triangle AMH</math> and <math>\triangle GMC</math> are similar because <math>\overline{AH} \parallel \overline{CG}</math>. Also, since <ma The triangles <math>CBN</math> and <math>CGH</math> are clearly similar with ratio <math>1:2</math>, hence <math>BN=2</math> and thus <math>AN=6</m

6 KB (867 words) - 00:17, 20 May 2023
2004 AMC 12B Problems/Problem 19
Above, <math>E,F,</math> and <math>G</math> are points of [[tangent (geometry)|tangency]]. By the Two Tangent Theorem, <math>BF = BE = 18</math> and <mat ...erline{FH} = 3x</math>.Triangles <math>OFH</math> and <math>BEH</math> are similar so <math>\frac{\overline{OF}}{\overline{BE}} = \frac{\overline{FH}}{\overl

3 KB (520 words) - 19:12, 20 November 2023
2007 AMC 10B Problems/Problem 11
<math>\triangle ADC</math> is similar to <math>\triangle ACR</math>, so <cmath>\frac {2\sqrt{2}}{3} = \frac {3}{2 [[Category:Introductory Geometry Problems]]

5 KB (851 words) - 22:02, 26 July 2021
2002 AMC 10A Problems/Problem 25
...F</math> is similar to triangle <math>BEC</math>, we can use properties of similar triangles and find that <math>DE = 12 \cdot \frac{5}{13} = \frac{60}{13}</m [[Category:Introductory Geometry Problems]]

8 KB (1,308 words) - 07:05, 19 December 2022
Isogonal conjugate
One can prove a similar theorem in the case <math>P</math> outside <math>\triangle ABC.</math> ...cated on the ray <math>OA_1</math> so that <math>\triangle OAA_1</math> is similar to <math>\triangle OA_2A</math>. Points <math>B_2</math> and <math>C_2</mat

54 KB (9,416 words) - 08:40, 18 April 2024
2008 AMC 12B Problems/Problem 11
...c 1 8</math> that of the entire cone (mountain). These cones are obviously similar so the radius and height of the small cone must be <math>\sqrt[3]{\frac1 8} [[Category:Introductory Geometry Problems]]

2 KB (251 words) - 22:12, 8 May 2021
2008 AIME I Problems/Problem 14
We let <math>AC=x</math>. From similar triangles, we have that <math>PC=\frac{x\sqrt{x^2+18x}}{x+9}</math> (Use Py ...nt on <math>x</math> is that it must be greater than <math>0</math>. Using similar triangles, we can deduce that <math>PA=\frac{9x}{x+9}</math>. Now, apply la

8 KB (1,333 words) - 00:18, 1 February 2024
2008 AIME II Problems/Problem 5
Similar to solution 1; Notice that it forms a right triangle. Remembering that the [[Category:Intermediate Geometry Problems]]

8 KB (1,338 words) - 23:15, 28 November 2023
2008 AIME II Problems/Problem 11
...eta) = \frac {3}{4}</math>. Therefore, <math>XC = \frac {64}{3}</math>. By similar reasoning in triangle <math>QBY</math>, we see that <math>BY = \frac {4r}{3 ...- (\overline{PZ})^2} = \sqrt{(16+r)^2-(16-r)^2} = 8\sqrt{r}</cmath> Using similar similarity as was done to find <math>\overline{BY}</math> we have <math>\fr

6 KB (1,065 words) - 20:12, 9 August 2022
1998 APMO Problems/Problem 4
...ath> are similar triangles, so <math>AFEN</math> and <math>ACBM</math> are similar figures. It follows that [[Category:Olympiad Geometry Problems]]

2 KB (345 words) - 18:42, 13 April 2008
1995 AHSME Problems/Problem 19
...e smaller triangle is <math>\sqrt{3}</math>. The ratio of the areas of two similar triangles is the square of the ratio of two corresponding side lengths, so [[Category:Introductory Geometry Problems]]

1 KB (200 words) - 20:58, 10 February 2019
Shoelace Theorem
...th>(x_1, y_1, z_1), (x_2, y_2, z_2),(x_3, y_3, z_3),(0,0,0)</math>. Does a similar formula hold for <math>n</math>Dimensional triangles for any <math>n</math> [[Category:Geometry]]

8 KB (1,358 words) - 15:32, 22 February 2024
2008 AMC 10A Problems/Problem 18
This solution is very similar to Solution 1, except instead of subtracting <math>a+b</math> from both sid [[Category:Introductory Geometry Problems]]

8 KB (1,339 words) - 14:15, 1 August 2022
2008 USAMO Problems/Problem 2
'''Lemma.''' <math>\triangle FEO</math> is directly similar to <math>\triangle NEM</math> ...rity. It is easy to see that they are oriented such that they are directly similar.

20 KB (3,565 words) - 11:54, 1 May 2024
2003 AMC 12A Problems/Problem 17
...find <math>XJ</math> and double it to get our answer. With some analytical geometry, we find that <math>XJ=\frac{8}{5}</math>, implying that <math>PK=\frac{16} Manipulating similar triangles gives us <math>PZ=\frac{16}{5}</math>

6 KB (1,026 words) - 22:35, 29 March 2023
1995 AHSME Problems/Problem 26
...triangle CFD</math> and <math>\triangle EOD</math> are [[Similar(geometry)|similar]]. [[Category:Introductory Geometry Problems]]

2 KB (359 words) - 20:01, 23 January 2017
How should I prepare?
...re for math contests. They cover a broad range of topics, from algebra to geometry to number theory to combinatorics and much much more. * Introduction to Geometry - [[Mathcounts]], [[AMC 8]], [[AMC 10]], [[AMC 12]], [[AIME]], [[HMMT]]

13 KB (1,926 words) - 11:22, 30 November 2023
2003 USAMO Problems/Problem 4
...</math> if and only if triangles <math>BCM</math> and <math>CDM</math> are similar, that is ...D = \angle BMC</math>, triangles <math>CMD</math> and <math>BMC</math> are similar. Consequently, <math>\angle MCD = \angle MBC</math>. This exact condition c

5 KB (820 words) - 02:39, 10 January 2023
2000 USAMO Problems/Problem 2
...espectively. Prove that all triangles in <math>S</math> are isosceles and similar to one another. ...e proved that all possible ABC are isosceles (as b = c), and that they are similar to a 5-5-8 triangle.

4 KB (703 words) - 18:40, 3 January 2019
2001 IMO Shortlist Problems/G1
...ther is on <math>AC</math>. Points <math>B_1,\ C_1</math> are defined in a similar way for inscribed squares with two vertices on sides <math>AC</math> and <m ...mplies that quadrilaterals <math>AEA_1F</math> and <math>ABA_2C</math> are similar. Therefore <math>\angle BAA_2=\angle EAA_1</math> and <math>\angle CAA_2=EA

3 KB (584 words) - 15:07, 12 December 2011
2001 IMO Shortlist Problems/G3
By similar reasoning, we can calculate <math>BG^2=(2c^2+2a^2-b^2)/9</math> and <math>C [[Category:Olympiad Geometry Problems]]

2 KB (430 words) - 18:13, 29 January 2019
Millennium Problems
...ort of mathematical research. However, specifically, the inspiration was a similar prize exactly a hundred years earlier. Paris had seen a similar event then, at the second International Congress of Mathematicians. The fam

13 KB (1,969 words) - 17:57, 22 February 2024
1984 IMO Problems/Problem 4
...D</math>. Then, since triangles <math>MNE</math> and <math>BCE</math> are similar, we have <math>\frac{CH}{r}=\frac{BG}{\frac{1}{2}AB}</math>. This gives an ...{\sin ECB}=\frac{EC}{EB}</math>. Since triangles <math>EDA,ECB</math> are similar, we have <math>\frac{EC}{EB}=\frac{CD}{AB}</math> and thus we have

4 KB (750 words) - 23:49, 29 January 2021
Carnot's Theorem
...ication of pythogorean theorem to triangles <math>PXB,PXC</math>. Now with similar relations for <math>Y</math> and <math>Z</math>, Carnot's theorem finishes [[Category:Geometry]]

4 KB (723 words) - 01:45, 18 February 2021
2000 USAMO Problems
...espectively. Prove that all triangles in <math>S</math> are isosceles and similar to one another. ...>k = 2, 3, \dots, 7,</math> <math>\omega_k</math> is externally [[tangent (geometry)|tangent]] to <math>\omega_{k - 1}</math> and passes through <math>A_k</mat

3 KB (495 words) - 19:02, 18 April 2014
2008 iTest Problems
...ua asks, "The battery shapes never change? Each year the new batteries are similar in shape - in all dimensions - to the batteries from previous years?" ...r she chooses <math>4</math> or <math>5</math> students at a time to tutor geometry each week:

71 KB (11,749 words) - 01:31, 2 November 2023
2008 iTest Problems/Problem 99
...ngle, we see that <math>[A_1A_{11}A_{12}] = \frac{1}{9}[AA_1A_2]</math>. A similar clipping about <math>A_2</math> gives <math>[A_2A_{21}A_{22}] = \frac{1}{9} [[Category:Olympiad Geometry Problems]]

4 KB (685 words) - 14:39, 7 October 2017
2007 AMC 12B Problems/Problem 25
Similar to solution 1, we allow [[Category:Introductory Geometry Problems]]

3 KB (470 words) - 19:46, 17 July 2023
1999 AHSME Problems/Problem 2
<math>\mathrm{(E) \ All\ equilateral\ triangles\ are\ similar\ to\ each\ other.} </math> ...es in an equilateral triangle are congruent, all equilateral triangles are similar by AAA similarity. Thus, <math>E</math> is true.

2 KB (259 words) - 14:28, 13 February 2019
2002 AMC 12A Problems/Problem 7
...mplify to get <math>\frac{r_1}{r_2}=\frac{2}{3}</math>. As all circles are similar to one another, the ratio of the areas is just the square of the ratios of [[Category:Introductory Geometry Problems]]

3 KB (492 words) - 14:46, 31 January 2024
2002 AMC 10A Problems/Problem 13
...is dropped (the big triangle and two smaller ones it gets split into) are similar! We take the triangle with hypotenuse <math>20</math>. The ratio of sides b [[Category:Introductory Geometry Problems]]

3 KB (395 words) - 13:22, 8 November 2021
2002 AMC 10A Problems/Problem 20
...l, triangles <math>\triangle GAD</math> and <math>\triangle HCD</math> are similar. Hence, <math>\frac{CH}{AG} = \frac{CD}{AD} = \frac{1}{3}</math>. ...l, triangles <math>\triangle GAF</math> and <math>\triangle JEF</math> are similar. Hence, <math>\frac{EJ}{AG} = \frac{EF}{AF} = \frac{1}{5}</math>. Therefore

3 KB (439 words) - 22:15, 9 June 2023
1972 USAMO Problems/Problem 5
...orresponding vertices. From the lemma, pentagon <math>A'B'C'D'E'</math> is similar to pentagon <math>ABCDE</math> with <math>AB = m(A'B')</math> and parallelo ...lograms, <math>BD' = EC' = (m - 1)a</math>. Triangle <math>EB'C'</math> is similar to triangle <math>ECB</math>, so <math>me = BC = \left(\frac{2m - 1}{m - 1}

4 KB (684 words) - 09:29, 26 March 2023
1960 IMO Problems/Problem 7
...h>h - x</math> must be the distance from <math>P</math> to the other base. Similar triangles give <math>\frac{x}{\frac{a}{2}} = \frac{\frac{c}{2}}{h - x}</mat [[Category:Olympiad Geometry Problems]]

2 KB (410 words) - 15:25, 23 March 2020
2000 AMC 10 Problems/Problem 5
(a) Triangles <math>ABP</math> and <math>MNP</math> are similar, and since <math>PM=\frac{1}{2}AP</math>, <math>MN=\frac{1}{2}AB</math>. [[Category:Introductory Geometry Problems]]

1 KB (242 words) - 01:24, 27 July 2023
2000 AMC 10 Problems/Problem 16
Extend line <math>\overline{DC}</math> as above. This creates two similar triangles whose side lengths have the ratio <math>5:4</math>. Therefore <m [[Category:Introductory Geometry Problems]]

7 KB (966 words) - 23:22, 31 July 2023
1995 USAMO Problems/Problem 3
...cated on the ray <math>OA_1</math> so that <math>\triangle OAA_1</math> is similar to <math>\triangle OA_2A</math>. Points <math>B_2</math> and <math>C_2</ma [[Category:Olympiad Geometry Problems]]

2 KB (364 words) - 01:42, 19 April 2024
2004 AMC 10B Problems/Problem 18
...he triangles <math>\triangle AFG</math> and <math>\triangle AEC</math> are similar, as they have all angles equal. Their ratio is <math>\frac {AF}{AE} = \frac ==Solution 3 (Coordinate Geometry)==

6 KB (904 words) - 12:54, 22 October 2023
2009 AMC 12A Problems/Problem 20
[[Category: Introductory Geometry Problems]] ...B = \angle DEC</math>, triangles <math>AEB</math> and <math>DEC</math> are similar. Their ratio is <math>\frac {AB}{CD} = \frac {3}{4}</math>. Since <math>AE

3 KB (543 words) - 21:09, 23 October 2023
2009 AMC 10A Problems/Problem 17
[[Category: Introductory Geometry Problems]] Triangle <math>EAB</math> is similar to <math>BAD</math>, as they have the same angles. Segment <math>BA</math>

4 KB (684 words) - 21:14, 23 October 2023
2009 AMC 10B Problems/Problem 18
== Solution 1 (Coordinate Geometry)== ...le, thus all their angles are equal, and therefore these two triangles are similar.

6 KB (930 words) - 22:14, 18 January 2024
2009 AMC 10B Problems/Problem 22
[[Category: Introductory Geometry Problems]] ...and <math>\triangle NQO</math> have the same angles and therefore they are similar. The ratio of their sides is <math>\frac{QM}{OQ} = \frac{\sqrt 5}1 = \sqrt

12 KB (1,868 words) - 03:36, 30 September 2023
2009 AIME I Problems/Problem 4
...2E_3}</math> where the points are in order from top to bottom. Clearly, by similar triangles, <math>BB_2 = \frac {1000}{17}MN</math> and <math>DF_1 = \frac {2 ==Solution 6(Coordinate Geometry)==

7 KB (1,117 words) - 00:23, 9 January 2023
2009 AIME I Problems/Problem 5
...which implies <math>AM||LP</math> and <math>\bigtriangleup{AMB}</math> is similar to <math>\bigtriangleup{LPB}</math> [[Category:Intermediate Geometry Problems]]

8 KB (1,224 words) - 19:52, 7 March 2024
2009 AIME I Problems/Problem 12
...ath>AD=\dfrac{12^{2}}{37}</math> and <math>DB=\dfrac{35^{2}}{37}</math> by similar triangles. Let <math>O</math> be the center of <math>\omega</math>; notice Since the ratios between corresponding lengths of two similar diagrams are equal, we can let <math>AD = 144, CD = 420</math> and <math>BD

12 KB (1,970 words) - 22:53, 22 January 2024
Radical axis
...ga_1</math>, for example, we adopt the convention that <math>P = M</math>; similar conventions hold for <math>\omega_2</math>. Power of a Point still holds in [[Category : Geometry]]

10 KB (1,797 words) - 02:05, 24 October 2023
2009 AIME II Problems/Problem 15
...>AO = \frac{1}{2}</math>. Since <math>CC'Y</math> and <math>AOY</math> are similar right triangles, we have <math>CY/AY = CC'/AO = \sin(t)</math>, and hence, [[Category: Intermediate Geometry Problems]]

11 KB (1,849 words) - 19:43, 2 January 2023
1965 IMO Problems/Problem 3
...vol APWX = k3(k+1)3 vol ABCD. The base of the prism WXPBYZ is BYZ which is similar to BCD with sides k/(k+1) times smaller and hence area k2(k+1)2 times small [[Category:Olympiad Geometry Problems]]

2 KB (309 words) - 10:52, 30 September 2022
2002 AMC 12A Problems/Problem 23
...ngle ABD</math> and <math>\triangle ACB</math> are [[Similarity (geometry)|similar]], so <math>\frac {16}{AB}=\frac {AB}{9} \Longrightarrow AB=12</math>. [[Category:Introductory Geometry Problems]]

6 KB (899 words) - 01:41, 5 July 2023
2010 AMC 12A Problems/Problem 8
== Solution 3 (Similar Triangles) == ...irc}</math> and <math>\angle ACF=\angle BAE</math>. Hence, triangle AEB is similar to triangle CFA. Since <math>AB=2AC</math>, <math>AE=2CF=2FE</math>, as tri

4 KB (621 words) - 15:12, 21 June 2023
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 (2,178 words) - 14:14, 11 September 2021
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
Alcumus
AoPS generates high-quality solutions for the students to learn and to do similar problems easily. *[[Geometry]]

3 KB (533 words) - 10:55, 7 February 2023
1950 AHSME Problems/Problem 47
...gle formed by taking away the rectangle and the two small portions left is similar to the big triangle, so the proportions of the heights is equal to the prop ...tions left is similar to the big triangle. Therefore the smaller triangle (similar to the big one) has an area of <math>\dfrac{2x(h-x)}{2}=x(h-x)</math>. Now

2 KB (353 words) - 16:50, 15 July 2023
2012 AMC 10A Problems/Problem 15
[[Category: Introductory Geometry Problems]] Triangle <math>EAB</math> is similar to triangle <math>EHI</math>; line <math>HI = 1/2</math>

6 KB (935 words) - 23:41, 13 September 2023
2012 AMC 12A Problems/Problem 16
==Solution 9 (Similar Triangles)== [[Category:Introductory Geometry Problems]]

9 KB (1,496 words) - 02:40, 2 October 2022
2012 AMC 12B Problems/Problem 17
...In other words, the longer leg is 3 times the shorter leg in any triangle similar to <math>\triangle AGB</math>. ...Triangles <math>\triangle EKM</math> and <math>\triangle MKF</math>, being similar to <math>\triangle AGB</math>, also have legs in a 1:3 ratio, therefore, <m

12 KB (2,183 words) - 21:05, 23 December 2023
University of Northern Colorado Math Contest
...ey for the Final Round of the contest in late January. The Final Round is similar in spirit to the First Round, but the problems are more challenging. Studen ...e students to exercise their creativity and ingenuity to solve problems in geometry, algebra, combinatorics, probability, and number theory. All the old contes

2 KB (336 words) - 15:09, 2 January 2020
2012 AMC 10B Problems/Problem 23
We can approach this problem similar to the solution above, by attacking it on the Cartesian plane. As mentioned [[Category: Introductory Geometry Problems]]

5 KB (851 words) - 11:25, 21 April 2024
2012 AIME I Problems/Problem 8
...is a right triangular pyramid. Note also that pyramid <math>MPBQ</math> is similar to <math>DPCN</math> with scale factor <math>.5</math> and thus the volume ...nd the volume <math>[DNCMQB]</math> is a sum of areas of cross sections of similar triangles running parallel to face <math>ABFE.</math> Let <math>x</math> be

10 KB (1,592 words) - 13:35, 4 April 2024
1968 IMO Problems/Problem 1
...ACB}{2} = \angle ABC</math>, so <math>ABC</math> and <math>ADB</math> are similar by AA Similarity. Hence, <math>c^2 = b(a+b)</math>. Then proceed as in Solu [[Category:Olympiad Geometry Problems]]

4 KB (729 words) - 08:47, 9 March 2019
2012 AIME II Problems/Problem 13
...retty much destroyed by complex plane geometry, which is similar to vector geometry only with the power of easy rotation. Place the triangle in the complex pl ...- 2 \alpha</math>. We can use all the congruent equilateral triangles in a similar manner obtaining:

13 KB (2,052 words) - 18:02, 5 February 2024
2012 AIME II Problems/Problem 15
...ive up after realizing that angle chasing won't work, you'd likely go in a similar approach to Solution 1 (below) or maybe be a bit more insightful and go wit [[Category:Intermediate Geometry Problems]]

13 KB (2,298 words) - 12:56, 10 September 2023
1988 USAMO Problems/Problem 4
...}</math>), so that <math>M</math> is the circumcenter of <math>AIC</math>. Similar results hold for <math>BIC</math> and <math>CIA</math>, and hence <math>O_c [[Category:Olympiad Geometry Problems]]

3 KB (504 words) - 19:25, 14 October 2021
1989 AHSME Problems/Problem 4
...m <math>A</math> and <math>B</math>; then the triangle <math>DBY</math> is similar to <math>DEF</math> but with corresponding sides of half the length. [[Category: Introductory Geometry Problems]]

2 KB (271 words) - 07:42, 22 October 2014
1978 USAMO Problems/Problem 2
.... In this case let AA' and BB' meet at O. Then triangles OAB and OA'B' are similar, so O must represent the same point. So assume A'B' is not parallel to AB. ...oordinate of O equals the y' coordinate of O. Similarly, XOX' and ZOZ' are similar, so OX/OZ = OX'/OZ', so the x-coordinate of O equals its x'-coordinate. In

4 KB (712 words) - 21:57, 12 November 2023
1978 USAMO Problems/Problem 4
...gles. Now, since <cmath>WX=2\cdot OW\sin\frac{\angle{WOX}}{2},</cmath> and similar for the other sides, we have that <math>WXYZ</math> is a regular tetrahedro [[Category:Olympiad Geometry Problems]]

2 KB (322 words) - 23:16, 18 July 2016
2013 AMC 12A Problems/Problem 9
...al triangles <math>\triangle BDE</math> and <math>\triangle EFC</math> are similar to <math>\triangle ABC</math>, and are therefore also isosceles triangles. [[Category:Introductory Geometry Problems]]

2 KB (352 words) - 16:25, 12 August 2023
2013 AMC 10A Problems/Problem 7
...ram of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least on Case <math>1</math>: The student chooses both algebra and geometry.

2 KB (379 words) - 13:06, 1 July 2023
2013 AMC 10A Problems/Problem 12
[[Category: Introductory Geometry Problems]] ...al triangles <math>\triangle BDE</math> and <math>\triangle EFC</math> are similar to <math>\triangle ABC</math>, and are therefore also isosceles triangles.

2 KB (376 words) - 11:09, 30 July 2022
2013 AMC 10B Problems/Problem 15
...hexagon can be broken into 6 small equilateral triangles, each of which is similar to the big equilateral triangle. The big triangle's area is 6 times the are [[Category:Introductory Geometry Problems]]

2 KB (240 words) - 12:18, 18 October 2022
2013 AMC 10B Problems/Problem 16
.../math>. Also, because <math>DE || AC</math>, <math>\triangle ABC</math> is similar to <math>\triangle{DBE}</math> with side length ratio <math>2:1</math>, so [[Category:Introductory Geometry Problems]]

5 KB (761 words) - 19:33, 11 January 2024
2013 AMC 12B Problems/Problem 13
...rithmetic progression. Triangles <math>ABD</math> and <math>DCB</math> are similar with <math>\angle DBA = \angle DCB</math> and <math>\angle ADB = \angle CBD For the sake of simplicity, lets rename the angles of each similar triangle. Let <math>\angle ADB = \angle CBD = \alpha</math>, <math>\angle D

3 KB (443 words) - 12:32, 8 January 2021
2013 AMC 12B Problems/Problem 19
[[Category: Introductory Geometry Problems]] ...}=\angle{ABF}</math>, so <math>\triangle ABF \sim \triangle ADE</math> are similar. In addition, <math>\triangle ADE \sim \triangle ACD</math>. We can easily

6 KB (1,004 words) - 22:38, 18 June 2023
2013 AMC 12B Problems/Problem 24
Furthermore, triangle <math>BCN</math> is similar to triangle <math>MCX</math>, so <math>BC/CM=CN/CX</math>, therefore <math> ...e have more similar triangles. In fact, going back to our original pair of similar triangles - <math>\triangle ANC</math> and <math>\triangle BXC</math> - giv

11 KB (1,876 words) - 00:08, 12 October 2023
1989 AHSME Problems/Problem 7
...can see that <math>\triangle ACH</math> and <math>\triangle MCN</math> are similar, implying <math>\overline{HN}=\overline{NC}</math>, implying that <math>\tr [[Category: Introductory Geometry Problems]]

2 KB (221 words) - 18:04, 21 October 2018
2014 AIME II Problems/Problem 11
Let <math>MN = x.</math> Meanwhile, since <math>\triangle R PM</math> is similar to <math>\triangle RCD</math> (angle, side, and side- <math>RP</math> and < [[Category:Intermediate Geometry Problems]]

6 KB (1,059 words) - 18:24, 20 January 2024
1967 AHSME Problems/Problem 20
Now we note that the picture is self-similar; if we erase the outer square, erase the outer circle, rotate the picture, [[Category:Introductory Geometry Problems]]

2 KB (390 words) - 01:40, 16 August 2023
Geometry Solutions
==Geometry Explained== This page is meant for explaining the most difficult problems in the geometry packet.

23 KB (3,182 words) - 12:30, 5 April 2014
1967 AHSME Problems/Problem 15
The difference in the areas of two similar triangles is <math>18</math> square feet, and the ratio of the larger area [[Category:Introductory Geometry Problems]]

1 KB (231 words) - 01:39, 16 August 2023
1967 IMO Problems/Problem 4
...o acute-angled triangles. Consider all triangles <math>ABC</math> that are similar to <math>\triangle A_1B_1C_1</math> (so that vertices <math>A_1</math>, <ma ...the triangles <math>ABC</math> circumscribed to <math>A_0B_0C_0</math> and similar to <math>A_1B_1C_1</math> by selecting <math>A</math> on <math>\mathcal C_A

2 KB (368 words) - 13:15, 29 January 2021
1962 AHSME Problems/Problem 16
...inches. Rectangle <math>R_2</math> with diagonal <math>15</math> inches is similar to <math>R_1</math>. Expressed in square inches the area of <math>R_2</math ...know that <math>\frac{b}a=\frac31=3</math>, because the two rectangles are similar. We also know that <math>a^2+b^2=15^2=225</math>. But <math>b=3a</math>, so

1 KB (170 words) - 22:16, 3 October 2014
2014 AMC 10A Problems/Problem 12
[[Category: Introductory Geometry Problems]] This solution is very similar to the one above, but instead we use the solutions to guide us. The interio

3 KB (482 words) - 11:50, 7 September 2021
2014 AMC 10A Problems/Problem 16
[[Category: Introductory Geometry Problems]] ...eight of <math>AFB</math>. Since <math>XHY</math> and <math>DHC</math> are similar for the same reasons as <math>XFY</math> and <math>AFB</math>, the height o

9 KB (1,411 words) - 19:51, 25 July 2023
2014 AMC 10A Problems/Problem 19
[[Category: Introductory Geometry Problems]] ...t is inside the cube with side length <math>3</math> be <math>x</math>. By similar triangles, <math>\dfrac{x}{3}=\dfrac{2\sqrt{33}}{10}</math>, giving <math>x

3 KB (377 words) - 21:23, 28 October 2023
2014 AMC 10A Problems/Problem 22
[[Category: Introductory Geometry Problems]] ==Solution 6 (Pure Euclidian Geometry)==

12 KB (1,821 words) - 18:16, 29 October 2023
1963 AHSME Problems/Problem 18
...y selection of <math>U</math>, as described, <math>\triangle EUM</math> is similar to: ...A</math>. This leads to prediction that <math>\triangle FEA</math> is the similar triangle as it shares an angle, and to prove this, we need to show that <ma

2 KB (410 words) - 21:18, 31 May 2020
1994 USAMO Problems/Problem 3
...he triangles <math>\triangle AEC</math> and <math>\triangle QED</math> are similar. Therefore, <math>\triangle AEC</math> and <math>\triangle QED</math> are similar, so <math>AC/CE=QD/DE</math>.

4 KB (747 words) - 08:02, 19 July 2016
2014 AMC 12A Problems/Problem 20
[[Category: Introductory Geometry Problems]] This problem is similar to [https://en.wikipedia.org/wiki/Fagnano%27s_problem Fagnano's Problem].

4 KB (665 words) - 04:35, 22 January 2024
2014 AMC 10B Problems/Problem 23
[[Category: Introductory Geometry Problems]] ...<math>r_1</math> and height <math>h</math>. The smaller right triangle is similar to the blue highlighted one in Solution <math>1</math>. Then <math>\frac{r_

6 KB (1,060 words) - 19:15, 11 August 2023
2014 AMC 12B Problems/Problem 21
...nt of <math>AB</math>. So <math>\triangle AJG</math>, along with all other similar triangles in the picture, is a 30-60-90 triangle, and we have <math>AG=\sqr ...>\triangle KEJ, \triangle JAG, \triangle GDH, \triangle HFK</math> are all similar. Using proportions and the pythagorean theorem, we find

8 KB (1,307 words) - 02:07, 1 January 2023
1964 AHSME Problems/Problem 35
A similar computation using <math>\triangle AEB</math> and <math>\triangle CEB</math> [[Category:Introductory Geometry Problems]]

6 KB (892 words) - 00:02, 12 July 2023
1964 AHSME Problems/Problem 29
...rtional, and the included angles are congruent, then the two triangles are similar by SAS congruence. Therefore, the third pair of sides must also be in the [[Category:Introductory Geometry Problems]]

2 KB (331 words) - 22:59, 29 June 2023
1993 AHSME Problems/Problem 27
...th> as the circle moves around the triangle. It turns out this triangle is similar to the <math>6-8-10</math> triangle (Proof: Realize that the slope of the l [[Category: Intermediate Geometry Problems]]

3 KB (420 words) - 11:10, 18 July 2020
1992 AHSME Problems/Problem 11
...ircle intersects <math>BC</math> as <math>E</math>. So <math>ABC</math> is similar to <math>DEC</math>. Since <math>DE</math> is the radius of the smaller cir [[Category:Introductory Geometry Problems]]

2 KB (324 words) - 12:02, 24 November 2016
1991 AHSME Problems/Problem 22
...the radius of the smaller circle be <math>y</math>, we can see that, using similar triangle, <math>x=2y</math>. In addition, the total hypotenuse of the large [[Category: Intermediate Geometry Problems]]

2 KB (295 words) - 19:09, 11 October 2016
1990 AHSME Problems/Problem 4
<math>DFE</math> and <math>AFB</math> are similar triangles, so <math>FD</math> is one quarter the length of the correspondin [[Category: Introductory Geometry Problems]]

1 KB (180 words) - 20:41, 2 July 2021
1990 AHSME Problems/Problem 28
...ic, we see that quadrilaterals <math>AXOW</math> and <math>OYCZ</math> are similar. [[Category: Intermediate Geometry Problems]]

2 KB (335 words) - 11:51, 5 October 2019
1952 AHSME Problems/Problem 49
== Alternative but very similar Solution == ...\frac{1}{3}K+\frac{1}{3}K+\frac{1}{3}K = K.</math> Then we can implement a similar but different area addition postulate to the first solution. It will be <ma

7 KB (1,136 words) - 10:01, 23 December 2023
2015 AMC 10B Problems/Problem 22
...de length. I chose to write it in terms of <math>FG</math> so we can apply similar triangles easily. To simplify the process lets write <math>FG</math> as <ma [[Category: Introductory Geometry Problems]]

7 KB (1,146 words) - 14:09, 13 April 2024
2007 iTest Problems/Problem 42
...line as seen in the diagram is <math>100\sqrt{2} - 100</math> feet. Using similar triangles, the height of the smaller triangle is <math>50\sqrt{2} - 50</mat [[Category:Intermediate Geometry Problems]]

3 KB (524 words) - 23:01, 17 June 2018
2007 IMO Problems/Problem 4
...s presents us <math>\dfrac{PK}{QL}=\dfrac{PK}{QD}=\dfrac{PC}{QC}</math> by similar triangles; now, we have only to prove <math>\dfrac{PC}{QC}=\dfrac{RQ}{RP}</ ...rea of <math>RPK.</math> Finding the area of <math>\triangle RQL</math> is similar.

8 KB (1,480 words) - 14:52, 5 August 2022
2014 IMO Problems/Problem 4
...ath>NED</math>, <math>MDF</math> and <math>FCP</math> must be two pairs of similar triangles. Therefore we must prove angles <math>CBM</math> and <math>ANC, A ...ac {BF}{FM}</math>, so triangles <math>BFM</math> and <math>NEC</math> are similar. So the angles <math>CBM</math> and <math>ANC, BCN</math> and <math>AMB</ma

13 KB (1,801 words) - 07:29, 1 May 2024
2012 UNCO Math Contest II Problems/Problem 7
...he centers of the circle and the point where the tangents meet, we get two similar triangles in ratio <math>1:3</math>. Let the hypotenuse of the smaller tria [[Category:Intermediate Geometry Problems]]

2 KB (336 words) - 12:57, 2 January 2020
2009 UNCO Math Contest II Problems/Problem 8
...he two parallel lines, we know that the two triangles with known areas are similar by angle-angle-angle similarity. Noticing that the areas of the triangles a [[Category:Intermediate Geometry Problems]]

1 KB (174 words) - 02:20, 29 January 2019
2007 UNCO Math Contest II Problems/Problem 9
...ve <math>BL = OM = \frac{BH}{3}</math>. This means the scale factor of the similar triangles <math>ABC</math> and <math>A_1B_1B</math> is <math>\frac {1}{3}</ [[Category:Intermediate Geometry Problems]]

3 KB (412 words) - 18:49, 29 January 2018
2006 UNCO Math Contest II Problems/Problem 5
In general, if <math>G</math> and <math>H</math> are similar triangles with sides <math>s_1</math> and <math>s_2</math> , then <math>\fr [[Category:Introductory Geometry Problems]]

1 KB (184 words) - 03:20, 13 January 2019
1988 AHSME Problems/Problem 2
Triangles <math>ABC</math> and <math>XYZ</math> are similar, with <math>A</math> corresponding to <math>X</math> and <math>B</math> to Since the triangles are similar, we know that the corresponding sides of the triangle are in ratio to each

950 bytes (142 words) - 06:21, 31 August 2015
1987 AHSME Problems/Problem 2
<math>\triangle DBE</math> is similar to <math>\triangle ABC</math> by AA, so <math>\overline{DE}</math> = 1 by s [[Category: Introductory Geometry Problems]]

1 KB (134 words) - 18:37, 23 April 2017
1986 AHSME Problems/Problem 16
is similar to <math>\triangle PCA</math>. The length of <math>PC</math> is [[Category: Intermediate Geometry Problems]]

1 KB (210 words) - 23:51, 10 February 2018
1986 AHSME Problems/Problem 19
...the first part of her route. For her <math>y</math>-coordinate, we can use similar logic to find that the coordinate is <math>\sqrt{3} + 0 - \frac{\sqrt{3}}{2 [[Category: Intermediate Geometry Problems]]

2 KB (240 words) - 20:45, 9 October 2017
1986 AHSME Problems/Problem 27
<math>ABE</math> and <math>DCE</math> are similar isosceles triangles. It remains to find the square of the ratio of their si [[Category: Intermediate Geometry Problems]]

1 KB (219 words) - 15:54, 2 August 2016
2015 AMC 10A Problems/Problem 21
Similar to solution 1, <math>\triangle CTD</math> is an isosceles right triangle. < [[Category: Introductory Geometry Problems]]

4 KB (574 words) - 07:18, 16 August 2023
AA similarity
...if two pairs of corresponding angles are congruent, then the triangles are similar. * [[Similarity (geometry)]]

915 bytes (147 words) - 01:18, 24 December 2022
2015 AMC 10B Problems/Problem 13
==Solution 2 (very similar to Solution 1)== [[Category: Introductory Geometry Problems]]

2 KB (315 words) - 17:14, 2 August 2022
2015 AIME I Problems/Problem 15
...o positions is equal to 10<math>w</math>:6<math>w</math> = 10:6. Through a similar argument, the areas between each set of vertical lines also maintains a rat ..., with <math>T'</math> the center of this circular cross section. Then, by similar triangles, <math>T'M' = \frac{3}{4} y</math> and thus <cmath>A'B' = 2A' M'

9 KB (1,407 words) - 19:37, 17 February 2024
2015 AIME I Problems/Problem 7
...th>, and triangles <math>\Delta BHJ</math> and <math>\Delta JFC</math> are similar to <math>\Delta EDC</math> since they are <math>1-2-\sqrt{5}</math> triangl ...ngth in terms of <math>x,y</math> since <math>AB=AN+NH+HB</math>. By using similar triangles as in the first part, we have

6 KB (1,105 words) - 21:02, 9 November 2023
2015 AIME I Problems/Problem 4
...erages.) Thus, by the Mean Geometry Theorem, <math>\triangle BMN</math> is similar to both <math>\triangle BAD</math> and <math>\triangle BEC</math>, which me ...>. Additionally, <math>MX = \frac{1}{2} (2\sqrt{3}) = \sqrt{3}</math> from similar triangles meaning we can now just do pythagorean theorem on right triangle

5 KB (852 words) - 00:12, 8 May 2024
2015 AIME I Problems/Problem 13
Similar to Solution <math>2</math>, so we use <math>\sin{2\theta}=2\sin\theta\cos\t [[Category:Intermediate Geometry Problems]]

9 KB (1,351 words) - 17:26, 16 January 2024
2015 AIME I Problems/Problem 11
...yields <math>AI=</math><math>\sqrt{x^2-2xy+64}</math>. <math>ADI</math> is similar to <math>AEB</math> by <math>AA</math>, so we [[Category: Intermediate Geometry Problems]]

5 KB (906 words) - 17:43, 27 September 2023
2015 AIME II Problems/Problem 7
Similar triangles can also solve the problem. ...Q</math> is parallel to <math>BC</math> meaning <math>\Delta APQ</math> is similar to <math>\Delta ABC</math>.

7 KB (1,180 words) - 14:08, 14 February 2023
2015 AIME II Problems/Problem 15
...a lot of time left in your mocking of this AIME, go ahead and use analytic geometry. ...s k=\frac{5}{3}</math>. We can do some more length chasing using triangles similar to <math>O_1BN</math> to get that <math>AK = AL = \frac{24}{15}</math>, <ma

31 KB (5,086 words) - 19:15, 20 December 2023

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