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• Pythagorean Theorem
  The '''Pythagorean Theorem''' states that for a [[right triangle]] with [[leg]]s of length <math>a</math> and <math>b</math> and [[hypotenus ...ath>\angle ACB</math>, and we use <math>[ABC]</math> to denote the area of triangle <math>ABC</math>.
  6 KB (978 words) - 12:02, 6 March 2025
• 2005 AMC 12A Problems/Problem 16
  dotfactor=3; pair O0=(9,9), O1=(1,1), O2=(3,1), O3=(1,3);
  2 KB (307 words) - 00:58, 18 November 2024
• 2016 AMC 10A Problems/Problem 11
  label("$4$",(8,3),dir(0)); <math>\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf
  8 KB (1,016 words) - 00:17, 31 December 2023
• 2016 AMC 10A Problems/Problem 16
  ...ing transformations will return <math>\triangle A''B''C''</math> to <math>\triangle ABC</math>?
  1 KB (235 words) - 14:52, 25 June 2023
• Inequality
  ...n particular, notice that although <math>3 > 2</math>, we must have <math>-3 < -2</math>. In particular, when multiplying or dividing by negative quant ..., in the example <math>x \ge \frac{3}{2}</math>, the value <math>x = \frac{3}{2}</math> satisfies the inequality because the inequality is nonstrict.
  12 KB (1,806 words) - 06:07, 19 June 2024
• 1954 AHSME Problems/Problem 46
  dotfactor=3; pair A=(-3*sqrt(3)/32,9/32), B=(3*sqrt(3)/32, 9/32), C=(0,9/16);
  3 KB (415 words) - 18:01, 24 May 2020
• 2015 IMO Problems
  <li> Show that for all integers <math>n\geq 3</math>, there exists a balanced set consisting of <math>n</math> points. </ <li> Determine all integers <math>n\geq 3</math> for which there exists a balanced centre-free set consisting of <mat
  4 KB (709 words) - 15:00, 1 June 2024
• Trivial Inequality
  ...</math> and <math>BC: AC=40: 41</math>. What is the largest area that this triangle can have? ([[1992 AIME Problems/Problem 13|AIME 1992]]) ...Find <math>r+s</math>. (Solution [[Problems Collection|here]] see problem 3 solution 1)
  3 KB (583 words) - 21:20, 2 August 2024
• Pigeonhole Principle
  Let <math>S = {1, 2, 3, \ldots, 2n}</math>. Show that if we choose <math>n+1</math> numbers from < === Example 3 ===
  11 KB (1,980 words) - 22:39, 19 February 2025
• Combination
  Consider the set of letters A, B, and C. There are <math>3!</math> different [[permutations]] of those letters. Since order doesn't ma ...that you choose the rest. This identity is also the reason why [[Pascal's Triangle]] is symmetrical.
  4 KB (638 words) - 21:55, 5 January 2025
• Cauchy-Schwarz Inequality
  === Proof 3 === * [http://www.mathlinks.ro/Forum/viewtopic.php?t=78687 (APMO 1991 #3)] Let <math>a_1</math>, <math>a_2</math>, <math>\cdots</math>, <math>a_n</m
  13 KB (2,048 words) - 15:28, 22 February 2024
• Menelaus' theorem
  ...ty]] of points on each of the three sides (extended when necessary) of a [[triangle]]. If line <math>PQ</math> intersecting <math>AB</math> on <math>\triangle ABC</math>, where <math>P</math> is on <math>BC</math>, <math>Q</math> is o
  5 KB (804 words) - 03:01, 12 June 2023
• Parabola
  draw(graph(f,(7+2*sqrt(13))/3,(7-2*sqrt(13))/3),red+linewidth(1)); ...th equation <math>y=x^2+bx+c</math> passes through the points (2,3) and (4,3). What is <math>c</math>?<br><br><math> \mathrm{(A) \ } 2\qquad \mathrm{(B
  3 KB (551 words) - 16:22, 13 September 2023
• Geometric inequality
  ===Triangle Inequality=== ...] says that the sum of the lengths of any two sides of a non[[degenerate]] triangle is greater than the length of the third side. This inequality is particular
  7 KB (1,300 words) - 01:11, 28 October 2024
• Ptolemy's theorem
  ...h> so <math>\angle ADP=\angle ABC</math>. Hence, <math>\triangle ABC \sim \triangle ADP</math> by AA similarity and <math>\frac{AB}{AD}=\frac{BC}{DP}\implies D ...gle CAD \implies \angle BAD=\angle CAP,</math> so <math>\triangle BAD\sim \triangle CAP.</math> This yields <math>\frac{AB}{AC}=\frac{BD}{CP}\implies CP=\frac{
  6 KB (922 words) - 17:34, 13 January 2025
• 1970 IMO Problems/Problem 6
  ...first <math>3</math> replaced by <math>4</math>. Hence at most <math>\frac{3}{4}</math> of the <math>10</math>, or <math>7.5</math>, can be acute, and h ...ts times <math>\frac{7}{\text{the number of subsets of 5 points containing 3 given points}}</math>. The total number of triangles is the same expression
  6 KB (1,054 words) - 18:09, 11 December 2024
• Induction
  ...rks for <math>n=1+1=2</math>, which in turn means it works for <math>n=2+1=3</math>, and so on. ...e. If a problem asks you to prove something for all integers greater than 3, you can use <math>n=4</math> as your base case instead. You might have to
  5 KB (768 words) - 00:59, 29 September 2024
• Triangle
  A '''triangle''' is a of [[polygon]] with three [[edge|sides]]. {{asy image|<asy>draw((0,1)--(2,0)--(3,2)--cycle);</asy>|right|A triangle}}
  4 KB (535 words) - 18:58, 31 January 2025
• Stewart's Theorem
  Given any [[triangle]] <math>\triangle ABC</math> with sides of length <math>a, b, c</math> and opposite [[vertex| ...gle ABD</math> at [[angle]] <math>\angle ADB</math> and in triangle <math>\triangle ACD</math> at angle <math>\angle CDA</math>, we get the following equations
  4 KB (709 words) - 19:22, 18 February 2025
• Jensen's Inequality
  ...riangle ABC</math>, we have <math>\sin{A}+\sin{B}+\sin{C}\leq \frac{3\sqrt{3}}{2}</math>. * Show that in any triangle <math>\triangle ABC</math> we have <math>\cos {A} \cos{B} \cos {C} \leq \frac{1}{8}</math>
  3 KB (628 words) - 06:29, 28 November 2024

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