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• Geometric inequality
  ...equality states that if <math>P</math> lies in <math>ABC</math> then <math>PA+PB+PC\ge 2(PD+PE+PF)</math> where <math>D, E, F</math> are the foot of the <b>Mordell's Lemma: </b> <math>PA\sin A\ge PE\sin C+PF\sin B</math>
  7 KB (1,300 words) - 00:11, 28 October 2024
• Ptolemy's theorem
  ...cribed in a circle. Point <math>P</math> is on this circle such that <math>AP \cdot CP = 56</math>, and <math>BP \cdot DP = 90</math>. What is the area o ...hat <math>P</math> is between <math>B</math> and <math>C</math>. Let <math>PA = a</math>, <math>PB = b</math>, <math>PC = C</math>, <math>PD = d</math>,
  7 KB (1,198 words) - 23:28, 19 September 2024
• Power of a Point Theorem
  ...ectively (this means that T is on the minor arc <math>AB</math>). If <math>AP = 20</math>, find the perimeter of <math>\triangle PQR</math>.
  5 KB (854 words) - 13:30, 30 June 2024
• 1988 AIME Problems/Problem 12
  <cmath>\frac{AP}{PA'}\frac{BP}{PB'}\frac{CP}{PC'} = 2 + \frac{AP}{PA'} + \frac{BP}{PB'} + \frac{CP}{PC'}</cmath> ...eva's Theorem identity (sometimes attributed to Gergonne): <math>\frac{AP}{PA'}=\frac{AC'}{C'B}+\frac{AB'}{B'C}</math>, and similarly for cevians <math>B
  4 KB (727 words) - 22:37, 7 March 2024
• 1990 AIME Problems/Problem 14
  ...angle <math>ABP^{}_{}</math> is cut out and removed, edges <math>\overline{AP}</math> and <math>\overline{BP}</math> are joined, and the figure is then c triple Pa;
  7 KB (1,086 words) - 07:16, 29 July 2023
• 1995 AIME Problems/Problem 12
  ...le OPA</math> is a <math>45-45-90</math> [[right triangle]], so <math>OP = AP = 1,</math> <math>OB = OA = \sqrt {2},</math> and <math>AB = \sqrt {4 - 2\s From the [[Law of Cosines]], <math>AC^{2} = AP^{2} + PC^{2} - 2(AP)(PC)\cos \theta,</math> so
  8 KB (1,172 words) - 20:57, 22 September 2022
• 1998 AIME Problems/Problem 11
  ...ne{BC},</math> and <math>\overline{CD},</math> respectively, so that <math>AP = 5, PB = 15, BQ = 15,</math> and <math>CR = 10.</math> What is the area o triple P=(5,0,0),Q=(20,0,15),R=(20,10,20),Pa=(15,20,20),Qa=(0,20,5),Ra=(0,10,0);
  7 KB (1,084 words) - 10:48, 13 August 2023
• 2001 AIME I Problems/Problem 7
  ...the intersection of all three [[angle bisector]]s. Draw the bisector <math>AP</math> to where it intersects <math>BC</math>, and name the intersection <m ...<math>P</math> has a weight of <math>63</math>, and the ratio of <math>FP:PA</math> is <math>20:43</math>. Therefore, the smaller similar triangle <math
  11 KB (1,829 words) - 05:57, 30 September 2024
• 2002 AIME II Problems/Problem 14
  ...f [[radius]] <math>19</math> with center <math>O</math> on <math>\overline{AP}</math> is drawn so that it is [[Tangent (geometry)|tangent]] to <math>\ove ...p of the circle to point <math>P</math>, with <math>x + 38</math> as <math>AP</math>. Given the perimeter is 152, subtracting the altitude yields the sem
  4 KB (658 words) - 18:15, 19 December 2021
• Quadratic reciprocity
  ...{}{}{a}{q}^p \zeta^{ap} = \sum_{a=0}^{q-1} \genfrac{(}{)}{}{}{a}{q} \zeta^{ap} . </cmath> ...\genfrac{(}{)}{}{}{p}{q} \sum_{a=0}^{q-1} \genfrac{(}{)}{}{}{pa}{q} \zeta^{ap} = \genfrac{(}{)}{}{}{a}{q} \tau_q . </cmath>
  7 KB (1,182 words) - 15:46, 28 April 2016
• Fermat point
  ...h> which has the minimum total distance to three [[vertices]] (i.e., <math>AP+BP+CP</math>). ...<math>BC</math>. Note <math>PB=P'B</math>, <math>PC=P'C</math>, and <math>PA>P'A</math>, so thus <math>P</math> is not the Fermat Point.
  4 KB (769 words) - 15:07, 29 December 2019
• British Flag Theorem
  ...math>P</math> is chosen inside [[rectangle]] <math>ABCD</math>, then <math>AP^{2}+CP^{2}=BP^{2}+DP^{2}</math>. The theorem is called the British flag the <cmath> \begin{align*}PA^2 &= AX^2+XP^2,\\ PB^2 &= BX^2+XP^2,\\ PC^2 &= CY^2+YP^2,\\ PD^2 &= DY^2+YP
  3 KB (425 words) - 23:14, 17 September 2024
• Isogonal conjugate
  ...t <math>P</math> lies on bisector of <math>\angle BAC</math> and <math>BD||AP.</math> ...</math> respectively. Denote by <math>p_a, p_b, p_c</math> the lines <math>PA</math>, <math>PB</math>, <math>PC</math>, respectively.
  54 KB (9,416 words) - 07:40, 18 April 2024
• 2008 AIME I Problems/Problem 14
  ...angle OTC</math>. Thus <math>\frac{AP}{OT} = \frac{CA}{CO} \Longrightarrow AP = \frac{9(x-9)}{x}</math>. By the [[Law of Cosines]] on <math>\triangle BAP <cmath>\begin{align*}BP^2 = AB^2 + AP^2 - 2 \cdot AB \cdot AP \cdot \cos \angle BAP \end{align*}</cmath>
  8 KB (1,339 words) - 12:29, 3 September 2024
• 2001 USAMO Problems/Problem 4
  ...that in this triangle the obtuse angle opposes the side congruent to <math>PA</math>. Prove that <math>\angle BAC</math> is acute. We know that <math>PB^2+PC^2 < PA^2</math> and we wish to prove that <math>AB^2 + AC^2 > BC^2</math>.
  8 KB (1,470 words) - 21:24, 18 June 2022
• 1992 USAMO Problems/Problem 4
  ...gle APB</math> is isosceles because the base angles are equal. Thus, <math>AP=BP</math>. Similarly, <math>A'P=B'P</math>. Thus, <math>AA'=BB'</math>. By ...tical angles. By power of point, <math>(AP)(A'P)=(BP)(B'P)\rightarrow\frac{AP}{B'P}=\frac{BP}{A'P}</math>
  5 KB (807 words) - 17:37, 25 June 2021
• 2000 AMC 10 Problems/Problem 5
  Points <math>M</math> and <math>N</math> are the midpoints of sides <math>PA</math> and <math>PB</math> of <math>\triangle PAB</math>. As <math>P</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>.
  2 KB (260 words) - 23:38, 14 July 2024
• 2001 AMC 12 Problems/Problem 23
  .../math>. When we divide the expression on the left by -p, we get <math>c-bp+ap^2-p^3</math>, so we can replace it in our original synthetic division equat We then want to synthetically divide <math>x^3+(a-p)x^2+(b-pa+p^2)x+\frac {d}{-k}</math> by the next factor, <math>(x-q)</math>. Using th
  7 KB (1,304 words) - 08:53, 5 October 2024
• 2009 AIME I Problems/Problem 4
  ...\overline{AC}</math> and <math>\overline{MN}</math>. Find <math>\frac {AC}{AP}</math>. <math>AP</math>(<math>AM</math> or <math>AN</math>) is <math>17x.</math>
  7 KB (1,117 words) - 23:23, 8 January 2023
• Radical axis
  ...ircle <math>\omega</math> at points A and B, then <math>|pow(P, \omega)| = PA * PB</math>. ...</math> is tangent to <math>(ABP)</math> and <math>(ACP)</math>, ray <math>AP</math> bisects <math>BC</math>
  12 KB (2,125 words) - 07:38, 23 May 2024

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