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• 2006 Indonesia MO Problems/Problem 5
  ...Also, note that <math>[BGA] = \frac13 [ABC]</math>, so <math>[APG] = \frac{AP}{AB} \cdot \frac13 [ABC]</math>. Similarly, <math>[AQG] = \frac{AQ}{AC} \c \frac{[BGM]}{[PAG]}+\frac{[CMG]}{[QGA]} &= \frac{\frac16 [ABC]}{\frac{AP}{AB} \cdot \frac13 [ABC]} + \frac{\frac16 [ABC]}{\frac{AQ}{AC} \cdot \frac1
  3 KB (489 words) - 18:48, 14 December 2019
• 2020 AMC 12A Problems/Problem 24
  ...there is a unique point <math>P</math> inside the triangle such that <math>AP=1</math>, <math>BP=\sqrt{3}</math>, and <math>CP=2</math>. What is <math>s< s^2 &= (AP)^2 + (CP)^2 - 2\cdot AP\cdot CP\cdot \cos{\angle{APC}} \\
  16 KB (2,509 words) - 17:49, 8 February 2024
• 2022 AMC 10A Problems
  ...C.</math> The line through <math>B</math> perpendicular to <math>\overline{AP}</math> intersects the line through <math>A</math> parallel to <math>\overl ...= CD.</math> There is a point <math>P</math> in the plane such that <math>PA=1, PB=2, PC=3,</math> and <math>PD=4.</math> What is <math>\tfrac{BC}{AD}?<
  18 KB (2,662 words) - 17:45, 5 August 2024
• 2021 AIME I Problems/Problem 6
  ...>, <math>DP=120\sqrt{2}</math>, and <math>GP=36\sqrt{7}</math>. Find <math>AP.</math> ...x^2+y^2+z^2)=512</math>, so <math>x^2+y^2+z^2=256</math>. This means <math>PA=16</math>. However, we scaled down everything by <math>12</math> so our ans
  4 KB (575 words) - 00:50, 12 January 2024
• 2021 AIME I Problems/Problem 9
  ...\sim \triangle QPA</math> and we have that <math>\frac{AQ}{AR} = \frac{PQ}{PA}</math> or that <math>\frac{18}{10} = \frac{QP}{15}</math>, which we can se ...<math>\frac{3\sqrt{2}}{4}</math>. From this, we see then that <cmath>AB = AP \cdot \frac{3\sqrt{2}}{4} = 15 \cdot \frac{3\sqrt{2}}{4} = \frac{45\sqrt{2}
  23 KB (3,640 words) - 18:16, 25 January 2024
• 2004 IMO Problems/Problem 5
  Prove that <math>ABCD</math> is a cyclic quadrilateral if and only if <math>AP = CP.</math> ...</math> is also on the perpendicular bisector of <math>AC</math>. So <math>PA=PC</math>.
  2 KB (336 words) - 22:28, 8 February 2024
• 2021 AMC 12B Problems/Problem 24
  ...\sqrt{AX^2 - XP^2} = 3\sqrt{x^2-1}</math>, and so<cmath>[ABCD] = 2[ABD] = AP\cdot BD = 3\sqrt{x^2-1}\cdot 8x = 24x\sqrt{x^2-1}=15</cmath>Solving this fo ...of each triangle, which are also the lengths of <math>QC</math> and <math>AP,</math> is <math>\tfrac{15}{2(x+3)}.</math> Suppose that <math>E = RS \cap
  11 KB (1,646 words) - 21:14, 28 May 2024
• 2021 IMO Problems/Problem 4
  ...ents we have, <math>AD = AM + MD = AP + ND</math>. So <math>AD + DT + XA = AP + ND + DT + XA = XP + NT</math>. <cmath>AD + DT + XA = AN+ND + TM – MD +XP-PA =</cmath>
  7 KB (1,196 words) - 10:30, 18 June 2023
• 2021 Fall AMC 12B Problems/Problem 22
  ...ath> is on the perpendicular bisector of segment <math>AB</math> and <math>PA \perp AC</math>. Let lines <math>OB</math> and <math>AP</math> intersect at point <math>D</math>.
  6 KB (943 words) - 00:41, 6 August 2023
• 2023 AIME I Problems/Problem 5
  ...on the circle circumscribing square <math>ABCD</math> that satisfies <math>PA \cdot PC = 56</math> and <math>PB \cdot PD = 90.</math> Find the area of <m ...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>,
  22 KB (3,480 words) - 19:33, 1 August 2024
• 2022 AMC 10A Problems/Problem 23
  ...= CD.</math> There is a point <math>P</math> in the plane such that <math>PA=1, PB=2, PC=3,</math> and <math>PD=4.</math> What is <math>\tfrac{BC}{AD}?< ...>. Under this reflection, <math>P^{\prime}A=PD=4</math>, <math>P^{\prime}D=PA=1</math>, <math>P^{\prime}C=PB=2</math>, and <math>P^{\prime}B=PC=3</math>.
  11 KB (1,794 words) - 15:36, 16 August 2024
• Steiner line
  <cmath>AP||BD \implies \frac {PD}{CD} = \frac {AB}{BC},</cmath> ...ly. Let points <math>P</math> and <math>Q</math> be such points that <math>PA = PB, PC = PD, QA = QD, QB = QC.</math>
  14 KB (2,381 words) - 12:07, 12 May 2024
• Gossard perspector
  Let <math>Pa</math> be a point such that <math>EPa</math> is parallel to <math>CP</math> Symilarly, <math>Pb: DPb||AP, FPb||CP, Pc: FPc||BP, EPc||AP.</math>
  18 KB (3,046 words) - 06:44, 19 January 2023
• 2023 AIME II Problems/Problem 3
  ...ath> such that <math>\angle PAB = \angle PBC = \angle PCA</math> and <math>AP = 10.</math> Find the area of <math>\triangle ABC.</math> ...iangle PBC</math> by AA Similarity. The ratio of similitude is <math>\frac{PA}{PB} = \frac{PB}{PC} = \frac{AB}{BC},</math> so <math>\frac{10}{PB} = \frac
  10 KB (1,443 words) - 01:42, 27 June 2024
• 2023 SSMO Speed Round Problems/Problem 5
  ...Find the sum of the maximum, <math>M</math>, and minimum values of <math>(PA)(PC)+(PB)(PD).</math> If you think there is no maximum, let <math>M=0.</mat ...c. From Ptolemy's Theorem, <math>(AD)(PP') = (PA)(P'D)+(PD)(P'A) \implies (PA)(PC)+(PB)(PD)=(AD)(CD)=6\cdot8 = 48,</math> meaning the answer is <math>48+
  749 bytes (122 words) - 14:20, 3 July 2023
• 2023 AMC 12A Problems/Problem 15
  ...math>) will have the same trig ratios. By proportion, the hypotenuse <math>AP</math> is <math>\frac{x}{100}(120) = \frac65 x</math>, so <math>\cos\theta .... Apply the Pythagoras theorem on <math>\triangle{ADP}</math> to get <math>AP = \sqrt{900 + x^2}</math>, which is also the length of every zigzag segment
  7 KB (1,105 words) - 11:17, 16 August 2024
• Power of a point theorem
  ...</math> intersect at a point <math> P </math> within a circle, then <math> AP\cdot BP=CP\cdot DP </math> ...at points <math> A </math> and <math> D </math> respectively, then <math> PA\cdot PB=PD\cdot PC </math>
  6 KB (999 words) - 16:44, 28 June 2024

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