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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 (1,296 words) - 14:22, 22 October 2023
• 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) - 20:39, 9 March 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 (827 words) - 17:30, 21 February 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) - 23:37, 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) - 08:16, 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) - 21:57, 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) - 11:48, 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
  9 KB (1,540 words) - 08:31, 1 December 2022
• 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) - 19:15, 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) - 16:46, 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) - 16:07, 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 (405 words) - 01:14, 3 December 2023
• 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) - 08:40, 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 (1,333 words) - 00:18, 1 February 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) - 22:24, 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) - 18:37, 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>.
  1 KB (242 words) - 01:24, 27 July 2023
• 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
  6 KB (1,035 words) - 09:18, 3 September 2023
• 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) - 00:23, 9 January 2023
• Radical axis
  ...ircle <math>\omega</math> at points A and B, then <math>|pow(P, \omega)| = PA * PB</math>. .../math>. Point <math>A</math> is located on <math>XY</math> such that <math>AP = 10</math> and <math>AQ =</math> <s>15</s> <math>12</math>. If the radius
  10 KB (1,797 words) - 02:05, 24 October 2023
• 2010 AIME II Problems/Problem 14
  ...le APC = 2\angle ACP</math> and <math>CP = 1</math>. The ratio <math>\frac{AP}{BP}</math> can be represented in the form <math>p + q\sqrt{r}</math>, wher <cmath>\frac {AP}{BP} = \frac {AO + OP}{BO - OP} = \frac {2 + \sqrt {2}}{2 - \sqrt {2}} = 3
  10 KB (1,507 words) - 00:31, 19 November 2023
• 2011 AIME II Problems/Problem 4
  ...<math>AC</math> and <math>BM</math>. The ratio of <math>CP</math> to <math>PA</math> can be expressed in the form <math>\dfrac{m}{n}</math>, where <math> ...by the Midline Theorem that <math>AP = PD'</math>. Thus, <cmath>\frac{CP}{PA} = \frac{1}{\frac{PD'}{PC}} = \frac{1}{1 - \frac{D'C}{PC}} = \frac{31}{20},
  6 KB (944 words) - 21:31, 14 January 2024
• 2011 AIME II Problems
  .../math> and the line <math>BM</math>. The ratio of <math>CP</math> to <math>PA</math> can be expressed in the form <math>\frac{m}{n}</math>, where <math>m ...>AB = 12</math> and <math>\angle O_1PO_2 = 120 ^{\circ}</math>, then <math>AP = \sqrt{a} + \sqrt{b}</math>, where <math>a</math> and <math>b</math> are p
  8 KB (1,301 words) - 08:43, 11 October 2020
• 2011 AIME II Problems/Problem 13
  ...= 12</math> and <math>\angle O_{1}PO_{2} = 120^{\circ}</math>, then <math>AP = \sqrt{a} + \sqrt{b}</math>, where <math>a</math> and <math>b</math> are p ...h> is <math>6\sqrt{2}</math>), or <math>QP=2\sqrt{6}</math> Finally, <math>AP=QP+AQ=2\sqrt{6}+6\sqrt{2}=\sqrt{24}+\sqrt{72} \Rightarrow \boxed{096}</math
  13 KB (2,055 words) - 05:25, 9 September 2022
• 1966 IMO Problems/Problem 3
  <cmath> = \sum \textrm{Distances from }P\textrm{ to faces of }WXYZ \leq PA + PB + PC + PD,</cmath> with equality only occurring when <math>AP</math>, <math>BP</math>, <math>CP</math>, and <math>DP</math> are perpendic
  2 KB (360 words) - 15:51, 11 December 2022
• 1996 AHSME Problems/Problem 11
  ...math> at point <math>P</math>, with radius <math>OP = 2</math>. Let <math>AP = PB = 1</math>, so that <math>P</math> is the midpoint of <math>AB</math>. Since <math>AP^2 + OP^2 = OA^2</math> by the [[Pythagorean Theorem]], we can find that <ma
  2 KB (266 words) - 14:07, 5 July 2013
• 2012 AIME I Problems/Problem 13
  ...0,\tfrac{\sqrt{3}}{2}x).</math> Let <math>P(a,b)</math> be such that <math>PA=3, PB=4</math> and <math>PC=5.</math> Then pair A = 3*dir(theta), Ap = rotate(150,O)*A, F=IP(c4,O--2*Ap), C=rotate(60,A)*F, E=rotate(60,A)*C, B=IP(c5,O--E), D=foot(C,O,E);
  11 KB (1,889 words) - 20:42, 25 January 2023
• 2012 USAMO Problems/Problem 5
  .../math>, <math>C'</math> be the points where the reflections of lines <math>PA</math>, <math>PB</math>, <math>PC</math> with respect to <math>\gamma</math <cmath>\frac{AB'}{\sin \angle APB'} = \frac{AP}{\sin \angle AB'P},</cmath>
  5 KB (767 words) - 22:32, 2 May 2023
• 2012 USAJMO Problems
  ...e{AB}</math> and <math>\overline{AC}</math>, respectively, such that <math>AP = AQ</math>. Let <math>S</math> and <math>R</math> be distinct points on s .../math>, <math>C'</math> be the points where the reflections of lines <math>PA</math>, <math>PB</math>, <math>PC</math> with respect to <math>\gamma</math
  3 KB (566 words) - 16:41, 5 August 2023
• 2013 AMC 12A Problems/Problem 23
  ...Point <math> P </math> is on <math> \overline{AC} </math> such that <math> AP = \sqrt{2} </math>. The square region bounded by <math> ABCD </math> is rot ...ength <math> \sqrt{6} + \sqrt{2} </math>. It must be that <math> \overline{AP} </math> divides the diagonal into two segments in the ratio <math>\sqrt{3}
  4 KB (603 words) - 16:51, 3 April 2020
• 2013 AIME I Problems/Problem 12
  <math>\frac{\frac{\sqrt{6} + \sqrt{2}}{4}}{1} = \frac{\frac{\sqrt{2}}{2}}{PA}</math> Simplifying, we find <math>PA = \sqrt{3} - 1</math>.
  9 KB (1,490 words) - 02:25, 2 May 2024
• 2013 USAMO Problems/Problem 6
  ...f triangles <math>PAB</math> and <math>PAC</math>, then <cmath>\left(\frac{PA}{XY}\right)^2+\frac{PB\cdot PC}{AB\cdot AC}=1.</cmath> ...at <math>Z</math>. Then obviously <math>Z</math> is the midpoint of <math>AP</math> and <math>AZ</math> is an altitude of triangle <math>A O_1 O_2</math
  4 KB (691 words) - 18:29, 10 May 2023
• 2013 USAMO Problems
  .../math>, <math>CPQ</math>, respectively. Given the fact that segment <math>AP</math> intersects <math>\omega_A</math>, <math>\omega_B</math>, <math>\omeg ...f triangles <math>PAB</math> and <math>PAC</math>, then <cmath>\left(\frac{PA}{XY}\right)^2+\frac{PB\cdot PC}{AB\cdot AC}=1.</cmath>
  3 KB (522 words) - 20:08, 30 April 2014
• Rotation
  ...Point <math>P</math> is inside <math>\triangle ABC</math>, such that <math>PA=11</math>, <math>PB=7</math>, and <math>PC=6</math>. Legs <math>\overline{A ...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<
  3 KB (432 words) - 23:22, 13 January 2021
• 1967 AHSME Problems/Problem 40
  ...ateral triangle <math>ABC</math> is a point <math>P</math> such that <math>PA=8</math>, <math>PB=6</math>, and <math>PC=10</math>. To the nearest intege <cmath>\angle P'AP = \angle PAB + \angle P'AB = \angle PAB + \angle PAC = 60^{\circ}.</cmath>
  7 KB (1,089 words) - 15:48, 11 November 2023
• 1993 AHSME Problems/Problem 25
  ...the points the same distance, we have <math>AA' = OO'</math>. Also, <math>AP = OP</math> by the original equilateral triangle. Therefore, by SAS congru Now, look at <math>\triangle A'PO'</math>. We have <math>PA' = PO'</math> from the above congruence. We also have the included angle <
  2 KB (375 words) - 19:13, 29 July 2019
• 1961 AHSME Problems
  ...respectively in <math>CB</math> and <math>AB</math> and such that <math>AC=AP=PQ=QB</math>. ...th>A</math> is between <math>P</math> and <math>B</math>). The ratio <math>PA:AB</math> is:
  18 KB (2,905 words) - 18:33, 5 April 2023
• 2016 AIME II Problems
  ...th> and <math>T</math> (other than <math>C</math>), respectively. If <math>AP=4,PQ=3,QB=6,BT=5,</math> and <math>AS=7</math>, then <math>ST=\frac{m}{n}</ ...ngle ABC</math> and are on opposite sides of the plane. Furthermore, <math>PA=PB=PC</math>, and <math>QA=QB=QC</math>, and the planes of <math>\triangle
  8 KB (1,312 words) - 21:16, 3 March 2021
• Mock Geometry AIME 2011 Problems/Problem 14
  ...athcal{E}</math> at <math>P</math>. Then the acute angles formed by <math>PA</math> and <math>PB</math> with respect to <math>\ell</math> are equal. (T ...hrough <math>Q</math>). Hence <math>P</math> is the point for which <math>AP+PB'</math> is minimized, and so <math>A</math>, <math>P</math>, <math>B'</m
  4 KB (657 words) - 11:40, 6 November 2016
• 2017 USAJMO Problems/Problem 3
  ...gle and let <math>P</math> be a point on its circumcircle. Let lines <math>PA</math> and <math>BC</math> intersect at <math>D</math>; let lines <math>PB< ...c{BD}{BC})=1.</math> Thus we have that <math>\frac{DK}{AK}=2\cdot\frac{DP}{AP}.</math>
  10 KB (1,829 words) - 03:13, 16 March 2022
• 2018 AMC 12A Problems/Problem 23
  ...egree measure of the acute angle formed by lines <math>MN</math> and <math>PA?</math> Let <math>P</math> be the origin, and <math>PA</math> lie on the <math>x</math>-axis.
  12 KB (1,878 words) - 22:11, 23 October 2021
• 2018 AMC 10B Problems/Problem 17
  ...</math> and <math>H</math> lie on <math>\overline{SP}</math> so that <math>AP=BQ<4</math> and the convex octagon <math>ABCDEFGH</math> is equilateral. Th Let <math>AP=BQ=x</math>. Then <math>AB=8-2x</math>.
  3 KB (581 words) - 18:54, 11 November 2023
• 2018 AIME II Problems/Problem 12
  Let <math>AP=x</math> and let <math>PC=\rho x</math>. Let <math>[ABP]=\Delta</math> and Let <math>AP=x</math> and let <math>PC=\rho x</math>. Let <math>[ABP]=\Delta</math> and
  18 KB (2,912 words) - 13:12, 24 January 2024
• 2018 AIME II Problems/Problem 14
  ...th> is tangent to <math>\omega</math> at <math>Y</math>. Assume that <math>AP = 3</math>, <math>PB = 4</math>, <math>AC = 8</math>, and <math>AQ = \dfrac ...\frac{\sin\alpha}{\sin\beta}.</cmath>It follows that <cmath>2 = \frac{AZ}{AP} + \frac{AZ}{AB} = \frac{AZ}3 + \frac{AZ}7,</cmath>implying <math>AZ = \tfr
  9 KB (1,518 words) - 14:32, 28 January 2024
• 2002 JBMO Problems/Problem 1
  ...the perpendicular from <math>C</math> to <math>PB</math>. Show that <math>PA + PB = 2 \cdot PD</math>. ...ath>E</math> be the foot of the perpendicular from <math>C</math> to <math>AP</math>. Then we have, <math>CEPD</math> is a cyclic quadrilateral with <mat
  1 KB (234 words) - 21:37, 8 December 2018
• 1988 IMO Problems
  ...B,P,C</math> are collinear and <math> BC</math> is perpendicular to <math> AP.</math>
  3 KB (499 words) - 13:29, 2 August 2021
• 2019 AIME I Problems
  ...>E</math> and <math>F</math> lie on <math>\overline{PR}</math>, with <math>PA=QB=QC=RD=RE=PF=5</math>. Find the area of hexagon <math>ABCDEF</math>. ...math>\omega</math> at <math>X</math> and <math>Y</math>. Assume that <math>AP=5</math>, <math>PB=3</math>, <math>XY=11</math>, and <math>PQ^2 = \tfrac{m}
  8 KB (1,331 words) - 06:57, 4 January 2021
• 2019 AIME I Problems/Problem 15
  ...math>\omega</math> at <math>X</math> and <math>Y</math>. Assume that <math>AP=5</math>, <math>PB=3</math>, <math>XY=11</math>, and <math>PQ^2 = \frac{m}{ By Power of a Point, <math>PX\cdot PY=PA\cdot PB=15</math>. Since <math>PX+PY=XY=11</math> and <math>XQ=11/2</math>,
  13 KB (2,252 words) - 11:32, 1 February 2024
• 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) - 02:08, 9 March 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
  8 KB (1,279 words) - 12:07, 28 October 2023
• 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>,
  19 KB (3,107 words) - 23:31, 17 January 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>.
  12 KB (1,806 words) - 12:52, 26 March 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
  9 KB (1,354 words) - 21:16, 26 January 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 (1,074 words) - 21:22, 20 November 2023
• 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 (1,010 words) - 02:38, 7 May 2024