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- ...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}} = 310 KB (1,507 words) - 00:31, 19 November 2023
- ...<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
- .../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 p8 KB (1,301 words) - 08:43, 11 October 2020
- ...= 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}</math13 KB (2,055 words) - 05:25, 9 September 2022
- <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 perpendic2 KB (360 words) - 15:51, 11 December 2022
- ...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 <ma2 KB (266 words) - 14:07, 5 July 2013
- ...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
- .../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
- ...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</math3 KB (566 words) - 16:41, 5 August 2023
- ...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
- <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
- ...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</math4 KB (691 words) - 18:29, 10 May 2023
- .../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
- ...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
- ...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
- ...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
- ...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
- ...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>\triangle8 KB (1,312 words) - 21:16, 3 March 2021
- ...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'</m4 KB (657 words) - 11:40, 6 November 2016
- ...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
