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- <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
- ...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
- ...</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
- 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> and18 KB (2,912 words) - 13:12, 24 January 2024
- ...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 = \tfr9 KB (1,518 words) - 14:32, 28 January 2024
- ...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 <mat1 KB (234 words) - 02:52, 6 June 2024
- ...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
- ...>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
- ...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
- ...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 \frac13 KB (489 words) - 18:48, 14 December 2019
- ...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
