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- ...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
- ...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+YP3 KB (405 words) - 01:14, 3 December 2023
- ...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
- ...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
- ...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
- ...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
- 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
- .../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 th6 KB (1,035 words) - 09:18, 3 September 2023
- ...\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
- ...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) - 08:38, 23 May 2024
- ...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
