Difference between revisions of "British Flag Theorem"

(Seriously, you guys need to learn asymptote.)
Line 2: Line 2:
pair A,B,C,D,P;
pair A,B,C,D,P;

Revision as of 16:15, 16 October 2007

The British flag theorem says that if a point P is chosen inside rectangle ABCD then $AP^{2}+PC^{2}=BP^{2}+DP^{2}$.

[asy] usepackage("amsmath"); size(200); pair A,B,C,D,P; A=(0,0); B=(200,0); C=(200,150); D=(0,150); P=(124,85); draw(A--B--C--D--cycle); label("<math>A</math>",A,(-1,0)); dot(A); label("<math>B</math>",B,(0,-1)); dot(B); label("<math>C</math>",C,(1,0)); dot(C); label("<math>D</math>",D,(0,1)); dot(D); dot(P); label("<math>P</math>",P,(1,1)); draw((0,85)--(200,85)); draw((124,0)--(124,150)); label("<math>w</math>",(124,0),(0,-1)); label("<math>x</math>",(200,85),(1,0)); label("<math>y</math>",(124,150),(0,1)); label("<math>z</math>",(0,85),(-1,0)); dot((124,0)); dot((200,85)); dot((124,150)); dot((0,85)); [/asy]

The theorem also applies to points outside the rectangle, although the proof is harder to visualize in this case.


In Figure 1, by the Pythagorean theorem, we have:

  • $AP^{2} = Aw^{2} + Az^{2}$
  • $PC^{2} = wB^{2} + zD^{2}$
  • $BP^{2} = wB^{2} + Az^{2}$
  • $PD^{2} = zD^{2} + Aw^{2}$


  • $AP^{2} + PC^{2} = Aw^{2} + Az^{2} + wB^{2} + zD^{2} = wB^{2} + Az^{2} + zD^{2} + Aw^{2} =\nolinebreak BP^{2} +\nolinebreak PD^{2}$

This article is a stub. Help us out by expanding it.

Invalid username
Login to AoPS