Difference between revisions of "British Flag Theorem"
m (added picture of flag) |
m |
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| Line 14: | Line 14: | ||
draw(C--P); | draw(C--P); | ||
draw(D--P); | draw(D--P); | ||
| − | label("A",A,(-1,0)); | + | label("$A$",A,(-1,0)); |
dot(A); | dot(A); | ||
| − | label("B",B,(0,-1)); | + | label("$B$",B,(0,-1)); |
dot(B); | dot(B); | ||
| − | label("C",C,(1,0)); | + | label("$C$",C,(1,0)); |
dot(C); | dot(C); | ||
| − | label("D",D,(0 | + | label("$D$",D,(-1,0)); |
dot(D); | dot(D); | ||
dot(P); | dot(P); | ||
| − | label("P",P, | + | label("$P$",P,NNE); |
draw((0,85)--(200,85)); | draw((0,85)--(200,85)); | ||
draw((124,0)--(124,150)); | draw((124,0)--(124,150)); | ||
Revision as of 21:51, 31 March 2014
The British flag theorem says that if a point P is chosen inside rectangle ABCD then 
. The theorem is called the British flag theorem due to the similarities between the British flag and a diagram of the points (shown below):
British Flag
![[asy] 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); draw(A--P); draw(B--P); draw(C--P); draw(D--P); label("$A$",A,(-1,0)); dot(A); label("$B$",B,(0,-1)); dot(B); label("$C$",C,(1,0)); dot(C); label("$D$",D,(-1,0)); dot(D); dot(P); label("$P$",P,NNE); draw((0,85)--(200,85)); draw((124,0)--(124,150)); label("$w$",(124,0),(0,-1)); label("$x$",(200,85),(1,0)); label("$y$",(124,150),(0,1)); label("$z$",(0,85),(-1,0)); dot((124,0)); dot((200,85)); dot((124,150)); dot((0,85)); [/asy]](http://latex.artofproblemsolving.com/3/b/4/3b47b62c5701b4cf4809728c92bed6867652d5ae.png)
The theorem also applies if the point 
is selected outside or on the boundary of the rectangle, although the proof is harder to visualize in this case.
Proof
In Figure 1, by the Pythagorean theorem, we have:
Therefore:
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