Difference between revisions of "British Flag Theorem"
(Seriously, you guys need to learn asymptote.) |
|||
| Line 1: | Line 1: | ||
The '''British flag theorem''' says that if a point P is chosen inside [[rectangle]] ABCD then <math>AP^{2}+PC^{2}=BP^{2}+DP^{2}</math>. | The '''British flag theorem''' says that if a point P is chosen inside [[rectangle]] ABCD then <math>AP^{2}+PC^{2}=BP^{2}+DP^{2}</math>. | ||
| − | + | <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); | ||
| + | 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 | + | The theorem also applies to points outside the rectangle, although the proof is harder to visualize in this case. |
== Proof == | == Proof == | ||
| Line 22: | Line 45: | ||
Therefore: | Therefore: | ||
| − | * <math>AP^{2} + PC^{2} = Aw^{2} + Az^{2} + wB^{2} + zD^{2} = wB^{2} + Az^{2} + zD^{2} + Aw^{2} = BP^{2} + PD^{2}</math> | + | *<math>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}</math> |
Revision as of 16:14, 16 October 2007
The British flag theorem says that if a point P is chosen inside rectangle ABCD then 
.
![[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); 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]](http://latex.artofproblemsolving.com/7/5/7/757bbb45146725af18d5a6d32cf80d67f7b55625.png)
The theorem also applies to points outside the rectangle, although the proof is harder to visualize in this case.
Proof
In Figure 1, by the Pythagorean theorem, we have:
Therefore:
This article is a stub. Help us out by expanding it.
