Difference between revisions of "British Flag Theorem"
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<cmath> \begin{align*}PA^2+PC^2 &= AX^2+XP^2+CY^2+YP^2,\\ PB^2+PD^2 &= BX^2+XP^2+DY^2+YP^2.\end{align*} </cmath> | <cmath> \begin{align*}PA^2+PC^2 &= AX^2+XP^2+CY^2+YP^2,\\ PB^2+PD^2 &= BX^2+XP^2+DY^2+YP^2.\end{align*} </cmath> | ||
From rectangles <math>AXYD</math> and <math>BXYC</math>, we have <math>AX = DY</math> and <math>BX = CY</math>, so the expressions above for <math>PA^2 + PC^2</math> and <math>PB^2 + PD^2</math> are equal, as desired. | From rectangles <math>AXYD</math> and <math>BXYC</math>, we have <math>AX = DY</math> and <math>BX = CY</math>, so the expressions above for <math>PA^2 + PC^2</math> and <math>PB^2 + PD^2</math> are equal, as desired. | ||
| + | |||
| + | ==Problems== | ||
| + | [http://artofproblemsolving.com/community/c3h579390 2014 MATHCOUNTS Chapter Sprint #29] | ||
[[Category:geometry]] | [[Category:geometry]] | ||
Revision as of 19:27, 11 October 2015
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):
![[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
Squares of lengths suggest right triangles. We build right triangles by drawing a line through perpendicular to two sides of the rectangle, as shown below. Both 
and 
are rectangles.
![[asy] pair A,B,C,D,P,X,Y; A = (0,0); B=(1,0); D = (0,0.7); C = B+D; P = (0.3,0.4); X = (0.3,0); Y=(0.3,0.7); draw(A--B--C--D--A--P--C); draw(X--Y); draw(B--P--D); draw(rightanglemark(P,X,A,1.5)); draw(rightanglemark(B,X,P,1.5)); draw(rightanglemark(P,Y,C,1.5)); draw(rightanglemark(D,Y,P,1.5)); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); label("$Y$",Y,N); label("$X$",X,S); label("$P$",P+(0,0.03),NE);[/asy]](http://latex.artofproblemsolving.com/2/f/c/2fca4691116928e07cf6da7092ac9fa993c61285.png)
Applying the Pythagorean Theorem to each of the four right triangles in the diagram, we have
So, we have
From rectangles and , we have 
and 
, so the expressions above for 
and 
are equal, as desired.
Problems
2014 MATHCOUNTS Chapter Sprint #29
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