Difference between revisions of "British Flag Theorem"
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==Problems== | ==Problems== | ||
| − | [http://artofproblemsolving.com/community/c3h579390 2014 MATHCOUNTS Chapter Sprint #29] | + | [http://artofproblemsolving.com/community/c3h579390 2014 MATHCOUNTS Chapter Sprint #29] \\ |
[https://artofproblemsolving.com/community/c4h2477234_distances_of_a_point_from_certices_of_a_square__2015_amq_concours_p5 2015 AMQ Concours #5] | [https://artofproblemsolving.com/community/c4h2477234_distances_of_a_point_from_certices_of_a_square__2015_amq_concours_p5 2015 AMQ Concours #5] | ||
[[Category:geometry]] | [[Category:geometry]] | ||
[[Category:Theorems]] | [[Category:Theorems]] | ||
Revision as of 10:52, 6 March 2021
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(300); 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/7/8/7/787e5bd8aa039bb5b512ba85ef3a0ce8e7160169.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.
