Difference between revisions of "British Flag Theorem"
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| − | + | In Euclidian Geometry, the '''British flag theorem''' states that if a point <math>P</math> is chosen inside [[rectangle]] <math>ABCD</math>, then <math>AP^{2}+CP^{2}=BP^{2}+DP^{2}</math>. The theorem is called the British flag theorem due to the similarities between the British flag and a diagram of the points (shown below): | |
[[File:UK.jpg|left|frame|British Flag]] | [[File:UK.jpg|left|frame|British Flag]] | ||
<asy> | <asy> | ||
| Line 36: | Line 36: | ||
</asy> | </asy> | ||
| − | + | This is also true when point <math>P</math> is located outside or on the boundary of <math>ABCD</math>, and even when <math>P</math> is located in a Euclidian space where <math>ABCD</math> is embedded. | |
== Proof == | == Proof == | ||
| − | + | We build right triangles by drawing a line through <math>P</math> perpendicular to two sides of the rectangle, as shown below. Both <math>AXYD</math> and <math>BXYC</math> are rectangles. | |
<asy> | <asy> | ||
pair A,B,C,D,P,X,Y; | pair A,B,C,D,P,X,Y; | ||
| Line 71: | Line 71: | ||
==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] 2005 mathcounts national target #7 |
[[Category:geometry]] | [[Category:geometry]] | ||
[[Category:Theorems]] | [[Category:Theorems]] | ||
Latest revision as of 01:14, 3 December 2023
In Euclidian Geometry, the British flag theorem states that if a point 
is chosen inside rectangle 
, 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)
This is also true when point is located outside or on the boundary of , and even when is located in a Euclidian space where is embedded.
Proof
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 \\ 2015 AMQ Concours #5 2005 mathcounts national target #7
