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]] | |
<asy> | <asy> | ||
− | size( | + | size(300); |
pair A,B,C,D,P; | pair A,B,C,D,P; | ||
A=(0,0); | A=(0,0); | ||
Line 10: | Line 10: | ||
P=(124,85); | P=(124,85); | ||
draw(A--B--C--D--cycle); | draw(A--B--C--D--cycle); | ||
− | label("A",A,(-1,0)); | + | draw(A--P); |
+ | draw(B--P); | ||
+ | draw(C--P); | ||
+ | draw(D--P); | ||
+ | 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)); | ||
− | label("w",(124,0),(0,-1)); | + | label("$w$",(124,0),(0,-1)); |
− | label("x",(200,85),(1,0)); | + | label("$x$",(200,85),(1,0)); |
− | label("y",(124,150),(0,1)); | + | label("$y$",(124,150),(0,1)); |
− | label("z",(0,85),(-1,0)); | + | label("$z$",(0,85),(-1,0)); |
dot((124,0)); | dot((124,0)); | ||
dot((200,85)); | dot((200,85)); | ||
Line 32: | 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> | |
− | + | 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> | ||
+ | Applying the Pythagorean Theorem to each of the four right triangles in the diagram, we have | ||
+ | <cmath> \begin{align*}PA^2 &= AX^2+XP^2,\\ PB^2 &= BX^2+XP^2,\\ PC^2 &= CY^2+YP^2,\\ PD^2 &= DY^2+YP^2.\end{align*} </cmath> | ||
+ | So, we have | ||
+ | <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. | ||
+ | ==Problems== | ||
+ | [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] 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):
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.
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