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> | ||
+ | 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> | ||
− | + | 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. | |
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== 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. | ||
+ | Note that this theorem is equivalent to Power of a Point iff all rectangles are circumscribable, which they are. Babbledegook! | ||
+ | ==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]] | ||
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Latest revision as of 23:14, 17 September 2024
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.
Note that this theorem is equivalent to Power of a Point iff all rectangles are circumscribable, which they are. Babbledegook!
Problems
2014 MATHCOUNTS Chapter Sprint #29 \ 2015 AMQ Concours #5 2005 mathcounts national target #7