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Difference between revisions of "British Flag Theorem"
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The '''British flag theorem''' says that if a point P is chosen inside [[rectangle]] ABCD then <math>AP^{2}+PC^{2}=BP^{2}+DP^{2}</math>.
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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):
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+
[[File:UK.jpg|left|frame|British Flag]]
 
<asy>
 
<asy>
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size(200);
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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);
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label("A",A,(-1,0));
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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));
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label("$B$",B,(0,-1));
 
dot(B);
 
dot(B);
−
label("C",C,(1,0));
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label("$C$",C,(1,0));
 
dot(C);
 
dot(C);
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label("D",D,(0,1));
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label("$D$",D,(-1,0));
 
dot(D);
 
dot(D);
 
dot(P);
 
dot(P);
−
label("P",P,(1,1));
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label("$P$",P,NNE);
 
draw((0,85)--(200,85));
 
draw((0,85)--(200,85));
 
draw((124,0)--(124,150));
 
draw((124,0)--(124,150));
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</asy>
 
</asy>
  
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The theorem also applies to points outside the rectangle, although the proof is harder to visualize in this case.
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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 ==
  
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In Figure 1, by the [[Pythagorean theorem]], we have:
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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.
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<asy>
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* <math>AP^{2} = Aw^{2} + Az^{2}</math>
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pair A,B,C,D,P,X,Y;
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* <math>PC^{2} = wB^{2} + zD^{2}</math>
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A = (0,0);
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* <math>BP^{2} = wB^{2} + Az^{2}</math>
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B=(1,0);
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* <math>PD^{2} = zD^{2} + Aw^{2}</math>
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D = (0,0.7);
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C = B+D;
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Therefore:
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P = (0.3,0.4);
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X = (0.3,0);
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*<math>AP^{2} + PC^{2} = Aw^{2} + Az^{2} + wB^{2} + zD^{2} = wB^{2} + Az^{2} + zD^{2} + Aw^{2} =\nolinebreak BP^{2} +\nolinebreak PD^{2}</math>
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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);
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label("$P$",P+(0,0.03),NE);</asy>
 +
Applying the Pythagorean Theorem to each of the four right triangles in the diagram, we have
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<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]]
−
 
−
{{stub}}
 
−
The above result i.e. theorem holds true even if the point P is selected on the boundary of rectangle or even outside the rectangle.
 

Latest revision as of 23:14, 17 September 2024

In Euclidian Geometry, the British flag theorem states that if a point $P$ is chosen inside rectangle $ABCD$, then $AP^{2}+CP^{2}=BP^{2}+DP^{2}$. The theorem is called the British flag theorem due to the similarities between the British flag and a diagram of the points (shown below):

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 $P$ is located outside or on the boundary of $ABCD$, and even when $P$ is located in a Euclidian space where $ABCD$ is embedded.

Proof

We build right triangles by drawing a line through $P$ perpendicular to two sides of the rectangle, as shown below. Both $AXYD$ and $BXYC$ 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 \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*} So, we have \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*} From rectangles $AXYD$ and $BXYC$, we have $AX = DY$ and $BX = CY$, so the expressions above for $PA^2 + PC^2$ and $PB^2 + PD^2$ 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

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