Difference between revisions of "British Flag Theorem"

m (Change Lines)
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== Proof ==
 
== Proof ==
  
In Figure 1, by the [[Pythagorean theorem]], we have:
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Squares of lengths suggest right triangles. 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>
* <math>AP^{2} = Aw^{2} + Az^{2}</math>
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pair A,B,C,D,P,X,Y;
* <math>PC^{2} = wB^{2} + zD^{2}</math>
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A = (0,0);
* <math>BP^{2} = wB^{2} + Az^{2}</math>
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B=(1,0);
* <math>PD^{2} = zD^{2} + Aw^{2}</math>
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D = (0,0.7);
 
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C = B+D;
Therefore:
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P = (0.3,0.4);
 
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X = (0.3,0);
*<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);
 
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draw(A--B--C--D--A--P--C);
 
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draw(X--Y);
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draw(B--P--D);
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draw(rightanglemark(P,X,A,1.5));
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draw(rightanglemark(B,X,P,1.5));
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draw(rightanglemark(P,Y,C,1.5));
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draw(rightanglemark(D,Y,P,1.5));
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label("$A$",A,SW);
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label("$B$",B,SE);
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label("$C$",C,NE);
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label("$D$",D,NW);
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label("$Y$",Y,N);
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label("$X$",X,S);
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label("$P$",P+(0,0.03),NE);</asy>
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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>
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So, we have
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<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.
 
[[Category:geometry]]
 
[[Category:geometry]]
  

Revision as of 13:48, 2 December 2014

The British flag theorem says 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(200); 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]

The theorem also applies if the point $P$ 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 $P$ 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.

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