During AMC testing, the AoPS Wiki is in read-only mode. Your account is not considered logged in on wiki pages and no edits can be made.

Difference between revisions of "British Flag Theorem"

(Seriously, you guys need to learn asymptote.)
m (Proof)
 
(20 intermediate revisions by 14 users not shown)
Line 1: Line 1:
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>.
+
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(200);
+
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("<math>A</math>",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("<math>B</math>",B,(0,-1));
+
label("$B$",B,(0,-1));
 
dot(B);
 
dot(B);
label("<math>C</math>",C,(1,0));
+
label("$C$",C,(1,0));
 
dot(C);
 
dot(C);
label("<math>D</math>",D,(0,1));
+
label("$D$",D,(-1,0));
 
dot(D);
 
dot(D);
 
dot(P);
 
dot(P);
label("<math>P</math>",P,(1,1));
+
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("<math>w</math>",(124,0),(0,-1));
+
label("$w$",(124,0),(0,-1));
label("<math>x</math>",(200,85),(1,0));
+
label("$x$",(200,85),(1,0));
label("<math>y</math>",(124,150),(0,1));
+
label("$y$",(124,150),(0,1));
label("<math>z</math>",(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>
  
The theorem also applies to points outside the rectangle, although the proof is harder to visualize in this case.
+
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 ==
  
In Figure 1, by the [[Pythagorean theorem]], we have:
+
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>
* <math>AP^{2} = Aw^{2} + Az^{2}</math>
+
pair A,B,C,D,P,X,Y;
* <math>PC^{2} = wB^{2} + zD^{2}</math>
+
A = (0,0);
* <math>BP^{2} = wB^{2} + Az^{2}</math>
+
B=(1,0);
* <math>PD^{2} = zD^{2} + Aw^{2}</math>
+
D = (0,0.7);
 
+
C = B+D;
Therefore:
+
P = (0.3,0.4);
 
+
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>
+
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]]
 
{{stub}}
 

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