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suppose there is a quadrilateral $ABCD$. any point $P$ which lies inside the quadrilateral satisfies $PA^2+PC^2 = PB^2+PD^2$. Should such a condition exist always in a rectangle or a square?.can there be any other quadrilateral in which such a point exists.

samarth srivastava
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  • Sidenote: The fact that $PA^2+PC^2 = PB^2+PD^2$ holds for all points $P$ in a rectangle $ABCD$ is known as the British Flag Theorem. http://en.wikipedia.org/wiki/British_flag_theorem – JimmyK4542 Jun 28 '14 at 06:48

3 Answers3

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This condition will always hold true in a rectangle. And since a square, by definition, is a rectangle, the condition will also hold true for the square.

Let us pick an arbitrary point P inside the quadrilateral, and let us create a line $\ell$ that passes through point P such that the line is perpendicular to side AB. Since AB and CD are parallel, it follows that $\ell$ is perpendicular to CD as well. Let's name the point of intersection of $\ell$ and AB point X, and $\ell$ and CD, point Y.

Instead of showing $ PA^2 + PC^2 = PB^2 + PD^2 $, I will show $$ PA^2 - PB^2 + PC^2 - PD^2 = 0 \\$$

Proof: \begin{align*} PA^2 &= PX^2 + XA^2 \\ PB^2 &= PX^2 + XB^2 \\ PC^2 &= PY^2 + YC^2 \\ PD^2 &= PY^2 + YD^2 \\ PA^2 - PB^2 + PC^2 - PD^2 &= XA^2 - XB^2 + YC^2 - YD^2 \\ PA^2 - PB^2 + PC^2 - PD^2 &= (XA^2 - YD^2) + (YC^2 - XB^2) \\ PA^2 - PB^2 + PC^2 - PD^2 &= 0 + 0 \\ PA^2 - PB^2 + PC^2 - PD^2 &= 0 \end{align*}

rjvani
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  • Is it really every allowed to denote a straight line as $L$, though? – user26486 Jun 26 '14 at 12:36
  • What do you mean? – rjvani Jun 26 '14 at 13:35
  • I mean, aren't straight lines only ever allowed to be defined either by a single lower case letter or by two capital letters? A capital letter in intself would instead define a point, and it couldn't really define anything else when you are talking about geometry (could, perhaps, define, e.g., a set in some cases, though, but not a line). – user26486 Jun 26 '14 at 13:46
  • Oh, yeah I guess so. I was typing quickly and didn't bother to LaTeX it, lol. I'll fix it – rjvani Jun 26 '14 at 13:49
  • Nice answer - I misread the question in the first place... – johannesvalks Jun 26 '14 at 14:56
  • @mathh: What's wrong with using capital letters for lines and small letters for points? As long as you define all your symbols you cannot be faulted for capitalization... – user21820 Jun 29 '14 at 10:08
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If $P=A$, then $$AC^2 = AB^2+AD^2,$$

if $P=C$, then $$AC^2 = BC^2+BD^2,$$

so $$AC^2 = AB^2+AD^2 = BC^2+CD^2.\tag{1}$$

Same way:

$$ AC^2 = AB^2+BC^2=CD^2+AD^2.\tag{2} $$

Hence (subtracting eq. $(1),(2)$) $$ AD^2-BC^2=-(AD^2-BC^2),\\ AB^2-CD^2=-(CD^2-AB^2),\\ $$ so $$AD=BC,\\AB=CD.\tag{3}$$

It is parallelogram (for now).

But $(1),(3) \Rightarrow$:

$$ AC^2=AB^2+BC^2, $$

so $\angle B = 90^\circ$. So, it is rectangle.

Here is the image that shows that "to be rectangle" is sufficient condition for such quadrilateral. ($\color{blue}{PA^2+PC^2} = \color{red}{PB^2+PD^2}$).

enter image description here

Oleg567
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Take $A = (0,0)$, $B = (b_x,0)$, $C = (c_x, c_y)$, $D = (d_x, d_y)$, $P = (x,y)$. Then $$PA^2 + PC^2 = PB^2 + PD^2 \quad\Leftrightarrow\quad \begin{array}{c}0 = -b_x^2 + c_x^2 + c_y^2 - d_x^2 - d_y^2 \\+\; 2x ( b_x - c_x + d_x ) + 2 y (-c_y + d_y )\end{array}$$ For the relation to be independent of $P$, we must have that the coefficient of $x$ and the coefficient of $y$ vanish, which in turn causes the "constant" term to vanish on its own: $$\begin{align} b_x - c_x + d_x &= 0 \qquad(1)\\[4pt] -c_y + d_y &= 0 \qquad(2)\\[4pt] -b_x^2 + c_x^2 +c_y^2 -d_x^2 - d_y^2 &= 0 \qquad (3) \end{align}$$ From $(2)$, we get $d_y = c_y$. Substituting this and $b_x = c_x-d_x$ (from $(1)$) into $(3)$ gives $$0 = -(c_x-d_x)^2+c_x^2-d_x^2 = 2 d_x (c_x - d_x )$$ Thus, $d_x = 0$, in which case $b_x = c_x$, and we see that $ABCD$ is a rectangle. On the other hand, we could have $c_x = d_x$ (so that $C=D$), in which case $b_x = 0$ (so that $A=B$); this makes $ABCD$ a "degenerate" rectangle in the form of a line segment.

Blue
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