British Flag Theorem
If \(ABCD\) is a rectangle and \(P\) is any point inside of it, then we have
\[ PA^2 + PC^2 = PB^2 + PD ^2 . \]
This can be proved by using the Pythagorean theorem and projecting point \(P\) to the 4 sides of the rectangle.
Drop perpendicular lines from the point \(P\) to the sides of the rectangle, meeting sides \(AB, BC, CD, AD\) at points \(W, X, Y, Z,\) respectively, as shown in the figure. These four points \(W, X, Y, Z\) form the vertices of an orthodiagonal quadrilateral. Applying the Pythagorean theorem to the right triangle \(AWP\) and observing that \(WP = AZ,\) it follows that
\[AP^{2} = AW^{2} + WP^{2} = AW^{2} + AZ^{2}.\]
By a similar argument, the squared lengths of the distances from \(P\) to the other three corners can be calculated as
\[\begin{align} PC^{2} &= WB^{2} + ZD^{2}\\ BP^{2} &= WB^{2} + AZ^{2}\\ PD^{2} &= ZD^{2} + AW^{2}. \end{align}\]
Therefore,
\[\begin{align} AP^{2} + PC^{2} &= \big(AW^{2} + AZ^{2}\big) + \big(WB^{2} + ZD^{2}\big) \\ &= \big(WB^{2} + AZ^{2}\big) + \big(ZD^{2} + AW^{2}\big) \\ &= BP^{2} + PD^{2}.\ _\square \end{align}\]
(Source: MATHCOUNT 2014)
Point \(E\) lies within rectangle \(ABCD\). If \(AE = 7\), \(BE = 5\), and \(CE = 8\), what is \(DE?\)
Let \(DE=x\). By the British flag theorem, we have
\[\begin{align} 7^2+8^2 &= 5^2+x^2\\ x^2&=88 \\ x&=2\sqrt{22}.\ _\square \end{align}\]
For the square \(ABCD\) with side length 4 units, denote \(P\) be a point on its incircle.
Compute \(PA^2+PB^2+PC^2+PD^2\).
Point \(E\) lies within rectangle \(ABCD,\) as shown below.
If the distances from the vertices to point \(E\) are all distinct integers, what is the least possible value of \(AE + BE + CE + DE?\)
