This is a paragraph.

An inequality in cyclic quadrilaterals

In a cyclic quadrilateral ABCD with diagonals AC=p and BD=q and circumradius R, the quantity 8R^2 is an upper bound for the sum of the squares of the diagonals; in other words: \boxed{p^2+q^2\leq 8R^2 .}

It is only when the parent quadrilateral is a rectangle that equality holds.

Easy proofs

Let ABCD be a cyclic quadrilateral with circumradius R and diagonals AC=p, BD=q. PROVE that p\sin A=q\sin B.

Rendered by QuickLaTeX.com

Since triangles ABC and ABD have the same circumradius R we have from the above diagram that p=2R\sin B and q=2R\sin A. Thus p\sin A=q\sin B=2R\sin A\sin B.

Let ABCD be a cyclic quadrilateral with circumradius R and diagonals AC=p, BD=q. PROVE that p\sin C=q\sin D.

Follows from the previous example, since the opposite angles at A and C, as well as at B and D, ensures that \sin A=\sin C and \sin B=\sin D.

Let ABCD be a cyclic quadrilateral with circumradius R and diagonals AC=p, BD=q. PROVE that p^2+q^2=4R^2(\sin^2A+\sin^2B).

We have p=2R\sin B, q=2R\sin A. Thus p^2+q^2=4R^2(\sin^2A+\sin^2B).

Let ABCD be a cyclic quadrilateral with circumradius R and diagonals AC=p, BD=q. Deduce that p^2+q^2\leq 8R^2.

Since p^2+q^2=4R^2(\sin^2A+\sin^2B) from the preceding example, and \sin^2A\leq 1,\sin^2B\leq 1, it follows that p^2+q^2\leq 8R^2.

Exclusive property

For rectangles only.

Let ABCD be a cyclic quadrilateral with circumradius R and diagonals AC=p, BD=q. If p^2+q^2=8R^2, PROVE that ABCD is a rectangle.

We always have p^2+q^2=4R^2(\sin^2A+\sin^2B). If the right side is now 8R^2, then we must have \sin^2A+\sin^2B=2. The latter equation can only be satisfied by taking \sin A=1 and \sin B=1. Similarly we have p^2+q^2=4R^2(\sin^2C+\sin^2D), and so \sin^2C+\sin^2D=2. We take \sin C=\sin D=1. Thus, in such a cyclic quadrilateral we have A=B=C=D=90^{\circ}.

Elusive password

Jump straight to the exercises at this stage — or read what follows for a little bit of distraction. As we promised previously, we’ll be bringing you occasional updates about our ongoing hide and seek — or, tug-of-war — game with some unknown bots. The game is exciting because we currently have the ascendancy.

Scene 1: Barrage by bots



Scene 2: Breathe for a bit

Is any of the above IP addresses is familiar?

Takeaway

Let ABCD be a convex cyclic quadrilateral with circumradius R and diagonals AC=p, BD=q. Then the following statements are equivalent:

  1. ABCD is a rectangle
  2. p^2+q^2=8R^2.

Task

  • (Special case) Let ABCD be a convex cyclic quadrilateral with vertices at A(1,4), B(0,0), C(6,-6), and D(3,6). PROVE that:
    1. diagonal AC bisects diagonal BD
    2. p^2+q^2=4R^2, where AC=p and BD=q
    3. p^2=2(R^2+x^2) and q^2=2(R^2-x^2), where x is the length of the Newton line.