![Rendered by QuickLaTeX.com ABCD](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-563addbbd3bac1a4b850c391dabc7878_l3.png)
![Rendered by QuickLaTeX.com AC=p](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-1b58a17267a80bbe3d12cbbb51a788f6_l3.png)
![Rendered by QuickLaTeX.com BD=q](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-fbe5c6a35a4bfb78c3d80e634cae961c_l3.png)
![Rendered by QuickLaTeX.com R](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-5cdb5b77a27c44c1d184bb37f903996e_l3.png)
![Rendered by QuickLaTeX.com 8R^2](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-1137934820793015f8243665154ffb20_l3.png)
![Rendered by QuickLaTeX.com \boxed{p^2+q^2\leq 8R^2 .}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-6719211c7fbe6157c23de7cae983be86_l3.png)
It is only when the parent quadrilateral is a rectangle that equality holds.
Easy proofs
![Rendered by QuickLaTeX.com ABCD](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-563addbbd3bac1a4b850c391dabc7878_l3.png)
![Rendered by QuickLaTeX.com R](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-5cdb5b77a27c44c1d184bb37f903996e_l3.png)
![Rendered by QuickLaTeX.com AC=p](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-1b58a17267a80bbe3d12cbbb51a788f6_l3.png)
![Rendered by QuickLaTeX.com BD=q](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-fbe5c6a35a4bfb78c3d80e634cae961c_l3.png)
![Rendered by QuickLaTeX.com p\sin A=q\sin B](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-7385a9a95d1561ebc92d9a426aed1763_l3.png)
Since triangles and
have the same circumradius
we have from the above diagram that
and
. Thus
.
![Rendered by QuickLaTeX.com ABCD](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-563addbbd3bac1a4b850c391dabc7878_l3.png)
![Rendered by QuickLaTeX.com R](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-5cdb5b77a27c44c1d184bb37f903996e_l3.png)
![Rendered by QuickLaTeX.com AC=p](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-1b58a17267a80bbe3d12cbbb51a788f6_l3.png)
![Rendered by QuickLaTeX.com BD=q](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-fbe5c6a35a4bfb78c3d80e634cae961c_l3.png)
![Rendered by QuickLaTeX.com p\sin C=q\sin D](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-1d83716b4c5796a9a506bf5571ab283a_l3.png)
Follows from the previous example, since the opposite angles at and
, as well as at
and
, ensures that
and
.
![Rendered by QuickLaTeX.com ABCD](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-563addbbd3bac1a4b850c391dabc7878_l3.png)
![Rendered by QuickLaTeX.com R](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-5cdb5b77a27c44c1d184bb37f903996e_l3.png)
![Rendered by QuickLaTeX.com AC=p](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-1b58a17267a80bbe3d12cbbb51a788f6_l3.png)
![Rendered by QuickLaTeX.com BD=q](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-fbe5c6a35a4bfb78c3d80e634cae961c_l3.png)
![Rendered by QuickLaTeX.com p^2+q^2=4R^2(\sin^2A+\sin^2B)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-658225c615e84168fa9799c2e2bd75ba_l3.png)
We have ,
. Thus
.
![Rendered by QuickLaTeX.com ABCD](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-563addbbd3bac1a4b850c391dabc7878_l3.png)
![Rendered by QuickLaTeX.com R](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-5cdb5b77a27c44c1d184bb37f903996e_l3.png)
![Rendered by QuickLaTeX.com AC=p](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-1b58a17267a80bbe3d12cbbb51a788f6_l3.png)
![Rendered by QuickLaTeX.com BD=q](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-fbe5c6a35a4bfb78c3d80e634cae961c_l3.png)
![Rendered by QuickLaTeX.com p^2+q^2\leq 8R^2](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-ab1728581c47387e455b6310b707aa58_l3.png)
Since from the preceding example, and
, it follows that
.
Exclusive property
For rectangles only.
![Rendered by QuickLaTeX.com ABCD](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-563addbbd3bac1a4b850c391dabc7878_l3.png)
![Rendered by QuickLaTeX.com R](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-5cdb5b77a27c44c1d184bb37f903996e_l3.png)
![Rendered by QuickLaTeX.com AC=p](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-1b58a17267a80bbe3d12cbbb51a788f6_l3.png)
![Rendered by QuickLaTeX.com BD=q](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-fbe5c6a35a4bfb78c3d80e634cae961c_l3.png)
![Rendered by QuickLaTeX.com p^2+q^2=8R^2](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-80accca7c528832df1877122c6443350_l3.png)
![Rendered by QuickLaTeX.com ABCD](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-563addbbd3bac1a4b850c391dabc7878_l3.png)
We always have . If the right side is now
, then we must have
. The latter equation can only be satisfied by taking
and
. Similarly we have
, and so
. We take
. Thus, in such a cyclic quadrilateral we have
.
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 familiar?
Takeaway
Let be a convex cyclic quadrilateral with circumradius
and diagonals
,
. Then the following statements are equivalent:
is a rectangle
.
Task
- (Special case) Let
be a convex cyclic quadrilateral with vertices at
,
,
, and
. PROVE that:
- diagonal
bisects diagonal
, where
and
and
, where
is the length of the Newton line.
- diagonal