- can be built from any non-isosceles right triangle
- contain a pair of
angles facing each other
- combine some properties of right kites with some properties of quasi-harmonic quadrilaterals.
Thus, they are special.
Partial proofs
Throughout we work with a non-isosceles right triangle in which
.
and
denote the circumcenter and nine-point center.
To see a basic construction of such a cyclic quadrilateral, take any non-isosceles right triangle in which
. Denote its circumcenter by
, its nine-point center by
, and its circumradius by
. Consider triangles
and
with circumcenters
and
. Let the segment
intersect the altitude from
at point
. Then the quadrilateral
, with
the foot of the altitude from
, is cyclic.
![Rendered by QuickLaTeX.com O_aO_b](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-6ac097388b6a909b3855c296dcc9bd1c_l3.png)
![Rendered by QuickLaTeX.com C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3238eb6669792857eccb924606e71d82_l3.png)
![Rendered by QuickLaTeX.com X](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-e53cc20b242aade793dbfe906ce70ee9_l3.png)
![Rendered by QuickLaTeX.com ONXF_c](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-dcc8f608b388301404302e935cd0e869_l3.png)
Consider the convex quadrilateral in the diagram below:
The interior angles at and
are both
, while the interior angles at
and
are
and
.
![Rendered by QuickLaTeX.com ONXF_c](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-dcc8f608b388301404302e935cd0e869_l3.png)
![Rendered by QuickLaTeX.com ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b5002f05fd90c80f81a9e0e9c845e02d_l3.png)
If the parent triangle is isosceles, the altitude coincides with the median
, in which case
is just a line.
![Rendered by QuickLaTeX.com ONXF_c](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-dcc8f608b388301404302e935cd0e869_l3.png)
Indeed, , and both are equal to
, where
is the circumradius of the parent triangle
.
![Rendered by QuickLaTeX.com O_b](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-e0cae4924a3b69bd1f44c7b6c0e51e28_l3.png)
![Rendered by QuickLaTeX.com O_a](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-081c1cefedcaab078975aa243526f793_l3.png)
![Rendered by QuickLaTeX.com AOC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-12d0a34023638d2bcdb04c5eef9f9f21_l3.png)
![Rendered by QuickLaTeX.com BOC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-afd9712c38ccd20b5761d4af1aa69f68_l3.png)
![Rendered by QuickLaTeX.com O_b,N,O_a](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-4cc49c03ca75d35c4d012bf80ffc2474_l3.png)
Since as stated, the nine-point center
is the midpoint of segment
. Triangles
and
both contain the segment
. Thus
are co-linear because they all lie on the right bisector of
.
Takeaway
Let be an acute triangle with circumcenter
and circumradius
. Further, let
and
be the diameters of the circumcircles of
and
. Then the following statements are equivalent:
is a right triangle in which
- the identity
holds.
In case the parent triangle is obtuse, we have the following analogous equivalence:
- the side-lengths
satisfy
- the identity
holds.
Task
- (Inverted squares) Let
be a right triangle in which
, the circumcenter is
, and the circumradius is
. PROVE that:
- the diameters
and
of the circles
and
are related to
via an inverted Pythagorean identity:
(1)
(The relation in (1) also holds when the parent right triangle is isosceles, but the quadrilateral
degenerates in that case. More on this later.)
- the diameters
- (Impossible situation) Let
be a right triangle in which
, and let
and
be as in the discussion.
- Find a condition under which a circle centred at
will be externally tangent to the circle
as in the diagram
- PROVE that if the circumcircles of
and
touch externally, then
.
- Deduce that it is impossible for the circumcircles of
and
to touch externally.
- Find a condition under which a circle centred at