This is a paragraph.

A special cyclic quadrilateral I

The class of cyclic quadrilaterals to be considered in this (and possibly next) post:

  • can be built from any non-isosceles right triangle
  • contain a pair of 90^{\circ} 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 ABC in which \angle C=90^{\circ}. O and N denote the circumcenter and nine-point center.

Explain how to construct a cyclic quadrilateral that exhibits the properties hinted at above.

To see a basic construction of such a cyclic quadrilateral, take any non-isosceles right triangle ABC in which \angle C=90^{\circ}. Denote its circumcenter by O, its nine-point center by N, and its circumradius by R. Consider triangles AOC and BOC with circumcenters O_b and O_a. Let the segment O_aO_b intersect the altitude from C at point X. Then the quadrilateral ONXF_c, with F_c the foot of the altitude from C, is cyclic.

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Suppose that the segment O_aO_b intersects the altitude through C at point X. PROVE that ONXF_c is a cyclic quadrilateral.

Consider the convex quadrilateral ONXF_c in the diagram below:

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The interior angles at N and F_c are both 90^{\circ}, while the interior angles at X and O are 2\angle B and 180^{\circ}-2\angle B.

In the cyclic quadrilateral ONXF_c above, explain why the parent right triangle ABC should be non-isosceles.

If the parent triangle is isosceles, the altitude CF_c coincides with the median CO, in which case ONXF_c is just a line.

In the cyclic quadrilateral ONXF_c above, PROVE that one side and one diagonal are equal.

Indeed, ON=NF_c, and both are equal to R/2, where R is the circumradius of the parent triangle ABC.

Let O_b and O_a denote the circumcenters of triangles AOC and BOC. PROVE that O_b,N,O_a are co-linear.

Since \angle C=90^{\circ} as stated, the nine-point center N is the midpoint of segment CO. Triangles AOC and BOC both contain the segment CO. Thus O_b,N,O_a are co-linear because they all lie on the right bisector of CO.


Let ABC be an acute triangle with circumcenter O and circumradius R. Further, let D_a and D_b be the diameters of the circumcircles of BOC and AOC. Then the following statements are equivalent:

  1. ABC is a right triangle in which \angle C=90^{\circ}
  2. the identity \frac{1}{R^2}=\frac{1}{D_a^2}+\frac{1}{D_b^2} holds.

In case the parent triangle ABC is obtuse, we have the following analogous equivalence:

  1. the side-lengths a,b,c satisfy (b^2-a^2)^2=(ac)^2+(cb)^2
  2. the identity \frac{1}{R^2}=\frac{1}{D_a^2}+\frac{1}{D_b^2} holds.


  • (Inverted squares) Let ABC be a right triangle in which \angle C=90^{\circ}, the circumcenter is O, and the circumradius is R. PROVE that:
    1. the diameters D_b and D_a of the circles AOC and BOC are related to R via an inverted Pythagorean identity:

      (1)   \begin{equation*} \frac{1}{D_a^2}+\frac{1}{D_b^2}=\frac{1}{R^2} \end{equation*}

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      (The relation in (1) also holds when the parent right triangle is isosceles, but the quadrilateral ONXF_c degenerates in that case. More on this later.)

  • (Impossible situation) Let ABC be a right triangle in which \angle C=90^{\circ}, and let O_a and O_b be as in the discussion.
    1. Find a condition under which a circle centred at O_b will be externally tangent to the circle BOC as in the diagram

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    2. PROVE that if the circumcircles of AOC and BOC touch externally, then c=a+b.
    3. Deduce that it is impossible for the circumcircles of AOC and BOC to touch externally.