This is a paragraph.

Quasi harmonic quadrilaterals I

Harmonic quadrilaterals are quadrilaterals that can be inscribed in a circle, with the additional property that the products of opposite sides are equal.

However, instead of requiring that the products of opposite sides are equal, we can stipulate that the products of consecutive sides are equal, and then call the resulting cyclic quadrilateral quasi harmonic.

Having this latter requirement (e.g. ab=cd in the diagram below) yields some nice lateral properties for quadrilaterals that satisfy them.

Rendered by QuickLaTeX.com

Rectangles.

Any rectangle can be inscribed in a circle, and the products of consecutive sides are equal (e.g. AB\times BC=CD\times DA below). Thus, rectangles are quasi-harmonic as per our definition. Note that a rectangle is not a harmonic quadrilateral, unless the rectangle is square.

Rendered by QuickLaTeX.com

Right kites and squares.

Right kites are both harmonic and quasi-harmonic. So are squares.

Rendered by QuickLaTeX.com

Consider quadrilateral ABCD with vertices at A(-7,7), B(-6,0), C(0,0), and D(2,4). Verify that this is a quasi-harmonic quadrilateral.

We need to show that it’s cyclic, and that the products of the lengths of consecutive sides are equal.

Rendered by QuickLaTeX.com

  • side-lengths: a=5\sqrt{2}, b=6, c=2\sqrt{5}, d=3\sqrt{10}
  • diagonals: AC=7\sqrt{2}, BD=4\sqrt{5}
  • Ptolemy’s theorem: AC\times BD=28\sqrt{10}, and ac+bd=10\sqrt{10}+18\sqrt{10}=28\sqrt{10}. Confirms ABCD is cyclic.
  • product of consecutive sides: ab=30\sqrt{2}=cd. Quasi harmonic.

Consider quadrilateral ABCD with vertices at A(1,4), B(0,0), C(6,-6), and D(3,6). Verify that this is a quasi-harmonic quadrilateral.

Rendered by QuickLaTeX.com

  • side-lengths: a=\sqrt{17}, b=6\sqrt{2}, c=3\sqrt{17}, d=2\sqrt{2}
  • diagonals: AC=5\sqrt{5}, BD=3\sqrt{5}
  • Ptolemy’s theorem: AC\times BD=75, and ac+bd=51+24=75. Confirms ABCD is cyclic.
  • product of consecutive sides: ab=6\sqrt{34}=cd. Quasi harmonic.

Consider quadrilateral ABCD with vertices at A\left(-\frac{12}{5},\frac{6}{5}\right), B(-3,6), C(1,8), and D(0,0). Verify that this is a quasi-harmonic quadrilateral.

Rendered by QuickLaTeX.com

  • side-lengths: AB=\frac{3}{5}\sqrt{65}, BC=2\sqrt{5}, CD=\sqrt{65}, DA=\frac{6}{5}\sqrt{5}
  • diagonals: AC=\frac{17}{5}\sqrt{5}, BD=3\sqrt{5}
  • Ptolemy’s theorem: AC\times BD=51, and AB\times CD+BC\times DA=39+12=51. Confirms ABCD is cyclic.
  • product of consecutive sides: AB\times BC=6\sqrt{13}=CD\times DA. Quasi harmonic.

Takeaway

Let ABCD be a convex cyclic quadrilateral with side-lengths AB=a, BC=b, CD=c, and DA=d. Then the following statements are equivalent:

  1. ab=cd
  2. diagonal AC bisects diagonal BD.

The longer diagonal in a quasi harmonic quadrilateral always bisects the shorter diagonal.

Task

  • (Bisection condition) Let ABCD be a convex cyclic quadrilateral with side-lengths AB=a, BC=b, CD=c, and DA=d. PROVE that:
    1. diagonal AC bisects diagonal BD if and only if ab=cd
    2. diagonal BD bisects diagonal AC if and only if ad=bc
    3. diagonals AC and BD bisect each other if and only if a=c and b=d (rectangle).
  • (Basic characteristics) Let ABCD be a convex cyclic quadrilateral in which the side-lengths are AB=a, BC=b, CD=c, DA=d, and the diagonals are AC=p, BD=q. If ab=cd (quasi-harmonic), PROVE that:
    1. 2p^2=a^2+b^2+c^2+d^2
    2. q^2=\frac{2(a^2+d^2)(b^2+c^2)}{a^2+b^2+c^2+d^2}.

Nine-point center equals a vertex III

Earlier we saw that the nine-point center of a triangle coincides with one of the vertices of the triangle precisely when the parent triangle is isosceles with an apex angle of 120^{\circ}.

Equivalently, two radii are parallel to two sides of the parent triangle, as will be shown in examples 1 and 2 below.

Everything we’ve been considering so far seems to work smoothly for the 30^{\circ}-30^{\circ}-120^{\circ}-triangle.

In \triangle ABC, let \angle A=\angle C=30^{\circ} and \angle B=120^{\circ}. PROVE that the radius through C is parallel to side AB and the radius through A is parallel to side BC.

Since the parent triangle is obtuse, its circumcenter O lies outside, opposite the obtuse angle. Thus OABC is a (convex) quadrilateral.

Rendered by QuickLaTeX.com

If R is the radius of the circumcircle of triangle ABC, then we have that R=a=BC and R=c=AB (see here). Since O is the circumcenter, we also have that OA=OC=R. Hence, OABC is a rhombus; in fact, a special rhombus with the property that the diagonal OB is equal to the equal sides. Every rhombus is a parallelogram so we conclude that radius OC is parallel to side AB and radius OA is parallel to side BC.

In \triangle ABC, suppose that the radius through C is parallel to side AB, and the radius through A is parallel to side BC. PROVE that \angle A=\angle C=30^{\circ} and \angle B=120^{\circ}.

Suppose that a triangle ABC with circumcenter O has the property that radius OA is parallel to side BC and radius OC is parallel to side AB. Such a triangle is necessarily obtuse, so that the circumcenter O lies outside the triangle and we again have a quadrilateral OABC. Join AC and let \angle CAB=\alpha^{\circ}. Since OC is parallel to AB, we have that \angle OCA=\alpha^{\circ}. Since O is the circumcenter, triangle AOC is isosceles and so \angle OAC=\angle OCA=\alpha^{\circ}. The situation is shown below, with the circumcircle included:

Rendered by QuickLaTeX.com

Since OA is parallel to CB, we have that \angle ACB=\angle OAC=\alpha^{\circ}. Thus, triangle ABC is isosceles; moreover, \angle ABC=180^{\circ}-2\alpha^{\circ}. By the inscribed angle theorem:

    \[180^{\circ}+2\alpha^{\circ}=2(180^{\circ}-2\alpha^{\circ})\implies \alpha=30\]

Thus, the interior angles of triangle ABC are \angle A=\angle C=30^{\circ} and \angle B=120^{\circ}.

Let a,b,c be the side-lengths of triangle ABC, whose area K satisfies K=\frac{1}{2}ac\cos A=\frac{1}{2}ac\cos C. PROVE that triangle ABC is the 30^{\circ}-30^{\circ}-120^{\circ} isosceles triangle.

Since the area satisfies \frac{1}{2}ac\cos A=\frac{1}{2}ac\cos C, we have that A=C.

Normally, K=\frac{1}{2}ac\sin B. And so we must have \sin B=\cos A. In turn, this implies

    \[B=90^{\circ}-A\quad\text{or}\quad 180^{\circ}-B= 90^{\circ}-A\]

B=90^{\circ}-A is not possible due to the fact that A=C and A+B+C=180^{\circ}. So we take 180^{\circ}-B= 90^{\circ}-A, which yields B=90^{\circ}+A. Using A+B+C=180^{\circ} we get

    \[A+(90^{\circ}+A)+A=180^{\circ}\implies A=30^{\circ}=C\]

Consequently, B=120^{\circ}.

Let O be the circumcenter of triangle ABC. If the convex quadrilateral OABC is a rhombus, PROVE that triangle ABC is the 30^{\circ}-30^{\circ}-120^{\circ} isosceles triangle.

No matter how the vertices are arranged, we’ll end up with a rhombus in which a diagonal is equal to the side-lengths. The interior angles of such a rhombus are 60^{\circ},60^{\circ},120^{\circ},120^{\circ}. Now just consider the triangle that remains when O is removed.

Let O be the circumcenter of triangle ABC. If the convex quadrilateral OABC is a rhombus, PROVE that the rhombus cannot be a square.

Consider various arrangements of the rhombus OABC, for example the one shown below:

Rendered by QuickLaTeX.com

Apply the inscribed angle theorem to see that it is not possible to have OABC be a square.

Takeaway

In any triangle ABC, let a,b,c be the side-lengths, H the orthocenter, O the circumcenter, and R the circumradius. Then the following statements are equivalent:

  1. a=c=R
  2. \angle A=\angle C=30^{\circ}
  3. the reflection of B over AC is O
  4. the area K satisfies K=\frac{1}{2}ac\cos A=\frac{1}{2}ac\cos C
  5. the reflection of A over BC is H and the reflection of C over AB is H

Task

  • (Aufbau) In triangle ABC, let a,b,c be the side-lengths, R the circumradius, O the circumcenter, N the nine-point center, and H the orthocenter. PROVE that the following statements are equivalent:
    1. B=N
    2. a=c=R
    3. \angle A=\angle C=30^{\circ}
    4. the reflection of B over AC is O
    5. the area K satisfies K=\frac{1}{2}ac\cos A=\frac{1}{2}ac\cos C
    6. the circle with diameter OH passes through vertices A and C
    7. radius OA is parallel to side CB and radius OC is parallel to side AB
    8. the reflection of A over BC is H and the reflection of C over AB is H
    9. AF_a is the geometric mean of BF_a and CF_a, and CF_c is the geometric mean of AF_c and BF_c. (F_a and F_c are the feet of the altitudes from A and C, respectively.)