Slopefessor Fr1day – Page 4 – High School Geometry

Following our previous discussion, we now consider quadrilaterals:

where a pair of interior opposite angles are both $90^{\circ}$
with one side and one diagonal having equal lengths.

For today’s purpose, let’s give the pseudonym special to cyclic quadrilaterals having the above two properties, and then prove that in any such special quadrilateral:

there’s a side whose length is twice the distance between the midpoints of the diagonals
the diagonals $p,q$ and the distance $x$ between the midpoints of the diagonals are related via
(1) $\begin{equation*}p^2=q^2+4x^2\end{equation*}$

Required knowledge

Euler’s quadrilateral formula is the main background needed in order to follow today’s discussion. It’s basically the equation

(2) $\begin{equation*}a^2+b^2+c^2+d^2=p^2+q^2+4x^2 \end{equation*}$

connecting the side-lengths $a,b,c,d$ of a quadrilateral with the diagonals $p,q$ and the distance $x$ between the midpoints of the diagonals.

Right kites

In a right kite, equation (2) simplifies to equation (1), that is $p^2=q^2+4x^2$ , as shown in our first example below.

Let $ABCD$

be a right kite with longer diagonal of length $p$

and shorter diagonal of length $q$

. If $x$

is the distance between the midpoints of the diagonals, PROVE that $\left(\frac{p}{2}\right)^2=\left(\frac{q}{2}\right)^2+x^2$

Consider the right kite shown below, with longer diagonal $AC=p$ , and shorter diagonal $BD=q$ .

Since $d=a, c=b$ in the kite shown above and $p^2=a^2+b^2$ because it’s a right kite, Euler’s quadrilateral formula (2) gives

$\begin{equation*} \begin{split} a^2+b^2+b^2+a^2&=p^2+q^2+4x^2\\ 2p^2&=p^2+q^2+4x^2\\ \therefore p^2&=q^2+4x^2 \end{split} \end{equation*}$

Let a cyclic quadrilateral $ABCD$

with side-lengths $a,b,c,d$

be special, in the sense of the two properties at the introduction. PROVE that $p^2=q^2+4x^2$

This example shows that apart from right kites, there are other (cyclic) quadrilaterals that satisfy equation (1), including the ones with the pseudonym special, which is our focus for today.

Such a special quadrilateral $ABCD$ contains two interior opposite angles measuring $90^{\circ}$ ; place them at $B$ and $D$ .

If we let diagonal $AC=p$ , then $p^2=a^2+b^2=c^2+d^2$ , and so by Euler’s quadrilateral formula (2):

$2p^2=p^2+q^2+4x^2\implies p^2=q^2+4x^2$

Let a cyclic quadrilateral $ABCD$

with side-lengths $a,b,c,d$

be special, in the sense of the two properties at the introduction. PROVE that there is a side whose length is twice the distance between the midpoints of the diagonals.

Using the same diagram/notation for the previous example, we have:

$c^2+d^2=p^2=q^2+4x^2$

Since one of the diagonals has the same length as one of the sides, it can’t be diagonal $AC$ because $AC$ is the diameter. So let’s suppose diagonal $BD=q$ as the same length as side $DA=d$ . Then the above equation gives $c=2x$ .

Rare kind

Consider a right kite $ABCD$

in which the interior angles at $A$

and $C$

are $120^{\circ}$

and $60^{\circ}$

. Show that $ABCD$

is special in today’s terms.

Diagonal $BD$ will now have the same length as sides $BC$ and $CD$ .

Show that the right kite in example 4 can be obtained from any $30^{\circ}-30^{\circ}-120^{\circ}$

isosceles triangle $ABC$

Suppose that $\angle B=120^{\circ}$ . Let $O$ and $H$ be the circumcenter and orthocenter of the parent triangle $ABC$ . Then the quadrilateral $OAHC$ is a particularly pleasant right kite.

Takeaway

Let $ABCD$ be a quadrilateral with side-lengths $a,b,c,d$ , diagonals $AC=p,BD=q$ , and $x$ the distance between the midpoints of the diagonals. If the interior angles at $B$ and $D$ are both $90^{\circ}$ , then the following statements are equivalent:

$c=2x$
$q=d$ .

Task

(Special property) Let be special, as per the definition at the introduction; specifically, let the interior opposite angles at and be both , and let diagonal have the same length as side . PROVE that:
1. the radius through $D$ is parallel to side $BC$
2. the orthocenter of triangle $BCD$ is a reflection of vertex $D$ over side $BC$ .

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.

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:

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 $CO$ .

Takeaway

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:

$ABC$ is a right triangle in which $\angle C=90^{\circ}$
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:

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

Task

(Inverted squares) Let be a right triangle in which , the circumcenter is , and the circumradius is . 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) $\begin{equation*} \frac{1}{D_a^2}+\frac{1}{D_b^2}=\frac{1}{R^2} \end{equation*}$
  
  (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 be a right triangle in which , and let and 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
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.

Author: Slopefessor Fr1day

A special cyclic quadrilateral II

Required knowledge

Right kites

Rare kind

Takeaway

Task

A special cyclic quadrilateral I

Partial proofs

Takeaway

Task