Krdytkyu

"Japanese theorem" is the name given to the following result (see Fig. 1 below) : if $A_1A_2A_3A_4$ is a cyclic quadrilateral, the incenters of the $4$ triangles $A_pA_qA_r$ with all triples ${p,q,r}$ are the vertices of a rectangle $R$.

Moreover, the medians of rectangle $R$ intersect circular arcs $A_{k}A_{k+1}$ (with $A_5:=A_1$) in their midpoints.

In fact, there is a second level of properties that is seldom given (see references and comments at the end of the text):

Let $frak{E}$ denote the set of the $4 times 3 = 12$ excenters of all triangles $A_pA_qA_r$:

• The convex hull of $frak{E}$ is a rectangle $R'$ with sides parallel to sides of rectangle $R$.


• The 8 other points of $frak{E}$ are situated on the sides of rectangle $R'$ at the intersection of the extended sides of rectangle $R$.

Fig. 1 : Rectangles R (blue vertices) and R' (red vertices). Midpoints of arcs $A_kA_{k+1}$ in black.

Fig. 2 enriches Fig. 1 with the four circles associated with triangle $A_1A_2A_3$ (its incircle and its three excircles) [Drawing all the incircles and excircles would give a too messy picture].

Fig. 2 : Incircle and the three excircles of triangle $A_1A_2A_3$.

Some supplementary properties exist. I will not give them here.

My questions dealing with this second part of the japanese theorem :

• is there is a simple proof of it ? The proof I have done is rather complicated. I have the feeling that there must be more direct proofs.




• are there known consequences of this rather surprizing result ?

References and comments :

• "The Penguin Dictionary of Curious and Interesting Geometry" by David Wells (Penguin, 1991) p. 43, a very interesting book, full of ill-known results:

(https://archive.org/details/ThePenguinDictionaryOfCuriousAndInterestingGeometry).

• The Wolfram corresponding article.




• Strangely, neither Wikipedia nor Cut-the-knot sites on "Japanese theorem" mention the second part.




• See also this site.




• Similar properties exist for tangential quadrilaterals. See for example (http://forumgeom.fau.edu/FG2011volume11/FG201108.pdf) cited in (https://en.wikipedia.org/wiki/Tangential_quadrilateral).

Not an answer, but an additional figure, showing cyclic quadrilateral $square ABCD$ and the various angle bisectors (and perpendiculars thereto) that define the incenters and excenters.

The incenter of $triangle BCD$ is labeled "$A^prime$", because $A$ is the excluded vertex of the quadrilateral. Likewise, the excenters of $triangle BCD$ feature "$A$", with subscripts indicating the vertex opposite the excenter in that triangle.

There are a few hard-to-see concurrences along the circle. For instance,

• $overline{AD_A} cap overline{DA_D}$ is the midpoint of $stackrel{frown}{BC}$. This is because each segment bisects an inscribed angle subtending that arc.


• $overline{A_BA_C}capoverline{D_BD_C}$ is the point diametrically opposite $overline{AD_A}capoverline{DA_D}$.

Here's another diagram, that arose from investigating an inverse:

Starting with rectangle $square WXYZ$, we choose points $P$, $Q$, $R$, $S$ and $P^prime$, $Q^prime$, $R^prime$, $S^prime$ (with primed and un-primed pairs being images under a symmetric reflection). The condition under which $overline{PQ^prime}$, $overline{R^prime Z}$, $overline{SX}$
to concur (at $A$) forces the corresponding concurrences at $B$, $C$, $D$; moreover, it forces $square ABCD$ to be cyclic.

Interestingly, there's no requirement that the "inner" rectangle lie inside the "outer" rectangle. A cyclic $square ABCD$ occurs no matter where $P$, $Q$, $R$, $S$ lie on the outer rectangle's edge-lines.

However, $square WXYZ$ is not the rectangle associated with $square ABCD$ by the second part of the Japanese Theorem; also, the "inner" rectangle isn't the rectangle of incenters from the first part. Those rectangles are quite distinct (the outer one is labeled $square W^prime X^prime Y^prime Z^prime$), although their edges are parallel to those of $square WXYZ$. (The "inner" rectangles do not necessarily share a center.)

Investigation continues ...

Here's a bit of a rebooted discussion of the "inverse" situation. (Throughout, I'll conveniently ignore various degeneracies, especially those that cause denominators to vanish.)

Let $square PQRS$ and $square P^prime Q^prime R^prime S^prime$ be "parallel" rectangles, with $P$ and $P^prime$ being "opposite" vertices as indicated in the figure. Also, let $P^prime$ project to points $P^{primeprime}$ and $P^{primeprimeprime}$ on the sides of $square PQRS$ as shown.

It's not hard to prove with coordinates that lines $overleftrightarrow{PP^prime}$, $overleftrightarrow{S^{primeprime}Q^prime}$, $overleftrightarrow{R^{primeprimeprime}S^prime}$ concur at a point, $P_circ$. (One can prove the concurrence using the trigonometric form of Ceva's Theorem in $triangle P^prime S^{primeprime} R^{primeprimeprime}$, but the details are a little messy.) Treating points as position vectors, one can write specifically that

$$P_circ = frac{P^prime;|square PR^prime| - P;|square P^prime R^prime|}{|square PR^prime| - |square P^prime R^prime| }$$
where $|square XY|$ denotes the area of the rectangle with diagonal $overline{XY}$. From this form, we deduce
$$frac{|overline{PP_circ}|}{|overline{PP^prime}|} = frac{|square PR^prime|}{|square PR^prime|-|square P^prime R^prime|} qquad
frac{|overline{P^prime P_circ}|}{|overline{PP^prime}|} = frac{|square P^prime R^prime|}{|square PR^prime|-|square P^prime R^prime|} qquad frac{|overline{P^prime P_circ}|}{|overline{PP_circ}|} = frac{|square P^prime R^prime|}{|square PR^prime|}$$

Likewise, we can get points-of-concurrency $Q_circ$, $R_circ$, $S_circ$. In general, these points are not concyclic. They are concyclic if $square P^prime Q^prime R^prime S^prime$ is centered horizontally, and/or vertically, with respect to $square PQRS$, or, more interestingly, if

$$|square PR^prime|,|square P^prime R| = |square PQRS|,|square P^prime Q^prime R^prime S^prime| = |square QS^prime|,|square Q^prime S| tag{$star$}$$
That is, the product of the areas of the inner and outer rectangles is equal to the product of the areas of either pair of "diagonal" rectangles.

(I don't have a clear geometric proof of this fact; I let Mathematica crunch through a coordinate argument.)

Intriguingly, condition $(star)$ has a very simple Ceva-like representation. If we define $p$, $q$, $r$, $s$ such that
$$overrightarrow{PP^{primeprime}} = p;overrightarrow{P^{primeprime}Q} qquad
overrightarrow{QQ^{primeprime}} = q;overrightarrow{Q^{primeprime}R} qquad overrightarrow{RR^{primeprime}} = r;overrightarrow{R^{primeprime}S} qquad overrightarrow{SS^{primeprime}} = s;overrightarrow{S^{primeprime}P}$$
then $(star)$ is equivalent to

$$p r + q s = pqrs tag{$starstar$}$$

(Typically, the Ceva ratio would be the reciprocal of the one I've written. Writing $overline{p}$ (etc) for that reciprocal, relation $(star)$ would be even simpler: $overline{p}overline{q}+overline{r}overline{s} = 1$. The relations below would be a little messier, however.)

Condition $(star)$ also implies the concurrence of $overleftrightarrow{QP^{primeprimeprime}}$, $overleftrightarrow{S^{primeprimeprime}R^{primeprime}}$, $overleftrightarrow{Q^{primeprime}S}$, with the common point being $P_circ$. (Importantly, mere horizontal/vertical centering of the rectangles does not guarantee this concurrence, so this property is not equivalent to $square P_circ Q_circ R_circ S_circ$ being cyclic.) This is how we get the configurations shown in the earlier part of this answer.

It's worth noting a few other relations involving the Ceva-like ratios. For instance,
$$frac{|square PQRS|}{|square PR^prime|} = (1+r)(1+s) qquad
frac{|square PQRS|}{|square P^prime R|} = (1+p)(1+q) qquad cdots$$
$$frac{|square PQRS|}{|square P^prime Q^prime R^prime S^prime|} = frac{(1+p)(1+q)(1+r)(1+s)}{(1-pr)(1-qs)} ;stackrel{(starstar)}{=}; (1+p)(1+q)(1+r)(1+s)$$
$$frac{|overline{P^prime P_circ}|}{|overline{PP_circ}|} = frac{(1-pr)(1-qs)}{(1+p)(1+q)};stackrel{(starstar)}{=}; frac{1}{(1+p)(1+q)}$$

Of course, no nowhere in here have we said anything about incenters or excenters or angle bisectors. We can see that the (extended) Japanese Theorem configuration is a special case of a $(starstar)$ configuration as follows:

Given cyclic $square P_circ Q_circ R_circ S_circ$, define
$$alpha := frac{1}{2}angle Q_circ P_circ R_circ quad
beta := frac{1}{2}angle R_circ Q_circ S_circ quad
gamma := frac{1}{2}angle S_circ R_circ P_circ quad
delta := frac{1}{2}angle P_circ S_circ Q_circ$$

(Note: $alpha+beta+gamma+delta = 90^circ$.) Let $P$, $Q$, $P^prime$ be excenters, opposite $P_circ$, $Q_circ$, $Q_circ$, for respective triangles $triangle P_circ Q_circ S_circ$, $triangle Q_circ P_circ R_circ$, $triangle Q_circ R_circ S_circ$. With $k$ the radius of the circle, one can calculate

$$|overline{PP^prime}| = 4 k sin(alpha+beta) sin(alpha+delta) qquad |overline{P^prime Q}| = 4 k cosalpha singamma$$
and likewise for various other excenters, so that
$$begin{align}
frac{1}{p}frac{1}{r}+frac{1}{q}frac{1}{s} &= frac{cosalphasingamma cdot cosgamma sinalpha + cosbetasindeltacdot cosdelta sinbeta}{sin(alpha+beta)sin(beta+gamma)sin(gamma+delta)sin(delta+alpha)} \[4pt]
&=frac{frac{1}{4}left(sin 2alpha sin 2gamma + sin 2beta sin 2deltaright)}{frac{1}{4}left(sin 2alpha sin 2gamma + sin 2beta sin 2deltaright)} \[4pt]
&=1
end{align}$$
satisfying $(starstar)$. (How $alpha$, $beta$, $gamma$, $delta$ relate to $p$, $q$, $r$, $s$ when $P$, $Q$, $R$, $S$, etc, are not excenters is not (yet) known.)