# Japanese theorem, in its extended form

"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 ?

• "The Penguin Dictionary of Curious and Interesting Geometry" by David Wells (Penguin, 1991) p. 43, a very interesting book, full of ill-known results:
• That's fascinating, all that! I'll definitely be saving this post! – AmbretteOrrisey Nov 22 '18 at 10:19
• Thanks. I am fascinated too by this harmony. More generally, I like to find, whenever possible, graphical representations of issues, even if it's not of a geometrical nature a priori. See for example the answer I gave recently : math.stackexchange.com/q/3006728 – Jean Marie Nov 22 '18 at 10:30
• I've been exploring that idea recently also: what I put about the solution of r"=r^(2n-1) being solved by r being the distance-from-origin of a point moving with constant speed along a generalised lemniscate r=(cos(nθ))^(1/n); and also the explanation of the integral of sec & the integral of sech being mutual inverses. – AmbretteOrrisey Nov 22 '18 at 10:44
• Any pointers about these generalised lemniscates ? – Jean Marie Nov 22 '18 at 10:56
• Yes! it's my post - the hektic oscillator one - I modified it only half an hour or so ago, so it won't be far down the list as well. Part of the exposition is an answer to my own question that I posted & modified as the answer(s) came to me incrementally. – AmbretteOrrisey Nov 22 '18 at 11:00

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

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 $P^{\prime\prime}$ and $P^{\prime\prime\prime}$ on the sides of $\square PQRS$ as shown.

It's not hard to prove with coordinates that lines $\overleftrightarrow{PP^\prime}$, $\overleftrightarrow{S^{\prime\prime}Q^\prime}$, $\overleftrightarrow{R^{\prime\prime\prime}S^\prime}$ concur at a point, $P_\circ$. (One can prove the concurrence using the trigonometric form of Ceva's Theorem in $\triangle P^\prime S^{\prime\prime} R^{\prime\prime\prime}$, 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 $$\overrightarrow{PP^{\prime\prime}} = p\;\overrightarrow{P^{\prime\prime}Q} \qquad \overrightarrow{QQ^{\prime\prime}} = q\;\overrightarrow{Q^{\prime\prime}R} \qquad \overrightarrow{RR^{\prime\prime}} = r\;\overrightarrow{R^{\prime\prime}S} \qquad \overrightarrow{SS^{\prime\prime}} = s\;\overrightarrow{S^{\prime\prime}P}$$ then $(\star)$ is equivalent to

$$p r + q s = pqrs \tag{\star\star}$$

(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 $\overleftrightarrow{QP^{\prime\prime\prime}}$, $\overleftrightarrow{S^{\prime\prime\prime}R^{\prime\prime}}$, $\overleftrightarrow{Q^{\prime\prime}S}$, 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, $$\frac{|\square PQRS|}{|\square PR^\prime|} = (1+r)(1+s) \qquad \frac{|\square PQRS|}{|\square P^\prime R|} = (1+p)(1+q) \qquad \cdots$$ $$\frac{|\square PQRS|}{|\square P^\prime Q^\prime R^\prime S^\prime|} = \frac{(1+p)(1+q)(1+r)(1+s)}{(1-pr)(1-qs)} \;\stackrel{(\star\star)}{=}\; (1+p)(1+q)(1+r)(1+s)$$ $$\frac{|\overline{P^\prime P_\circ}|}{|\overline{PP_\circ}|} = \frac{(1-pr)(1-qs)}{(1+p)(1+q)}\;\stackrel{(\star\star)}{=}\; \frac{1}{(1+p)(1+q)}$$

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 $(\star\star)$ 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

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

• Thank you very much for these very interesting and deep investigations. Your second part, a reciprocal point of view, is typically in the spirit of what I desire to understand... Besides, 2 little questions : 1) What are points (A),(B),(C),(D) in your first figure ? Which graphical software do you use for your figures ? – Jean Marie Jan 21 '18 at 10:04
• 1) $(A)$, $(B)$, $(C)$, $(D)$ are meant to indicate that the segments actually continue to endpoints $A$, $B$, $C$, $D$. (The diagram was getting muddled in the middle, and I didn't want to make it worse. Then again, that "$(A)$" stuff isn't very intuitive. shrug) 2) I use GeoGebra for my images. Specifically, GeoGebra "Classic". (I'm not terribly fond of the user interface of the new GeoGebra Geometry app.) – Blue Jan 21 '18 at 16:57
• Thanks for your answers. – Jean Marie Jan 21 '18 at 18:32
• @JeanMarie: Thanks for the checkmark! I have a few more "inverse" observations to add when I get some time. – Blue Jan 23 '18 at 6:44
• I will appreciate it very much! Thanks again for your investigations. – Jean Marie Jan 23 '18 at 6:47