# Under what conditions will the rectangle of the Japanese theorem be a square?

In geometry, the Japanese theorem for cyclic quadrilaterals states that the centers of the incircles of certain triangles inside a cyclic quadrilateral are vertices of a rectangle.

Question. Under what conditions will the $M_1M_2M_3M_4$ rectangle be a square?

• Rather than starting with an inscribed quadrilateral, constructing the Japanese rectangle, and trying to figure out what properties ensure it's square, can you start with a square and workout the most general way to construct a corresponding inscribed quadrilateral? – user14972 Dec 31 '14 at 12:46

Some very ugly symbol-bashing led me to this relation:

$$\sin\alpha \sin\gamma = \sin \beta \sin\delta$$

Edit. Proof (without much symbol-bashing at all!) begins with a lemma about inscribed and circumscribed circles.

Lemma. Let $I$ be the incenter of $\triangle ABC$, and let $A^\prime$ be the point where $\overrightarrow{AI}$ meets the circumcircle of $\triangle ABC$. (Note that $A^\prime$ is the midpoint of $\stackrel{\frown}{BC}$.) Then $$|\overline{A^\prime I}| = |\overline{A^\prime B}| = |\overline{A^\prime C}| = 2 r \sin\frac12 \angle A$$ where $r$ is the circumradius.

Proof of Lemma. The bisectors of $\angle A$ and $\angle B$ (and $\angle C$) meet at $I$, so we can write $$\alpha := \angle BAA^\prime = \angle CAA^\prime = \frac12\angle A \quad\text{and}\quad \beta := \angle ABI = \angle CBI = \frac12\angle B$$

• The Exterior Angle Theorem applied at vertex $I$ of $\triangle ABI$ implies that $\angle BIA^\prime = \alpha + \beta$.

• The Inscribed Angle Theorem implies that $\angle A^\prime B C = \angle A^\prime A C = \alpha$, as both angles subtend the common arc $\stackrel{\frown}{A^\prime C}$. Therefore, $\angle IBA^\prime = \alpha + \beta$.

• Consequently, $\triangle IBA^\prime$ is isosceles with base $IB$.

• The Law of Sines implies that $|\overline{A^\prime B}| = 2 r \sin\alpha$, so this must also be the length of $\overline{A^\prime I}$.

Lemma in hand, we return to the original problem, with cyclic quadrilateral $\square ABCD$ determining incircles with centers $W$, $X$, $Y$, $Z$ as shown.

With $M$ the midpoint of $\stackrel{\frown}{BC}$, we observe that rays $\overrightarrow{AX}$ and $\overrightarrow{DY}$ converge at $M$. By the Lemma, with $r$ the radius of the circle, $$|\overline{MX}| = 2 r \sin\alpha = |\overline{MY}|$$

Moreover, since inscribed angles $\angle ACD$ and $\angle AMD$ subtend $\stackrel{\frown}{AD}$, they are congruent, so that the vertex angle of isosceles triangle $\triangle XMY$ has measure $2\gamma$. Therefore,

$$|\overline{XY}| = 2 r \sin\alpha \sin\gamma \qquad\text{and, likewise,}\qquad |\overline{WX}| = 2 r \sin\beta \sin\delta$$ with $\beta$ and $\delta$ as in the first diagram above. Note that, by symmetry, $|\overline{WZ}| = 2r\sin\alpha\sin\gamma$ and $|\overline{YZ}| = 2r\sin\beta\sin\delta$. We have almost proven the Japanese theorem itself, since $\square WXYZ$ is now at least a parallelogram. Taking rectangle-ness for granted, we see that $\square WXYZ$ is more specifically a square if and only if two adjacent sides are congruent; this is precisely the condition $$\sin\alpha\;\sin\gamma \;=\; \sin\beta\;\sin\delta$$ as claimed. $\square$

• Correct me is I'm wrong, but from you derivation it seems to me that the condition $\sin\alpha\;\sin\gamma \;=\; \sin\beta\;\sin\delta$ is necessary, but is it also sufficient? So is this an equivalence? – user153012 Jan 8 '15 at 13:21
• I've shown that, in any circumscribed quad with $\alpha$, $\beta$, $\gamma$, $\delta$ and $W$, $X$, $Y$, $Z$ as marked, we have $|\overline{XY}| = |\overline{WZ}| = 2 r \sin\alpha\sin\gamma$ and $|\overline{WX}| = |\overline{YZ}| = 2 r \sin\beta\sin\delta$. That's just what those lengths are in terms of the angles (and the circumradius $r$). The Japanese Theorem guarantees that $\square WXYZ$ is a rectangle. Congruent adjacent sides is then necessary and sufficient for the rectangle to be a square; the sine relation simply restates the congruence condition, so there is logical equivalence. – Blue Jan 8 '15 at 13:40