Two Squares In Circle


Two squares in circle


Squares $ABCD\;$ and $EFGH\;$ are inscribed in a circle, forming in intersection an octagon $MNPQRSTU.$

Two squares in circle - problem

Prove that $[MNPQRSTU]\ge 2(\sqrt{2}-1)[ABCD],\;$ where $[\mathcal{X}]\;$ denotes the area of figure $\mathcal{X}.$

Proof 1

WLOG, assume that the squares have side $1.\;$ If $x=[MNPQRSTU],\;$ then $2x=2-[\text{the eight triangles}].\;$ The minor arcs ${AB}$ and ${EF}$ are equal, so that $AB=EF,\;$ making $AEBF\;$ an isosceles trapezoid. It follows that $\Delta NAF\;$ is isosceles and $NA=NF.\;$ Since triangles $NFP\;$ and $NAM\;$ are similar, we deduce that they are in fact congruent. Thus all eight triangles are congruent.

We denote $AN=a\sin t,\;$ $\displaystyle 0\lt t\lt\frac{\pi}{2}.$ This implies that $NP=a\;$ and $PB=a\cos t\;$ from which $a=\displaystyle\frac{1}{1+\sin t+\cos t}.\;$ Thus

$[\text{the eight triangles}]=\displaystyle\frac{4\sin t\cos t}{(1+\sin t+\cos t)^2}.\;$

If $\sin t+\cos t=y,\;$ $2\sin t\cos t=y^2-1\;$ and $a\lt y\le\sqrt{2}.$ From here,

$\displaystyle\frac{4\sin t\cos t}{(1+\sin t+\cos t)^2}=\frac{2(y^2-1)}{(y+1)^2}=\frac{2(y-1)}{y+1}\le\frac{2(\sqrt{2}-1)}{\sqrt{2}+1}$

so that $[\text{the eight triangles}]\le 2(3-2\sqrt{2}),\;$ $2x\ge 2-2(3-2\sqrt{2}),\;$ or $x\ge 2(\sqrt{2}-1).$

Proof 2

By symmetry, the four cut-off triangles of either square are equal. By the Carpets Theorem, the sum of the ares of the cut-off triangles of one of the squares equals to the sum of the areas of the others. It follows that all eight triangles are equal.

Let $x=AM=ME,\;$ $y=EU=UD,\;$ $MU=z.\;$ WLOG, assume $x+y+z=1.\;$ By the Pythagorean Theorem, $x^2+y^2=z^2=(1-x-y)^2,$ which is equivalent to $\displaystyle (1-x)(1-y)=\frac{1}{2}.\;$ By the AM-GM inequality,

$\displaystyle \frac{(1-x)+(1-y)}{2}\ge\sqrt{(1-x)(1-y)}=\frac{1}{\sqrt{2}},$

or, $x+y\le 2-\sqrt{2}.$ On the other hand, $\displaystyle (1-x)(1-y)=\frac{1}{2}\;$ is equivalent to $\displaystyle xy=x+y-\frac{1}{2},\;$ implying that $\displaystyle xy\le\frac{3}{2}-\sqrt{2}.$ But $xy=2[\text{one of the triangles}].\;$ So that $\displaystyle [\text{the eight triangles}]\le 2(3-2\sqrt{2}),\;$ and, subsequently, $[MNPQRSTU]\ge 2(\sqrt{2}-1).$

Note that the minimum is achieved when $x=y.$


The problem which is due to Miguel Ochoa Sanchez has been posted by Leo Giugiuc at the CutTheKnotMath facebook page along with a solution (Proof 1) by Dan Sitaru and Leo Giugiuc.


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