Angles between vectors of center of two incircles

Question

I have two two incircle between rectangle and two quadrilateral circlein. It's possible to determine exact value of $\phi,$ angles between vectors of center of two circles.

Answer 1

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Place the origin $O$ at the lower left-hand corner, and let the positive $x,y$-axes be the the natural choices based on the diagram.

Let $r$ be the radius of the red circles.

Let $P,Q$ be the centers of the circles of radii $2,1$, respectively.

Then we can express the coordinates of $P,Q$ as $P=(x,2)$ and $Q=(x,r-1)$, where $x$ is an unknown.

Computing $|OQ|$ in two ways, we get the equation $$\sqrt{x^2+(r-1)^2}=r+1$$ hence $$x^2+(r-1)^2=(r+1)^2\tag{eq1}$$ Computing $|OP|$ in two ways, we get the equation $$\sqrt{x^2+2^2}=r-2$$ hence $$x^2+4=(r-2)^2\tag{eq2}$$ From $(\text{eq}1){\,-\,}(\text{eq}2)$, we get $r=8$.

Plugging $r=8$ into $(\text{eq}2)$, we get $x=4\sqrt{2}$.

Using the known values of $r$ and $x$, the distance formula yields \begin{align*} |OP|&=6\\[4pt] |OQ|&=9\\[4pt] |PQ|&=5\\[4pt] \end{align*} hence by the law of cosines, $\cos(\large{\phi})={\large{\frac{23}{27}}}$, so we get $\large{\phi}=\cos^{-1}\bigl({\large{\frac{23}{27}}}\bigr)$.

Answer 2

There are constraints that $A$ and $B$ must fullfill, i.e., the following system of equations coming from Pythagoras theorem applied to certain right triangles :

$$\begin{cases}(B-2)^2+2^2=(A/2)^2\\(B-1)^2+(A/2)^2=(B+1)^2\end{cases}$$

giving $A=8 \sqrt{2}$ and $B=8$.

If now we take equations as in the partial solution you gave (a good idea) :

$$\tan(\theta) = \dfrac{2}{A/2} = \dfrac{\sqrt{2}}{4}\tag{1}$$

and

$$\tan(\phi + \theta) = \dfrac{B-1}{A/2} = \dfrac{7}{4 \sqrt{2}}\tag{2}$$

Equation (2) can also be written :

$$\dfrac{\tan(\phi) + \tan(\theta)}{1-\tan(\phi)\tan(\theta)} = \dfrac{7\sqrt{2}}{8}\tag{3}$$

Let $T:=\tan(\phi)$. (3) is equivalent to :

$$\dfrac{T + \sqrt{2}/4}{1-T\sqrt{2}/4} = \dfrac{7\sqrt{2}}{8}\tag{4}$$

giving $T=\dfrac{10 \sqrt{2}}{23}$. Therefore

$\phi=\arctan(\dfrac{10 \sqrt{2}}{23})\approx 31.5863 \ \text{degrees}.$

Answer 3

Using Pythagoras theorem is not difficult to find that the radius $B$ of red arcs is $8$. It follows that $PQ=6$, $PR=9$ and $QR=5$. Apply now the cosine rule to triangle $PQR$ to find $$ \cos\phi={23\over27}. $$

Answer 4

Let $|AB|=|CD|=a$,$|BC|=|AD|=b$, $|O_1E|=r_1$, $|O_2F|=r_2$, $\angle O_1AO_2=\phi$, $\angle O_1AF=\alpha$ $\angle O_2AF=\beta$.

Then \begin{align} \tan\alpha&=\frac{b-r_1}{a/2} \tag{1}\label{1} ,\\ \sin\alpha&=\frac{b-r_1}{b+r_1} ,\\ \tan\alpha&=\frac{\sin\alpha}{\sqrt{1-\sin^2\alpha}} =\frac{b-r_1}{b+r_1} \left/ \sqrt{1-\Big( \frac{b-r_1}{b+r_1} \Big)^2} \right. =\tfrac12\,\frac{b-r_1}{\sqrt{b\,r_1}} \tag{2}\label{2} . \end{align}

From \eqref{1}$=$\eqref{2} it follows \begin{align} a&=4\,\sqrt{b\,r_1} \tag{3}\label{3} . \end{align}

Similarly, \begin{align} \tan\beta&=\frac{2r_2}a \tag{4}\label{4} ,\\ \sin\beta&=\frac{r_2}{b-r_2} ,\\ \tan\beta&= \frac{\sin\beta}{\sqrt{1-\sin^2\beta}} =\frac{r_2}{\sqrt{b\,(b-2\,r_2)}} \tag{5}\label{5} . \end{align}

From \eqref{4}$=$\eqref{5}: \begin{align} a&=2\,\sqrt{b\,(b-2\,r_2)} \tag{6}\label{6} , \end{align}

and from \eqref{3}$=$\eqref{6} we have

\begin{align} b&=4r_1+2r_2 ,\\ a&=4\,\sqrt{r_1\,(4\,r_1+2\,r_2)} . \end{align}

\begin{align} \tan\phi&=\tan(\alpha-\beta) =\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta} =\frac{2(3r_1+r_2)\,\sqrt{2\,r_1\,(2\,r_1+r_2)}}{16\,r_1^2+11\,r_1\,r_2+2\,r_2^2} \end{align}

For $r_1=1$, $r_2=2$ we have

\begin{align} a&=8\,\sqrt2\approx 11.31370850 ,\\ b&=8 ,\\ \phi&=\arctan\Big(\frac{10\,\sqrt2}{23} \Big) \approx 31.586338^\circ . \end{align}