Geometry Problem with isobarycenter and isoceles triangle

Question

$XYZ$ a triangle with $XY=YZ.$

$D$ middle of $XY,$ $G$ isobarycenter of $XYZ.$

a circle $O,$ with center $D, XY$ are on $O.$

a circle $O',$ with center $G,$ $X$ and $Z$ are on $O'.$

1/the circumference O=O',

Given the above information, what is the value of the angle XYZ?

Answer 1

Following is an alternate derivation of the formula using more trigonometry.

Let $E$ be the midpoint of $XZ$, let $\theta$ be the angle $\measuredangle GXE$ and $\varphi$ be the angle $\measuredangle XYZ$. If one scale the whole figure such that $|GX| = 1$, then using the fact $\measuredangle XEG$ is a right angle, we have:

$$|XE| = \cos\theta\quad\text{ and }\quad |GE| = \sin\theta$$

Since $G$ is the isobarycenter, $|YE| = 3|GE| = 3\sin\theta$, This implies

$$\tan\frac{\varphi}{2} = \frac{|XE|}{|YE|} = \frac{1}{3\tan\theta}$$

The condition $|GX| = |DX| = |DY|$ implies $|XY| = 2$. Since $|XY|^2 = |XE|^2 + |YE|^2$, we get:

$$2^2 = \cos^2\theta + (3\sin\theta)^2 = 1 + 8\sin\theta^{\,2} \implies \sin\theta = \sqrt{\frac38} \implies \tan\theta = \sqrt{\frac35}$$ As a result, we get $$\varphi = 2 \tan^{-1}\left(\frac{1}{3\sqrt{\frac35}}\right) = 2\tan^{-1}\sqrt{\frac{5}{27}}$$

Answer 2

The angle for which the radii agree is approximately $46.6°$.

Assume coordinates:

\begin{align*} X &= \begin{pmatrix}1\\0\end{pmatrix} & Y &= \begin{pmatrix}0\\y\end{pmatrix} & Z &= \begin{pmatrix}-1\\0\end{pmatrix} \end{align*}

Compared to the image and common coordinate systems, this is flipped along the $y$ axis. Now you get

\begin{align*} D &= \begin{pmatrix}\frac12\\[1ex]\frac y2\end{pmatrix} & G &= \begin{pmatrix}0\\[1ex]\frac y3\end{pmatrix} \\ \lvert D,X\rvert^2 &= \left(\frac12\right)^2+\left(\frac y2\right)^2 = \frac{1+y^2}4 & \lvert G,X\rvert^2 &= 1 + \left(\frac y3\right)^2 = \frac{9+y^2}9 \end{align*}

Setting these equal gives you

\begin{align*} \frac{1+y^2}4 &= \frac{9+y^2}9 \\ 9(1+y^2)&=4(9+y^2) \\ 9+9y^2&=36+4y^2 \\ 5y^2 &= 27 \\ y^2&=\frac{27}5 \\ y&=\sqrt{\frac{27}5} \end{align*}

Now let $\alpha$ denote the angle you are looking for. The triangle $XYC$ will be a right triangle, with one angle $\frac\alpha2$. one of its legs is $1$, the other $y$. From this you get

\begin{align*} \tan\frac\alpha2 &= \frac1y = \sqrt{\frac5{27}} \\ \frac\alpha2 &= \operatorname{atan}\sqrt{\frac5{27}} \\ \alpha &= 2\operatorname{atan}\sqrt{\frac5{27}} \approx 46.6° \end{align*}