, 2024

We join $B$ to $E$ and $A$ to $D$.

Since $MC$ is tangent to the circles
with centres $A$ and $B$ at $D$ and $E$, respectively, then $AD$ and $BE$ are perpendicular to $MC$.

Since the radius of the circle with centre $B$ is $3$, then $AB =
3$and$BE =3$.

Since the radius of the circle with centre $A$ is $4$, then $AD =
4$and$AT = 4$.

Let $CB = x$ and $MT = y$.

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We note that $\triangle CEB$,
$\triangle CDA$ and $\triangle CTM$ are all similar, since they
are right-angled at $E$, $D$ and $T$, respectively, and share a common angle
at $C$.

Since $\triangle CEB$ and $\triangle CDA$ are similar, then $\dfrac{CB}{CA} = \dfrac{BE}{AD}$ and so
$\dfrac{x}{x+3} = \dfrac{3}{4}$ which
gives $4x = 3x + 9$ and so $x = 9$.

By the Pythagorean Theorem, $CE = $\sqrt{CB^2}$
- BE^2} = $\sqrt{9^2}$ - 3^2} = $\sqrt{72} = 6\sqrt{2}\$.$

Since $\triangle CEB$ and $\triangle CTM$ are similar, then $\dfrac{BE}{CE} = \dfrac{MT}{CT}$ and so
$\$\dfrac{3}{6\sqrt{2}} = \dfrac{y}{9+3+4}$$which gives$y =
$\text{16}$ 3}{6 ${\sqrt{2}} = \dfrac{8}{\sqrt{2}} = 4\sqrt{2}\$.$

Finally, the area of $\$\triangle$
MNC$is equal to$$\dfrac{1}{2} \cdot MN \cdot$ CT$.

If we joined $B$ to $G$, we would see that $\triangle CEB$ is congruent to $\triangle CGB$ (each is right-angled, they
have a common hypotenuse, and $BE =
BG)$. This means that$$\angle$ BCE =
$\angle$ BCG$, which in turn means that$MT = TN$.

Since $MT = TN$, then $MN = 2 \cdot 4\sqrt{2} = 8 \sqrt{2}$ and
so the area of $\triangle MNC$ is
$\$\dfrac{1}{2} \cdot 8\sqrt{2} \cdot$
16$or$$64\sqrt{2}\$.$
First, we note that $$\log_3 z = \dfrac{\log_{10} z}{\log_{10} 3} = \dfrac{2 \log_{10} z}{2 \log_{10} 3} = \dfrac{\log_{10}(z^2)}{\log_{10}(3^2)} = \dfrac{\log_{10}(z^2)}{\log_{10} 9} = \log_9 (z^2)$$ Similarly,
$\log_4 y = \log_{16}(y^2)$ and $\log_5 x = \log_{25}(x^2)$.

We also note from the original system of equations that $x > 0$ and $y > 0$ and $z > 0$.

Therefore, we can re-write the original system of equations as $$\begin{align*} \log_9 x + \log_9 y + \log_9 (z^2) & = 2 \\ \log_{16} x + \log_{16}(y^2) + \log_{16} z & = 1 \\ \log_{25}(x^2) + \log_{25} y + \log_{25} z & = 0\end{align*}$$ Using logarithm rules, this is equivalent to the
system $$\begin{align*} \log_9 (xyz^2) & = 2 \\ \log_{16} (xy^2z) & = 1 \\ \log_{25} (x^2yz) & = 0 \end{align*}$$ and to the system
$$\begin{align*} xyz^2 & = 9^2 = 81 \\ xy^2z & = 16^1 = 16 \\ x^2yz & = 25^0 = 1\end{align*}$$ Multiplying these three
equations together, we obtain $x^4 y^4 z^4 =
1296$and so$(xyz)^4 =
6^4$.

Thus, $xyz = 6$.

Since $xyz^2 = 81$ and $xyz = 6$, then $z = $\dfrac{xyz^2}{xyz} = \dfrac{81}{6} = \dfrac{27}{2}\$.$

Similarly, $y = $\dfrac{xy^2z}{xyz} = \dfrac{16}{6} = \dfrac{8}{3}$$and$x
= $\dfrac{x^2yz}{xyz} = \dfrac{1}{6}\$.$

Therefore, $(x,y,z) = $\left(\dfrac{1}{6}, \dfrac{8}{3}, \dfrac{27}{2}\right)\$.$

We can check by substitution that this triple does satisfy the original
system of equations.