, 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 $MT = y$. [[IMAGE0]] 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

$\text{9 x + 9 y + 9 (z 2) = 2 16 x + 16 (y 2) + 16 z = 1 25 (x 2) + 25 y + 25 z = 0}$ Using logarithm rules, this is equivalent to the system

$\text{9 (xyz 2) = 2 16 (xy 2z) = 1 25 (x 2yz) = 0}$$and to the system$$\text{xyz 2 = 9 2 = 81 xy 2z = 16 1 = 16 x 2yz = 25 0 = 1}$ 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.