A triangle $$ABC$$ with $$AC=20$$ is inscribed in a circle $$\omega$$. A tangent $$t$$ to $$\omega$$ is drawn through $$B$$. The distance of $$t$$ from $$A$$ is $$25$$ and that from $$C$$ is $$16$$. If $$S$$ denotes the area of the triangle $$ABC$$, find the largest integer not exceeding $$S/20$$.

Let the tangent line at $$B$$ be the $$y$$-axis.
Take $$B$$ as the origin and measure $$x$$ to the right, $$y$$ upward.
Because the radius drawn to the point of contact is perpendicular to the tangent, the centre $$O$$ of the circum-circle lies somewhere on the positive $$x$$-axis. Suppose its coordinates are $$\left(R,0\right)$$, where $$R$$ is the circum-radius.

The circle is therefore
$$\left(x-R\right)^2+y^2=R^2 \qquad -(1)$$

The perpendicular distance of a point from the $$y$$-axis is the absolute value of its $$x$$-coordinate. Hence we can place the remaining vertices as

$$A\,(25,\,y_A),\qquad C\,(16,\,y_C)$$

(both to the right of the tangent so that the squared quantities below are non-negative).

Substituting $$A$$ and $$C$$ in $$-(1)$$ gives

$$\begin{aligned} (25-R)^2+y_A^{\,2}&=R^2 \;\Longrightarrow\; y_A^{\,2}=50R-625 \quad -(2)\\ (16-R)^2+y_C^{\,2}&=R^2 \;\Longrightarrow\; y_C^{\,2}=32R-256 \quad -(3) \end{aligned}$$

The given side length $$AC=20$$ yields

$$\begin{aligned} 20^2 &= (25-16)^2+\left(y_A-y_C\right)^2\\ 400 &= 9^2+\left(y_A-y_C\right)^2\\ \left(y_A-y_C\right)^2 &= 319 \qquad -(4) \end{aligned}$$

Write

$$u = y_A,\qquad v = y_C,$$

so that from $$-(2)$$ and $$-(3)$$

$$u^2 = 50R-625,\qquad v^2 = 32R-256 \qquad -(5)$$

Equation $$-(4)$$ gives

$$u^2+v^2-2uv = 319 \quad\Longrightarrow\quad 50R-625+32R-256-2uv=319.$$

Simplifying,

$$82R-2uv = 1200 \;\Longrightarrow\; 41R-uv = 600 \qquad -(6)$$

Next, compute the area of $$\triangle ABC$$ with coordinates $$A(25,u),\,\, B(0,0),\,\, C(16,v).$$ Using the determinant (vector cross-product) formula,

$$\begin{aligned} 2S &= \Bigl|(x_C-x_A)(y_B-y_A)-(y_C-y_A)(x_B-x_A)\Bigr|\\ &= \Bigl|(-9)(-u)- (v-u)(-25)\Bigr|\\ &= \bigl|-16u+25v\bigr|. \end{aligned}$$

Hence

$$S=\dfrac{1}{2}\,\bigl|-16u+25v\bigr| \qquad -(7)$$

Square the bracket in $$-(7)$$ and use $$-(5),(6)$$:

$$\begin{aligned} (-16u+25v)^2 &= 256u^2 + 625v^2 -800uv\\ &= 256(50R-625)+625(32R-256)-800(41R-600)\\ &= (12800R-160000)+(20000R-160000)-(32800R-480000)\\ &= 160000. \end{aligned}$$

Thus

$$\bigl|-16u+25v\bigr| = 400,$$

and from $$-(7)$$

$$S=\dfrac{400}{2}=200.$$

Finally,

$$\dfrac{S}{20} = \dfrac{200}{20}=10,$$

so the largest integer not exceeding $$S/20$$ is

10.