Key Concepts and Formulas

• Equilateral Triangle Properties: In an equilateral triangle, the centroid, orthocenter, circumcenter, and incenter coincide.
• Centroid Formula: The centroid of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by $\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$.
• Distance Formula: The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Step-by-Step Solution

Step 1: Understand the Given Information and Calculate the Side Length

We are given an equilateral triangle $ABC$ with vertices $A(\sin t, -\cos t)$, $B(\cos t, \sin t)$, and $C(a, b)$. Since it's equilateral, all sides have equal length. We calculate the square of the length of side $AB$:

$AB^2 = (\cos t - \sin t)^2 + (\sin t - (-\cos t))^2$ $AB^2 = (\cos t - \sin t)^2 + (\sin t + \cos t)^2$ Expanding the terms: $AB^2 = (\cos^2 t - 2\sin t \cos t + \sin^2 t) + (\sin^2 t + 2\sin t \cos t + \cos^2 t)$ Using the identity $\sin^2 t + \cos^2 t = 1$: $AB^2 = (1 - 2\sin t \cos t) + (1 + 2\sin t \cos t)$ $AB^2 = 2$ Thus, $AB^2 = BC^2 = CA^2 = 2$.

Step 2: Express the Orthocenter Coordinates using the Centroid Formula

Let $H(h, k)$ be the orthocenter of triangle $ABC$. Since $ABC$ is equilateral, the orthocenter coincides with the centroid. The centroid coordinates are: $h = \frac{\sin t + \cos t + a}{3} \quad \ldots (1)$ $k = \frac{-\cos t + \sin t + b}{3} \quad \ldots (2)$

Step 3: Derive a Relationship for $a$ and $b$

From equations (1) and (2), we express $a$ and $b$: From (1): $3h = \sin t + \cos t + a \implies a = 3h - (\sin t + \cos t)$ From (2): $3k = -\cos t + \sin t + b \implies b = 3k - (\sin t - \cos t)$

Rearranging these equations: $3h - a = \sin t + \cos t \quad \ldots (3)$ $3k - b = \sin t - \cos t \quad \ldots (4)$

Squaring both equations (3) and (4) and adding them: $(3h - a)^2 = (\sin t + \cos t)^2 = \sin^2 t + 2\sin t \cos t + \cos^2 t = 1 + 2\sin t \cos t$ $(3k - b)^2 = (\sin t - \cos t)^2 = \sin^2 t - 2\sin t \cos t + \cos^2 t = 1 - 2\sin t \cos t$ Adding the squared equations: $(3h - a)^2 + (3k - b)^2 = (1 + 2\sin t \cos t) + (1 - 2\sin t \cos t)$ $(3h - a)^2 + (3k - b)^2 = 2 \quad \ldots (5)$ This equation describes a relationship between the orthocenter $H(h,k)$ and the coordinates of vertex $C(a,b)$.

Step 4: Use the Given Locus of the Orthocenter to Find $a$ and $b$

The problem states that the orthocenter $H(h,k)$ lies on a circle with center $\left(1, \frac{1}{3}\right)$. Thus, $(h-1)^2 + (k-\frac{1}{3})^2 = R^2$.

From equation (5): $(3h - a)^2 + (3k - b)^2 = 2$. Rewriting this equation: $9(h - \frac{a}{3})^2 + 9(k - \frac{b}{3})^2 = 2$ $(h - \frac{a}{3})^2 + (k - \frac{b}{3})^2 = \frac{2}{9}$

The locus of the orthocenter is a circle with center $(\frac{a}{3}, \frac{b}{3})$. We are given that the orthocenter lies on a circle with center $(1, \frac{1}{3})$. For these to be the same circle, we must have: $\frac{a}{3} = 1 \implies a = 3$ $\frac{b}{3} = \frac{1}{3} \implies b = 1$

Step 5: Calculate $a^2 - b^2$

We have $a = 3$ and $b = 1$. Therefore, $a^2 - b^2 = 3^2 - 1^2 = 9 - 1 = 8$

Common Mistakes & Tips

• Remember that the orthocenter and centroid coincide only for equilateral triangles.
• When eliminating trigonometric functions, squaring and adding is a useful technique.
• Be careful with algebraic manipulations to avoid errors.

Summary

We used the properties of equilateral triangles and the given information about the orthocenter's locus to find the coordinates of vertex $C(a, b)$. We equated the center of the derived locus of the orthocenter with the given center of the circle to solve for $a$ and $b$. Finally, we calculated $a^2 - b^2$.

The final answer is \boxed{8}, which corresponds to option (B).