Find the locus of the vertex of a triangle whose, sum of base angles is fixed.

Answer

To find the locus of the vertex of a triangle whose sum of the base angles is fixed, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Triangle**:
Let triangle \( ABC \) have \( AB \) as the base. Let \( \angle A \) be the vertex angle and \( \angle B \) and \( \angle C \) be the base angles. We know that the sum of the base angles \( \angle B + \angle C \) is fixed.

2. **Setting Up the Coordinate System**:
Place the triangle in a Cartesian coordinate system. Let \( A(h, k) \) be the vertex opposite the base \( BC \), and let points \( B \) and \( C \) lie on the x-axis. Assume \( B \) is at \( (x_1, 0) \) and \( C \) is at \( (x_2, 0) \).

3. **Using Angle Relationships**:
Since \( \angle B + \angle C \) is fixed, we can denote it as \( \theta_1 + \theta_2 = C \) (a constant). This means that \( \angle A \) can be expressed as \( \angle A = 180^\circ - C \).

4. **Using Tangent Function**:
We can express the tangents of the angles in terms of the coordinates:
- \( \tan(\theta_1) = \frac{k}{h - x_1} \)
- \( \tan(\theta_2) = \frac{k}{h - x_2} \)

5. **Relating the Angles**:
Using the tangent addition formula:
\[ \tan(\theta_1 + \theta_2) = \frac{\tan(\theta_1) + \tan(\theta_2)}{1 - \tan(\theta_1)\tan(\theta_2)} \]
Since \( \theta_1 + \theta_2 = C \), we can set:
\[ \tan(C) = \frac{\tan(\theta_1) + \tan(\theta_2)}{1 - \tan(\theta_1)\tan(\theta_2)} \]

6. **Substituting Values**:
Substitute the expressions for \( \tan(\theta_1) \) and \( \tan(\theta_2) \):
\[ \tan(C) = \frac{\frac{k}{h - x_1} + \frac{k}{h - x_2}}{1 - \frac{k^2}{(h - x_1)(h - x_2)}} \]

7. **Simplifying the Equation**:
Cross-multiply and simplify to find a relationship between \( h \) and \( k \). This will lead to a quadratic equation in terms of \( h \) and \( k \).

8. **Identifying the Locus**:
The resulting equation will be of the form \( A(h^2) + B(k^2) + C(h) + D(k) + E = 0 \), which represents a parabola.

9. **Final Result**:
Thus, the locus of the vertex \( A(h, k) \) of the triangle, where the sum of the base angles is fixed, is a parabola.