Find the maximum area of the ellipse that can be inscribed in an isoceles triangles of area A and having one axis lying along the perpendicular from the vertex of the triangles to its base.

Answer

To find the maximum area of an ellipse inscribed in an isosceles triangle with a given area \( A \), we can follow these steps:

### Step 1: Understand the Geometry
We have an isosceles triangle with a base along the x-axis and a vertex at the top. Let the coordinates of the triangle be \( (0, b) \), \( (-a, 0) \), and \( (a, 0) \). The area of the triangle can be expressed as:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2a \times b = ab \]
Given that the area is \( A \), we have:
\[ ab = A \quad \text{(1)} \]

### Step 2: Define the Ellipse
The ellipse is inscribed in the triangle with its major axis along the y-axis. Let the semi-major axis be \( \alpha \) and the semi-minor axis be \( \beta \). The area of the ellipse is given by:
\[ \text{Area of ellipse} = \pi \alpha \beta \quad \text{(2)} \]

### Step 3: Relate \( \alpha \) and \( \beta \) to the Triangle
From the geometry, we know that the ellipse will touch the sides of the triangle. The triangle's height is \( b \) and the base is \( 2a \). The relationship between \( \alpha \) and \( \beta \) can be derived from the triangle's dimensions. The ellipse's axes must fit within the triangle, leading to the constraints:
\[ \alpha \leq b \quad \text{and} \quad \beta \leq a \quad \text{(3)} \]

### Step 4: Substitute \( b \) from Equation (1)
From equation (1), we can express \( b \) in terms of \( a \):
\[ b = \frac{A}{a} \quad \text{(4)} \]
Substituting this into the inequality gives:
\[ \alpha \leq \frac{A}{a} \]

### Step 5: Express \( \beta \) in terms of \( \alpha \)
Using the relationship from the triangle, we can express \( \beta \) in terms of \( \alpha \):
\[ \beta = \frac{A}{2\alpha} \quad \text{(5)} \]

### Step 6: Substitute into the Area of the Ellipse
Substituting equations (4) and (5) into the area of the ellipse:
\[ \text{Area of ellipse} = \pi \alpha \left(\frac{A}{2\alpha}\right) = \frac{\pi A}{2} \quad \text{(6)} \]

### Step 7: Find Maximum Area
To maximize the area of the ellipse, we need to ensure that the constraints from the triangle are satisfied. The maximum area occurs when \( \alpha \) and \( \beta \) are at their respective maximum values, leading to:
\[ \text{Maximum Area} = \frac{\pi A}{2} \]

### Conclusion
Thus, the maximum area of the ellipse that can be inscribed in the isosceles triangle of area \( A \) is:
\[ \text{Maximum Area} = \frac{\pi A}{2} \]

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