ABC is an isosceles triangle inscribed in a circle of radius r , if AB = AC and h is the altitude from A to BC . If the ` triangleABC` has perimeter P and ` triangle ` is area then ` lim_( h to 0) 512 r Delta/p^(3)` equals

Answer

To solve the problem, we need to find the limit:

\[ \lim_{h \to 0} \frac{512r \Delta}{P^3} \]

where \( \Delta \) is the area of triangle ABC and \( P \) is its perimeter.

### Step 1: Understand the Geometry
Given that triangle ABC is isosceles with \( AB = AC \) and inscribed in a circle of radius \( r \), we denote the altitude from \( A \) to \( BC \) as \( h \). The base \( BC \) will be denoted as \( b \).

### Step 2: Find the Length of \( BC \)
Using the properties of the circle and the triangle, we can find \( BC \). The altitude \( h \) divides \( BC \) into two equal segments, and we can denote the midpoint of \( BC \) as \( D \).

Using the Pythagorean theorem in triangle \( OBD \) (where \( O \) is the center of the circle):

\[ r^2 = h^2 + BD^2 \]

Let \( BD = x \). Therefore, we have:

\[ BD^2 = r^2 - h^2 \quad \Rightarrow \quad x = \sqrt{r^2 - h^2} \]

Since \( BC = 2 \times BD \):

\[ b = 2\sqrt{r^2 - h^2} \]

### Step 3: Find the Length of \( AB \) and \( AC \)
Using triangle \( ABD \):

\[ AB^2 = AD^2 + BD^2 \quad \Rightarrow \quad AB^2 = h^2 + (r^2 - h^2) = r^2 \]

Thus,

\[ AB = AC = r \]

### Step 4: Calculate the Perimeter \( P \)
The perimeter \( P \) of triangle ABC is given by:

\[ P = AB + AC + BC = r + r + 2\sqrt{r^2 - h^2} = 2r + 2\sqrt{r^2 - h^2} \]

### Step 5: Calculate the Area \( \Delta \)
The area \( \Delta \) of triangle ABC can be calculated as:

\[ \Delta = \frac{1}{2} \times BC \times h = \frac{1}{2} \times 2\sqrt{r^2 - h^2} \times h = h\sqrt{r^2 - h^2} \]

### Step 6: Substitute \( \Delta \) and \( P \) into the Limit
Now we substitute \( \Delta \) and \( P \) into the limit:

\[ \lim_{h \to 0} \frac{512r \Delta}{P^3} = \lim_{h \to 0} \frac{512r \cdot h\sqrt{r^2 - h^2}}{(2r + 2\sqrt{r^2 - h^2})^3} \]

### Step 7: Simplify the Expression
As \( h \to 0 \):

- \( \sqrt{r^2 - h^2} \to r \)
- \( P \to 2r + 2r = 4r \)

Thus,

\[ P^3 \to (4r)^3 = 64r^3 \]

Now substituting back into the limit:

\[ \lim_{h \to 0} \frac{512r \cdot h \cdot r}{64r^3} = \lim_{h \to 0} \frac{512hr}{64r^3} = \lim_{h \to 0} \frac{8h}{r^2} \]

As \( h \to 0 \), this limit approaches \( 0 \).

### Final Result
Thus, the final answer is:

\[ \lim_{h \to 0} \frac{512r \Delta}{P^3} = 0 \]