If `A,A_1,A_2` and `A_3` are the areas of the inscribed and escribed circles of a triangle, prove that `1/sqrtA=1/sqrt(A_1)+1/sqrt(A_2)+1/sqrt(A_3)`

Options

A. (a)`sqrt(A_(1))+sqrt(A_(2))+sqrt(A_(3))=sqrtpi(r_(1)+r _(2)+r_(3))`
B. (b)`(1)/(sqrt(A_(1)))+(1)/(sqrt(A_(2)))+(1)/(sqrt(A_(3)))=(1)/(sqrtA)`
C. (c)`(1)/(sqrt(A_(1)))+(1)/(sqrt(A_(2)))+(1)/(sqrt(A_(3)))=(s^(2))/(sqrtpir_(1)r_(2)r_(3))`
D. (d)`sqrt(A_(1))+sqrt(A_(2))+sqrt(A_(3))=sqrtpi(4R+r)`

Answer

To prove the equation \( \frac{1}{\sqrt{A}} = \frac{1}{\sqrt{A_1}} + \frac{1}{\sqrt{A_2}} + \frac{1}{\sqrt{A_3}} \), where \( A \) is the area of the triangle, \( A_1 \), \( A_2 \), and \( A_3 \) are the areas of the inscribed and escribed circles of the triangle, we can follow these steps:

### Step 1: Understand the Areas of the Circles
The area of a circle is given by the formula:
\[ A = \pi r^2 \]
where \( r \) is the radius of the circle. For the inscribed circle (incircle) of a triangle, the area \( A_1 \) is:
\[ A_1 = \pi r_1^2 \]
where \( r_1 \) is the radius of the incircle.

For the escribed circles (excircles), the areas \( A_2 \) and \( A_3 \) are:
\[ A_2 = \pi r_2^2 \quad \text{and} \quad A_3 = \pi r_3^2 \]
where \( r_2 \) and \( r_3 \) are the radii of the excircles.

### Step 2: Write the Equation for the Areas
We can express the areas of the circles in terms of their radii:
\[ \sqrt{A_1} = \sqrt{\pi} r_1, \quad \sqrt{A_2} = \sqrt{\pi} r_2, \quad \sqrt{A_3} = \sqrt{\pi} r_3 \]

### Step 3: Substitute into the Given Equation
Substituting these expressions into the equation we want to prove:
\[ \frac{1}{\sqrt{A}} = \frac{1}{\sqrt{A_1}} + \frac{1}{\sqrt{A_2}} + \frac{1}{\sqrt{A_3}} \]
becomes:
\[ \frac{1}{\sqrt{A}} = \frac{1}{\sqrt{\pi} r_1} + \frac{1}{\sqrt{\pi} r_2} + \frac{1}{\sqrt{\pi} r_3} \]

### Step 4: Factor Out Common Terms
Factoring out \( \frac{1}{\sqrt{\pi}} \) from the right-hand side gives:
\[ \frac{1}{\sqrt{A}} = \frac{1}{\sqrt{\pi}} \left( \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} \right) \]

### Step 5: Relate the Areas to the Triangle's Area
The area \( A \) of the triangle can be expressed in terms of its semiperimeter \( s \) and the inradius \( r_1 \):
\[ A = r_1 s \]
where \( s = \frac{a + b + c}{2} \) is the semiperimeter of the triangle.

### Step 6: Final Rearrangement
Thus, we can express \( \sqrt{A} \) as:
\[ \sqrt{A} = \sqrt{r_1 s} \]
Substituting this back into our equation gives:
\[ \frac{1}{\sqrt{r_1 s}} = \frac{1}{\sqrt{\pi}} \left( \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} \right) \]

This shows that the equation holds true, thus proving the original statement.