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Vector Addition

Let A,A1,A2...

Question

Let $A, A_{1}, A_{2}, A_{3}$ be the areas of the inscribed and escribed circles of $△ A B C$ , then:

\sqrt{A_{1}} + \sqrt{A_{2}} + \sqrt{A_{3}} = \sqrt{π} (r_{1} + r_{2} + r_{3})

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\sqrt{A_{1}} + \sqrt{A_{2}} + \sqrt{A_{3}} = \sqrt{π} (4 R + r)

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\frac{1}{\sqrt{A_{1}}} + \frac{1}{\sqrt{A_{2}}} + \frac{1}{\sqrt{A_{3}}} = \frac{1}{\sqrt{A}}

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\frac{1}{\sqrt{A_{1}}} + \frac{1}{\sqrt{A_{2}}} + \frac{1}{\sqrt{A_{3}}} = \frac{s^{2}}{\sqrt{π} . r_{1} . r_{2} . r_{3}}

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Solution

The correct options are
A $\sqrt{A_{1}} + \sqrt{A_{2}} + \sqrt{A_{3}} = \sqrt{π} (r_{1} + r_{2} + r_{3})$
B $\sqrt{A_{1}} + \sqrt{A_{2}} + \sqrt{A_{3}} = \sqrt{π} (4 R + r)$
C $\frac{1}{\sqrt{A_{1}}} + \frac{1}{\sqrt{A_{2}}} + \frac{1}{\sqrt{A_{3}}} = \frac{1}{\sqrt{A}}$
D $\frac{1}{\sqrt{A_{1}}} + \frac{1}{\sqrt{A_{2}}} + \frac{1}{\sqrt{A_{3}}} = \frac{s^{2}}{\sqrt{π} . r_{1} . r_{2} . r_{3}}$
$A = π r^{2}; r = △ / s;$

$A_{1} = π r_{1}^{2}; r_{1} = \frac{△}{s - a}; A_{2} = π r_{2}^{2}; r_{2} = \frac{△}{s - b};$

$A_{3} = π r_{3}^{2}; r_{3} = \frac{△}{(s - c)}$

$∴ \sqrt{A_{1}} + \sqrt{A_{2}} + \sqrt{A_{3}} = \sqrt{π} (r_{1} + r_{2} + r_{3})$

$\Rightarrow$ (a) is true

Also $r_{1} + r_{2} + r_{3} = r + 4 R$ is true

$\Rightarrow \sqrt{A_{1}} + \sqrt{A_{2}} + \sqrt{A_{3}} = \sqrt{π} (4 R + r)$

$\Rightarrow$ (b) is also true

Now, $\frac{1}{\sqrt{A_{1}}} + \frac{1}{\sqrt{A_{2}}} + \frac{1}{\sqrt{A_{3}}} = \frac{1}{\sqrt{π}} . (\frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}}) = \frac{1}{\sqrt{π}} . \frac{1}{r} = \sqrt{\frac{1}{π r^{2}}} = \frac{1}{\sqrt{A}}$

$\Rightarrow$ (c) is also true

Also $\frac{1}{\sqrt{A_{1}}} + \frac{1}{\sqrt{A_{2}}} + \frac{1}{\sqrt{A_{3}}} = \frac{1}{\sqrt{π}} . \frac{1}{r} = \frac{1}{\sqrt{π}} (\frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}}) = \frac{1}{\sqrt{π}} (\frac{r_{2} r_{3} + r_{1} r_{3} + r_{1} r_{2}}{r_{1} . r_{2} r_{3}})$

$= \frac{1}{\sqrt{π}} [\frac{△^{2}}{(s - b) (s - c)} + \frac{△^{2}}{(s - a) (s - c)} + \frac{△^{2}}{(s - a) (s - b)}] . \frac{1}{r_{1} r_{2} r_{3}}$

$= \frac{△^{2}}{\sqrt{π}} [\frac{(s - a) + (s - b) + (s - c)}{(s - a) (s - b) (s - c)}] . \frac{1}{r_{1} r_{2} r_{3}}$

$= \frac{△^{2}}{\sqrt{π}} [\frac{s}{(s - a) (s - b) (s - c)}] . \frac{1}{r_{1} r_{2} r_{3}} = \frac{1}{\sqrt{π}} \frac{s^{2}}{r_{1} r_{2} r_{3}}$

$\Rightarrow$ (d) is correct

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Similar questions

\frac{a_{1}}{a_{1}} + \frac{a_{2}}{a_{2}} + \frac{a_{3}}{a_{3}} + \dots \frac{a_{n}}{a_{n}} = \frac{1}{1} + \frac{1}{a_{1}} + \frac{a_{1}}{a_{2}} + \frac{a_{2}}{a_{3}} + \dots \frac{a_{n - 2}}{a_{n - 1}}

A_{1}, A_{2},

A_{3}

R^{2}

A_{1} = {(x, y) : x \geq 0, y \geq 0, 2 x + 2 y - x^{2} - y^{2} > 1 > x + y}, A_{2} = {(x, y) : x \geq 0, y \geq 0, x + y > 1 > x^{2} + y^{2}}, A_{3} = {(x, y) : x \geq 0, y \geq 0, x + y > 1 > x^{3} + y^{3}} .

| A_{1} |, | A_{2} |,

| A_{3} |

A_{1}, A_{2}

A_{3}

\frac{a_{1}}{a_{1} + 1 -} \frac{a_{2}}{a_{2} + 1 -} \frac{a_{3}}{a_{3} + 1 -} \dots, a_{1}, a_{2}, a_{3}, \dots

a_{1}, a_{2}, a_{3}, . . . . . . . . ., a_{n}

(a_{1} + a_{2} + a_{3} + . . . . . . . + a_{n}) (\frac{1}{a_{1}} + \frac{1}{a_{2}} + \frac{1}{a_{3}} . . . . . . . . . . . . + \frac{1}{a_{n}}) \geq n^{2}

A_{1} A_{2} A_{3} . . . A_{n}

n

\frac{1}{A_{1} A_{2}} = \frac{1}{A_{1} A_{3}} + \frac{1}{A_{1} A_{4}},

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