Montrer-que-4r-2-a-2-b-2-c-2-d-2-

Question Number 187142 by a.lgnaoui last updated on 14/Feb/23

M o n t r e r q u e :

4 r^{2} = a^{2} + b^{2} + c^{2} + d^{2}

Commented by a.lgnaoui last updated on 14/Feb/23

Commented by mahdipoor last updated on 14/Feb/23

C (O, r) : x^{2} + y^{2} = r^{2}

P = (i, j) {\begin{cases} l i n e C P D : x = i \\ l i n e A P B : y = j \end{cases}

A = (- \sqrt{r^{2} - j^{2}}, j) B = (+ \sqrt{r^{2} - j^{2}}, j)

C = (i, \sqrt{r^{2} - i^{2}}) D = (i, - \sqrt{r^{2} - i^{2}})

a^{2} + b^{2} + c^{2} + d^{2} =∣ A P ∣^{2} + ∣ A B ∣^{2} + \dots =

[{(i - (- \sqrt{r^{2} - j^{2}}))}^{2} + {(j - j)}^{2}] + [{(i - \sqrt{r^{2} - j^{2}})}^{2} + {(j - j)}^{2}]

+ \dots = 2 i^{2} + 2 (r^{2} - j^{2}) + 2 j^{2} + 2 (r^{2} - i^{2}) = 4 r^{2}

Answered by mr W last updated on 14/Feb/23

{(b - \frac{a + b}{2})}^{2} + {(\frac{c + d}{2})}^{2} = r^{2}

\Rightarrow {(a - b)}^{2} + {(c + d)}^{2} = 4 r^{2} \dots (i)

{(d - \frac{c + d}{2})}^{2} + {(\frac{a + b}{2})}^{2} = r^{2}

\Rightarrow {(c - d)}^{2} + {(a + b)}^{2} = 4 r^{2} \dots (i i)

(i) - (i i) :

4 c d - 4 a b = 0 \Rightarrow c d = a b

(i) :

a^{2} + b^{2} - 2 a b + c^{2} + d^{2} + 2 c d = 4 r^{2}

\Rightarrow a^{2} + b^{2} + c^{2} + d^{2} = 4 r^{2} ✓

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