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Question Number 111799 by ajfour last updated on 05/Sep/20

Commented by mr W last updated on 05/Sep/20

$s i r, n o t p o s s i b l e : A = B = C$
Commented by ajfour last updated on 05/Sep/20

$I f r e g i o n s A, B, C h a v e e q u a l a r e a s,$ $f i n d h, k .$
Commented by ajfour last updated on 05/Sep/20

$y = {(x - h)}^{2}$ $A = B = C i f$ $k = \frac{h^{2}}{4}$ $e x a m p l e : h = 2, k = 1$ $A + B = \frac{h^{3}}{3} = \frac{8}{3}$ $C = 2 \times 1 - 2 (\frac{1}{3}) = \frac{4}{3}$ $B = 1 \times 1 + \frac{1}{3} = \frac{4}{3}$
Commented by mr W last updated on 05/Sep/20

$i m a d e a m i s t a k e . i n f a c t {i t}^{'} s t r u e$ $f o r a n y h = 2 \sqrt{k} .$
	Answered by 1549442205PVT last updated on 05/Sep/20

	Answered by mr W last updated on 05/Sep/20

	$C = \frac{2}{3} \times k \times 2 \sqrt{k} = \frac{4 k \sqrt{k}}{3}$ $B = (h - \sqrt{k}) k + \frac{1}{3} \times k \sqrt{k} = h k - \frac{2 k \sqrt{k}}{3}$ $A + B = \frac{h^{3}}{3}$ $B = C :$ $h k - \frac{2 k \sqrt{k}}{3} = \frac{4 k \sqrt{k}}{3}$ $\Rightarrow h = 2 \sqrt{k}$ $A + B = 2 C :$ $\frac{h^{3}}{3} = 2 \times \frac{4 k \sqrt{k}}{3}$ $h^{3} = 8 k \sqrt{k}$ $\Rightarrow h = 2 \sqrt{k}$ $i . e . f o r a n y h = 2 \sqrt{k} w e a l w a y s h a v e$ $A = B = C .$
	Commented by ajfour last updated on 05/Sep/20

	$Y e s! T h a n k s S i r .$
	Answered by 1549442205PVT last updated on 05/Sep/20

	$Suppose F (- a, 0), A (\sqrt{k}, 0), B (- \sqrt{k}, 0)$ $A = \int_{- a}^{- \sqrt{k}} (x^{2} - k) dx = \frac{x^{3}}{3} - kx ∣_{- a}^{- \sqrt{k}} = \frac{- k \sqrt{k}}{3} + \frac{a^{3}}{3} + k \sqrt{k} - a (1)$ $B = \int_{- \sqrt{k}}^{0} x^{2} dx + k (a - \sqrt{k}) = \frac{x^{3}}{3} ∣_{- \sqrt{k}}^{0} + k (a - \sqrt{k})$ $= \frac{k \sqrt{k}}{3} + k (a - \sqrt{k}) = ka - \frac{2 \sqrt{k}}{3} (2)$ $C = \int_{- \sqrt{k}}^{\sqrt{k}} (k - x^{2}) dx = (kx - \frac{x^{3}}{3}) ∣_{- \sqrt{k}}^{\sqrt{k}}$ $= 2 k \sqrt{k} - 2 \frac{k \sqrt{k}}{3} = \frac{4 k \sqrt{k}}{3} (3)$ $B = C \Leftrightarrow \frac{4 k \sqrt{k}}{3} = \frac{3 ka - 2 \sqrt{k}}{3}$ $\Leftrightarrow 4 k = 3 a \sqrt{k} - 2 \Leftrightarrow 4 k - 3 a \sqrt{k} + 2 = 0 (4)$ $A = C \Leftrightarrow a^{3} + 2 k \sqrt{k} - 3 a = 4 k \sqrt{k}$ $\Leftrightarrow a^{3} - 3 a = 2 k \sqrt{k} (5)$ $(4) \Rightarrow a = \frac{4 k + 2}{3 \sqrt{k}} . Replace into (5) we get$ ${64 k}^{3} + {96 k}^{2} + 48 k + 8 - 9 k (12 k + 6) = 27 k \sqrt{k} . 2 k \sqrt{k}$ $\Leftrightarrow {10 k}^{3} - {12 k}^{2} - 6 k + 8 = 0 .$ $\Leftrightarrow {(k - 1)}^{2} (10 k + 8) = 0$ $\Leftrightarrow k - 1 = 0 \Leftrightarrow k = 1, h = a = 2$ $Thus A = B = C \Leftrightarrow k = 1, h = 2$
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