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Question Number 145202 by mnjuly1970 last updated on 03/Jul/21

Commented by loveineq last updated on 03/Jul/21

$$\mathrm{Wow},\mathrm{it}\mathrm{likes}\mathrm{a}\mathrm{therom}.$$

Answered by ajfour last updated on 03/Jul/21

$${(x+3p)}^2+y^2=a^2$$$${(x+p)}^2+y^2=b^2$$$${(x-p)}^2+y^2=c^2$$$${(x+3p)}^2+y^2=d^2$$$$a^2+3c^2=4x^2+12p^2+4y^2$$$$d^2+3b^2=4x^2+12p^2+4y^2$$$$hence$$$$a^2+3c^2=d^2+3b^2$$

Commented by mnjuly1970 last updated on 03/Jul/21

$$thankyousomuchmrajfor$$

Answered by mr W last updated on 03/Jul/21

$$\frac{x^2+b^2-a^2}{2xb}=-\frac{x^2+b^2-c^2}{2xb}$$$$\Rightarrow 2x^2=a^2+c^2-2b^2...\left(i\right)$$$$similarly$$$$\Rightarrow 2x^2=b^2+d^2-2c^2...\left(ii\right)$$$$a^2+c^2-2b^2=b^2+d^2-2c^2$$$$\Rightarrow a^2+3c^2=d^2+3b^2$$

Commented by mnjuly1970 last updated on 03/Jul/21

$$thankalotmrW,verynice$$$$solution...$$

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