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Question Number 123040 by ajfour last updated on 22/Nov/20

Commented by ajfour last updated on 22/Nov/20

$$Findmaximumsidelengthof$$$$equilateral△ABC,intermsof$$$$p,q,andr;theverticesof$$$$whichlierespectivelyonthree$$$$concentriccirclesofradii$$$$p<q<r.AssumevertexAisas$$$$shownonoutermostcircleandon$$$$verticalaxes.$$

Commented by mr W last updated on 22/Nov/20

$$wegettwotriangleswithsidelength$$$$s=\sqrt{\frac{p^2+q^2+r^2\pm \sqrt{3\delta }}{2}}with$$$$\delta =(p+q+r)(-p+q+r)(p-q+r)(p+q-r)$$

Commented by ajfour last updated on 23/Nov/20

Commented by mr W last updated on 23/Nov/20

$$great!$$

Answered by mr W last updated on 22/Nov/20

Commented by ajfour last updated on 23/Nov/20

$$Thanksalotsir.$$

Commented by mr W last updated on 22/Nov/20

$$B(-\frac{s}{2},0)$$$$C(\frac{s}{2},0)$$$$A(0,\frac{\sqrt{3}s}{2})$$$$G(h,k)say$$$$h^2+{(k-\frac{\sqrt{3}s}{2})}^2=p^2...\left(i\right)$$$${(h+\frac{s}{2})}^2+k^2=q^2...\left(ii\right)$$$${(h-\frac{s}{2})}^2+k^2=r^2...\left(iii\right)$$$$\left(ii\right)-\left(iii\right):$$$$\Rightarrow 2hs=q^2-r^2$$$$\left(ii\right)-\left(i\right):$$$$hs+\frac{s^2}{4}+\sqrt{3}ks-\frac{3s^2}{4}=q^2-p^2$$$$\Rightarrow 2\sqrt{3}ks=s^2+q^2+r^2-2p^2$$$$into\left(iii\right):$$$$3{(q^2-r^2-s^2)}^2+{(s^2+q^2+r^2-2p^2)}^2=12r^2{s}^2$$$$s^4-(p^2+q^2+r^2)s^2+p^4+q^4+r^4-p^2{q}^2-q^2{r}^2-r^2{p}^2=0$$$$s^2=\frac{p^2+q^2+r^2\pm \sqrt{{(p^2+q^2+r^2)}^2-4(p^4+q^4+r^4-p^2{q}^2-q^2{r}^2-r^2{p}^2)}}{2}$$$$=\frac{p^2+q^2+r^2\pm \sqrt{6(p^2{q}^2+q^2{r}^2+r^2{p}^2)-3(p^4+q^4+r^4)}}{2}$$$$=\frac{p^2+q^2+r^2\pm \sqrt{3(p+q+r)(-p+q+r)(p-q+r)(p+q-r)}}{2}$$$$\Rightarrow s=\sqrt{\frac{p^2+q^2+r^2\pm \sqrt{3(p+q+r)(-p+q+r)(p-q+r)(p+q-r)}}{2}}$$

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