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Question Number 52911 by ajfour last updated on 15/Jan/19

Commented by ajfour last updated on 15/Jan/19

$$letforexample:q=5,p=7,r=4.$$

Commented by MJS last updated on 16/Jan/19

$$\mathrm{not}\mathrm{sure}\mathrm{if}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{getting}\mathrm{easier},\mathrm{but}\mathrm{there}\mathrm{are}$$$$\mathrm{also}\mathrm{pairs}\mathrm{of}\mathrm{triangles}\mathrm{with}\mathrm{equal}\mathrm{areas}$$$$△bcp\equiv △cqr=\frac{bc}{2}$$$$△abq\equiv △apr=\frac{ab}{2}$$

Answered by mr W last updated on 15/Jan/19

$$a^2+b^2=q^2$$$$b^2+c^2=p^2$$$$b^2+{(c-a)}^2=r^2$$$$$$$$c^2-{(c-a)}^2=p^2-r^2$$$$2ac-a^2=p^2-r^2$$$$\Rightarrow c=\frac{a^2+p^2-r^2}{2a}$$$$q^2-a^2+{\left(\frac{a^2+p^2-r^2}{2a}\right)}^2=p^2$$$$a^2-{\left(\frac{a^2+p^2-r^2}{2a}\right)}^2=q^2-p^2$$$$3a^4+2(p^2+r^2-2q^2)a^2-{(p^2-r^2)}^2=0$$$$a^2=\frac{\sqrt{{(p^2+r^2-2q^2)}^2+3{(p^2-r^2)}^2}-(p^2+r^2-2q^2)}{3}$$$$\Rightarrow a=\sqrt{\frac{\sqrt{{(p^2+r^2-2q^2)}^2+3{(p^2-r^2)}^2}-(p^2+r^2-2q^2)}{3}}$$$$\Rightarrow c=...$$$$\Rightarrow b=\sqrt{q^2-a^2}=...$$$$$$$$example:p=7,q=5,r=4$$$$\Rightarrow a=\sqrt{\frac{\sqrt{{(49+16-50)}^2+3{(49-16)}^2}-(49+16-50)}{3}}$$$$\Rightarrow a=\sqrt{\frac{6\sqrt{97}-15}{3}}\approx 3.834$$$$...$$

Commented by ajfour last updated on 15/Jan/19

$$ThankyouSir,thiswastoughforme.$$

Answered by MJS last updated on 16/Jan/19

$$a=\sqrt{\frac{-p^2+2q^2-r^2}{3}+\frac{2\sqrt{p^4+q^4+r^4-(p^2{q}^2+p^2{r}^2+q^2{r}^2)}}{3}}$$$$b=\sqrt{\frac{p^2+q^2+r^2}{3}-\frac{2\sqrt{p^4+q^4+r^4-(p^2{q}^2+p^2{r}^2+q^2{r}^2)}}{3}}$$$$c=\sqrt{\frac{2p^2-q^2-r^2}{3}+\frac{2\sqrt{p^4+q^4+r^4-(p^2{q}^2+p^2{r}^2+q^2{r}^2)}}{3}}$$$$\mathrm{with}p=7q=5r=4\mathrm{I}\mathrm{get}$$$$a=\sqrt{-5+2\sqrt{97}}\approx 3.83376$$$$b=\sqrt{30-2\sqrt{97}}\approx 3.20972$$$$c=\sqrt{19+2\sqrt{97}}\approx 6.22075$$

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