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Question Number 217358 by mnjuly1970 last updated on 11/Mar/25

Answered by mr W last updated on 12/Mar/25

$$let\mathrm{\Delta }=areaoftriangle\mathrm{\Delta }ABC$$$$wehave$$$$\mathrm{\Delta }=\frac{\sqrt{2(a^2{b}^2+b^2{c}^2+c^2{a}^2)-(a^4+b^4+c^4)}}{4}$$$$\mathrm{\Delta }=\frac{ca\sin B}{2}=\frac{ab\sin C}{2}=\frac{bc\sin A}{2}$$$$$$$$\frac{xc\sin \omega }{2}+\frac{ya\sin \omega }{2}+\frac{zb\sin \omega }{2}=\mathrm{\Delta }$$$$\Rightarrow (xc+ya+zb)\sin \omega =2\mathrm{\Delta }...\left(i\right)$$$$$$$$y^2=x^2+c^2-2xc\cos \omega$$$$z^2=y^2+a^2-2ya\cos \omega$$$$x^2=z^2+b^2-2zb\cos \omega$$$$\Rightarrow (xc+ya+zb)\cos \omega =\frac{a^2+b^2+c^2}{2}...\left(ii\right)$$$$\left(i\right)/\left(ii\right):$$$$\Rightarrow \tan \omega =\frac{4\mathrm{\Delta }}{a^2+b^2+c^2}$$$$\sin \omega =\frac{4\mathrm{\Delta }}{\sqrt{{\left(4\mathrm{\Delta }\right)}^2+{(a^2+b^2+c^2)}^2}}$$$$=\frac{4\mathrm{\Delta }}{\sqrt{2(a^2{b}^2+b^2{c}^2+c^2{a}^2)-(a^4+b^4+c^4)+a^4+b^4+c^4+2(a^2{b}^2+b^2{c}^2+c^2{a}^2)}}$$$$=\frac{2\mathrm{\Delta }}{\sqrt{a^2{b}^2+b^2{c}^2+c^2{a}^2}}$$$$\Rightarrow \frac{\sin \omega }{2\mathrm{\Delta }}=\frac{1}{\sqrt{a^2{b}^2+b^2{c}^2+c^2{a}^2}}$$$$$$$$in\mathrm{\Delta }PABwehave$$$$\frac{y}{\sin \omega }=\frac{c}{\sin (\omega +B-\omega )}=\frac{c}{\sin B}=\frac{c^{2}a}{ca\sin B}=\frac{c^{2}a}{2\mathrm{\Delta }}$$$$\Rightarrow y=\frac{c^{2}a\sin \omega }{2\mathrm{\Delta }}=\frac{c^{2}a}{\sqrt{a^2{b}^2+b^2{c}^2+c^2{a}^2}}$$$$similarly$$$$z=\frac{a^{2}b\sin \omega }{2\mathrm{\Delta }}=\frac{a^{2}b}{\sqrt{a^2{b}^2+b^2{c}^2+c^2{a}^2}}$$$$x=\frac{b^{2}c\sin \omega }{2\mathrm{\Delta }}=\frac{b^{2}c}{\sqrt{a^2{b}^2+b^2{c}^2+c^2{a}^2}}$$$$\Rightarrow x+y+z=\frac{a^{2}b+b^{2}c+c^{2}a}{\sqrt{a^2{b}^2+b^2{c}^2+c^2{a}^2}}✓$$

Commented by mnjuly1970 last updated on 12/Mar/25

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