Question Number 223626 by fantastic last updated on 01/Aug/25
$$\begin{cases}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}=\mathrm{25}}\\{{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +{yz}=\mathrm{49}}\\{{z}^{\mathrm{2}} +{x}^{\mathrm{2}} +{zx}=\mathrm{64}}\end{cases} \\ $$$$\left({x}+{y}+{z}\right)^{\mathrm{2}} −\mathrm{100}=?? \\ $$
Commented by Frix last updated on 01/Aug/25
$$\mathrm{Show}\:\mathrm{it}\:\mathrm{please}. \\ $$
Commented by Frix last updated on 01/Aug/25
$$\mathrm{2}\:\mathrm{answers}:\:−\mathrm{91}\:\mathrm{or}\:+\mathrm{29} \\ $$
Commented by fantastic last updated on 01/Aug/25
$$ \\ $$
Commented by Frix last updated on 01/Aug/25
$$\mathrm{We}\:\mathrm{can}\:\mathrm{simply}\:\mathrm{solve}\:\mathrm{it} \\ $$$${x}=\mathrm{0}\wedge{y}=\pm\mathrm{5}\wedge{z}=\mp\mathrm{8} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:−\mathrm{91} \\ $$$${x}=\pm\frac{\mathrm{40}}{\:\sqrt{\mathrm{129}}}\wedge{y}=\pm\frac{\mathrm{25}}{\:\sqrt{\mathrm{129}}}\wedge{z}=\pm\frac{\mathrm{64}}{\:\sqrt{\mathrm{129}}} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{29} \\ $$
Commented by fantastic last updated on 01/Aug/25
$${Sir}\:{i}\:{have}\:{another}\:{method} \\ $$
Commented by Frix last updated on 01/Aug/25
$$\mathrm{We}\:\mathrm{can}\:\mathrm{always}\:\mathrm{solve}\:\mathrm{it},\:\mathrm{even}\:\mathrm{if}\:\mathrm{there}'\mathrm{s}\:\mathrm{no} \\ $$$$\mathrm{triangle},\:\mathrm{but}\:\mathrm{not}\:\mathrm{always}\:\mathrm{in}\:\mathbb{R}: \\ $$$$\begin{cases}{{x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} =\mathrm{1}}\\{{y}^{\mathrm{2}} +{yz}+{z}^{\mathrm{2}} =\mathrm{2}}\\{{x}^{\mathrm{2}} +{xz}+{z}^{\mathrm{2}} =\mathrm{6}}\end{cases} \\ $$$$\left(\mathrm{3}\right)−\left(\mathrm{2}\right)\:\Rightarrow\:{z}=−\frac{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} −\mathrm{4}}{{x}−{y}} \\ $$$$\mathrm{Inserting}\:\mathrm{in}\:\left(\mathrm{2}\right)\:\mathrm{or}\:\left(\mathrm{3}\right)\:\mathrm{and}\:\mathrm{transforming} \\ $$$${y}^{\mathrm{4}} −{xy}×{y}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} −{x}\left({x}^{\mathrm{2}} −\mathrm{8}\right){y}+{x}^{\mathrm{4}} −\mathrm{10}{x}^{\mathrm{2}} +\mathrm{16}=\mathrm{0} \\ $$$$\mathrm{Using}\:\left(\mathrm{1}\right): \\ $$$${y}^{\mathrm{2}} =−{x}^{\mathrm{2}} −{xy}+\mathrm{1} \\ $$$${y}^{\mathrm{4}} =\left({x}^{\mathrm{2}} +{xy}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{Inserting}\:\mathrm{above}\:\mathrm{leads}\:\mathrm{to} \\ $$$$\mathrm{3}{xy}−\mathrm{12}{x}^{\mathrm{2}} +\mathrm{19}=\mathrm{0}\:\Rightarrow\:{y}=\frac{\mathrm{12}{x}^{\mathrm{2}} −\mathrm{19}}{\mathrm{3}{x}} \\ $$$$\mathrm{Inserting}\:\mathrm{in}\:\left(\mathrm{1}\right)\:\Rightarrow \\ $$$${x}^{\mathrm{4}} −\frac{\mathrm{58}}{\mathrm{21}}{x}^{\mathrm{2}} +\frac{\mathrm{361}}{\mathrm{189}}=\mathrm{0} \\ $$$$… \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{12}}}{\mathrm{12}}\mathrm{i} \\ $$$$ \\ $$$$\mathrm{Same}\:\mathrm{method}\:\mathrm{with}\:\mathrm{constants}\:\mathrm{9},\:\mathrm{16},\:\mathrm{49} \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{37} \\ $$
Answered by fantastic last updated on 01/Aug/25
Commented by fantastic last updated on 01/Aug/25
![x^2 +y^2 +xy=25 or x^2 +y^2 −2xy×cos 120^0 =25[∵cos 120^0 =−(1/2)] similarly we will get y^2 +z^2 −2yzcos 120^0 =7^2 z^2 +x^2 −2zxcos 120^0 =8^2 we can make a triangle out of these the big triangles area=those 3 triangles total area ∴10(√3)=((√3)/4)(xy+yz+zx) ∴xy+yz+zx=40 combining those three equation we will get 2x^2 +2y^2 +2z^2 +xy+yz+zx=25+49+64=138 or 2(x^2 +y^2 +z^2 )=138−40=98 x^2 +y^2 +z^2 =49 (x+y+z)^2 =x^2 +y^2 +z^2 +2(xy+yz+zx) ⇒49+2×40=49+80=129 So (x+y+z)^2 −100=129−100=29✓](https://www.tinkutara.com/question/Q223646.png)
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}=\mathrm{25} \\ $$$${or}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{xy}×\mathrm{cos}\:\mathrm{120}^{\mathrm{0}} =\mathrm{25}\left[\because\mathrm{cos}\:\mathrm{120}^{\mathrm{0}} =−\frac{\mathrm{1}}{\mathrm{2}}\right] \\ $$$${similarly}\:{we}\:{will}\:{get} \\ $$$${y}^{\mathrm{2}} +{z}^{\mathrm{2}} −\mathrm{2}{yz}\mathrm{cos}\:\mathrm{120}^{\mathrm{0}} =\mathrm{7}^{\mathrm{2}} \\ $$$${z}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{2}{zx}\mathrm{cos}\:\mathrm{120}^{\mathrm{0}} =\mathrm{8}^{\mathrm{2}} \\ $$$${we}\:{can}\:{make}\:{a}\:{triangle}\:{out}\:{of}\:{these} \\ $$$${the}\:{big}\:{triangles}\:{area}={those}\:\mathrm{3}\:{triangles}\:{total}\:{area} \\ $$$$\therefore\mathrm{10}\sqrt{\mathrm{3}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\left({xy}+{yz}+{zx}\right) \\ $$$$\therefore{xy}+{yz}+{zx}=\mathrm{40} \\ $$$${combining}\:{those}\:{three}\:{equation}\:{we}\:{will}\:{get} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} +\mathrm{2}{z}^{\mathrm{2}} +{xy}+{yz}+{zx}=\mathrm{25}+\mathrm{49}+\mathrm{64}=\mathrm{138} \\ $$$${or}\:\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)=\mathrm{138}−\mathrm{40}=\mathrm{98} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{49} \\ $$$$\left({x}+{y}+{z}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +\mathrm{2}\left({xy}+{yz}+{zx}\right) \\ $$$$\Rightarrow\mathrm{49}+\mathrm{2}×\mathrm{40}=\mathrm{49}+\mathrm{80}=\mathrm{129} \\ $$$${So}\:\left({x}+{y}+{z}\right)^{\mathrm{2}} −\mathrm{100}=\mathrm{129}−\mathrm{100}=\mathrm{29}\checkmark \\ $$
Commented by Frix last updated on 01/Aug/25
$$\mathrm{Yes}\:\mathrm{but}\:\mathrm{you}\:\mathrm{lost}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{solution}. \\ $$
Commented by mr W last updated on 02/Aug/25
$${x},{y},{z}\:{are}\:{distances}\:{from}\:{Fermat}\: \\ $$$${point}\:{to}\:{the}\:{vertexes}\:{of}\:{a}\:{triangle}. \\ $$$${see}\:{Q}\mathrm{164174} \\ $$
Commented by fantastic last updated on 01/Aug/25
$${there}\:{must}\:{be}\:{a}\:{way} \\ $$