Question Number 15567 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 11/Jun/17
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 11/Jun/17
$${ABCD}\:\&\:{MNPQ},{are}\:{squares}. \\ $$$${show}\:{that}: \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={c}^{\mathrm{2}} +{d}^{\mathrm{2}} \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 13/Jun/17
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 13/Jun/17
$${a}^{\mathrm{2}} ={s}^{\mathrm{2}} +\left({p}+{n}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} ,{b}^{\mathrm{2}} ={q}^{\mathrm{2}} +\left({r}+{n}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${c}^{\mathrm{2}} ={p}^{\mathrm{2}} +\left({q}+{n}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} ,{d}^{\mathrm{2}} ={r}^{\mathrm{2}} +\left({s}+{n}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={s}^{\mathrm{2}} +{q}^{\mathrm{2}} +{p}^{\mathrm{2}} +{r}^{\mathrm{2}} +{n}^{\mathrm{2}} +\mathrm{2}{n}\sqrt{\mathrm{2}}\left({p}+{r}\right)\left({i}\right) \\ $$$${c}^{\mathrm{2}} +{d}^{\mathrm{2}} ={s}^{\mathrm{2}} +{q}^{\mathrm{2}} +{p}^{\mathrm{2}} +{r}^{\mathrm{2}} +{n}^{\mathrm{2}} +\mathrm{2}{n}\sqrt{\mathrm{2}}\left({q}+{s}\right)\left({ii}\right) \\ $$$${q}+{s}++{n}\sqrt{\mathrm{2}}={m},{p}+{r}+{n}\sqrt{\mathrm{2}}={m} \\ $$$$\Rightarrow{p}+{r}={q}+{s} \\ $$$$\left({i}\right)−\left({ii}\right)\Rightarrow\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−\left({c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} =\boldsymbol{{c}}^{\mathrm{2}} +\boldsymbol{{d}}^{\mathrm{2}} \:\:\:\:\:\:\:\:.\blacksquare \\ $$
Answered by ajfour last updated on 12/Jun/17
Commented by ajfour last updated on 12/Jun/17
$$\:{let}\:\:\:\boldsymbol{{s}}\mathrm{cos}\:\theta=\boldsymbol{{s}}_{\boldsymbol{{x}}} ,\:\:\boldsymbol{{s}}\mathrm{sin}\:\theta=\boldsymbol{{s}}_{\boldsymbol{{y}}} \\ $$$$\:\:\boldsymbol{{l}}\:{is}\:{the}\:{side}\:{of}\:{biggest}\:{square} \\ $$$${all}\:{components}\:{of}\:{lengths}\:\boldsymbol{{a}},\:\boldsymbol{{b}}, \\ $$$$\:\boldsymbol{{c}},\:\boldsymbol{{d}},\:{and}\:\boldsymbol{{s}}\:{are}\:{positive}. \\ $$$$\:\:{from}\:{figure}\:{above}, \\ $$$$\:\:\:\:\:{c}_{{x}} ={b}_{{x}} +{s}_{{y}} \:\:\:;\:\:\:{c}_{{y}} ={a}_{{y}} −{s}_{{y}} \\ $$$$\:\:\:\:\:{d}_{{x}} ={a}_{{x}} +{s}_{{y}} \:\:\:;\:\:\:{d}_{{y}} ={b}_{{y}} −{s}_{{y}} \\ $$$$\:\:\boldsymbol{{c}}^{\mathrm{2}} +\boldsymbol{{d}}^{\mathrm{2}} =\boldsymbol{{c}}_{\boldsymbol{{x}}} ^{\mathrm{2}} +\boldsymbol{{c}}_{\boldsymbol{{y}}} ^{\mathrm{2}} +\boldsymbol{{d}}_{\boldsymbol{{x}}} ^{\mathrm{2}} +\boldsymbol{{d}}_{\boldsymbol{{y}}} ^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\left(\boldsymbol{{b}}_{\boldsymbol{{x}}} +\boldsymbol{{s}}_{\boldsymbol{{y}}} \right)^{\mathrm{2}} +\left(\boldsymbol{{a}}_{\boldsymbol{{y}}} −\boldsymbol{{s}}_{\boldsymbol{{y}}} \right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:+\left(\boldsymbol{{a}}_{\boldsymbol{{x}}} +\boldsymbol{{s}}_{\boldsymbol{{y}}} \right)^{\mathrm{2}} +\left(\boldsymbol{{b}}_{\boldsymbol{{y}}} −\boldsymbol{{s}}_{\boldsymbol{{y}}} \right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:=\:\left(\boldsymbol{{a}}_{\boldsymbol{{x}}} ^{\mathrm{2}} +\boldsymbol{{a}}_{\boldsymbol{{y}}} ^{\mathrm{2}} \right)+\left(\boldsymbol{{b}}_{\boldsymbol{{x}}} ^{\mathrm{2}} +\boldsymbol{{b}}_{\boldsymbol{{y}}} ^{\mathrm{2}} \right)+\mathrm{4}\boldsymbol{{s}}_{\boldsymbol{{y}}} ^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:+\mathrm{2}\boldsymbol{{s}}_{\boldsymbol{{y}}} \left[\left(\boldsymbol{{a}}_{\boldsymbol{{x}}} +\boldsymbol{{b}}_{\boldsymbol{{x}}} \right)−\left(\boldsymbol{{a}}_{\boldsymbol{{y}}} +\boldsymbol{{b}}_{\boldsymbol{{y}}} \right)\right] \\ $$$$\:\:\:{Now}\:\:\left(\boldsymbol{{a}}_{\boldsymbol{{x}}} +\boldsymbol{{b}}_{\boldsymbol{{x}}} \right)=\boldsymbol{{l}}−\left(\boldsymbol{{s}}_{\boldsymbol{{x}}} +\boldsymbol{{s}}_{\boldsymbol{{y}}} \right) \\ $$$$\:\:{and}\:\:\:\:\:\left(\boldsymbol{{a}}_{\boldsymbol{{y}}} +\boldsymbol{{b}}_{\boldsymbol{{y}}} \right)=\boldsymbol{{l}}−\left(\boldsymbol{{s}}_{\boldsymbol{{x}}} −\boldsymbol{{s}}_{\boldsymbol{{y}}} \right) \\ $$$$\Rightarrow\:\left(\boldsymbol{{a}}_{\boldsymbol{{x}}} +\boldsymbol{{b}}_{\boldsymbol{{x}}} \right)−\left(\boldsymbol{{a}}_{\boldsymbol{{y}}} +\boldsymbol{{b}}_{\boldsymbol{{y}}} \right)=\:−\mathrm{2}\boldsymbol{{s}}_{\boldsymbol{{y}}} \\ $$$${continuing},\:{we}\:{have} \\ $$$$\:\:\:\:\:\boldsymbol{{c}}^{\mathrm{2}} +\boldsymbol{{d}}^{\mathrm{2}} \:=\:\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{{s}}_{\boldsymbol{{y}}} ^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{2}\boldsymbol{{s}}_{\boldsymbol{{y}}} \left(−\mathrm{2}\boldsymbol{{s}}_{\boldsymbol{{y}}} \right) \\ $$$${or}\:\:\:\:\:\boldsymbol{{c}}^{\mathrm{2}} +\boldsymbol{{d}}^{\mathrm{2}} \:=\:\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} \:. \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 12/Jun/17
$${excellent}!\:{mr}\:{Ajcour}.{thank}\:{you}\:{very}\: \\ $$$${much}. \\ $$
Commented by ajfour last updated on 12/Jun/17
$${thank}\:{you}\:{sir},\:{i}\:{shall}\:{yet}\:{look}\:{for} \\ $$$$\:{a}\:{shorter}\:{way}.. \\ $$
Answered by ajfour last updated on 12/Jun/17
Commented by ajfour last updated on 12/Jun/17