Find-the-side-of-a-triangle-if-the-distances-from-an-arbitrary-point-inside-a-regular-triangle-to-its-vertices-are-m-n-and-k-Help-please-

Question Number 208872 by Ismoiljon_008 last updated on 26/Jun/24

$$ \\ $$$$\:\:\:{Find}\:{the}\:{side}\:{of}\:{a}\:{triangle}\:{if}\:{the}\:{distances} \\ $$$$\:\:\:{from}\:{an}\:{arbitrary}\:{point}\:{inside}\:{a}\:{regular}\:{triangle}\: \\ $$$$\:\:\:{to}\:{its}\:{vertices}\:{are}\:{m},\:{n}\:{and}\:{k}. \\ $$$$\:\:{Help}\:{please} \\ $$

Answered by mr W last updated on 26/Jun/24

Commented by mr W last updated on 26/Jun/24

B(−(s/2),0) C((s/2),0) A(0,(((√3)s)/2)) G(h,k) say h^2 +(k−(((√3)s)/2))^2 =p^2 ...(i) (h+(s/2))^2 +k^2 =q^2 ...(ii) (h−(s/2))^2 +k^2 =r^2 ...(iii) (ii)−(iii): ⇒2hs=q^2 −r^2 (ii)−(i): hs+(s^2 /4)+(√3)ks−((3s^2 )/4)=q^2 −p^2 ⇒2(√3)ks=s^2 +q^2 +r^2 −2p^2 into (iii): 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 =((p^2 +q^2 +r^2 ±(√((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) =((p^2 +q^2 +r^2 ±(√(6(p^2 q^2 +q^2 r^2 +r^2 p^2 )−3(p^4 +q^4 +r^4 ))))/2) =((p^2 +q^2 +r^2 ±(√(3(p+q+r)(−p+q+r)(p−q+r)(p+q−r))))/2) ⇒s=(√((p^2 +q^2 +r^2 ±(√(3(p+q+r)(−p+q+r)(p−q+r)(p+q−r))))/2)) +: point inside the triangle −: point outside the triangle

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

Commented by Tawa11 last updated on 26/Jun/24