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Question-83471




Question Number 83471 by mr W last updated on 02/Mar/20
Commented by mr W last updated on 02/Mar/20
The distances from a point M to the  vertices of a given triangle with  side lengthes a, b, c are p, q, r  respectively.  Find the correlation between theses  distances.
$${The}\:{distances}\:{from}\:{a}\:{point}\:{M}\:{to}\:{the} \\ $$$${vertices}\:{of}\:{a}\:{given}\:{triangle}\:{with} \\ $$$${side}\:{lengthes}\:\boldsymbol{{a}},\:\boldsymbol{{b}},\:\boldsymbol{{c}}\:{are}\:\boldsymbol{{p}},\:\boldsymbol{{q}},\:\boldsymbol{{r}} \\ $$$${respectively}. \\ $$$${Find}\:{the}\:{correlation}\:{between}\:{theses} \\ $$$${distances}. \\ $$
Answered by mr W last updated on 03/Mar/20
If the point M is “above” the triangle  ABC, then we have a pyramid. The  volume of the pyramid is:  (see Q40469)
$${If}\:{the}\:{point}\:{M}\:{is}\:“{above}''\:{the}\:{triangle} \\ $$$${ABC},\:{then}\:{we}\:{have}\:{a}\:{pyramid}.\:{The} \\ $$$${volume}\:{of}\:{the}\:{pyramid}\:{is}: \\ $$$$\left({see}\:{Q}\mathrm{40469}\right) \\ $$
Commented by mr W last updated on 03/Mar/20
Commented by mr W last updated on 03/Mar/20
since the point M is coplanar with  the triangle ABC, the volume is 0.  therefore  a^2 p^2 f_a +b^2 q^2 f_b +c^2 r^2 f_c −δ=0  this is an equation which the  distances p, q, r must satisfy.
$${since}\:{the}\:{point}\:{M}\:{is}\:{coplanar}\:{with} \\ $$$${the}\:{triangle}\:{ABC},\:{the}\:{volume}\:{is}\:\mathrm{0}. \\ $$$${therefore} \\ $$$${a}^{\mathrm{2}} {p}^{\mathrm{2}} {f}_{{a}} +{b}^{\mathrm{2}} {q}^{\mathrm{2}} {f}_{{b}} +{c}^{\mathrm{2}} {r}^{\mathrm{2}} {f}_{{c}} −\delta=\mathrm{0} \\ $$$${this}\:{is}\:{an}\:{equation}\:{which}\:{the} \\ $$$${distances}\:{p},\:{q},\:{r}\:{must}\:{satisfy}. \\ $$
Answered by mr W last updated on 03/Mar/20
let α=∠BMC, β=∠CMA, γ=∠AMB  cos α=((q^2 +r^2 −a^2 )/(2qr))  cos β=((r^2 +p^2 −b^2 )/(2rp))  cos γ=((p^2 +q^2 −c^2 )/(2pq))  α+β=2π−γ  cos (α+β)=cos γ  ((q^2 +r^2 −a^2 )/(2qr))×((r^2 +p^2 −b^2 )/(2rp))−((√(4q^2 r^2 −(q^2 +r^2 −a^2 )^2 ))/(2qr))×((√(4r^2 p^2 −(r^2 +p^2 −b^2 )^2 ))/(2rp))=((p^2 +q^2 −c^2 )/(2pq))  ⇒(q^2 +r^2 −a^2 )(r^2 +p^2 −b^2 )−(√([4q^2 r^2 −(q^2 +r^2 −a^2 )^2 ][4r^2 p^2 −(r^2 +p^2 −b^2 )^2 ]))=2r^2 (p^2 +q^2 −c^2 )
$${let}\:\alpha=\angle{BMC},\:\beta=\angle{CMA},\:\gamma=\angle{AMB} \\ $$$$\mathrm{cos}\:\alpha=\frac{{q}^{\mathrm{2}} +{r}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{qr}} \\ $$$$\mathrm{cos}\:\beta=\frac{{r}^{\mathrm{2}} +{p}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{rp}} \\ $$$$\mathrm{cos}\:\gamma=\frac{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{pq}} \\ $$$$\alpha+\beta=\mathrm{2}\pi−\gamma \\ $$$$\mathrm{cos}\:\left(\alpha+\beta\right)=\mathrm{cos}\:\gamma \\ $$$$\frac{{q}^{\mathrm{2}} +{r}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{qr}}×\frac{{r}^{\mathrm{2}} +{p}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{rp}}−\frac{\sqrt{\mathrm{4}{q}^{\mathrm{2}} {r}^{\mathrm{2}} −\left({q}^{\mathrm{2}} +{r}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} }}{\mathrm{2}{qr}}×\frac{\sqrt{\mathrm{4}{r}^{\mathrm{2}} {p}^{\mathrm{2}} −\left({r}^{\mathrm{2}} +{p}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} }}{\mathrm{2}{rp}}=\frac{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{pq}} \\ $$$$\Rightarrow\left({q}^{\mathrm{2}} +{r}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)\left({r}^{\mathrm{2}} +{p}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)−\sqrt{\left[\mathrm{4}{q}^{\mathrm{2}} {r}^{\mathrm{2}} −\left({q}^{\mathrm{2}} +{r}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} \right]\left[\mathrm{4}{r}^{\mathrm{2}} {p}^{\mathrm{2}} −\left({r}^{\mathrm{2}} +{p}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} \right]}=\mathrm{2}{r}^{\mathrm{2}} \left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} −{c}^{\mathrm{2}} \right) \\ $$

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