Question Number 161015 by mr W last updated on 10/Dec/21
Commented by mr W last updated on 11/Dec/21
$${The}\:{distances}\:{from}\:{the}\:{orthocenter} \\ $$$${to}\:{the}\:{vertexes}\:{of}\:{a}\:{triangle}\:{are} \\ $$$$\boldsymbol{{p}},\:\boldsymbol{{q}},\:\boldsymbol{{r}}\:{respectively}.\:{Find}\:{the}\:{side} \\ $$$${lengthes}\:{of}\:{the}\:{triangle}. \\ $$
Answered by mr W last updated on 11/Dec/21
Commented by mr W last updated on 14/Dec/21
![(p/v)=(q/u) ⇒pu=qv (p/w)=(r/u) ⇒pu=rw ⇒ determinant (((pu=qv=rw)))=(1/k), say ⇒u=(1/(pk)), v=(1/(qk)), w=(1/(rk)) say the area of triangle ABC is Δ and the side lengthes are a, b, c. Δ=(1/2)a(p+u)=(1/2)b(q+v)=(1/2)c(r+w) Δ=(1/2)(au+bv+cw) Δ=(1/2)(((2Δu)/(p+u))+((2Δv)/(q+v))+((2Δw)/(r+w))) ⇒ determinant ((((u/(p+u))+(v/(q+v))+(w/(r+w))=1))) (1/((p/u)+1))+(1/((q/v)+1))+(1/((r/w)+1))=1 (1/(p^2 k+1))+(1/(q^2 k+1))+(1/(r^2 k+1))=1 (p^2 k+1)(q^2 k+1)+(q^2 k+1)(r^2 k+1)+(r^2 k+1)(p^2 k+1)=(p^2 k+1)(q^2 k+1)(r^2 k+1) (p^2 q^2 +q^2 r^2 +r^2 p^2 )k^2 +2(p^2 +q^2 +r^2 )k+3=p^2 q^2 r^2 k^3 +(p^2 q^2 +q^2 r^2 +r^2 p^2 )k^2 +(p^2 +q^2 +r^2 )k+1 p^2 q^2 r^2 k^3 −(p^2 +q^2 +r^2 )k−2=0 determinant (((k^3 −((p^2 +q^2 +r^2 )/(p^2 q^2 r^2 )) k−(2/(p^2 q^2 r^2 ))=0))) (1/(p^4 q^4 r^4 ))−(1/(27))(((p^2 +q^2 +r^2 )/(p^2 q^2 r^2 )))^3 ≤0 ⇒3 real roots generally two of the three roots are suitable. determinant (((k=(2/(pqr))(√((p^2 +q^2 +r^2 )/3)) sin {((2𝛑)/3)−(1/3) sin^(−1) [pqr((3/(p^2 +q^2 +r^2 )))^(3/2) ]})),((k=−(2/(pqr))(√((p^2 +q^2 +r^2 )/3)) sin {(𝛑/3)−(1/3) sin^(−1) [pqr((3/(p^2 +q^2 +r^2 )))^(3/2) ]}))) k>0 for the case that the orthocenter lies inside the triangle and k<0 for the case that it lies outside. with k we can get: 𝚫=(1/( (√(((1/(p+(1/(pk))))+(1/(q+(1/(qk))))+(1/(r+(1/(rk)))))(−(1/(p+(1/(pk))))+(1/(q+(1/(qk))))+(1/(r+(1/(rk)))))((1/(p+(1/(pk))))−(1/(q+(1/(qk))))+(1/(r+(1/(rk)))))((1/(p+(1/(pk))))+(1/(q+(1/(qk))))−(1/(r+(1/(rk))))))))) a=((2𝚫)/(p+(1/(pk)))) b=((2𝚫)/(q+(1/(qk)))) c=((2𝚫)/(r+(1/(rk)))) example 1: p=7, q=6, r=5 ⇒a≈9.79245055 or 1.56227610 ⇒b≈10.43513717 or 3.92946646 ⇒c≈10.94952454 or 5.14205276 (see first diagram) example 2: p=q=r=2 ⇒a=b=c≈3.46410162 (=2(√3))](https://www.tinkutara.com/question/Q161081.png)
$$\frac{{p}}{{v}}=\frac{{q}}{{u}}\:\Rightarrow{pu}={qv} \\ $$$$\frac{{p}}{{w}}=\frac{{r}}{{u}}\:\Rightarrow{pu}={rw} \\ $$$$\Rightarrow\begin{array}{|c|}{\boldsymbol{{pu}}=\boldsymbol{{qv}}=\boldsymbol{{rw}}}\\\hline\end{array}=\frac{\mathrm{1}}{{k}},\:{say} \\ $$$$\Rightarrow{u}=\frac{\mathrm{1}}{{pk}},\:{v}=\frac{\mathrm{1}}{{qk}},\:{w}=\frac{\mathrm{1}}{{rk}} \\ $$$$ \\ $$$${say}\:{the}\:{area}\:{of}\:{triangle}\:{ABC}\:{is}\:\Delta \\ $$$${and}\:{the}\:{side}\:{lengthes}\:{are}\:{a},\:{b},\:{c}. \\ $$$$\Delta=\frac{\mathrm{1}}{\mathrm{2}}{a}\left({p}+{u}\right)=\frac{\mathrm{1}}{\mathrm{2}}{b}\left({q}+{v}\right)=\frac{\mathrm{1}}{\mathrm{2}}{c}\left({r}+{w}\right) \\ $$$$\Delta=\frac{\mathrm{1}}{\mathrm{2}}\left({au}+{bv}+{cw}\right) \\ $$$$\Delta=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{2}\Delta{u}}{{p}+{u}}+\frac{\mathrm{2}\Delta{v}}{{q}+{v}}+\frac{\mathrm{2}\Delta{w}}{{r}+{w}}\right) \\ $$$$\Rightarrow\begin{array}{|c|}{\frac{\boldsymbol{{u}}}{\boldsymbol{{p}}+\boldsymbol{{u}}}+\frac{\boldsymbol{{v}}}{\boldsymbol{{q}}+\boldsymbol{{v}}}+\frac{\boldsymbol{{w}}}{\boldsymbol{{r}}+\boldsymbol{{w}}}=\mathrm{1}}\\\hline\end{array} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\frac{{p}}{{u}}+\mathrm{1}}+\frac{\mathrm{1}}{\frac{{q}}{{v}}+\mathrm{1}}+\frac{\mathrm{1}}{\frac{{r}}{{w}}+\mathrm{1}}=\mathrm{1} \\ $$$$\frac{\mathrm{1}}{{p}^{\mathrm{2}} {k}+\mathrm{1}}+\frac{\mathrm{1}}{{q}^{\mathrm{2}} {k}+\mathrm{1}}+\frac{\mathrm{1}}{{r}^{\mathrm{2}} {k}+\mathrm{1}}=\mathrm{1} \\ $$$$\left({p}^{\mathrm{2}} {k}+\mathrm{1}\right)\left({q}^{\mathrm{2}} {k}+\mathrm{1}\right)+\left({q}^{\mathrm{2}} {k}+\mathrm{1}\right)\left({r}^{\mathrm{2}} {k}+\mathrm{1}\right)+\left({r}^{\mathrm{2}} {k}+\mathrm{1}\right)\left({p}^{\mathrm{2}} {k}+\mathrm{1}\right)=\left({p}^{\mathrm{2}} {k}+\mathrm{1}\right)\left({q}^{\mathrm{2}} {k}+\mathrm{1}\right)\left({r}^{\mathrm{2}} {k}+\mathrm{1}\right) \\ $$$$\left({p}^{\mathrm{2}} {q}^{\mathrm{2}} +{q}^{\mathrm{2}} {r}^{\mathrm{2}} +{r}^{\mathrm{2}} {p}^{\mathrm{2}} \right){k}^{\mathrm{2}} +\mathrm{2}\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} \right){k}+\mathrm{3}={p}^{\mathrm{2}} {q}^{\mathrm{2}} {r}^{\mathrm{2}} {k}^{\mathrm{3}} +\left({p}^{\mathrm{2}} {q}^{\mathrm{2}} +{q}^{\mathrm{2}} {r}^{\mathrm{2}} +{r}^{\mathrm{2}} {p}^{\mathrm{2}} \right){k}^{\mathrm{2}} +\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} \right){k}+\mathrm{1} \\ $$$${p}^{\mathrm{2}} {q}^{\mathrm{2}} {r}^{\mathrm{2}} {k}^{\mathrm{3}} −\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} \right){k}−\mathrm{2}=\mathrm{0} \\ $$$$\begin{array}{|c|}{{k}^{\mathrm{3}} −\frac{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} }{{p}^{\mathrm{2}} {q}^{\mathrm{2}} {r}^{\mathrm{2}} }\:{k}−\frac{\mathrm{2}}{{p}^{\mathrm{2}} {q}^{\mathrm{2}} {r}^{\mathrm{2}} }=\mathrm{0}}\\\hline\end{array} \\ $$$$\frac{\mathrm{1}}{{p}^{\mathrm{4}} {q}^{\mathrm{4}} {r}^{\mathrm{4}} }−\frac{\mathrm{1}}{\mathrm{27}}\left(\frac{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} }{{p}^{\mathrm{2}} {q}^{\mathrm{2}} {r}^{\mathrm{2}} }\right)^{\mathrm{3}} \leqslant\mathrm{0}\:\Rightarrow\mathrm{3}\:{real}\:{roots} \\ $$$$ \\ $$$${generally}\:{two}\:{of}\:{the}\:{three}\:{roots}\:{are} \\ $$$${suitable}. \\ $$$$\begin{array}{|c|c|}{\boldsymbol{{k}}=\frac{\mathrm{2}}{\boldsymbol{{pqr}}}\sqrt{\frac{\boldsymbol{{p}}^{\mathrm{2}} +\boldsymbol{{q}}^{\mathrm{2}} +\boldsymbol{{r}}^{\mathrm{2}} }{\mathrm{3}}}\:\boldsymbol{\mathrm{sin}}\:\left\{\frac{\mathrm{2}\boldsymbol{\pi}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\:\boldsymbol{\mathrm{sin}}^{−\mathrm{1}} \:\left[\boldsymbol{{pqr}}\left(\frac{\mathrm{3}}{\boldsymbol{{p}}^{\mathrm{2}} +\boldsymbol{{q}}^{\mathrm{2}} +\boldsymbol{{r}}^{\mathrm{2}} }\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]\right\}}\\{\boldsymbol{{k}}=−\frac{\mathrm{2}}{\boldsymbol{{pqr}}}\sqrt{\frac{\boldsymbol{{p}}^{\mathrm{2}} +\boldsymbol{{q}}^{\mathrm{2}} +\boldsymbol{{r}}^{\mathrm{2}} }{\mathrm{3}}}\:\boldsymbol{\mathrm{sin}}\:\left\{\frac{\boldsymbol{\pi}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\:\boldsymbol{\mathrm{sin}}^{−\mathrm{1}} \:\left[\boldsymbol{{pqr}}\left(\frac{\mathrm{3}}{\boldsymbol{{p}}^{\mathrm{2}} +\boldsymbol{{q}}^{\mathrm{2}} +\boldsymbol{{r}}^{\mathrm{2}} }\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]\right\}}\\\hline\end{array} \\ $$$${k}>\mathrm{0}\:{for}\:{the}\:{case}\:{that}\:{the}\:{orthocenter} \\ $$$${lies}\:{inside}\:{the}\:{triangle}\:{and}\:{k}<\mathrm{0}\:{for} \\ $$$${the}\:{case}\:{that}\:{it}\:{lies}\:{outside}. \\ $$$$ \\ $$$${with}\:{k}\:{we}\:{can}\:{get}: \\ $$$$\boldsymbol{\Delta}=\frac{\mathrm{1}}{\:\sqrt{\left(\frac{\mathrm{1}}{\boldsymbol{{p}}+\frac{\mathrm{1}}{\boldsymbol{{pk}}}}+\frac{\mathrm{1}}{\boldsymbol{{q}}+\frac{\mathrm{1}}{\boldsymbol{{qk}}}}+\frac{\mathrm{1}}{\boldsymbol{{r}}+\frac{\mathrm{1}}{\boldsymbol{{rk}}}}\right)\left(−\frac{\mathrm{1}}{\boldsymbol{{p}}+\frac{\mathrm{1}}{\boldsymbol{{pk}}}}+\frac{\mathrm{1}}{\boldsymbol{{q}}+\frac{\mathrm{1}}{\boldsymbol{{qk}}}}+\frac{\mathrm{1}}{\boldsymbol{{r}}+\frac{\mathrm{1}}{\boldsymbol{{rk}}}}\right)\left(\frac{\mathrm{1}}{\boldsymbol{{p}}+\frac{\mathrm{1}}{\boldsymbol{{pk}}}}−\frac{\mathrm{1}}{\boldsymbol{{q}}+\frac{\mathrm{1}}{\boldsymbol{{qk}}}}+\frac{\mathrm{1}}{\boldsymbol{{r}}+\frac{\mathrm{1}}{\boldsymbol{{rk}}}}\right)\left(\frac{\mathrm{1}}{\boldsymbol{{p}}+\frac{\mathrm{1}}{\boldsymbol{{pk}}}}+\frac{\mathrm{1}}{\boldsymbol{{q}}+\frac{\mathrm{1}}{\boldsymbol{{qk}}}}−\frac{\mathrm{1}}{\boldsymbol{{r}}+\frac{\mathrm{1}}{\boldsymbol{{rk}}}}\right)}} \\ $$$$\boldsymbol{{a}}=\frac{\mathrm{2}\boldsymbol{\Delta}}{\boldsymbol{{p}}+\frac{\mathrm{1}}{\boldsymbol{{pk}}}} \\ $$$$\boldsymbol{{b}}=\frac{\mathrm{2}\boldsymbol{\Delta}}{\boldsymbol{{q}}+\frac{\mathrm{1}}{\boldsymbol{{qk}}}} \\ $$$$\boldsymbol{{c}}=\frac{\mathrm{2}\boldsymbol{\Delta}}{\boldsymbol{{r}}+\frac{\mathrm{1}}{\boldsymbol{{rk}}}} \\ $$$$ \\ $$$$\underline{{example}\:\mathrm{1}:\:{p}=\mathrm{7},\:{q}=\mathrm{6},\:{r}=\mathrm{5}} \\ $$$$\Rightarrow{a}\approx\mathrm{9}.\mathrm{79245055}\:{or}\:\mathrm{1}.\mathrm{56227610} \\ $$$$\Rightarrow{b}\approx\mathrm{10}.\mathrm{43513717}\:{or}\:\mathrm{3}.\mathrm{92946646} \\ $$$$\Rightarrow{c}\approx\mathrm{10}.\mathrm{94952454}\:{or}\:\mathrm{5}.\mathrm{14205276} \\ $$$$\left({see}\:{first}\:{diagram}\right) \\ $$$$\underline{{example}\:\mathrm{2}:\:{p}={q}={r}=\mathrm{2}} \\ $$$$\Rightarrow{a}={b}={c}\approx\mathrm{3}.\mathrm{46410162}\:\left(=\mathrm{2}\sqrt{\mathrm{3}}\right) \\ $$
Commented by Tawa11 last updated on 11/Dec/21
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by mr W last updated on 12/Dec/21
Commented by mr W last updated on 14/Dec/21
Commented by mr W last updated on 14/Dec/21
Commented by mr W last updated on 14/Dec/21
Commented by mr W last updated on 14/Dec/21
Answered by mr W last updated on 27/Dec/21
$${according}\:{to}\:{Q}\mathrm{162071} \\ $$$${we}\:{can}\:{also}\:{calculate}\:{the}\:{side}\:{lengthes} \\ $$$${using}\:{following}\:{formulas}: \\ $$$$ \\ $$$$\begin{array}{|c|c|}{\frac{\mathrm{1}}{{k}}=\frac{\mathrm{2}}{{pqr}}\sqrt{\frac{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} }{\mathrm{3}}}\:\mathrm{sin}\:\left\{\frac{\pi}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{sin}^{−\mathrm{1}} \:{pqr}\left(\frac{\mathrm{3}}{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} }\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right\}}\\{\frac{\mathrm{1}}{{k}}=−\frac{\mathrm{2}}{{pqr}}\sqrt{\frac{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} }{\mathrm{3}}}\:\mathrm{sin}\:\left\{\frac{\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{sin}^{−\mathrm{1}} \:{pqr}\left(\frac{\mathrm{3}}{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} }\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right\}}\\\hline\end{array} \\ $$$$ \\ $$$${a}=\sqrt{{q}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{k}} \\ $$$${b}=\sqrt{{r}^{\mathrm{2}} +{p}^{\mathrm{2}} −\mathrm{2}{k}} \\ $$$${c}=\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} −\mathrm{2}{k}} \\ $$