Question Number 161912 by mr W last updated on 24/Dec/21
Commented by mr W last updated on 24/Dec/21
Commented by mr W last updated on 24/Dec/21
$${it}\:{is}\:{to}\:{find}\:{a}\:{triangle}\:{whose}\:{sides} \\ $$$${tangent}\:{the}\:{three}\:{circles}\:{with}\:{radii} \\ $$$${p},\:{q},\:{r}\:{respectively}. \\ $$
Commented by mr W last updated on 24/Dec/21
$${if}\:{the}\:{distances}\:{from}\:{the}\:{orthocenter} \\ $$$${of}\:{a}\:{triangle}\:{to}\:{its}\:{sides}\:{are}\:{p},\:{q},\:{r} \\ $$$${respectively},\:{find}\:{the}\:{side}\:{lengthes} \\ $$$${of}\:{the}\:{triangle}. \\ $$
Commented by henderson last updated on 24/Dec/21
$$\mathrm{very}\:\mathrm{interesting}... \\ $$$$\mathrm{i}\:\mathrm{want}\:\mathrm{to}\:\mathrm{know}\:\mathrm{it}\:! \\ $$
Commented by mr W last updated on 25/Dec/21
Commented by mr W last updated on 25/Dec/21
$${generally}\:{we}\:{can}\:{find}\:{two}\:{triangles} \\ $$$${under}\:{the}\:{given}\:{condition}. \\ $$$${red}\:{triangle}\:{with}\:{orthocenter}\:{inside} \\ $$$${blue}\:{triangle}\:{with}\:{orthocenter}\:{outside} \\ $$
Answered by mr W last updated on 25/Dec/21
Commented by mr W last updated on 26/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. (1/2)a(p+u)=(1/2)b(q+v)=(1/2)c(r+w)=Δ (1/2)(au+bv+cw)=2Δ (1/2)(((2Δu)/(p+u))+((2Δv)/(q+v))+((2Δw)/(r+w)))=2Δ ⇒ determinant ((((u/(p+u))+(v/(q+v))+(w/(r+w))=2))) (1/((p/u)+1))+(1/((q/v)+1))+(1/((r/w)+1))=2 (1/(p^2 k+1))+(1/(q^2 k+1))+(1/(r^2 k+1))=2 (p^2 k+1)(q^2 k+1)+(q^2 k+1)(r^2 k+1)+(r^2 k+1)(p^2 k+1)=2(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=2p^2 q^2 r^2 k^3 +2(p^2 q^2 +q^2 r^2 +r^2 p^2 )k^2 +2(p^2 +q^2 +r^2 )k+2 2p^2 q^2 r^2 k^3 +(p^2 q^2 +q^2 r^2 +r^2 p^2 )k^2 −1=0 determinant ((((1/k^3 )−(p^2 q^2 +q^2 r^2 +r^2 p^2 ) (1/k)−2(pqr)^2 =0))) (pqr)^4 −(((p^2 q^2 +q^2 r^2 +r^2 p^2 )/3))^3 ≤0 ⇒three real roots (1/k_n )=2(√((p^2 q^2 +q^2 r^2 +r^2 p^2 )/3)) sin {((2nπ)/3)−(1/3) sin^(−1) [(pqr)^2 ((3/(p^2 q^2 +q^2 r^2 +r^2 p^2 )))^(3/2) ]} (n=0,1,2) generally two (i.e. n=1,2) of the three roots are suitable: determinant ((((1/k_1 )=2(√((p^2 q^2 +q^2 r^2 +r^2 p^2 )/3)) sin {(π/3)+(1/3) sin^(−1) [(pqr)^2 ((3/(p^2 q^2 +q^2 r^2 +r^2 p^2 )))^(3/2) ]})),(((1/k_2 )=−2(√((p^2 q^2 +q^2 r^2 +r^2 p^2 )/3)) sin {(π/3)−(1/3) sin^(−1) [(pqr)^2 ((3/(p^2 q^2 +q^2 r^2 +r^2 p^2 )))^(3/2) ]}))) the positive value is for the case that the orthocenter lies inside the triangle and the negative value for the case that it lies outside the triangle. with k we 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: p=10, q=6, r=4 a≈28.522248 or 15.173152 b≈26.202978 or 12.655577 c≈20.945175 or 4.829686](Q161982.png)
$$\frac{{p}}{{v}}=\frac{{q}}{{u}}\:\Rightarrow{pu}={qv} \\ $$$$\frac{{p}}{{w}}=\frac{{r}}{{u}}\:\Rightarrow{pu}={rw} \\ $$$$\Rightarrow\begin{array}{|c|}{\boldsymbol{{pu}}=\boldsymbol{{qv}}=\boldsymbol{{rw}}}\\\hline\end{array}=\frac{\mathrm{1}}{{k}},\:{say} \\ $$$$\Rightarrow{u}=\frac{\mathrm{1}}{{pk}},\:{v}=\frac{\mathrm{1}}{{qk}},\:{w}=\frac{\mathrm{1}}{{rk}} \\ $$$$ \\ $$$${say}\:{the}\:{area}\:{of}\:{triangle}\:{ABC}\:{is}\:\Delta \\ $$$${and}\:{the}\:{side}\:{lengthes}\:{are}\:{a},\:{b},\:{c}. \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{a}\left({p}+{u}\right)=\frac{\mathrm{1}}{\mathrm{2}}{b}\left({q}+{v}\right)=\frac{\mathrm{1}}{\mathrm{2}}{c}\left({r}+{w}\right)=\Delta \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left({au}+{bv}+{cw}\right)=\mathrm{2}\Delta \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{2}\Delta{u}}{{p}+{u}}+\frac{\mathrm{2}\Delta{v}}{{q}+{v}}+\frac{\mathrm{2}\Delta{w}}{{r}+{w}}\right)=\mathrm{2}\Delta \\ $$$$\Rightarrow\begin{array}{|c|}{\frac{\boldsymbol{{u}}}{\boldsymbol{{p}}+\boldsymbol{{u}}}+\frac{\boldsymbol{{v}}}{\boldsymbol{{q}}+\boldsymbol{{v}}}+\frac{\boldsymbol{{w}}}{\boldsymbol{{r}}+\boldsymbol{{w}}}=\mathrm{2}}\\\hline\end{array} \\ $$$$\frac{\mathrm{1}}{\frac{{p}}{{u}}+\mathrm{1}}+\frac{\mathrm{1}}{\frac{{q}}{{v}}+\mathrm{1}}+\frac{\mathrm{1}}{\frac{{r}}{{w}}+\mathrm{1}}=\mathrm{2} \\ $$$$\frac{\mathrm{1}}{{p}^{\mathrm{2}} {k}+\mathrm{1}}+\frac{\mathrm{1}}{{q}^{\mathrm{2}} {k}+\mathrm{1}}+\frac{\mathrm{1}}{{r}^{\mathrm{2}} {k}+\mathrm{1}}=\mathrm{2} \\ $$$$\left({p}^{\mathrm{2}} {k}+\mathrm{1}\right)\left({q}^{\mathrm{2}} {k}+\mathrm{1}\right)+\left({q}^{\mathrm{2}} {k}+\mathrm{1}\right)\left({r}^{\mathrm{2}} {k}+\mathrm{1}\right)+\left({r}^{\mathrm{2}} {k}+\mathrm{1}\right)\left({p}^{\mathrm{2}} {k}+\mathrm{1}\right)=\mathrm{2}\left({p}^{\mathrm{2}} {k}+\mathrm{1}\right)\left({q}^{\mathrm{2}} {k}+\mathrm{1}\right)\left({r}^{\mathrm{2}} {k}+\mathrm{1}\right) \\ $$$$\left({p}^{\mathrm{2}} {q}^{\mathrm{2}} +{q}^{\mathrm{2}} {r}^{\mathrm{2}} +{r}^{\mathrm{2}} {p}^{\mathrm{2}} \right){k}^{\mathrm{2}} +\mathrm{2}\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} \right){k}+\mathrm{3}=\mathrm{2}{p}^{\mathrm{2}} {q}^{\mathrm{2}} {r}^{\mathrm{2}} {k}^{\mathrm{3}} +\mathrm{2}\left({p}^{\mathrm{2}} {q}^{\mathrm{2}} +{q}^{\mathrm{2}} {r}^{\mathrm{2}} +{r}^{\mathrm{2}} {p}^{\mathrm{2}} \right){k}^{\mathrm{2}} +\mathrm{2}\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} \right){k}+\mathrm{2} \\ $$$$\mathrm{2}{p}^{\mathrm{2}} {q}^{\mathrm{2}} {r}^{\mathrm{2}} {k}^{\mathrm{3}} +\left({p}^{\mathrm{2}} {q}^{\mathrm{2}} +{q}^{\mathrm{2}} {r}^{\mathrm{2}} +{r}^{\mathrm{2}} {p}^{\mathrm{2}} \right){k}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\begin{array}{|c|}{\frac{\mathrm{1}}{{k}^{\mathrm{3}} }−\left({p}^{\mathrm{2}} {q}^{\mathrm{2}} +{q}^{\mathrm{2}} {r}^{\mathrm{2}} +{r}^{\mathrm{2}} {p}^{\mathrm{2}} \right)\:\frac{\mathrm{1}}{{k}}−\mathrm{2}\left({pqr}\right)^{\mathrm{2}} =\mathrm{0}}\\\hline\end{array} \\ $$$$\left({pqr}\right)^{\mathrm{4}} −\left(\frac{{p}^{\mathrm{2}} {q}^{\mathrm{2}} +{q}^{\mathrm{2}} {r}^{\mathrm{2}} +{r}^{\mathrm{2}} {p}^{\mathrm{2}} }{\mathrm{3}}\right)^{\mathrm{3}} \leqslant\mathrm{0}\:\Rightarrow{three}\:{real}\:{roots} \\ $$$$\frac{\mathrm{1}}{{k}_{{n}} }=\mathrm{2}\sqrt{\frac{{p}^{\mathrm{2}} {q}^{\mathrm{2}} +{q}^{\mathrm{2}} {r}^{\mathrm{2}} +{r}^{\mathrm{2}} {p}^{\mathrm{2}} }{\mathrm{3}}}\:\mathrm{sin}\:\left\{\frac{\mathrm{2}{n}\pi}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{sin}^{−\mathrm{1}} \left[\left({pqr}\right)^{\mathrm{2}} \left(\frac{\mathrm{3}}{{p}^{\mathrm{2}} {q}^{\mathrm{2}} +{q}^{\mathrm{2}} {r}^{\mathrm{2}} +{r}^{\mathrm{2}} {p}^{\mathrm{2}} }\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]\right\}\:\left({n}=\mathrm{0},\mathrm{1},\mathrm{2}\right) \\ $$$$ \\ $$$${generally}\:{two}\:\left({i}.{e}.\:{n}=\mathrm{1},\mathrm{2}\right)\:{of}\:{the}\:{three}\: \\ $$$${roots}\:{are}\:{suitable}: \\ $$$$\begin{array}{|c|c|}{\frac{\mathrm{1}}{{k}_{\mathrm{1}} }=\mathrm{2}\sqrt{\frac{{p}^{\mathrm{2}} {q}^{\mathrm{2}} +{q}^{\mathrm{2}} {r}^{\mathrm{2}} +{r}^{\mathrm{2}} {p}^{\mathrm{2}} }{\mathrm{3}}}\:\mathrm{sin}\:\left\{\frac{\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{sin}^{−\mathrm{1}} \left[\left({pqr}\right)^{\mathrm{2}} \left(\frac{\mathrm{3}}{{p}^{\mathrm{2}} {q}^{\mathrm{2}} +{q}^{\mathrm{2}} {r}^{\mathrm{2}} +{r}^{\mathrm{2}} {p}^{\mathrm{2}} }\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]\right\}}\\{\frac{\mathrm{1}}{{k}_{\mathrm{2}} }=−\mathrm{2}\sqrt{\frac{{p}^{\mathrm{2}} {q}^{\mathrm{2}} +{q}^{\mathrm{2}} {r}^{\mathrm{2}} +{r}^{\mathrm{2}} {p}^{\mathrm{2}} }{\mathrm{3}}}\:\mathrm{sin}\:\left\{\frac{\pi}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{sin}^{−\mathrm{1}} \left[\left({pqr}\right)^{\mathrm{2}} \left(\frac{\mathrm{3}}{{p}^{\mathrm{2}} {q}^{\mathrm{2}} +{q}^{\mathrm{2}} {r}^{\mathrm{2}} +{r}^{\mathrm{2}} {p}^{\mathrm{2}} }\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]\right\}}\\\hline\end{array} \\ $$$${the}\:{positive}\:{value}\:{is}\:{for}\:{the}\:{case}\:{that} \\ $$$${the}\:{orthocenter}\:{lies}\:{inside}\:{the}\:{triangle} \\ $$$${and}\:{the}\:{negative}\:{value}\:{for}\:{the}\:{case}\: \\ $$$${that}\:{it}\:{lies}\:{outside}\:{the}\:{triangle}. \\ $$$$ \\ $$$${with}\:{k}\:{we}\:{get} \\ $$$$\boldsymbol{\Delta}=\frac{\mathrm{1}}{\:\sqrt{\left(\frac{\mathrm{1}}{\boldsymbol{{p}}+\frac{\mathrm{1}}{\boldsymbol{{pk}}}}+\frac{\mathrm{1}}{\boldsymbol{{q}}+\frac{\mathrm{1}}{\boldsymbol{{qk}}}}+\frac{\mathrm{1}}{\boldsymbol{{r}}+\frac{\mathrm{1}}{\boldsymbol{{rk}}}}\right)\left(−\frac{\mathrm{1}}{\boldsymbol{{p}}+\frac{\mathrm{1}}{\boldsymbol{{pk}}}}+\frac{\mathrm{1}}{\boldsymbol{{q}}+\frac{\mathrm{1}}{\boldsymbol{{qk}}}}+\frac{\mathrm{1}}{\boldsymbol{{r}}+\frac{\mathrm{1}}{\boldsymbol{{rk}}}}\right)\left(\frac{\mathrm{1}}{\boldsymbol{{p}}+\frac{\mathrm{1}}{\boldsymbol{{pk}}}}−\frac{\mathrm{1}}{\boldsymbol{{q}}+\frac{\mathrm{1}}{\boldsymbol{{qk}}}}+\frac{\mathrm{1}}{\boldsymbol{{r}}+\frac{\mathrm{1}}{\boldsymbol{{rk}}}}\right)\left(\frac{\mathrm{1}}{\boldsymbol{{p}}+\frac{\mathrm{1}}{\boldsymbol{{pk}}}}+\frac{\mathrm{1}}{\boldsymbol{{q}}+\frac{\mathrm{1}}{\boldsymbol{{qk}}}}−\frac{\mathrm{1}}{\boldsymbol{{r}}+\frac{\mathrm{1}}{\boldsymbol{{rk}}}}\right)}} \\ $$$$\boldsymbol{{a}}=\frac{\mathrm{2}\boldsymbol{\Delta}}{\boldsymbol{{p}}+\frac{\mathrm{1}}{\boldsymbol{{pk}}}},\:\boldsymbol{{b}}=\frac{\mathrm{2}\boldsymbol{\Delta}}{\boldsymbol{{q}}+\frac{\mathrm{1}}{\boldsymbol{{qk}}}},\:\boldsymbol{{c}}=\frac{\mathrm{2}\boldsymbol{\Delta}}{\boldsymbol{{r}}+\frac{\mathrm{1}}{\boldsymbol{{rk}}}} \\ $$$$ \\ $$$$\underline{{example}:\:{p}=\mathrm{10},\:{q}=\mathrm{6},\:{r}=\mathrm{4}} \\ $$$${a}\approx\mathrm{28}.\mathrm{522248}\:{or}\:\mathrm{15}.\mathrm{173152} \\ $$$${b}\approx\mathrm{26}.\mathrm{202978}\:{or}\:\mathrm{12}.\mathrm{655577} \\ $$$${c}\approx\mathrm{20}.\mathrm{945175}\:{or}\:\mathrm{4}.\mathrm{829686} \\ $$
Commented by mr W last updated on 25/Dec/21
Commented by mr W last updated on 25/Dec/21
Commented by mr W last updated on 25/Dec/21
Commented by Ar Brandon last updated on 25/Dec/21
$$\mathrm{What}'\mathrm{s}\:\mathrm{all}\:\mathrm{this},\:\mathrm{Sir} \\ $$$$ \\ $$🙁☹️
Commented by mr W last updated on 25/Dec/21
$${how}\:{to}\:{find}\:{the}\:{triangle}\:{if}\:{you}\:{only} \\ $$$${know}\:{the}\:{distances}\:{from}\:{its}\: \\ $$$${orthocenter}\:{to}\:{its}\:{three}\:{sides}. \\ $$
Commented by Ar Brandon last updated on 25/Dec/21
$$ \\ $$😭
Commented by Tawa11 last updated on 25/Dec/21
$$\mathrm{Wow},\:\mathrm{great}\:\mathrm{sir}. \\ $$