Question-224249

Question Number 224249 by mnjuly1970 last updated on 29/Aug/25

Commented by mnjuly1970 last updated on 29/Aug/25

$${please}\:{prove}\:\:\underline{\underbrace{\lesseqgtr}} \\ $$

Commented by mr W last updated on 30/Aug/25

ΔABC mustn′t be equilateral. for any triangle ABC we have cot α=cot A+cot B+cot C. proof see below.

$$\Delta{ABC}\:{mustn}'{t}\:{be}\:{equilateral}. \\ $$$${for}\:{any}\:{triangle}\:{ABC}\:{we}\:{have} \\ $$$$\mathrm{cot}\:\alpha=\mathrm{cot}\:{A}+\mathrm{cot}\:{B}+\mathrm{cot}\:{C}. \\ $$$${proof}\:{see}\:{below}. \\ $$

Commented by TonyCWX last updated on 30/Aug/25

$${Fun}\:{fact},\:{that}\:{point}\:{is}\:{known}\:{as}\:{the}\:{Brocard}\:{point}. \\ $$

Answered by Mathwiz last updated on 29/Aug/25

Answered by mr W last updated on 30/Aug/25

Commented by mr W last updated on 30/Aug/25

ΔABD=((qc sin α)/2) ΔBCD=((ra sin α)/2) ΔCAD=((pb sin α)/2) say Δ=area of ΔABC Δ=(((ra+pb+qc)sin α)/2) ⇒(ra+pb+qc)sin α=2Δ ...(i) p^2 =q^2 +c^2 −2qc cos α q^2 =r^2 +a^2 −2ra cos α r^2 =p^2 +b^2 −2pb cos α ⇒(ra+qb+qc) cos θ=((a^2 +b^2 +c^2 )/2) ...(ii) (ii)/(i): cot α=((a^2 +b^2 +c^2 )/(4Δ)) ...(I) Δ=((ab sin C)/2)=((ca sin B)/2)=((ab sin C)/2) ⇒sin A=((2Δ)/(bc)) cos A=((b^2 +c^2 −a^2 )/(2bc)) cot A=((b^2 +c^2 −a^2 )/(2bc))×((bc)/(2Δ))=((b^2 +c^2 −a^2 )/(4Δ)) similarly cot B=((c^2 +a^2 −b^2 )/(4Δ)) cot C=((a^2 +b^2 −c^2 )/(4Δ)) ⇒cot A+cot B+cot C=((a^2 +b^2 +c^2 )/(4Δ)) ...(I) therefore cot α=cot A+cot B+cot C ✓

$$\Delta{ABD}=\frac{{qc}\:\mathrm{sin}\:\alpha}{\mathrm{2}} \\ $$$$\Delta{BCD}=\frac{{ra}\:\mathrm{sin}\:\alpha}{\mathrm{2}} \\ $$$$\Delta{CAD}=\frac{{pb}\:\mathrm{sin}\:\alpha}{\mathrm{2}} \\ $$$${say}\:\Delta={area}\:{of}\:\Delta{ABC} \\ $$$$\Delta=\frac{\left({ra}+{pb}+{qc}\right)\mathrm{sin}\:\alpha}{\mathrm{2}} \\ $$$$\Rightarrow\left({ra}+{pb}+{qc}\right)\mathrm{sin}\:\alpha=\mathrm{2}\Delta\:\:\:…\left({i}\right) \\ $$$${p}^{\mathrm{2}} ={q}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{qc}\:\mathrm{cos}\:\alpha \\ $$$${q}^{\mathrm{2}} ={r}^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{2}{ra}\:\mathrm{cos}\:\alpha \\ $$$${r}^{\mathrm{2}} ={p}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{pb}\:\mathrm{cos}\:\alpha \\ $$$$\Rightarrow\left({ra}+{qb}+{qc}\right)\:\mathrm{cos}\:\theta=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}}\:\:\:…\left({ii}\right) \\ $$$$\left({ii}\right)/\left({i}\right): \\ $$$$\mathrm{cot}\:\alpha=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{4}\Delta}\:\:\:…\left({I}\right) \\ $$$$\Delta=\frac{{ab}\:\mathrm{sin}\:{C}}{\mathrm{2}}=\frac{{ca}\:\mathrm{sin}\:{B}}{\mathrm{2}}=\frac{{ab}\:\mathrm{sin}\:{C}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sin}\:{A}=\frac{\mathrm{2}\Delta}{{bc}} \\ $$$$\mathrm{cos}\:{A}=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}} \\ $$$$\mathrm{cot}\:{A}=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}}×\frac{{bc}}{\mathrm{2}\Delta}=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{4}\Delta} \\ $$$${similarly} \\ $$$$\mathrm{cot}\:{B}=\frac{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{4}\Delta} \\ $$$$\mathrm{cot}\:{C}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{4}\Delta} \\ $$$$\Rightarrow\mathrm{cot}\:{A}+\mathrm{cot}\:{B}+\mathrm{cot}\:{C}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{4}\Delta}\:\:\:…\left({I}\right) \\ $$$${therefore} \\ $$$$\mathrm{cot}\:\alpha=\mathrm{cot}\:{A}+\mathrm{cot}\:{B}+\mathrm{cot}\:{C}\:\:\checkmark \\ $$

Commented by mnjuly1970 last updated on 31/Aug/25

$${mercey}\:{Sir}\:{W}\:\:\underline{\underbrace{\lesseqgtr}} \\ $$

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