Question Number 204944 by mr W last updated on 02/Mar/24
Commented by mr W last updated on 03/Mar/24
$${the}\:{distances}\:{from}\:{a}\:{point}\:{inside} \\ $$$${a}\:{triangle}\:{to}\:{its}\:{vertexes}\:{are}\:{p},\:{q},\:{r} \\ $$$${respectively}.\:{find}\:{the}\:{maximum} \\ $$$${area}\:{of}\:{the}\:{triangle}\:{and}\:{its}\:{sides}. \\ $$
Answered by mr W last updated on 03/Mar/24
Commented by mr W last updated on 03/Mar/24
![α+β+γ=2π Δ=area of triangle Δ=((qr sin α)/2)+((rp sin β)/2)+((pq sin γ)/2) Δ=((pqr)/2)(((sin α)/p)+((sin β)/q)+((sin γ)/r))=((pqr)/2)Φ with Φ=((sin α)/p)+((sin β)/q)+((sin γ)/r) Δ_(max) ⇔Φ_(max) F=((sin α)/p)+((sin β)/q)+((sin γ)/r)−λ(α+β+γ−2π) (∂F/∂α)=((cos α)/p)−λ=0 ⇒cos α=pλ similarly cos β=qλ cos γ=rλ we see r cos β=q cos γ=qrλ, that means AS⊥BC. similarly BS⊥AC, CS⊥AB. that means S is the orthocenter of the triangle. cos (α+β)=cos (2π−γ)=cos γ pqλ^2 −(√((1−p^2 λ^2 )(1−q^2 λ^2 )))=rλ pqλ^2 −rλ=(√((1−p^2 λ^2 )(1−q^2 λ^2 ))) p^2 q^2 λ^4 −2pqrλ^3 +r^2 λ^2 =(1−p^2 λ^2 )(1−q^2 λ^2 ) p^2 q^2 λ^4 −2pqrλ^3 +r^2 λ^2 =1−(p^2 +q^2 )λ^2 +p^2 q^2 λ^4 ⇒(1/λ^3 )−((p^2 +q^2 +r^2 )/λ)+2pqr=0 ⇒(1/λ)=−2(√((p^2 +q^2 +r^2 )/3)) sin [(π/3)+(1/3)sin^(−1) ((pqr)/((((p^2 +q^2 +r^2 )/3))^(3/2) ))] ⇒λ=−(1/(2(√((p^2 +q^2 +r^2 )/3)) sin [(π/3)+(1/3)sin^(−1) ((pqr)/((((p^2 +q^2 +r^2 )/3))^(3/2) ))])) sin α=(√(1−p^2 λ^2 )) sin β=(√(1−q^2 λ^2 )) sin γ=(√(1−r^2 λ^2 )) ⇒Δ_(max) =((pqr)/2)((√((1/p^2 )−λ^2 ))+(√((1/q^2 )−λ^2 ))+(√((1/r^2 )−λ^2 ))) a=(√(q^2 +r^2 −2qr cos α))=(√(q^2 +r^2 −2pqrλ)) similarly b=(√(r^2 +p^2 −2pqrλ)) c=(√(p^2 +q^2 −2pqrλ))](https://www.tinkutara.com/question/Q204950.png)
$$\alpha+\beta+\gamma=\mathrm{2}\pi \\ $$$$\Delta={area}\:{of}\:{triangle} \\ $$$$\Delta=\frac{{qr}\:\mathrm{sin}\:\alpha}{\mathrm{2}}+\frac{{rp}\:\mathrm{sin}\:\beta}{\mathrm{2}}+\frac{{pq}\:\mathrm{sin}\:\gamma}{\mathrm{2}} \\ $$$$\Delta=\frac{{pqr}}{\mathrm{2}}\left(\frac{\mathrm{sin}\:\alpha}{{p}}+\frac{\mathrm{sin}\:\beta}{{q}}+\frac{\mathrm{sin}\:\gamma}{{r}}\right)=\frac{{pqr}}{\mathrm{2}}\Phi \\ $$$${with}\:\Phi=\frac{\mathrm{sin}\:\alpha}{{p}}+\frac{\mathrm{sin}\:\beta}{{q}}+\frac{\mathrm{sin}\:\gamma}{{r}} \\ $$$$\Delta_{{max}} \Leftrightarrow\Phi_{{max}} \\ $$$${F}=\frac{\mathrm{sin}\:\alpha}{{p}}+\frac{\mathrm{sin}\:\beta}{{q}}+\frac{\mathrm{sin}\:\gamma}{{r}}−\lambda\left(\alpha+\beta+\gamma−\mathrm{2}\pi\right) \\ $$$$\frac{\partial{F}}{\partial\alpha}=\frac{\mathrm{cos}\:\alpha}{{p}}−\lambda=\mathrm{0}\:\Rightarrow\mathrm{cos}\:\alpha={p}\lambda \\ $$$${similarly} \\ $$$$\mathrm{cos}\:\beta={q}\lambda \\ $$$$\mathrm{cos}\:\gamma={r}\lambda \\ $$$$ \\ $$$${we}\:{see}\:{r}\:\mathrm{cos}\:\beta={q}\:\mathrm{cos}\:\gamma={qr}\lambda,\:{that}\: \\ $$$${means}\:{AS}\bot{BC}.\: \\ $$$${similarly}\:{BS}\bot{AC},\:{CS}\bot{AB}. \\ $$$${that}\:{means}\:{S}\:{is}\:{the}\:{orthocenter}\:{of} \\ $$$${the}\:{triangle}. \\ $$$$ \\ $$$$\mathrm{cos}\:\left(\alpha+\beta\right)=\mathrm{cos}\:\left(\mathrm{2}\pi−\gamma\right)=\mathrm{cos}\:\gamma \\ $$$${pq}\lambda^{\mathrm{2}} −\sqrt{\left(\mathrm{1}−{p}^{\mathrm{2}} \lambda^{\mathrm{2}} \right)\left(\mathrm{1}−{q}^{\mathrm{2}} \lambda^{\mathrm{2}} \right)}={r}\lambda \\ $$$${pq}\lambda^{\mathrm{2}} −{r}\lambda=\sqrt{\left(\mathrm{1}−{p}^{\mathrm{2}} \lambda^{\mathrm{2}} \right)\left(\mathrm{1}−{q}^{\mathrm{2}} \lambda^{\mathrm{2}} \right)} \\ $$$${p}^{\mathrm{2}} {q}^{\mathrm{2}} \lambda^{\mathrm{4}} −\mathrm{2}{pqr}\lambda^{\mathrm{3}} +{r}^{\mathrm{2}} \lambda^{\mathrm{2}} =\left(\mathrm{1}−{p}^{\mathrm{2}} \lambda^{\mathrm{2}} \right)\left(\mathrm{1}−{q}^{\mathrm{2}} \lambda^{\mathrm{2}} \right) \\ $$$${p}^{\mathrm{2}} {q}^{\mathrm{2}} \lambda^{\mathrm{4}} −\mathrm{2}{pqr}\lambda^{\mathrm{3}} +{r}^{\mathrm{2}} \lambda^{\mathrm{2}} =\mathrm{1}−\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)\lambda^{\mathrm{2}} +{p}^{\mathrm{2}} {q}^{\mathrm{2}} \lambda^{\mathrm{4}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\lambda^{\mathrm{3}} }−\frac{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} }{\lambda}+\mathrm{2}{pqr}=\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\lambda}=−\mathrm{2}\sqrt{\frac{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} }{\mathrm{3}}}\:\mathrm{sin}\:\left[\frac{\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{{pqr}}{\left(\frac{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} }{\mathrm{3}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\right] \\ $$$$\Rightarrow\lambda=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\frac{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} }{\mathrm{3}}}\:\mathrm{sin}\:\left[\frac{\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{{pqr}}{\left(\frac{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} }{\mathrm{3}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\right]} \\ $$$$\mathrm{sin}\:\alpha=\sqrt{\mathrm{1}−{p}^{\mathrm{2}} \lambda^{\mathrm{2}} } \\ $$$$\mathrm{sin}\:\beta=\sqrt{\mathrm{1}−{q}^{\mathrm{2}} \lambda^{\mathrm{2}} } \\ $$$$\mathrm{sin}\:\gamma=\sqrt{\mathrm{1}−{r}^{\mathrm{2}} \lambda^{\mathrm{2}} } \\ $$$$\Rightarrow\Delta_{{max}} =\frac{{pqr}}{\mathrm{2}}\left(\sqrt{\frac{\mathrm{1}}{{p}^{\mathrm{2}} }−\lambda^{\mathrm{2}} }+\sqrt{\frac{\mathrm{1}}{{q}^{\mathrm{2}} }−\lambda^{\mathrm{2}} }+\sqrt{\frac{\mathrm{1}}{{r}^{\mathrm{2}} }−\lambda^{\mathrm{2}} }\right) \\ $$$$ \\ $$$${a}=\sqrt{{q}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{qr}\:\mathrm{cos}\:\alpha}=\sqrt{{q}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{pqr}\lambda} \\ $$$${similarly} \\ $$$${b}=\sqrt{{r}^{\mathrm{2}} +{p}^{\mathrm{2}} −\mathrm{2}{pqr}\lambda} \\ $$$${c}=\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} −\mathrm{2}{pqr}\lambda} \\ $$
Commented by mr W last updated on 03/Mar/24
$${example}:\:{p}=\mathrm{2},\:{q}=\mathrm{3},\:{r}=\mathrm{5} \\ $$
Commented by mr W last updated on 03/Mar/24
Commented by mr W last updated on 04/Mar/24
![with λ=(1/(2(√((p^2 +q^2 +r^2 )/3)) sin [(π/3)−(1/3)sin^(−1) ((pqr)/((((p^2 +q^2 +r^2 )/3))^(3/2) ))])) we get the minimum area. in this case the orthocenter lies outside the triangle.](https://www.tinkutara.com/question/Q204953.png)
$${with}\:\lambda=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\frac{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} }{\mathrm{3}}}\:\mathrm{sin}\:\left[\frac{\pi}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{{pqr}}{\left(\frac{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} }{\mathrm{3}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\right]} \\ $$$${we}\:{get}\:{the}\:{minimum}\:{area}.\:{in}\:{this} \\ $$$${case}\:{the}\:{orthocenter}\:{lies}\:{outside} \\ $$$${the}\:{triangle}. \\ $$
Commented by mr W last updated on 03/Mar/24
