Esther Hoffman

2022-05-03

Find all points P such that $\frac{BC}{P{H}_{A}}+\frac{AC}{P{H}_{B}}+\frac{AB}{P{H}_{C}}$ is minimum.

Genevieve James

Beginner2022-05-04Added 12 answers

Step 1

Let:

$P{H}_{B}=y$

and altitudes from A, B and C as ${h}_{a}$ , ${h}_{b}$ , and ${h}_{c}$ for sides a, b and c respectively.It can be shown that:

$\frac{x}{{h}_{a}}+\frac{y}{{h}_{b}}+\frac{z}{{h}_{c}}=1$

Or:

$\frac{a}{x}a{h}_{a}+\frac{b}{y}b{h}_{b}+\frac{c}{z}c{h}_{c}=1$

$a{h}_{a}=b{h}_{b}=c{h}_{c}=2S$

where S is the area of triangle, so we have to minimize $f(x,y,z)=A=\frac{a}{x}+\frac{b}{y}+\frac{c}{z}$

subjected to $g(x,y,z)=ax+by+cz-2S$

We use Lagrange multiplier; we have:

$\mathcal{L}(x,y,z,\lambda )=f(x,y,z)+\lambda g(x,y,z)$

Taking partial derivative we get:

$-\frac{a}{{x}^{2}}-\frac{b}{{y}^{2}}-\frac{c}{{z}^{2}}+\lambda (a+b+c)=0$

which gives:

$a{\textstyle (}\lambda -\frac{1}{{x}^{2}}{\textstyle )}+b{\textstyle (}\lambda -\frac{1}{{y}^{2}}{\textstyle )}+c{\textstyle (}\lambda -\frac{1}{{z}^{2}}{\textstyle )}=0$

which gives:

$(}\lambda -\frac{1}{{x}^{2}}{\textstyle )}={\textstyle (}\lambda -\frac{1}{{y}^{2}}{\textstyle )}={\textstyle (}\lambda -\frac{1}{{z}^{2}}{\textstyle )}=0$

Or:

$x=y=z$

Or we may put $\lambda =\frac{1}{x}$ , $\lambda =\frac{1}{y}$ and $\lambda =\frac{1}{z}$ in g(x, y, z) and get:

$\frac{a}{\lambda}+\frac{b}{\lambda}+\frac{c}{\lambda}=2S\Rightarrow \lambda =\frac{a+b+c}{2S}$

putting this in f(x, y, z) we get:

$f(x,y,z)=A=\frac{(a+b+c{)}^{2}}{2S}$

which is minimum of A.

Let:

$P{H}_{B}=y$

and altitudes from A, B and C as ${h}_{a}$ , ${h}_{b}$ , and ${h}_{c}$ for sides a, b and c respectively.It can be shown that:

$\frac{x}{{h}_{a}}+\frac{y}{{h}_{b}}+\frac{z}{{h}_{c}}=1$

Or:

$\frac{a}{x}a{h}_{a}+\frac{b}{y}b{h}_{b}+\frac{c}{z}c{h}_{c}=1$

$a{h}_{a}=b{h}_{b}=c{h}_{c}=2S$

where S is the area of triangle, so we have to minimize $f(x,y,z)=A=\frac{a}{x}+\frac{b}{y}+\frac{c}{z}$

subjected to $g(x,y,z)=ax+by+cz-2S$

We use Lagrange multiplier; we have:

$\mathcal{L}(x,y,z,\lambda )=f(x,y,z)+\lambda g(x,y,z)$

Taking partial derivative we get:

$-\frac{a}{{x}^{2}}-\frac{b}{{y}^{2}}-\frac{c}{{z}^{2}}+\lambda (a+b+c)=0$

which gives:

$a{\textstyle (}\lambda -\frac{1}{{x}^{2}}{\textstyle )}+b{\textstyle (}\lambda -\frac{1}{{y}^{2}}{\textstyle )}+c{\textstyle (}\lambda -\frac{1}{{z}^{2}}{\textstyle )}=0$

which gives:

$(}\lambda -\frac{1}{{x}^{2}}{\textstyle )}={\textstyle (}\lambda -\frac{1}{{y}^{2}}{\textstyle )}={\textstyle (}\lambda -\frac{1}{{z}^{2}}{\textstyle )}=0$

Or:

$x=y=z$

Or we may put $\lambda =\frac{1}{x}$ , $\lambda =\frac{1}{y}$ and $\lambda =\frac{1}{z}$ in g(x, y, z) and get:

$\frac{a}{\lambda}+\frac{b}{\lambda}+\frac{c}{\lambda}=2S\Rightarrow \lambda =\frac{a+b+c}{2S}$

putting this in f(x, y, z) we get:

$f(x,y,z)=A=\frac{(a+b+c{)}^{2}}{2S}$

which is minimum of A.

timbreoizy

Beginner2022-05-05Added 15 answers

Step 1

Consider the areas of the triangles ${S}_{A}=(PBC)\text{},\text{}{S}_{B}=(PAC)\text{},\text{}{S}_{C}=(PAB),S=(ABC)$ . Note that ${S}_{A}+{S}_{B}+{S}_{C}=S$ and that the quantity that we seek to minimize is

$Q=\frac{BC}{P{H}_{A}}+\frac{AC}{P{H}_{B}}+\frac{BA}{P{H}_{C}}=\frac{{a}^{2}}{2{S}_{A}}+\frac{{b}^{2}}{2{S}_{B}}+\frac{{c}^{2}}{2{S}_{C}}$

and the problem can be recast as the minimization of the quantity $Q({S}_{A},{S}_{B},{S}_{C})$ under the constraint $h({S}_{A},{S}_{B},{S}_{C})={S}_{A}+{S}_{B}+{S}_{C}=S$ (why can there not be more constraints?) This problem can be solved in various ways but the most elegant one is to use the Cauchy-Schwarz inequality (how?) to conclude that

$\frac{{a}^{2}}{2{S}_{A}}+\frac{{b}^{2}}{2{S}_{B}}+\frac{{c}^{2}}{2{S}_{C}}\ge \frac{(a+b+c{)}^{2}}{2S}$

The equality holds only when $P{H}_{A}=P{H}_{B}=P{H}_{C}$ (why?). This is obeyed by the incenter of the triangle. This solution is also unique, because $Q-\lambda h$ is convex whenever the areas of the triangles are positive (which is trivially true in the interior of ABC).

Consider the areas of the triangles ${S}_{A}=(PBC)\text{},\text{}{S}_{B}=(PAC)\text{},\text{}{S}_{C}=(PAB),S=(ABC)$ . Note that ${S}_{A}+{S}_{B}+{S}_{C}=S$ and that the quantity that we seek to minimize is

$Q=\frac{BC}{P{H}_{A}}+\frac{AC}{P{H}_{B}}+\frac{BA}{P{H}_{C}}=\frac{{a}^{2}}{2{S}_{A}}+\frac{{b}^{2}}{2{S}_{B}}+\frac{{c}^{2}}{2{S}_{C}}$

and the problem can be recast as the minimization of the quantity $Q({S}_{A},{S}_{B},{S}_{C})$ under the constraint $h({S}_{A},{S}_{B},{S}_{C})={S}_{A}+{S}_{B}+{S}_{C}=S$ (why can there not be more constraints?) This problem can be solved in various ways but the most elegant one is to use the Cauchy-Schwarz inequality (how?) to conclude that

$\frac{{a}^{2}}{2{S}_{A}}+\frac{{b}^{2}}{2{S}_{B}}+\frac{{c}^{2}}{2{S}_{C}}\ge \frac{(a+b+c{)}^{2}}{2S}$

The equality holds only when $P{H}_{A}=P{H}_{B}=P{H}_{C}$ (why?). This is obeyed by the incenter of the triangle. This solution is also unique, because $Q-\lambda h$ is convex whenever the areas of the triangles are positive (which is trivially true in the interior of ABC).

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