Find all points P such that $\frac{BC}{PH_A}+\frac{AC}{PH_B}+\frac{AB}{PH_C}$ is minimum

Question

ABC is any triangle, we choose P, an interior point. The point P is projected orthogonally on the lines (BC),(AC) and (AB) respectively in $H_A$, $H_B$ and $H_C$. Find all points P such that :

$$ \frac{BC}{PH_A}+\frac{AC}{PH_B}+\frac{AB}{PH_C} $$

is minimum.

Can you help me to have a start please? I have been looking with the areas of small triangles but I don't know how to do it.

regards

Answer 1

Let:

$PH_A=x$, $PH_B=y$, $PH_C=z$

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:

$$\frac ax{ah_a}+\frac by{bh_b}+\frac cz{ch_c}=1$$

$ah_a=bh_b=ch_c=2S$

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:

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

which gives:

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

Or:

$$x=y=z$$

Or we may put $\lambda =\frac 1x$, $\lambda =\frac 1y$ and $\lambda =\frac 1z$ 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.

Answer 2

Here is a few hints towards one way to solve the problem:

Consider the areas of the triangles $S_A=(PBC)~,~S_B=(PAC)~,~S_C=(PAB), S=(ABC)$. Note that $S_A+S_B+S_C=S$ and that the quantity that we seek to minimize is

$$Q=\frac{BC}{PH_A}+\frac{AC}{PH_B}+\frac{BA}{PH_C}=\frac{a^2}{2S_A}+\frac{b^2}{2S_B}+\frac{c^2}{2S_C}$$

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

$$\frac{a^2}{2S_A}+\frac{b^2}{2S_B}+\frac{c^2}{2S_C}\geq\frac{(a+b+c)^2}{2S}$$

The equality holds only when $PH_A=PH_B=PH_C$ (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 $ABC$).