P is a point inside triangle ABC, and D,E,F are the legs of the altitudes from P to BC,CA,AB

Question

P is a point inside triangle ABC, and D,E,F are leg of the heights from P to BC,CA,AB. Find all points like P so that $\frac{BC}{PD}+\frac{CA}{PE}+\frac{AB}{PF}$ has minimum value

It is an Olympiad problem

Answer 1

$$Min(\frac{BC}{PD}+\frac{CA}{PE}+\frac{AB}{PF})=?$$

$$\frac{BC}{PD}+\frac{CA}{PE}+\frac{AB}{PF}=(\frac{BD}{PD}+\frac{DC}{PD})+(\frac{CE}{PE}+\frac{AE}{PE})+(\frac{AF}{PF}+\frac{FB}{PF})$$

$$=Cot(B_2)+Cot(C_1)+Cot(C_2)+Cot(A_1)+Cot(A_2)+Cot(B_1)$$ $$Min((Cot(A_1)+Cot(A_2))+(Cot(B_1)+Cot(B_2))+(Cot(C_1)+Cot(C_2)))=Min((Cot(A_1)+Cot(A_2))+Min((Cot(B_1)+Cot(B_2))+Min((Cot(C_1)+Cot(C_2))$$ Because: $$(constant)\angle A=\angle A_1+\angle A_2$$ $$(constant)\angle B=\angle B_1+\angle B_2$$ $$(constant)\angle C=\angle C_1+\angle C_2$$ $Cot(A_1)+Cot(A_2)$ is minimum when $\angle A_1=\angle A_2$ and similarly $\angle B_1=\angle B_2$ , $\angle C_1=\angle C_2$
So P is intersection of bisectors of Angles A,B,C (P is Center of inscribed circle)

Answer 2

Let's denote altitude from A, B and C as $h_a$, $h_b$ and $h_c$ respectively.Let:

$M=\frac{BC}{PD}+\frac{AC}{PE}+\frac{AB}{PF}$

If $S$ is the area of triangle, we can write:

$$M=\frac{2S}{PD\times h_a}+\frac{2S}{PE\times h_b}+\frac{2S}{PF\times h_c}$$

$$M=2S\big(\frac{1}{PD\times h_a}+\frac{1}{PE\times h_b}+\frac{1}{PF\times h_c}\big)$$

$2S$ is constant, therefore we must minimize:

$$\frac{M}{2S}=\frac{1}{PD\times h_a}+\frac{1}{PE\times h_b}+\frac{1}{PF\times h_c}$$

$$\frac{M}{2S}=\frac{h_bh_c(PE\times PF)+h_ah_c(PD\times PF)+h_ah_b(PD\times PE)}{h_ah_bh_c(PD\times PE\times PF)}$$

In any triangle if P is a point inside we have:

$$\frac{PD}{h_a}+\frac{PE}{h_b}+\frac{PF}{h_c}=1$$

This relation can easily be found using relation between area of triangle with ares of three triangles their altitudes are PD, PE and PF. Now we rewrite this relation as:

$$\frac{PD^2}{h_a\times PD}+\frac{PE^2}{h_b\times PE}+\frac{PF^2}{h_c\times PF}=1$$

Now by doing operation on this relation(summing three fractions) and compare it with RHS of $\frac{M}{2S}$, for M to be minimum we must have:

$\frac{M}{2S}=\big(\frac{1}{PD^2}+\frac{1}{PE^2}+\frac{1}{PF^2}\big)$

This relation are symmetric for $PD$, $PE$ and $PF$, that means they can be equal.The geometric interpretation of this is if P is the center of inscribed circle (the intersection of bisectors of angles A, B and C) then M is minimum. Hence minimum of $M$ is $2S$.You can check this on figure; $M=10.9$ for $P_1$ which is the center of inscribed circle. Other values are $(12.9), (15.3), (58.77),(11.94)$.