Maximization of function in 3 variables

Question

If $( x,y,z)$ be the lengths of perpendiculars from any interior point P of a triangle $ABC$ on sides $BC,CA$ and $AB$ respectively then find the minimum value of :

$$ x^2+ y^2 + z^2 $$ The sides of triangle being $a,b,c$.

I thought of using Lagrange's method of multipliers but am not able to find another function in terms of $x,y,z$ and $a,b,c$

Any help will be appreciated. Thanks.

You need to be able to identify parameters and constraints. How would you represent a point $P$ in the triangle? How would $x^2 + y^2 + z^2$ be calculated once a representation of $P$ is given? Are there constraints on the representation of $P$? — Tunococ, Aug 01 '13 at 09:43
@Tunococ that is where the problem lies...I am not able to understand how to find a relationship between the coordinates of the point $P$ , the variables $x$, $y$, $z$ and the constants $a$, $b$, $c$? — Sheetal Sarin, Aug 01 '13 at 10:48
You may have to assume something about how the triangle is represented. For example, you can assume that they vertices of the triangle are three points in $\mathbb R^2$ ordered counterclockwise (with respect to the interior). That means a point in a triangle would be on the left of all vectors $AB$, $BC$ and $CA$. (This is just one example to start with.) — Tunococ, Aug 01 '13 at 11:08

score 1 · Answer 1 · answered Aug 01 '13 at 15:30

Here is a solution without the method of Lagrange multipliers. Note that $ax+by+cz=2S$ where $S$ is the area of the triangle. Therefore, by Cauchy-Schwarz inequality we have $$\frac{4S^2}{a^2+b^2+c^2}= \frac{(ax+by+cz)^2}{a^2+b^2+c^2}\leq x^2+y^2+z^2.$$ The equality holds when $x:y:z=a:b:c$. The corresponding point is the radical center of the three Apollonian circles associated with the triangle.

score 0 · Answer 2 · edited Apr 13 '17 at 12:19

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As in $ x^2 + y^2 + z^2 $ all are squares , sum of squares is minimum when all terms (for which squares are taken) are equal .( refer How to prove the sum of squares is minimum?) . So the point should be the orthocentre of triangle with equal altitudes . if altitude be 'h'. distance from orthocentre to base is $h/2$ , then minimum is $3(h^2)/4$ . height can be expresses in terms of area which is in terms of sides . so a relation is obtained as you wanted .

edited Apr 13 '17 at 12:19

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answered Aug 01 '13 at 12:07

Harish Kayarohanam

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I think you misunderstand the questions. there is no equal altitude. – chenbai Aug 01 '13 at 12:14
Ya , it is asked when the sum of squares(unequal) will become minimum . so they will become minimum when they are equal . I don't understand what is wrong . Please clarify. If so I can delete the answer . Thanks – Harish Kayarohanam Aug 01 '13 at 12:15
it doesn't ask when a,b,c are equal or not. a.b,c should be parameters. – chenbai Aug 01 '13 at 12:20
But I haven;t made any such assumptions . solving it resulted in such a result . – Harish Kayarohanam Aug 01 '13 at 12:26

chenbai · Answer 3 · 2013-08-02T00:05:00.123

like Tunococ said,you can setup a triangle by your self first:

let $A(0,u),B(v,0),C(w,0), u>0,v>0,w<v$ then $a= v-w,b=\sqrt{u^2+w^2},c=\sqrt{u^2+v^2}$

let point $P(p,q)$, then $x^2=q^2, y^2=\dfrac{(up+wq-uw)^2}{u^2+w^2},z^2=\dfrac{(up+vq-uv)^2}{u^2+v^2}$,since P is in the $\triangle ABC$,so $q\ge 0, up+wq-uw \le 0,up+vq-uv \le 0$

$u,v,w$ can be written in $f(a,b,c)$ so the question be come to find minimum of $Q(p,q)=x^2+y^2+z^2$ ,now you can go further.

Edit: it has a direct way as SB pointed out:

$ax+by+cz=2S$ so it is more easy to do.

Maximization of function in 3 variables

3 Answers3