# Find the minimum value of the sum of the squares of the distances from M to the lines AB, AC and BC

The problem

Let ABCA'B'C' be a regular triangular prism with base edge $$AB = 2 \sqrt{3}$$ cm and height AA' = 1 cm. If M is a point in the plane of the triangle A'B'C' 0 , then the minimum value of the sum of the squares of the distances from M to the lines AB, AC and BC is:

my idea

the drawing

As you can see we drew $$MM' \perp (ABC)$$ and from M' we drew the perpendiculars $$M'X \perp AB, M'Y\perp BC, M'Z \perp AC$$ and by the theorem of the 3 perpendiculars we get that $$MX \perp AB, MY\perp BC, MZ \perp AC$$

By the theorem of Pythagoras, we realize that for $$MX^2+MY^2+ MZ^2$$ to be minimal, $$M'X^2+M'Y^2+ M'Z^2$$ must

By the inequality of means we get that the minimum happens when $$M'X=M'Y=M'Z$$ so basically when $$MX=MY=MZ$$

This means that $$M'$$ is the centre of the circumscribed circle of the triangle $$ZXY$$

I don't know what to do next! I hope one of you can help me! Thanks!

• You are right. I erase this reference. Commented Jul 19 at 17:03
• Here is a good reference ; see also here Commented Jul 19 at 17:11

## 1 Answer

As you said, the sum of the three squared distances, by Pythagorean theorem are

$$f = 3 (1)^2 + (M'X)^2 + (M'Y)^2 + (M'Z)^2$$

Now we know that these altitudes satisfy

$$\frac{1}{2} ( 2 \sqrt{3} ) ( M'X + M'Y + M'Z ) = \text{AREA} = \dfrac{ \sqrt{3} (2 \sqrt{3} )^2 }{4} = 3 \sqrt{3}$$

Thus

$$M'X + M'Y + M'Z = 3$$

So our problem becomes as follows:

$$\text{Minimize } (x^2 + y^2 + z^2) \hspace{5pt} \text{subject to } x + y + z = 3$$

From Cauchy-Schwarz, if we take $$u = [x, y, z]$$ and $$v = [1, 1, 1]$$ then we know that

$$( u \cdot v )^2 \le \|u\|^2 \|v \|^2$$

Substituting $$u$$ and $$v$$ gives

$$(x + y + z )^2 \le (x^2 + y^2 + z^2) (1^2 + 1^2 + 1^2)$$

That is

$$3^2 \le 3 (x^2 + y^2 + z^2)$$

Therefore,

$$x^2 + y^2 + z^2 \ge 3$$

Hence,

$$f \ge 3 + 3 = 6$$