# Inequality problem

1. Dec 23, 2005

### recon

In $$\triangle ABC, AB=4, BC=5, and AC=7$$. Point X is in the interior of the triangle such that $$AX^2 + BX^2 + CX^2$$ is a minimum. What is X, and what is the value of this expression?

2. Dec 23, 2005

### Tide

This looks like a homework problem. What have you tried so far?

3. Dec 23, 2005

### recon

By the triangle inequality,

AX + CX >= 7
CX + BX >= 5
BX + AX >= 4

Therefore, AX + BX + CX >= 8

By the Cauchy-Schwarz inequality,

(AX^2 + BX^2 + CX^2)*(1^2 + 1^2 + 1^2) >= (AX + BX + CX)^2 >= 64

4. Dec 23, 2005

### Tide

If

$$f(x, y) = \sum_{i = 1}^{3}\left [(x-x_i)^2 + (y-y_i)^2\right]$$

can you find the value(s) of x and y that minimize the function?

5. Dec 23, 2005

### recon

It's not homework. I saw the problem somewhere on the web, and am trying to solve it using my mediocre knowledge of inequalities. I'm only in a normal A-Level math course, so almost none of the problems we get involve euclidean geometry.

Hmm, actually I think it may be solvable using the triangle inequality only. I think X may be at point A.

6. Dec 23, 2005

### Tide

Then test it! The method I proposed places that point at the center of the triangle. Compare the values of f(x, y) at both points.

P.S. I haven't tested whether the point I found is a minimum or a maximum!

7. Dec 27, 2005

### recon

What do you mean by the 'centre' of the triangle?

I found the problem in a paper about inequalities, so I've been trying to use the inequalities given in the paper to solve the problem. The inequalities in the paper were the AM-GM-HM inequality, the Cauchy-Schwarz inequality, the Rearrangement inequality and the Triangle Inequality. I've given up trying to solve the problem with inequalities.

Instead, I've approached the problem geometrically, by embedding the triangle in a cartesian coordinate system with point A at the origin and AB lying on the positive y-axis, and letting X = (x,y).

Using routine coordinate geometry (cosine rule & pythagorean theorem), I am able to identify C as having coordinates $$(2\sqrt{6},5)$$. It follows from this that

$$AX^2 + BX^2 + CX^2 =& x^2 + y^2 + x^2 + (4-y)^2 + (2\sqrt{6} - x)^2 + (5 - y)^2 = 3(x - \frac{2\sqrt{6}}3)^2 + 3(y-3)^2 + 30$$

The Minimum occurs at $$X = (\frac{2\sqrt{6}}3, 3)$$

Last edited: Dec 27, 2005
8. Dec 27, 2005

### maverick6664

yeah, and notice it's the center of gravity of the triangle

Last edited: Dec 27, 2005
9. Dec 27, 2005

### HallsofIvy

Staff Emeritus
It's at the centroid. A triangle does not have mass and so does not have a "center of gravity"!

(I am waging a hopeless war against using physics terms in mathematics.)

10. Dec 27, 2005

### maverick6664

oh sorry, in japanese the literal translation is "center of gravity."

but centroid also has the meaning of "center of mass"

Last edited: Dec 27, 2005
11. Dec 27, 2005

### roger

but hallofisy mathworld uses those terms centre of mass / gravity as well

12. Dec 27, 2005

### HallsofIvy

Staff Emeritus
I said it was a hopeless war!

By the way, if the triangle is in a coordinate system, then the coordinates of the centroid is the average of the coordinates of the three vertices.

That only works, in general, for triangles. For more complicated figures, such as a quadrilateral, you can disect the figure into triangles but then you need to "weight" the average by the area of each triangle.