Coordinate Geometry: Triangle Vertices & Point P Movement in OAB Plane

[denian]

the vertices of a triangle are 0, and the points A (a, 0) and B (0,a) where a>0.
the point P moves in the plane OAB such that OP square + AP square + BP square = k square, where k is a positive constant. show that k mst be bigger than (2/3)a X square root of 3


i tried. but can't even know how to start. any hints?

 

[HallsofIvy]

Let P be (x,y). The OP square= x²+ y², AP square= (x-a)²+ y², and BP square= x²+ (y-a)².
Now, what is OP square + AP square+ BP square. Keep in mind that x² and y² must be positive.

 

[M98Ranger]

To start, let's draw a diagram to visualize the situation. We have a triangle with vertices at (0,0), A(a,0), and B(0,a). The point P can move anywhere in the plane OAB, and we have the condition that the sum of the squares of the distances from P to O, A, and B is equal to a constant k. Let's label the distance from P to O as r, the distance from P to A as s, and the distance from P to B as t.

Now, we can use the Pythagorean theorem to write the equation as:

r^2 + s^2 + t^2 = k^2

We also know that the distance between any two points in the plane can be found using the distance formula:

d = √((x2-x1)^2 + (y2-y1)^2)

Using this formula, we can write the distances r, s, and t in terms of the coordinates of P:

r = √((0-x)^2 + (0-y)^2) = √(x^2 + y^2)

s = √((a-x)^2 + (0-y)^2) = √((a-x)^2 + y^2)

t = √((0-x)^2 + (a-y)^2) = √(x^2 + (a-y)^2)

Substituting these values into our original equation, we get:

√(x^2 + y^2)^2 + √((a-x)^2 + y^2)^2 + √(x^2 + (a-y)^2)^2 = k^2

Simplifying, we get:

x^2 + y^2 + (a-x)^2 + y^2 + x^2 + (a-y)^2 = k^2

Expanding the squares and simplifying further, we get:

3x^2 + 3y^2 - 2ax - 2ay + 2a^2 = k^2

Now, we can use the fact that a>0 to rewrite this equation as:

3x^2 + 3y^2 - 2ax - 2ay + a^2 + a^2 + a^2 = k^2

Factoring out the common factor of 3, we get:

3(x