AX^2*CX^2+BX^2*DX^2=((x+1)^2+(y+1)^2)*((x-1)^2+(y-1)^2)+((x+1)^2+(y-1)^2)*((x-1)^2+(y+1)^2)=10, because x^2+y^2=1.

Peter Breuer, for every radius r>0, the sum AX2*CX2+BX2*DX2 is the same for every point X of the circle. Hence, we can calculate the sum by placing X at any of the four common points of the circle and the square, since there each of the two terms of the sum is calculated very easily, and is equal to 5r4. This means that the sum is equal to 10r4, for every r>0. Replacing 1 and i with r and ir, respectively, your result 2x2x*2 + 8 reads 2x2x*2 +8r4, where xx*=r2.

A note regarding the question statement. In mathematics, to inscribe means to draw on the inside of, just touching but never crossing the sides. See, for instance, https://www.mathsisfun.com/definitions/inscribe.html. So, in this case, it is not the square that is inscribed in the circle; it is the circle that is inscribed in the square.

Incircle, yes. https://mathworld.wolfram.com/Incircle.html.

The circumcircle is a triangle's circumscribed circle, i.e., the unique circle that passes through each of the triangle's three vertices. https://mathworld.wolfram.com/Circumcircle.html and https://en.wikipedia.org/wiki/Circumcircle.

I think a geometrical proof of the distance AX2∙CX2+BX2∙DX2 being the same for every point of the circle would be, if nothing else, worth pursuing.

Peter Breuer It's great! Your research deserves respect and admiration!

SOLVE THE PROBLEM! The same geometric structure as above, considering only that the radius of the circle is an arbitrary R>0. Show that for any position of point X on the circle the sum

AX2+BX2+CX2+DX2 depends only on R( i.e. the sum is constant when X moves on the circle).

REMARK To clarify, here, regarding the problem posted by Liudmyla, the circle is inscribed in the square.

My question(problem) about the sum is valid also if we consider the square inscribed in a circle with radius R. Is the problem of Liudmyla valid too, if we consider the square inscribed in a circle?

Regarding Dinu's problem.

We choose the two coordinate axes as in the first answer.

(i) The circle is inscribed in the square.

We have D=(R,-R), C=(R,R), B=(-R,R), and A=(-R,-R). Also, X=(x,y) such that x2+y2=R2. Consequently, AX2+BX2+CX2+DX2=(x+R)2+(y+R)2+(x+R)2+(y-R)2+(x-R)2+(y-R)2+(x-R)2+(y+R)2=2[(x+R)2+(x-R)2]+2[(y+R)2+(y-R)2]=2(2x2+2R2)+2(2y2+2R2)=4(x2+y2)+8R2=12R2.

(ii) The square is inscribed in the circle.

It is easy to see that the angle formed by the line segment OD and the radius that is drawn in the figure is 45 degrees. As a result, each of the previous coordinates of the points A, B, C, and D is multiplied by sqrt(2)/2. Consequently, AX2+BX2+CX2+DX2=(x+sqrt(2)/2*R)2+(y+sqrt(2)/2*R)2+(x+sqrt(2)/2*R)2+(y-sqrt(2)/2*R)2+(x-sqrt(2)/2*R)2+(y-sqrt(2)/2*R)2+(x-sqrt(2)/2*R)2+(y+sqrt(2)/2*R)2=2[(x+sqrt(2)/2*R)2+(x-sqrt(2)/2*R)2]+2[(y+sqrt(2)/2*R)2+(y-sqrt(2)/2*R)2]=2(2x2+R2)+2(2y2+R2)=4(x2+y2)+4R2=8R2.

Note that replacing the radius R by sqrt(2)/2*R into the expression 4(x2+y2)+8R2 that gives the sum AX2+BX2+CX2+DX2 in the case where the circle is inscribed in the square yields the expression 4(x2+y2)+4R2 that gives the sum AX2+BX2+CX2+DX2 in the case where the square is inscribed in the circle.

Likewise, replacing R by sqrt(2)/2*R into the expression 2(x2+y2)2 +8R4 that gives the sum AX2*CX2+BX2*DX2 in the case where the circle is inscribed in the square yields 2(x2+y2)2 +2R4=4R4, which must be the sum AX2*CX2+BX2*DX2 in the case where the square is inscribed in the circle.

Spiros Konstantogiannis Dear Spiros, your solutions are valid, but you made the same mistake as Sergey Gladun( probably because the statement of Liudmyla's problem is not so clear, i.e. who is inscribed, square ABCD in the cicrle or circle in the square?). Fortunately both problems( my problem and Liudmyla's problem) hold in both cases( square inscribed in circle, circle insribed in the square).

Therefore, in the case The circle is inscribed in the square the center O of the circle is the origin and D( (R/2)(sqrt(2)); -(R/2)(sqrt(2))), ...

All solutions of a problem from plane geometry based on analytic geometry can be shortened using geometrical interpretation of complex numbers, as in the Peter Breuer 's solution.

In my next message, pure geometrical solutions for my problem.

I will use the following result, called sometimes "The Median theorem": If S is the middle point of segment YZ and X is a point wich is not collinear with Y and Z, then

4XS2=2(XY2+XZ2)-YZ2

CASE 1 - M( I denote the point with M, not X) moves on the circle C(O,R) inscribed in the square ABCD.

O is the middle point of AC and BD and we apply Median theorem in triangles MAC and MBD:

4MO2=2(MA2+MC2)-AC2 => 4R2=2(MA2+MC2)-8R2 =>MA2+MC2=6R2

Similarly MB2+MD2=6R2, then MA2+MB2+MC2+MD2=12R2

CASE 2 - M moves on the circle C(O,R) and the square ABCD is inscribed in the circle C(O,R)

Angles AMC and BMD are right angles and we can apply Phytagora's theorem:

MA2+MC2=AC2=4R2=BD2=MB2+MD2 =>MA2+MB2+MC2+MD2=8R2

Clearly R from case 1 is not the same with R from case 2 !!!

Dear Dinu, you write "Therefore, in the case The circle is inscribed in the square the center O of the circle is the origin and D( (R/2)(sqrt(2)); -(R/2)(sqrt(2))), ...". If the circle is inscribed in the square, then the circle is inside the square. As a result, the point D is outside the circle, since it is a vertex of the square. However, the point ((R/2)(sqrt(2)),-(R/2)(sqrt(2))) is on the circle, since ((R/2)(sqrt(2)))2+(-(R/2)(sqrt(2)))2=R2. Do you mean that the square is inscribed in the circle or am I missing something?

Spiros Konstantogiannis Sorry Spiros, no mistake in your proofs! Me, I have confused positions in the comment about your proofs, so it's my mistake! Sorry again! Sorry to Sergey Gladun too!

I think my confusion was generated by above statement of Liudmyla: "In the square ABCD inscribed in a circle with center O and radius 1, let X be an arbitrary point within the inscribed circle." But it's not a problem, all is understandable, it's more important that Liudmyla's questions are nice and interesting, inviting to think on beautiful plane geometry problems!

It works slightly simpler with the (cos t, sin t) for the point on the circle. The real question is why is this quantity constant? Does represent something? And for Liudmyla: how did you figured this out? Was it just a random computation and you saw that it is constant or you knew it will be constant by some other reason?

I think I figured it out. Say that for a > 0 we have the square with vertices A(-a, - a), B (- a, a), C (a, a), D (a, - a). Then for A > 8 a^4 the locus of all points P such that PA^2 PC^2 + PB^2 PD^2 = A is the circle of center the origin and radius square root of (A - 8 a^4).

With a change of variable can be generalize to every square, the locus being the circle with center in the center of the square and the radius the square root of (A - side^4)

Above I forgot to divide by 2 under the square root

Spiros Konstantogiannis

The expression

ax2*CX2+BX2*DX2 holds true for both the inscribed circle within a square and the circumscribed circle around the square, with an arbitrary point

X on the circle. However, there exists another set of problems involving the sum AX2+BX2+CX2+DX2 with a circle and a square, which can be easily and quickly solved using the property of the median.

I can provide solutions for all of them if you find it intriguing.

To Liudmyla

As I mentioned above, it is true that the expression is constant for all circles

with center in the center of the square. Different constant depending on the constant. Inscribed or excribed is just a particular case.

Liudmyla Hetmanenko, it would be interesting to know the details if they are different from those given by Dinu Teodorescu.

Dear colleagues, I suggest you look at the following generalization of this task. See pdf-file.

Dear colleagues and fans of geometry, I suggest you familiarize yourself with the following general theorem. See pdf-file.

Sergey E. Gladun Very nice your generalizations, and solutions using complex numbers also!

Dinu Teodorescu Thank you very much for taking the time to look at the theorem.

Sergey Evgenjevich Gladun: Nice work. You should make a note with both.

Gabriel T. Prajitura Spiros Konstantogiannis Thank you very much for taking the time to review both theorems.

Please see the following file.