Skip to main content
MathOverflow

Return to Question

Post Closed as "Not suitable for this site" by abx, Robert Bryant, Desiderius Severus, Alex M., YCor
added 63 characters in body
Source Link
jcdornano
jcdornano
  • 469
  • 2
  • 15

Let $f: P^4 \to \mathbb R^+$, $(A,B,C,D)\mapsto (AB-BC)^2+(BC-CD)^2+(CD-DA)^2+(AC-BD)^2$. Where $AB$ is the distance betwin $A$ and $B$.

Let $A$, $B$, $C$, $D$ be four points of the plane $P$, respectively in $S_A$, $S_B$, $S_C$, $S_D$ that are areas delimited by squares. We also suppose that no point is the corner of its square—but it can be on the egde.

Suppose that $f(ABCD)>0$$f((A,B,C,D))>0$ (then $ABCD$ is not a square), does there always exist $A'\in S_A$, $B'\in S_B$, $C'\in S_C$, $D'\in S_D$ such that $f(A'B'C'D')<f(ABCD)$$f((A',B',C',D'))<f((A,B,C,D))$?

Let $f: P^4 \to \mathbb R^+$, $(A,B,C,D)\mapsto (AB-BC)^2+(BC-CD)^2+(CD-DA)^2+(AC-BD)^2$.

Let $A$, $B$, $C$, $D$ be four points of the plane $P$, respectively in $S_A$, $S_B$, $S_C$, $S_D$ that are areas delimited by squares. We also suppose that no point is the corner of its square—but it can be on the egde.

Suppose that $f(ABCD)>0$ (then $ABCD$ is not a square), does there always exist $A'\in S_A$, $B'\in S_B$, $C'\in S_C$, $D'\in S_D$ such that $f(A'B'C'D')<f(ABCD)$?

Let $f: P^4 \to \mathbb R^+$, $(A,B,C,D)\mapsto (AB-BC)^2+(BC-CD)^2+(CD-DA)^2+(AC-BD)^2$. Where $AB$ is the distance betwin $A$ and $B$.

Let $A$, $B$, $C$, $D$ be four points of the plane $P$, respectively in $S_A$, $S_B$, $S_C$, $S_D$ that are areas delimited by squares. We also suppose that no point is the corner of its square—but it can be on the egde.

Suppose that $f((A,B,C,D))>0$ (then $ABCD$ is not a square), does there always exist $A'\in S_A$, $B'\in S_B$, $C'\in S_C$, $D'\in S_D$ such that $f((A',B',C',D'))<f((A,B,C,D))$?

Spelling; changed quantification so that it seems better to match the statement
Source Link
LSpice
LSpice
  • 12.9k
  • 4
  • 45
  • 69

Can anyevery quadrangle thatwhose corners are in little squares be more square-like when we aloudallow corners to move in the little squares

Let $A,B,C,D$$f: P^4 \to \mathbb R^+$, $(A,B,C,D)\mapsto (AB-BC)^2+(BC-CD)^2+(CD-DA)^2+(AC-BD)^2$.

Let $A$, $B$, $C$, $D$ be four points of the planplane $P$, respectivallyrespectively in $S_A, S_B, S_C,S_D$$S_A$, $S_B$, $S_C$, $S_D$ that are areas delimited by squares. We also suppose that no point is the corner of his square - butits square—but it can be on the egde.

Let $f: P^4 \to \mathbb R^+$, $(A,B,C,D)\mapsto (AB-BC)^2+(BC-CD)^2+(CD-DA)^2+(AC-BD)^2$

Suppose that $f(ABCD)>0$ (then $ABCD$ is not a square), does there always exist $A'\in S_A, B'\in S_B, C'\in S_C,D'\in S_D$$A'\in S_A$, $B'\in S_B$, $C'\in S_C$, $D'\in S_D$ such that $f(A'B'C'D')<f(ABCD)$  ?

Can any quadrangle that corners are in little squares be more square-like when we aloud corners to move in the little squares

Let $A,B,C,D$ be four points of the plan $P$, respectivally in $S_A, S_B, S_C,S_D$ that are areas delimited by squares. We also suppose that no point is the corner of his square - but it can be on the egde.

Let $f: P^4 \to \mathbb R^+$, $(A,B,C,D)\mapsto (AB-BC)^2+(BC-CD)^2+(CD-DA)^2+(AC-BD)^2$

Suppose that $f(ABCD)>0$ (then $ABCD$ is not a square), does there always exist $A'\in S_A, B'\in S_B, C'\in S_C,D'\in S_D$ such that $f(A'B'C'D')<f(ABCD)$  ?

Can every quadrangle whose corners are in little squares be more square-like when we allow corners to move in the little squares

Let $f: P^4 \to \mathbb R^+$, $(A,B,C,D)\mapsto (AB-BC)^2+(BC-CD)^2+(CD-DA)^2+(AC-BD)^2$.

Let $A$, $B$, $C$, $D$ be four points of the plane $P$, respectively in $S_A$, $S_B$, $S_C$, $S_D$ that are areas delimited by squares. We also suppose that no point is the corner of its square—but it can be on the egde.

Suppose that $f(ABCD)>0$ (then $ABCD$ is not a square), does there always exist $A'\in S_A$, $B'\in S_B$, $C'\in S_C$, $D'\in S_D$ such that $f(A'B'C'D')<f(ABCD)$?

added 7 characters in body
Source Link
jcdornano
jcdornano
  • 469
  • 2
  • 15

Let $A,B,C,D$ be four points of the plan $P$, respectivally in $S_A, S_B, S_C,S_D$ that are areas delimited by squares. We also suppose that no point is the corner of his square - but it can be on the egde.

Let $f: P^4 \to \mathbb R^+$, $(A,B,C,D)\mapsto (AB-BC)^2+(BC-CD)^2+(CD-DA)^2+(AC-BD)^2$

Suppose that $f(ABCD)>0$ (then $ABCD$ is not a square), does there always exist $A'\in S_A, B'\in S_B, C'\in S_C,D'\in S_D$ such that $f(A'B'C'D')<f(ABCD)$ ?

Let $A,B,C,D$ be four points of the plan $P$, respectivally in $S_A, S_B, S_C,S_D$ that are areas delimited by squares. We also suppose that no point is the corner of his square - but it can be on the egde.

Let $f: P^4 \to \mathbb R^+$, $(A,B,C,D)\mapsto (AB-BC)^2+(BC-CD)^2+(CD-DA)^2+(AC-BD)^2$

Suppose that $f(ABCD)>0$ (then $ABCD$ is not a square), does there exist $A'\in S_A, B'\in S_B, C'\in S_C,D'\in S_D$ such that $f(A'B'C'D')<f(ABCD)$ ?

Let $A,B,C,D$ be four points of the plan $P$, respectivally in $S_A, S_B, S_C,S_D$ that are areas delimited by squares. We also suppose that no point is the corner of his square - but it can be on the egde.

Let $f: P^4 \to \mathbb R^+$, $(A,B,C,D)\mapsto (AB-BC)^2+(BC-CD)^2+(CD-DA)^2+(AC-BD)^2$

Suppose that $f(ABCD)>0$ (then $ABCD$ is not a square), does there always exist $A'\in S_A, B'\in S_B, C'\in S_C,D'\in S_D$ such that $f(A'B'C'D')<f(ABCD)$ ?

Source Link
jcdornano
jcdornano
  • 469
  • 2
  • 15