Let P 1 and P 2 be two convex quadrilaterals such that P 1 ≠ P 2 and A r e a ( P 1 ) ≥ A r e a ( P 2 ). Is it true that it is not possible that all sides and diagonals of P 1 are shorter than the corresponding sides and diagonals of P 2 ?Suppose that all four sides and one diagonal of P 1 are shorter than the corresponding sides and diagonal of P 2 . Then, as it is explained in the answer to my older question, the two triangles which triangulate P 2 should be "flatter" (otherwise it's not possible A r e a ( P 1 ) ≥ A r e a ( P 2 )). From this it should follow that the other diagonal of P 1 is longer than the corresponding diagonal of P 2 .Do you think it's correct? Can you help me formalizing it?

Question

Let       P    1   and       P    2   be two convex quadrilaterals such that       P    1    ≠      P    2   and   A  r  e  a  (      P    1    )  ≥  A  r  e  a  (      P    2    ). Is it true that it is not possible that all sides and diagonals of       P    1   are shorter than the corresponding sides and diagonals of       P    2  ?Suppose that all four sides and one diagonal of       P    1   are shorter than the corresponding sides and diagonal of       P    2  . Then, as it is explained in the answer to my older question, the two triangles which triangulate       P    2   should be &quot;flatter&quot; (otherwise it&#039;s not possible   A  r  e  a  (      P    1    )  ≥  A  r  e  a  (      P    2    )). From this it should follow that the other diagonal of       P    1   is longer than the corresponding diagonal of       P    2  .Do you think it&#039;s correct? Can you help me formalizing it?

Domenigmh · Accepted Answer

It is possible that   A  r  e  a  (      P    1    )  ≥  A  r  e  a  (      P    2    ) while each side and each diagonal of       P    1   is strictly smaller than the corresponding one in       P    2  .Take for example       P    1   to be a square of side   1. Take       P    2   to be an isosceles trapezoid with the small base and the non-parallel sides all equal to 2  2. Flatten the trapezoid while keeping those three sides at the same length   1. In the limit the large base tends to   3  ⋅  2  =  6, the diagonals tend to   2  ⋅  2  =  4  , and the area tends to   0  , yet all the sides and diagonals of       P    2   are strictly larger than those of       P    1  . Even stronger, any side or diagonal of       P    2   is strictly larger than any other side or diagonal of       P    1  .

Let P_1 and P_2 be two convex quadrilaterals such that P_1≠P_2 and Area(P_1)≥Area(P_2). Is it true that it is not possible that all sides and diagonals of P_1 are shorter than the corresponding sides and diagonals of P_2?

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