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?

sailorlyts14eh 2022-09-04 Answered
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 .
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Answers (1)

Domenigmh
Answered 2022-09-05 Author has 7 answers
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 .
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