BMAD Posted December 1, 2022 Report Share Posted December 1, 2022 (edited) For any Convex Quadrilateral, show that the ratio of the Area to its Perimeter^2 is always <1/16, bonus points if you can show that it holds for concave quadrilaterals (not squares). Edited December 1, 2022 by BMAD Quote Link to comment Share on other sites More sharing options...

0 EventHorizon Posted February 28 Report Share Posted February 28 Was there an easier method? Spoiler For all non-square convex quadrilaterals, area < 1/16 * perimeter^2 Case 1: Not all side lengths are equal Let a convex quadrilateral (Q1) consist of side lengths s1, s2, s3,and s4, where s1 is consecutive to s2, s2 to s3, s3 to s4, and s4 to s1. Since the quadrilateral is convex, the diagonals cross inside the shape. Let s1≠s2. Split the quadrilateral into two triangles consisting of sides lengths s1, s2, and d (length of the diagonal) for the first triangle (T1A) and s3, s4, and d for other triangle (T2A). Construct a new triangle, T1B, with side lengths (s1+s2)/2, (s1+s2)/2, and d. By Lemma 2 and since s1≠s2, the area of T1A is less than the area of T1B. Similarly, construct triangle T2B with side lengths (s3+s4)/2, (s3+s4)/2, and d. By Lemma 2, the area of T2A is less than or equal to the area of T2B. Connect T1B and T2B along the side of length d to create quadrilateral Q2. Q2 has the same perimeter as Q1, and the area of Q2 is greater than the area of Q1. Now split Q2 along the diagonal which is not the diagonal shared by T1B and T2B. Let the length of this diagonal be d2. This will create triangle T3A with side lengths (s1+s2)/2, (s3+s4)/2, and d2 and triangle T4A with the same side lengths. Construct T3B, T4B, and ultimately Q3 by the same process as before. Q3 has the same perimeter as Q2, and the area of Q3 is greater than or equal to the area of Q2. Q3 has four sides of length (s1+s2+s3+s4)/4, so Q3 is a rhombus. The area of a rhombus is base times height. The base is just a side length, and the height is the side length times the sine of one of the angles. So the area is ((s1+s2+s3+s4)/4)^2 * sin Z, where Z is one of the angles of the rhombus. Let Q4 be a square with four sides of length (s1+s2+s3+s4)/4. The area of Q4 is ((s1+s2+s3+s4)/4)^2 = (1/16)*(s1+s2+s3+s4)^2. Q4 has the same perimeter as Q3, and the area of Q4 is greater than or equal to the area of Q3 since sin Z <= 1. So Area(Q1) < Area(Q2) <= Area(Q3) <= Area(Q4) = (1/16)*(s1+s2+s3+s4)^2. So the area of Q1 is less than one sixteenth of its perimeter squared. Case 2: All side lengths are equal. If all side lengths are equal, but the quadrilateral is not a square, then it is a rhombus. The sine value of all rhombus angles are the same, and since it is not a square, this sine value is less than 1. Let Z be an angle measure in the rhombus. The area is then side^2 * sin Z < side^2 = (perimeter/4)^2 = (1/16)perimeter^2. Lemma 1: The product of two unequal non-negative real numbers is less than their average squared. Let a,b be unequal non-negative real numbers. ab = (((a+b)/2) + ((a-b)/2)) * (((a+b)/2) - ((a-b)/2)) = ((a+b)/2)^2 - ((a-b)/2)^2 < ((a+b)/2)^2, since ((a-b)/2)^2 is positive. Lemma 2: A triangle with side lengths a, b, and c has an area less than the area of a triangle with side lengths (a+b)/2, (a+b)/2, c, if a≠b. Equal areas if a=b. Case 1: a≠b Let triangle T1 have side lengths a, b, and c, where a≠b. Let triangle T2 have side lengths (a+b)/2, (a+b)/2, and c. The semiperimeter of both triangles s = (a+b+c)/2. Using Heron's formula, the area of T1 is sqrt( s(s-a)(s-b)(s-c) ). Similarly, the area of T2 is sqrt( s(s-(a+b)/2)(s-(a+b)/2)(s-c) ). Notice that all of s, (s-a), (s-b), (s-c), and (s-(a+b)/2) are non-negative. The average of (s-a) and (s-b) is (s-(a+b)/2). By Lemma 1, (s-a)(s-b) < (s-(a+b)/2)(s-(a+b)/2) Therefore s(s-a)(s-b)(s-c) < s(s-(a+b)/2)(s-(a+b)/2)(s-c). And since those are both non-negative reals, sqrt( s(s-a)(s-b)(s-c) ) < sqrt( s(s-(a+b)/2)(s-(a+b)/2)(s-c) ) So the area of T1 is less than the area of T2. Case 2: a=b If a=b, then both triangles have the same side lengths. Therefore they have the same areas. Concave Spoiler Given a concave quadrilateral Q0. Let V1 be the vertex of Q0 that has an interior angle greater than 180 degrees. Let the vertices V2 and V4 be the vertices connected to V1, and V3 be the remaining vertex. Mirror V1 across the line connecting V2 and V4 to point P1. Construct quadrilateral Q1 by connecting the points P1, V2, V3, V4. Q1 has the same perimeter as Q0, and the area of Q1 is greater than the area of Q0. Above was shown that the area of a convex non-square quadrilateral was less than one sixteenth of its perimeter squared, so Area(Q0) < Area(Q1) < (1/16)Perimeter(Q1)^2 = (1/16)Perimeter(Q0)^2. So the area of Q0 is less than one sixteenth of its perimeter squared. If V1 and V3 share the same point and Q1 happens to end up a square, the area of Q0 is 0 which is less than one sixteenth its perimeter. Quote Link to comment Share on other sites More sharing options...

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## BMAD

For any Convex Quadrilateral, show that the ratio of the Area to its Perimeter^2 is always <1/16, bonus points if you can show that it holds for concave quadrilaterals (not squares).

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