Let P be a convex n-gon on the plane. For k=bar{1,n} define a_k as the length of k-th side of P and d_k as the length of projection of P onto the line containing k-th side of the polygon P. Prove that 2<a_1/d_1+a_2/d_2+...+a_n/d_n leq 4.

Barrett Osborn

Barrett Osborn

Answered question

2022-11-06

Inequality for convex polygons
Let P be a convex n-gon on the plane. For k = 1 , n ¯ define a k as the length of k-th side of P and d k as the length of projection of P onto the line containing k-th side of the polygon P. Prove that
2 < a 1 d 1 + a 2 d 2 + + a n d n 4.
Firstly, let us prove the first inequality. Indeed, if p is the perimeter of the polygon P, then it's clear that 2 d k < p for all k { 1 , 2 , , n }. Hence, we obtain
a 1 d 1 + a 2 d 2 + + a n d n > 2 ( a 1 + a 2 + + a n ) p = 2 p p = 2 ,
as desired.
Now, for the second part note that equality holds if, for example, P is a rectangle, so the second inequality is sharp. For polygon P denote f ( P ) := a 1 d 1 + a 2 d 2 + + a n d n .
Then, it can be shown that if P′ is polygon which is centrally symmetric to P, then the Minkowski sum Q = P + P satisfy the following equality f ( Q ) = f ( P ) .
Thus, it's sufficient to prove the inequality for Q, i. e. for centrally symmetric polygons (it's well-known that P + P is a centrally symmetric polygon). However, it's quite unclear how to continue this approach.
So, is there any way to end this solution?

Answer & Explanation

Nkgopotsev1g

Nkgopotsev1g

Beginner2022-11-07Added 15 answers

Step 1
As was mentioned before it's sufficient to prove the inequality for centrally symmetric polygons. So, suppose that n = 2 m and P = A 1 A 2 A 2 m and let O be center of the symmetry of P. Let h k be the length of projection of P onto line which is perepndicular to k-th side of polygon P. Denote the area of polygon P as S. Then, due to convexity of P we have d k h k S for every k { 1 , 2 , , 2 m }.
Step 2
Hence, k = 1 2 m a k d k k = 1 2 m a k h k S .
Now note that a k h k is equal to the area of parallelogram B i B i + 1 B i + m B i + m + 1 , but S ( B i B i + 1 B i + m B i + m + 1 ) = 4 S ( O B i B i + 1 ) (the corresponding four triangles has the same area). Thus, k = 1 2 m a k d k k = 1 2 m a k h k S = k = 1 2 m 4 S ( O B i B i + 1 ) S = 4 S S = 4 ,

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