Problem 12204
(American Mathematical Monthly, Vol.127, October 2020) Proposed by F. Visescu (Romania).
Prove that the absolute value of the sum of the cosines of the four angles in a convex quadrilateral is less than1/2.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Let α1, α2, α3, α4be the four angles of the convex quadrilateral, we are going to show that
−1 2 <
4
X
k=1
cos(αk) < 1 2.
We have that α1+ α2+ α3+ α4 = 2π and we may assume that 0 < α1 ≤ α2 ≤ α3 ≤ α4 < π. It suffices to show that
4
X
k=1
cos(αk) <1
2. (1)
Indeed let βi = π − αi for i = 1, 2, 3, 4, then β1+ β2+ β3+ β4= 2π, 0 < β4≤ β3 ≤ β2 ≤ β1< π, and
−
4
X
k=1
cos(αk) =
4
X
k=1
cos(βk) <1
2 =⇒ −1 2 <
4
X
k=1
cos(αk) In order to prove (1) we distinguish four cases.
1) If α4≤ π/2 then α1= α2= α3= α4= π/2 andP4
k=1cos(αk) = 0 < 1/2.
2) If α3≤ π/2 < α4, since x → cos(x) is concave in [0, π/2], then by Jensen inequality,
4
X
k=1
cos(αk) ≤ 3 cos α1+ α2+ α3
3
+ cos(α4) = 3 cos 2π − α4 3
+ cos(α4)
< 3 cos 2π − π 3
+ cos(π) = 3
2 − 1 = 1 2 because α4→ 3 cos 2π−α3 4 + cos(α4) is stricly increasing in [π/2, π].
3) If α2≤ π/2 < α3then, again by Jensen inequality,
4
X
k=1
cos(αk) ≤ 2 cos α1+ α2
2
+ cos(α3) + cos(α4) = 2 cos 2π − (α3+ α4) 2
+ cos(α3) + cos(α4)
= −2 cos α3+ α4
2
+ cos(α3) + cos(α4) < −2 cos α3+ π 2
+ cos(α3) − 1
< −2 cos π/2 + π 2
+ 0 − 1 =√
2 − 1 < 1 2
because, since x → cos(x) is convex in [π/2, π] it follows that α4→ −2 cos α3+α2 4+cos(α3)+cos(α4) is strictly increasing in [π/2, π], and α3→ −2 cos α32+π + cos(α3) − 1 is strictly decreasing [π/2, π].
4) If α1≤ π/2 < α2then
4
X
k=1
cos(αk) = cos(α2) + cos(α3) + cos(α4) + cos(α2+ α3+ α4)
= 2 cos α2+ α3
2
cos α2− α3 2
+ 2 cos 2α4+ α2+ α3
2
cos α2+ α3
2
= 4 cos α2+ α3
2
cos α3+ α4
2
cos α4+ α2
2
< 0 < 1 2
because (α2+ α3)/2, (α3+ α4)/2, (α4+ α2)/2 ∈ (π/2, π). Remark. 1/2 is the best constant: consider a convex quadrilateral with α1= α2= α3= π/3 + t and α4= π − 3t then lim
t→0+ 4
X
k=1
cos(αk) = 1/2.
