For 1 ≤ k ≤ 4, let hk denote the distance from thek-th vertex to the plane of the opposite face

Problem 11783

(American Mathematical Monthly, Vol.121, June-July 2014) Proposed by Zhang Yun (China).

Given a tetrahedron, letr denote the radius of its inscribed sphere. For 1 ≤ k ≤ 4, let h^k denote the distance from thek-th vertex to the plane of the opposite face. Prove that

4

X

k=1

h^k−r h^k+ r ≥ 12

5 .

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

The volume of the tetrahedron is given by h^kA^k

3 =rS 3

where A^k is the area of the face of opposite to the k-th vertex and S =P4

k=1A^k is the surface area of the tetrahedron. Hence h^k= r/t^k with t^k= A^k/S ∈ (0, 1) and

4

X

k=1

h^k−r h^k+ r =

4

X

k=1

f (t^k) ≥ 4f

4

X

k=1

t^k

!

= 4f 1 4



=12 5

where f (t) = (1 − t)/(1 + t) is a convex function in [0, +∞).