SOLUTIONS

No problem is ever permanently closed. Any comments, new solutions, or new insights on old problems are always welcomed by the editor.

17. Proposed by Stanley Rabinowitz, Westford, Massachusetts.

Let ABCD be an isosceles tetrahedron. (An isosceles tetrahedron is a tetrahedron whose opposite edges are equal.) Denote the dihedral angle at edge AB by ^{}A̸B. Prove that

\frac{AB}{sin^{}A̸B} = \frac{AC}{sin^{}A̸C} = \frac{AD}{sin^{}A̸D} .

Solution by the proposer.

Let AH be the altitude from A to face BCD and let AE be the altitude from A to

BC in face ABC. Then triangle AHE is a right triangle with ^{}A̸EH = ^{}B̸C. Hence

sin^{}B̸C = AH/AE. But AH = 3V/K, where V is the volume of the tetrahedron and K is

the area of any face; and AE = 2K/BC. Therefore sin^{}B̸C = 3V (BC)/2K^{2} and hence

\frac{sin^{}B̸C}{BC}

is constant. Since BC was arbitrary, this is true for all six edges of the tetrahedron.

18. Proposed by Jayanthi Ganapathy, University of Wisconsin–Oshkosh, Oshkosh, Wisconsin.

Let f be differentiable on an interval of the form (M, ∞). Suppose

lim_{x→∞}(f(x) + xf′(x)) = α ,

where α is finite. Prove

lim_{x→∞} f(x) and  lim_{x→∞} f′(x)

exist and evaluate these limits.

Solution I by Robert Doucette (student), University of Iowa, Iowa City, Iowa. Let g : (M, ∞) → R be defined by g(x) = xf(x). The hypothesis becomes

lim_{x→∞} g′(x) = α .

Let ϵ > 0. We may choose x_{0} > max {M, 0} such that x > x_{0} implies

|g′(x) − α| < \frac{ϵ}{4} .

Next choose x_{1} > x_{0} such that

x_{1}(\frac{ϵ}{4}) > x_{0}(|α| + \frac{ϵ}{4}) and |g(x_{0})| .

Suppose x > x_{1}. By the Mean Value Theorem, there is a c > x_{0} such that

g(x) − g(x_{0}) = g′(c)(x − x_{0}) .

With this

|\frac{g(x) − g(x_{0})}{x} − α| ≤ |g′(c) − α| + |g′(c)|\frac{x_{0}}{x}

< \frac{ϵ}{4} + (|α| + \frac{ϵ}{4})\frac{x_{0}}{x_{1}}

< \frac{ϵ}{2} .

Therefore

|f(x) − α| = |\frac{g(x)}{x} − α| ≤ |\frac{g(x) − g(x_{0})}{x} − α| + \frac{|g(x_{0})|}{x}

< \frac{ϵ}{2} + \frac{|g(x_{0})|}{x_{1}}

< ϵ .

This shows that

lim_{x→∞} f(x) = α .

Also

lim_{x→∞} f′(x) = lim_{x→∞} \frac{g′(x) − f(x)}{x} = 0 .

Solution II by Don Redmond, Southern Illinois University, Carbondale, Illinois.

We use two lemmas that can be found in G.H. Hardy, A Course in Pure Mathematics, 10th ed., p. 280.

\underline{Lemma 1}. Suppose that

lim_{x→+∞} ϕ′(x) = a ,

where a is finite. If a ≠ 0, then

lim_{x→+∞} \frac{ϕ(x)}{ax} = 1

and if a = 0, then

lim_{x→+∞} \frac{ϕ(x)}{x} = 0 .

\underline{Lemma 2}. If

lim_{x→+∞} ψ(x) = a ,

where a is finite, then

lim_{x→+∞} ψ′(x) = 0 .

In the problem note that

f(x) + xf′(x) = (xf(x))′ .

Thus the statement of the problem is that

lim_{x→+∞}(xf(x))′ = α ,

where α is finite. By Lemma 1 we see that, if α ≠ 0

lim_{x→+∞} \frac{xf(x)}{αx} = 1 ,

that is,

lim_{x→+∞} f(x) = α ;

and if α = 0, then

lim_{x→+∞} \frac{xf(x)}{x} = 0 ,

that is,

lim_{x→+∞} f(x) = 0 .

Thus we see that

lim_{x→+∞} f(x) = α ,

where α is finite, and so, by Lemma 2, we see that

lim_{x→+∞} f′(x) = 0 .

Solution III by the proposer.

Since

f(x) = f(x) · \frac{x}{x}

we see by L’Hôpital’s Rule that

lim_{x→∞} f(x) = lim_{x→∞} \frac{xf(x)}{x}

= lim_{x→∞} \frac{f(x) + xf′(x)}{1}

= α .

Hence

lim_{x→∞} f(x) = α

and

lim_{x→∞} xf′(x) = 0

so that

lim_{x→∞} f′(x) = lim_{x→∞} xf′(x) · \frac{1}{x} = 0 .

If the condition

lim_{x→∞}(f(x) + xf′(x)) = α

is changed to

lim_{x→∞}(kf(x) + f′(x)) = α , k > 0 ,

the conclusion about the existence of limits still holds. Follow the same argument more or less except use the fact that

f(x) = f(x) · \frac{e^{kx}}{e^{kx}}

and apply L’Hôpital’s Rule.

19. Proposed by Russell Euler, Northwest Missouri State University, Maryville, Missouri.

Prove that triangle ABC is isosceles if and only if

tan(A − B) + tan(B − C) = tan(A − C) .

Solution by Stanley Rabinowitz, Westford, Massachusetts.

With the given information, the following statements are equivalent:

tan[(A − B) + (B − C)] = \frac{tan(A − B) + tan(B − C)}{1 − tan(A − B) tan(B − C)}
tan(A − C) = \frac{tan(A − C)}{1 − tan(A − B) tan(B − C)}
tan(A − C) − tan(A − C) tan(A − B) tan(B − C) = tan(A − C)
tan(A − C) tan(A − B) tan(B − C) = 0
tan(A − C) = 0 or tan(A − B) = 0 or tan(B − C) = 0 .

Hence, since A, B, C are angles of a triangle, so that −π < A − B < π, etc., we must

have A − C = 0 or A − B = 0 or B − C = 0, i.e. △ABC is isosceles.

Note that we did not actually need the fact that A + B + C = π.

Also solved by Donald Fuller, Gainesville College, Gainesville, Georgia; Jayanthi Ganapathy, University of Wisconsin – Oshkosh, Oshkosh, Wisconsin; Bob Prielipp, University of Wisconsin – Oshkosh, Oshkosh, Wisconsin; Donald Skow, University of Texas – Pan American, Edinburg, Texas; and the proposer.

20. Proposed by Troy Hicks, University of Missouri–Rolla, Rolla, Missouri.

The useful lemma given below is part of the folklore of topology. A non–elementary proof was given in [1] and the lemma was used to prove the following theorem. A Tychonoff space X is separable and metrizable if and only if every compatible uniformity on X contains a compatible totally bounded uniformity. Then applications of this result were given to group actions, compactifications, and proximity classes of uniformities. It seems appropriate to have an elementary proof in the literature. Supply one.

\underline{Lemma}. Let (X, U) be a uniform space. Then for any closed subset A of X and any point z ∈ X − A, there is a uniformily continuous function f : (X, U) → [0, 1] such that f(z) = 0 and f(A) = 1.

\underline{Reference}

1. P. L. Sharma and T. L. Hicks, “Uniformities on Separable Metrizable Spaces,” Math. Japonica, 25, No. 6 (1980), 677–680.

Solution by the proposer.

The following two results can be found in standard texts. They will be used without

proof.

(a) Every uniformity can be derived from the family G(U) of its uniformily continuous

pseudo–metrices; that is

{V_{ρ,r} : ρ ∈ G(U) and r > 0}

is a base for U where

V_{ρ,r} = {(x, y) : ρ(x, y) < r} .

(b) f : (X, U) → (Y, V) is uniformily continuous if and only if for every d ∈ G(V) and each ϵ > 0, there exists ρ ∈ G(U) and δ > 0 such that, if ρ(x, y) < δ, then d(f(x), f(y)) < ϵ.

\underline{Proof}. From (a), there exists ρ ∈ G(U) and r > 0 such that

V_{ρ,r}[z] = {y : ρ(z, y) < r} ⊂ X − A .

Define f by

f(y) = min {1, ρ(z, y)/r} .

Observe that f(z) = 0 and f(A) = 1. We show that f is uniformily continuous. Given 0 < ϵ < 1, we may assume 0 < r < ϵ. From (b), it suffices to prove that ρ(x, y) < rϵ = δ implies |f(x) − f(y)| < ϵ.

\underline{Case 1}. x, y, ∈ X − V_{ρ,r}[z]. Then f(x) = f(y) = 1.

\underline{Case 2}. x, y, ∈ V_{ρ,r}[z]. Then
|f(x) − f(y)| = |ρ(z, x)/r − ρ(z, y)/r| = ρ(x, y)/r < rϵ/r = ϵ .
\underline{Case 3}. x ∈ V_{ρ,r}[z] and y ∉ V_{ρ,r}[z]. Then
r ≤ ρ(y, z) ≤ ρ(y, x) + ρ(x, z)

or

r − ρ(x, z) ≤ ρ(y, x) < rϵ .

Now

|f(y) − f(x)| = r/r − ρ(x, z)/r < rϵ/r = ϵ .