PROBLEMS           Vol. 102(3), March 2002

ð   4719: Proposed by: Richard L. Francis, Cape Girardeau, MO

If x, y and z are distinct real numbers such that x + y + z = xyz, then the triple x, y, z is decimally perfect. Show that the tangent of the angles of a scalene oblique triangle are decimally perfect.

 ð   4720: Proposed by: Michael Brozinsky, Central Islip, NY

Let A and B be adjacent vertices of square ABCD. If point E is interior to the square, show that ÜAEB> 45¡.

 ð   4721: Proposed by: Joseâ Luis Diâaz-Barrero, Barcelona, Spain

Find all prime numbers p such that (a) 23p + 4 is a perfect square; (b) l3p + 1 is a perfect square.

 ð   4722: Proposed by: Richard L. Francis, Cape Girardeau, MO

Show if two perfect numbers are consecutive integers that the even one must be the smaller.

á    4723: Find n if:

á   4724: Find the shortest distance from the origin to the hyperbola x² + 8xy + 7x² = 225.

Solutions 

4686: Suppose that f is a continuous function defined on all x in [-1000, 10001. Also suppose that: f(0) = 0, f(2 ÷ x) = f(2 + x) and f(7-x) = f(7 + x). What is the minimum number of zeros for the function f on the given domain?

 Solution:Paul M. Harms, North Newton, KS.

If x = x₁ is a zero of f, the condition f(2 ÷ x) = f(2 + x) gives a zero of f which is the point on the x ÷ axis symmetric to x = x₁ about x = 2. Similarly for f(7 ÷ x) = f(x+7). Starting with x = 0 as a zero of f and alternating symmetric conditions, first using f(2 ÷ x) = f(2 + x) to find another zero of f, (x = 4), we obtain the zeros

x  = 0,4, 10, -6, 20, -16,30, -26,..., 1000, -996,·. Starting with x = 0 as a zero and alternating the ãsymmeticä condition on f (x-7) = f (7 + x), we obtain the zeros x = 0,14, -10,24, -20,34, -30,·,994, -1000·. The 200 positive zeros of f in [-1000,1000] are x = 4, 10, 14,20, 24,30,..., 994, 1000 and the 200 negative zeros of f in [-1000,1000] are

x  = -6, -10, -16, -20. ÷ 26,·, -996, -1000. The total number of zeros of f from above is 401, including x = 0. This is the minimum number of zeros for f.

 Also solved by: N.J. Kuenzi, Oshkosh,WI.

 4687: True or False? Substantiate your answer.

a) iⁱ is a real number (i =… -1,i² = 1.)

b)     It is possible to construct using only Euclidean tools, a regular polygon with 80 sides.

c)    x⁴ + y⁴ can be factored over the complex numbers:

x⁴ + y⁴ = (x² - iy²)(x² ÷ iy²) = (x² - iy²)(x+…(i)y)(x- …(i)y) where i=…-1; i² = 1. Moreover, x⁴ + y⁴ can be factored over the reals. 

 d)         Although  ,for no n₀ is  an integer.

 e)     The graphs of two distinct quadratic equations cannot have more than four points of intersection.

 Solution by Paul M. Harms, North Newton, KS.

a) True;

b) True; Gauss proved that a regular polygon having a prime number of sides of the form 2²ⁿ +1 can be constructed with Euclidean tools. Since a 5 sided regular polygon can be constructed so can a regular polygon of 5(2ⁿ) sides were n is a positive integer. 80 = 5 (2¹⁶).

c) True/False; x⁴ + y⁴ = (x² + y² +…2xy) (x² + y² +…2xy) but generally, ãto factorä means to ãto factor into the same formä; the unstated part of the problem is: can z⁴ + y⁴ be factored into the product of binomials. If this werenât implicitly understood, then prime numbers and every real number can be factored: p = …p…p. (Commentary added by editor.)

d) Trivially; if n₀ =1,  is an integer, but for all integers n₀ >1, is not an integer.

e) False; Consider:

The points (-1, 6), (3, -2), (1, 1), (0, 4) and (2, 0) satisfy the system. (Commentary added by editor.)

 4688: Let AB be the diameter of a circle with center 0 and radius R. Let CD be a chord of the circle intersecting AB at E, such that ÜOED= 45¡. Prove that is independent of the location of E. I.e., prove that is a constant.

 Solution by Michael Brozinsky, Central Islip, NY.

Let CE = a, DE = b, EQ = d and note that ÜOEC = 135¡. From DEDO and DEOC we have by the law of cosines:

(r² - b² - d²)²-2b²d² -(r²- a²-d²)² ÷ 2a²d². Expansion and simplification gives: (b² - a²)(b²+ a²)= 2r²(b² ÷ a²), and so is a constant. (The case that E is O, i.e., a = b follows trivially.)

 Also solved by: Irwin Feinstein, Wilmette, IL; Paul M. Harms,

North Newton, KS; N.J. Kuenzi, Oshkosh,WI and Michael

Brozinsky (who also submitted a second solution different from the one above.)

4689: Compute:

 Solution by William T. Bailey, Blacksburg, VA.

Let x= 111, 111,111,111 then

     = x = = = 3x +1     So,  = 333,333,333,334. 

Also solved by: Paul M. Harms, North Newton, KS; N.J. Kuenzi, Oshkosh, WI; Scott H. Brown, Montgomery, AL.

4690: Show that every third degree polynomial is symmetric with respect to its point of inflection. (Think of a solution which is not straight forward computation.)

Solution 1) by Michael Brozinsky, Central Islip, NY.

If f(x) is a third degree polynomial with inflection point at x = τ, we have by Taylorâs formula that

  F(τ +x) = F(τ) + Fâ(τ)x + Fäâ(τ) x³/6

since Fä( τ) = 0 and derivatives of order greater than three vanish. Hence, replacing x by öx  we have

Solution 2) by Paul M. Harms, North Newton, KS.

It can be shown that the graph of every polynomial of the form x³ + mx = y is symmetric around the origin, its point of inflection; adding n to the left hand side of the equation simply raises the graph n units; its inflection point is now at (0,n). So, every polynomial of the form x³  + mx + n = y is symmetric around its inflection point (0, n). Every third degree polynomial of the form ax³ + bx²  +cx + d = y can be transformed into one of the form w³ + mw +n = y by letting x = w-b/3a. This transformation simply slides the graph of ax³ + bx² + cx + d = y so that its point of inflection is on the y axis. Hence, the graph of every third degree polynomial of the form y= ax³  +bx² +cx + d is symmetric around its point of inflection.

 4691: Given ΔABC. Let the interior angle bisectors of A and B meet the sides BC and AC in points D and E respectively, and let the bisector of the external angle at C meet side AB extended at F. Prove that the points D, E and F are collinear.

 Solution by Michael Brozinsky, Central Islip, NY.

Without loss of generality place ΔABC, with opposite sides respectively a, b and c, in the coordinate plane such that vertex A is at the origin, vertex C has coordinates (b, 0) and vertex B has coordinates (g, h). Let the angle bisectors of ÜA and ÜC intersect BC and AB at E and D respectively, and let the bisector of the external angle at B intersect the extension of AC at F:(f, 0). By the angle bisector theorem  and . As a result, the coordinates of D, E, and F are:

, ,

the slopes of DE and FE simplify to h. ; hence the points D, E and F are collinear.