SOLUTIONS

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45. [1992, 88] Proposed by Mohammad K. Azarian, University of Evansville, Evansville, Indiana.

Let α, β, and γ be the direction angles of a vector R. Without using Lagrange multipliers, show that

cos α + cos β + cos γ − 2 cos α cos β cos γ ≤ \sqrt{2}.

Solution by the proposer.

If we let

cos α = \frac{a}{\sqrt{2}}, cos β = \frac{b}{\sqrt{2}}, cos γ = \frac{c}{\sqrt{2}},

then we need to show that

a + b + c − abc ≤ 2.

Thus, it is enough to prove that

4(4 − (a + b + c − abc)^{2}) ≥ 0.

But, using

a^{2} + b^{2} + c^{2} = 2(cos^{2} α + cos^{2} β + cos^{2} γ) = 2,

we have

\begin{aligned}4(4 − (a + b + c − abc)^{2}) & = (2 − 2ab)(2 − 2ac)(2 − 2bc) + 4(abc)^{2} \\ & = (c^{2} + (a − b)^{2})(b^{2} + (a − c)^{2})(a^{2} + (b − c)^{2}) + 4(abc)^{2}. \end{aligned}

This completes the proof. Note that equality holds if R is perpendicular to one of the axes and is making a 45-degree angle with each of the other two axes.

46. [1992, 88] Proposed by Robert E. Kennedy and Curtis Cooper, Central Missouri State University, Warrensburg, Missouri.

What is the leading digit of 2^{googol}?

Solution by Kevin Robinson, Messiah College, Grantham, Pennsylvania and the proposers.

The leading digit of 2^{googol} is 2. To see this, we observe that for any positive integer n,

2^{n} = 10^{⌊n log 2⌋} · 10^{{n log 2}},

where ⌊·⌋ denotes the greatest integer function, {·} denotes the fractional part operator, and log 2 denotes the base 10 logarithm of 2. By the above identity, we have the leading digit of 2^{n} is the units digit of 10^{{n log 2}}. Using the computer algebra system DERIVE, the first 120 digits of log 2 are

0.30102999566398119521

37388947244930267681 89881462108541310427 46112710818927442450 94869272521181861720 40684477191430995379 .

Therefore,

10^{{googol log 2}} = 10^{{10^{100} log 2}} = 10^{.40684...} = 2….

Thus, the leading digit of 2^{googol} is 2.

One incorrect solution was also received.

47. [1992, 88] Proposed by Russell Euler, Northwest Missouri State University, Maryville, Missouri.

Find all the solutions of

(x − 1)x(x + 1)(x + 2) = −1.

Solution I by Bob Prielipp, University of Wisconsin - Oshkosh, Oshkosh, Wisconsin; Robert L. Doucette, McNeese State University, Lake Charles, Louisiana; Kevin Robinson, Messiah College, Grantham, Pennsylvania; Joseph E. Chance, University of Texas - Pan American, Edinburg, Texas; Frank Flanigan, San Jose State University, San Jose, California; Kandasamy Muthuvel, University of Wisconsin - Oshkosh, Oshkosh, Wisconsin; Hertag T. Freitag, Roanoke, Virginia; J. Sriskandarajah, University of Wisconsin Center, Richland Center, Wisconsin and the proposer.

We may construct the following collection of equivalent equalities.

(x − 1)x(x + 1)(x + 2) = −1
x^{4} + 2x^{3} − x^{2} − 2x + 1 = 0
(x^{2} + x − 1)^{2} = 0

x = \frac{−1 + \sqrt{5}}{2} (double root) or x = \frac{−1 − \sqrt{5}}{2} (double root).

Solution II by Mohammad K. Azarian, University of Evansville, Evansville, Indiana and J. Sriskandarajah, University of Wisconsin Center, Richland Center, Wisconsin.

From the substitution

x = \frac{u − 1}{2},

we obtain

(x − 1)x(x + 1)(x + 2) = (\frac{u − 3}{2})(\frac{u − 1}{2})(\frac{u + 1}{2})(\frac{u + 3}{2}) = −1.

This implies that

u^{4} − 10u^{2} + 25 = 0.

Thus, u^{2} = 5 is a double root. Hence, u = ±\sqrt{5}. Therefore,

x = \frac{±\sqrt{5} − 1}{2},

where both of these roots are of multiplicity two.

Generalized Solution I by Joseph E. Chance and Donald P. Skow (jointly), University of Texas - Pan American, Edinburg, Texas.

Let k be an integer. Find all the solutions of

(x − k)(x − k + 1)(x − k + 2)(x − k + 3) = −1.

Let

r = (x − k + 1)(x − k + 2).

Then

r = (x − k)^{2} + 3(x − k) + 2.

Hence,

r − 2 = (x − k)^{2} + 3(x − k) = (x − k)(x − k + 3).

Thus, the original equation becomes

r(r − 2) = −1.

This implies r = 1. Hence,

(x − k)^{2} + 3(x − k) + 1 = 0

and so

x − k = \frac{−3 ± \sqrt{5}}{2}.

Therefore,

x = \frac{−3 ± \sqrt{5} + 2k}{2} .

If k = 1, we have the solution to problem 47.

Generalized Solution II by Frank Flanigan, San Jose State University, San Jose, California.

Let

P (x) = (x + a_{1})(x + a_{2})(x + b_{1})(x + b_{2}) + k.

Suppose a_{1}, a_{2}, b_{1}, b_{2}, and k are elements of a field F. Assume

a_{1} + a_{2} = b_{1} + b_{2} = s.

Now, if a = a_{1}a_{2}, b = b_{1}b_{2} and y = x^{2} + sx, then

P (x) = (x^{2} + sx + a)(x^{2} + sx + b) + k
= (y + a)(y + b) + k
= y^{2} + (a + b)y + (ab + k)
= (y − σ)(y − τ)
= (x^{2} + sx − σ)(x^{2} + sx − τ)

It is important that the roots σ, τ of the quadratic in y will lie in the original scalar field F of characteristic ≠ 2 if and only if the discriminant

△ = △_{y} = (a − b)^{2} − 4k

is a square in F. This is decisive for the irreducibility of P (x) in the ring F [x] of polynomials.

Problem 47 can now be solved by taking F = ℝ and a_{1} = 0, a_{2} = 1, b_{1} = −1, b_{2} = 2, and k = 1.

Also solved by N. J. Kuenzi, University of Wisconsin - Oshkosh, Oshkosh, Wisconsin; J. Sriskandarajah, University of Wisconsin Center, Richland Center, Wisconsin (second solution) and Donald P. Skow, University of Texas - Pan American, Edinburg, Texas. Frank Flanigan commented that his general solution to P (x) = 0 can be extended to the case when P (x) involves the product of six, rather than four, suitable linear factors.

48. [1992, 89] Proposed by Alvin Beltramo (student), Central Missouri State University, Warrensburg, Missouri.

A standard deck of 52 cards is shuffled and then the cards are displayed, one at a time. Before each card is displayed, a person with a perfect memory guesses what each card is. How many cards can this person expect to guess correctly?

Solution I by Jenhua Tao, Central Missouri State University, Warrensburg, Missouri; Robert L. Doucette, McNeese State University, Lake Charles, Louisiana; K. L. D. Gunawardena, University of Wisconsin - Oshkosh, Oshkosh, Wisconsin; N. J. Kuenzi, University of Wisconsin - Oshkosh, Oshkosh, Wisconsin and Kevin Robinson, Messiah College, Grantham, Pennsylvania.

Define X_{i} to be the Bernoulli random variable for the ith card, i.e., X_{i} = 0 if the person guesses the ith card incorrectly and X_{i} = 1 if the person guesses the ith card correctly. Then

X = X_{1} + X_{2} + ⋯ + X_{52}

is the number of cards this person guesses correctly. Thus,

E(X) = E(X_{1}) + E(X_{2}) + ⋯ + E(X_{52}).

Since this person has perfect memory,

P (X_{i} = 1) = \frac{1}{52 − i + 1}

and

E(X_{i}) = \frac{1}{52 − i + 1}.

So,

E(X) = \frac{1}{52} + \frac{1}{51} + \frac{1}{50} + ⋯ + 1 ≐ 4.538.

Generalized Solution I by K. L. D. Gunawardena, University of Wisconsin - Oshkosh, Oshkosh, Wisconsin.

Let X_{i} = 1 if the ith card is guessed correctly and X_{i} = 0 otherwise. Then

S = ∑_{i=1}^{52} X_{i}

denotes the total number of cards guessed correctly. Now suppose the person does not have a perfect memory, but he can remember the last k cards displayed prior to the displaying of the ith card. Thus,

P (X_{i} = 1) = \{ \begin{matrix}\frac{1}{53−i}, & 1 ≤ i ≤ k; \\ \frac{1}{52−k} , & k + 1 ≤ i ≤ 52. \end{matrix}

Therefore,

E(S) = ∑_{i=1}^{k} \frac{1}{53 − i} + ∑_{i=k+1}^{52} \frac{1}{52 − k} = 1 + ∑_{i=1}^{k} \frac{1}{53 − i}.

Note that k = 51 corresponds to a person with a perfect memory.

Generalized Solution II by the proposer. Suppose there are n cards, numbered 1 to n in the deck. Then the expected number of correct guesses by a person with a perfect memory is

∑_{k=1}^{n} \frac{1}{k} .

But,

∑_{k=1}^{n} \frac{1}{k} = log n + γ + O(\frac{1}{n}),

where log is the natural logarithm and γ is Euler’s constant. Thus, the expected number of correct guesses for a person with perfect memory, in a deck of n cards numbered 1 to n is asymptotic to log n.

Also solved by the proposer. One incorrect solution was also received. Kevin Robinson and the proposer noted that the exact value is

\frac{14063600165435720745359}{3099044504245996706400}.