The Elements, Book I (Prop I - V)

The Thirteen Books of the Elements

Euclid


Book I

Reference edition: Sir Thomas Heath, Johan Ludvig Heiberg
 The University Press, 1908

Free Google Ebook

Proposition I:  Given finite straight line AB, to construct an equilateral triangle.

Proposition II:  To place at a given point a straight line equal to a given straight line.

Given point A and straight line BC,

Construct equilateral triangle ΔABD [Prop. I]

Construct the circle with center B, and radius BC.

Extend the straight line DB to intersect the circle.  Call the point of intersection F
BC ≅ BF

Construct the circle with center D and radius DF.

Extend straight line DA to intersect this new circle.  Call the point of intersection G:

Now, DG ≅ DF

and AD ≅ BD
And therefore DG − AD ≅ DF − BD
But DG − AD = AG
and DF − BD = BF
Hence AG ≅ BF.

Proposition III:  Given two unequal straight lines, to cut off from the greater a straight line equal to the lesser.

Given straight line AB (the greater) and straight line C (the lesser),

Use an endpoint of C to draw a circle of radius C.

Connect A to the endpoint and build an equilateral triangle. (Prop I)
Extend the side of the triangle to the circle. (Prop II)

Draw a second circle with center at D and extend DA to intersect the circle. (Prop II)

Draw a third circle with center A which also intersects this point:

The intersection of this last circle with AB is labeled γ.
Aγ ≅ C and the construction is complete.

Proposition IV: Side-Angle-Side Congruence

If you can find more than one way to draw the segment BC, let me know.

Proposition V: In an isosceles triangle, the base angles are equal.  If the equal sides are extended in straight lines, the angles under the base are equal.

Given  AB

Construct AC = AB and isosceles triangle ΔABC.

Construct AC = AB and isosceles triangle ΔABC.
Extend AB  to arbitrary point F.
Place  CG = BF  [Prop II, or by definition with the compass]

Connect   FC, GB
     ΔFCA ≅ ΔGBA  [Prop IV, common angle ∠A]
The corresponding angles and sides are then equal, namely
     FC = GB,    ∠CFA ≅ ∠BGA

Then ΔFCB ≅ ΔGBC and
     ∠FBC = ∠GCB,   (the angles under the base)
     ∠BCF = ∠CBG
And  ∠ABC  = ∠ABG −∠CBG  = ∠ACF − ∠BCF  = ∠ACB (the base angles of the isosceles triangle)
Which were the things to be proved.

{Next −−>}

____________________________________
There are hundreds of propositions remaining of which I will post a selection, including only a restatement of the Proposition and a completed diagram.

Kirkman's Schoolgirl Problem

Glad Kirkman knows he's the one with the problem.

In a boarding school there are fifteen schoolgirls who always take their daily walks in rows of threes.  How can it be arranged so that each schoolgirl walks in the same row with every other schoolgirl exactly once a week?  

[Heinrich Dörrie; 100 Great Problems of Elementary Mathematics] 

Kirkman originally posed the problem in 1850, in the Lady's and Gentleman's Diary, a yearly repository of mathematical puzzles and bad poetry about them.

Boxy McGee has found the general solution, and two particular solutions:

A superabundance of solutions!

The puzzle blocks are general; the cards and the hanging pentagon are particular solutions.  Now, about solving it.  I need a mental picture. Let me try the question again.

In a certain town of G---- the Abbess is strict.  The 15 girls are left alone only for their daily walks, which they may take in any of five gardens and woods.  Three girls may visit any one location at a time.  When all the girls go out, they make five trios, and each trio visits a different wood. 
The girls may walk with whomever they wish, but they must submit a schedule to the Abbess in advance.
It happens that Sister Agnès has made an arrangement with Guillot, a strong and beautiful young man whose face was well known and liked by all the girls.  She devised the following ruse.  Every day of the week, Guillot will wait in a certain kingcup field, where the girls often dally out of sight of the Abbess.  Here, he will trade places with Agnès.  He will take her habit, and number of other things she has meant to give him for some time.  She will take his day-clothes, and a number of other things he has earnestly tried to give her, and enjoy an afternoon in town.  He will then take her place, and two girls will enjoy his company in the woods.  On the return trip, the girls will stop at the same spot, and the two will again trade places.  
In this way, for the duration of the week Agnès will have the pleasure of Guillot's company twice a day, going and coming.  In addition, one of the five rows of girls, being now two girls and Guillot, will enjoy a number of hours in a secluded wood.
Now, the girls are jealous of their prize, and eager.  None wishes to be left out, and they cannot decide who shall strip him of trousers twice in the kingcups, and who shall be contented to walk with their young Adonis, and un-dress him in the woods.
Agnès is a sly fox.  See what she has done.  The week prior, she submitted a special schedule to the Abbess in which all 15 girls go out every day of the week. Then, into a hat she placed the fifteen names of the girls, each on a separate slip of paper.  Out of the hat, a single name will be drawn at random.  This name is the one who Guillot will replace for the duration of the week, who shall have him twice daily in kingcup fields.
Agnès' claim to the girls is this: she had arranged the schedule so that no matter which name is to be drawn out of the hat,  each of the remaining fourteen girls will walk in the same row as Guillot exactly once during the week.
To which the girls agree, and it is done, and they are done, there is a great deal of having and doing, and the Old Sister is never more satisfied with the fervor and sincerity of the pleading cries of her young lambs, which fill the valley for seven gold days in of spring.
Eight of the girls become pregnant that winter, are mercifully beaten and thrown into the streets.

The question remains. How did Agnès do it?

THE SOLUTION

Notes:
1) This is not a problem of combinatorics. How many ways are there? is a different question.
2)  This is not a system of 15 linear equations.  5 girls, 3 rows, 7 days, no repetitions are all constraints which limit the solution.  How many independent variables are there? is part of the question.
3) Call Day 1 Sunday.  Each girl may return to the same position the following Sunday.
4) Say each girl walks a path with seven stops and the last stop returns to the first.  But then we need six more girls to fill the remaining stops.  Can the girls be divided up this way? (Yes. For this problem it works very well.  The solution space is cyclic, and very small).

I: SUBSTITUTION

The problem is the same if a stranger takes the place of any one of the girls.  To me, this is the crux.  We can impose a series of limitations without changing the question.  It is surprisingly difficult to check the restrictions for consistency.  Does it throw out any valid solutions?  Substitution is not necessary.  But if the first constraint is, 'we may choose an arbitrary girl as a fixed frame of reference',  this can be double-checked by a simple substitution: can Guillot take the place of any one of the girls, chosen at random?  Yes.

So, set a frame of reference, and move the other girls around it. Choose a symbol for this person, put it somewhere obvious, and don't move it.  Dörrie uses *, and puts it in the middle of a row.  I choose row 1, and refer to her as the Joker.  Hence,

(1)    Row 1 will always be    − *  −

II. PARTITION; SUITS

Each day, two different girls walk with the Joker in row 1.  A total of seven girls will pass through Row 1 on her left, and seven on her right.  Hand each of these girls a playing card.  Girls on the left are given Diamonds.  Girls on the right, Spades.  Hand out the Ace through 7 of each suit, in order:
     On Sunday, give out the Aces.
     On Monday, give out the 2's.
     On Tuesday, the 3's, and so on.
Now, consider the first day, Sunday.  We have

(2)     Row 1, Sunday:     ◊A * ♠A

Arrangement of Suits:

12 girls hold among them the 2-7 of Diamonds and Spades, arranged in 4 rows.  What can we say about them?  There are four cases:
              1                 2                 3                  4    
          ◊  *  ♠          ◊   *  ♠          ◊  *   ♠          ◊  *  ♠
          ◊  ◊  ◊          ◊  ◊  ◊          ◊  ◊  ♠          ◊  ◊  ♠ 
          ◊  ◊  ◊          ◊  ◊  ♠          ◊  ◊  ♠          ◊  ◊  ♠  
          ♠  ♠  ♠          ◊  ♠  ♠          ◊  ◊  ♠          ◊  ♠  ♠ 
          ♠  ♠  ♠          ♠  ♠  ♠          ♠  ♠  ♠          ◊  ♠  ♠ 
Any other permutation can be turned into one of these four by changing the order of the rows and rearranging the cards within a row, which do not affect the solution.  Each girl occupies a different position every day: a Diamond will visit every square on the graph labeled with a diamond.  Each day, each girl walks with 2 others.  In seven days, she has 7*2=14 partners.  They must be:
     The joker,
     the 6 other girls in the same suit
     the 7 girls in the opposite suit.


Case 1:  A Diamond walks with
  - the Joker and 1 ♠
  - (2 rows) *(2 ◊ per row) * (3 visits to each row) = 12 ◊
And likewise for a spade.  Throw out Case 1.

Case 2:  The two suits still do not meet often enough.  Throw out Case 2.

Case 3:  A Diamond walks with
      - the Joker and 1 ♠
      - (3 rows) * (1 ♠  per row) * (2 visits) = 6 ♠
   And in those same rows
      - (3 rows) * (1 ◊  per row) * (2 visits) = 6 ◊
A Spade walks with
      - the Joker and 1 ◊
      - (3 rows) *  (2 ◊ per row)  *  (1 visit) = 6 ◊
      - (1 row)  *  (2  ♠ per row) * (3 visits) = 6 ♠
 Case 3 Satisfies the conditions.

Case 4: Girls in the same suit do not meet often enough. Opposite suits meet too often.  Throw out Case 4.

• Case 3 is the only valid case.  We must have

It it, alas, easier to arrange the remaining information in a satisfactory way using

III. MODULO ARITHMETIC

We need to count by sevens.  Gauss to the rescue.
DEFINITION:  A is congruent to B modulo 7 if the difference A−B is divisible by seven.
          A   ≡  B  mod 7     if      (A−B)  =  7n ,       n any integer.

Do not stare directly into the abstraction!  This is how we count days:  7, 14, 21... days from now it will be the same day of the week as today, which is 0 days away:
          7 ≡  14  ≡  21 ...  ≡  0 mod 7
The definition arises on its own in the context of the problem.

                    Sun     Mon    Tues     Wed      Thur      Fri       Sat      Sun 

                      1         2           3           4           5          6          7      8≡1

Beginning with Sunday, assign the numbers 1-7 to the days of the week.  Say today is Sunday. In two days it will be Tuesday.  Five days ago it was also Tuesday.  Counting days is a circle:

Pick any position on the circle, x.  Add clockwise, subtract counterclockwise.  The following hold no matter where we start:

(3)     Adding or subtracting a multiple of seven lands on the same position, x.
               x + 7n ≡ x       (Any multiple of seven days away is the same day of the week)
          Adding 1 is equivalent to subtracting 6
               x + 1 ≡ x − 6     (Tomorrow is the same day of the week as six days ago)
          Adding 2 is equivalent to subtracting 5, and so on
               x + 2 ≡ x − 5
               x + 3 ≡ x − 4
               x + 4 ≡ x − 3
               x + 5 ≡ x − 2
               x + 6 ≡ x − 1
Where '≡' is read 'equivalent to' or 'congruent to'.  Here, it means "the same day of the week as".

We can also count the distance between any pair of days.  Consider the following diagrams:

 The Red block covers a pair of numbers 1 step apart.  The green pair are 2 numbers apart, and the blue  are 3 apart.  What can we say about them?
      • Any pair of days we choose can be covered by one of the three pieces.  Invariably,
           A difference of ±1 and ±6 refer to the Red piece
           A difference of  ±2 and ±5  refer to Green
           A difference of  ±3 and ±4  refer to Blue
     •  The list of congruences (3) becomes
             x − 1 ≡ x + 6,        x − 6 ≡ x + 1
             x − 2 ≡ x + 5,        x − 5 ≡ x + 2
             x − 3 ≡ x + 4,        x − 4 ≡ x + 3
Each piece is uniquely identified by one of the three differences ±1, ±2, ±3.

IV. DISTANCE

DEFINITION: The distance between a pair of girls in a row is one of the three differences ±1, ±2, ±3.

Each day, the girls trade places: Each moves to the next lowest number in her suit.  The ◊A takes the place of the ◊7, who takes the place of ◊6, ... and the ◊2 moves to the place vacated by ◊A.  The girls walk in a loop  ◊A. . . ◊7.  How can we add this loop to the diagram?

Consider the following arrangement:

         Sunday       Monday         Tuesday        Wednesday

        A   *    ♠         ◊2   *    ♠        ◊3  *     ♠         ◊4   *    ♠
       ◊2  ◊4   ♠        ◊3  ◊5   ♠        ◊4  ◊6   ♠        ◊5  ◊7   ♠

       ◊3  ◊5   ♠        ◊4  ◊6   ♠        ◊5  ◊7   ♠        ◊6   A   ♠     . . .
       ◊6  ◊7   ♠        ◊7   A   ♠         A  ◊2   ♠         ◊2  ◊3   ♠
        ♠    ♠    ♠        ♠    ♠    ♠         ♠    ♠    ♠          ♠    ♠    ♠

The Diamonds in rows 2 and 3 are the same distance apart:
     ◊4 − ◊2 =  ◊5 − ◊3 = 2.  In other words:
 The highlighted rows are repeated from the day before.  If the distance between a pair of diamonds in a row is x.  Each girl walks in that row twice, once with the girl x steps in front of her and x steps behind her.

• Example: The expression 5 ±2  refers to both 3 and 7.  ◊5 appears...-in row 2 on Monday and Wednesday, with the ◊3 and the ◊7, and-in row 3 on Sunday and Tuesday, again with the ◊3 and the ◊7.

The pair of diamonds in each row must be a different distance apart.

V. DIAMONDS

• Rule: If we can arrange the 6 cards ◊2, ◊3, ◊4, ◊5, ◊6, ◊7  so that the cards in rows 2, 3 and 4 are ±1, ±2 and ±3 steps apart, respectively, then the seven Diamonds will walk with one another only once.  Thus we have

Any way the Red, Green, and Blue pieces can be assembled, is a valid solution.  Remembering that one of the places is fixed by the Ace,

...one last constraint arises for the Diamonds: if the Red piece is placed next to the Ace, it is no longer possible to fit both the Green and Blue pieces. This leaves only three possible locations for the Red piece: (◊3 ◊4), (◊4 ◊5), (◊5 ◊6) each of which uniquely determines the Green and Blue pieces.

         ◊3 ◊4            ◊4 ◊5           ◊6 ◊5 
         ◊5 ◊7            ◊7 ◊2           ◊4 ◊2  
         ◊6 ◊2            ◊3 ◊6           ◊3 ◊7

If we count backward in the third case: 7-6-5-4-3-2, we visit the same places, in the same order, as counting forward in the first. Paths I and III are identical. This is a result of the rule, a difference of ±1 and ± 6 refers to the same row; a difference of ±2 and ±5, & c.  
For the numbers 1-7, reversing the direction requires an offset:
     x' ≡  2 − x
Which turns path 1 in to path 3:
           ◊3 ◊4          ◊(-1) ◊(-2)              ◊6 ◊5 
   2 −   ◊5 ◊7   =     ◊(-3) ◊(-5)      ≡     ◊4 ◊2  
           ◊6 ◊2          ◊(-4) ◊( 0)              ◊3 ◊7

The second case is its own mirror image.  Reversing the direction has no effect.  Dressing them up in smart woolen vests does.  It keeps them warm.  Thus we have for the Diamonds

• The arrangement of diamonds in rows 2, 3, 4:
             ◊4 ◊i
             ◊7  ◊j
             ◊6 ◊k
  Case  I:       i, j, k = (3, 5, 2)
  Case II*:    i, j, k = (5, 2, 3)

VI. SPADES

Again, we are only concerned with the values 2-7.  The Ace has already been used, the inequality x ≠ 1, for any Spade x, is a constraint on all that follows.  Labeling the remaining six spades B - G, we have

ROW 5:
All of the Spades meet each other in the fifth row, so the distances between all 3 girls E, F, G must be different.

    E − F ≡ ±1     (E and F are 1 step apart)

    F − G ≡ ±2    (F and G are 2 steps apart)

    E − G ≡ ±3    (E and G are 3 steps apart)
With three girls in the same suit, we must resolve the ± sign.  For simplicity, I have arranged E, F, and G so that the equations are only satisfied if the three differences are all the same sign.  This is easier to see with a diagram:

E, F, G are clockwise: 2, 3, 5

Flip the piece over and E, F, G are counterclockwise, in order.  

Thus we have two possibilities for the last row:

(E, F, G)  ≡ (E, E+1, E+3)   or  (E, E−1, E−3)

ROWS 2-4:

In addition, a Spade must meet six different diamonds when visiting B, C, D: the differences between the three girls in rows 2-4 must all be different. Consider Case I:
     ◊4 ◊3  B
     ◊7 ◊5  C
     ◊6 ◊2  D
The six differences, B−◊4,  B−◊3,  C−◊7,  C−◊5,  D−◊6,  D−◊2  must all be mutually incongruent, modulo 7.  We can rewrite the inequalities
     B−x ≠ C−y              C ≠ B − x + y 
     C−y ≠ D−z      as     D ≠ C − y + z 
     B−x ≠ D−z              D ≠ B − x + z 
where x, y, z are Diamonds in rows 2, 3, and 4 respectively.  In addition, a spade and a diamond in the same row cannot have the same number, because they already meet in the first row with the Joker.  So we add the inqualities
     B ≠ x,   C ≠ y,   D ≠ z

Proceeding in this way... {Click to show the details}

      C ≠ B+2     (C ≠ B − ◊3 + ◊5)
      C ≠ B+4     (C ≠ B − ◊3 + ◊7)
      C ≠ B+1     (C ≠ B − ◊4 + ◊5)
      C ≠ B+3     (C ≠ B − ◊4 + ◊7)
      C ≠ B         
(a)  C ≡ {B+5,  B+6}

      D ≠ C+4     (D ≠ C − ◊5 + ◊2)   (and  -3 ≡ 4)
      D ≠ C+1     (D ≠ C − ◊5 + ◊6)
      D ≠ C+2     (D ≠ C − ◊7 + ◊2)        (-5 ≡ 2)
      D ≠ C+6     (D ≠ C − ◊7 + ◊6)        (-1 ≡ 6)
      D ≠ C         
(b)  D ≡ {C+3,  C+5}

      D ≠ B+6     (D ≠ B − ◊3 + ◊2)        (-1 ≡ 6)
      D ≠ B+3     (D ≠ B − ◊3 + ◊6)
      D ≠ B+5     (D ≠ B − ◊4 + ◊2)        (-2 ≡ 5)
      D ≠ B+2     (D ≠ B − ◊4 + ◊6)
      D ≠ B         
(c)  D ≡ {B+4,  B+1}

When C ≡ B+5                      When C ≡ B+6
    (b)  D ≡ {B+1, B+3}            (b)  D ≡ {B+2, B+4}
    (c)  D ≡ {B+4,  B+1}           (c)  D ≡ {B+4,  B+1}
          D ≡ B+1                               D ≡ B+4

Hence,

For a given value of B, we have:
       (B, C, D) ≡ (B, B+5, B+1) and  (B, B+6, B+4).
With the constraints  B ≠ x,   C ≠ y,   D ≠ z:
       B ≠ 1, 3, 4;  C ≠ 1, 5, 7;  D ≠ 1, 6, 2

Test the remaining possible values of B:  2, 5, 6, 7.
      (B, B+5, B+1):         (B, B+6, B+4):
      (2,  2+5, 2+1)           (2,  2+6, 2+4) 
      (5,  5+5, 5+1)           (5, 5+6,  5+4)        ≡ (6, 4, 7) and (7, 6, 4)
      (6,  6+5, 6+1)           (6,  6+6, 6+4) 
      (7,  7+5, 7+1)           (7,  7+6, 7+4)
Using only the numbers 4, 6, 7.  The other triplet (2, 3, 5) satisfies the equations for E, F, G.  Hence,

SPADES, Case I:
     (B, C, D) ≡ (6, 4 ,7) and (7, 6, 4)
     (E, F, G)  ≡ (2, 3, 5)

Following the same procedure, we obtain for  Case  II: 

     (B, C, D)  ≡  (B, B+1, B+3) or  (B, B+6, B+4)
     (B, C, D)  ≡  (2, 3, 5) and (7, 6, 4)
There is only one solution here. (2, 3, 5), and (7, 6, 4) are the same path, traveling in opposite directions.  If we choose (B, C, D) ≡ (7, 6, 4) the values for (E, F, G) are identical for both Case I and II.  Hence, we have

SPADES, Case II*:
     (B, C, D)  ≡ (7, 6, 4)
     (E, F, G)   ≡ (2, 3, 5)

PART SEVEN: THE REVENGE

At this point, a crucial question arises.  Why are you still reading this?  I accept the precipitous descent: I have been ruined by mathematics. I am a slut for math.  I wake early and nuzzle her skirttails.  If you are reading these words, perhaps you, too, have a problem.

oh hell. on with the solution.

The problem is now solved.  The general solution is:

with only three possible arrangements, in two cases:

Case I:

     i, j, k ≡ (3, 5, 2)

     B, C, D ≡ (6, 4, 7) and (7, 6, 4)

Case II*:

      i, j, k ≡ (5, 2, 3)

     B, C, D ≡ (7, 6, 4) 

where, in the first case, if one suit proceeds in the reverse direction, the other must as well;

whereas in the second case, Diamonds and Spades may proceed in either direction, independently of one another.  Thus the solution reduces to the question how to arrange for arbitrary girls i, j, and  k to meet once each with  {B, C, and D}.   Of course, Kirkman's way of asking is more prettily dressed.

Capturing the Slope of e^x

The exponential inequality from Cauchy's mean theorem
            
can be applied  to the intriguing pair of functions
            
to determine their limiting values using ordinary algebra.  As x becomes infinitely large, the two functions become closer and closer together.  They never meet for any finite value of x, but converge to the same number.  φ(x) approaches from below, always increasing, and Φ(x) from above, always decreasing. The limit is the Euler number, e.

This is the 12th problem in Heinrich Dörrie's 100 Great Problems of Elementary Mathematics.  Along the way, the following inequality is established:
            
which he uses as the basis for solving the next problem:

 (#13): Transform the exponential function e^x into a progression in terms of powers of x.
A charming inequality (marked (1) below) appears in the introductory steps.  Begin with
            
Now consider an arbitrary number a, and a second number  A = a + δ.   δ is also arbitrary. We establish only that A > a, and the difference between the two is δ.


The next two steps are a roundabout way of implying the Calculus.  Set u = δ, and then u = −δ. This gives
            
Multiply the first expressions by e^a and the second by e^A.
            
A bit of rearranging gives
            
where the value (e^A − e^a)/δ can now be sandwiched between two inequalities:

There are two points, a and A.  A being the greater of the two,  e^a < e^A, which are the leftmost and rightmost terms.   The term in the middle is the slope of the line connecting these two points: rise/run.  We can also make δ as small as we like.  In fact, we can make it 0.
Division by zero is indeterminate, and so it is undefined.  In this case, we can define it.  If δ is zero, then
e^A = e^a+0 = e^a
In other words, we can shrink the distance between the leftmost and rightmost terms until they are the same. The middle term remains trapped between the two, and so at δ = 0, its value, too, must be e^a.

So what does expression (1) mean?  The value of the function e^x at any point x is identical to the slope of the tangent line at that point, for every value of x.  It says,

"e^x is its own derivative."

Cauchy's Mean Theorem

The geometric mean of n positive numbers is less than the arithmetic mean.

(Except when all n numbers are the same, in which case the two means are equal)

 ■ The Arithmetic Mean  —  Add n numbers together, and divide by n:
          
■ The Geometric mean —   Multiply n numbers, and take the nth root:
          

In symbols, Cauchy's mean theorem states

The equal condition can be omitted as long as a, b, c, ... n are not all identical.

This is problem #10 in 100 Great Problems of Elementary Mathematics. Dörrie holds the sum constant and examines the product, so for practice I'll go in the reverse direction.  Suppose we have two positive numbers x and y, with x > y.   If the product xy is kept constant, what happens to the two means as x and y are brought closer together?
Introduce a new number a which satisfies

The two new numbers,  (x/a) and (ay) are sandwiched between x and y.  Their product is constant and the same as xy:
                 
What values of a satisfy (2)?  Evaluating each inequality sign separately gives the range of a:


According to Cauchy's theorem, two properties should be satisfied by a.   First, I should always have
(4.1)             ½(x/a+ay) > √xy,
except when x/a = ay.  Second when a =  √ x/y, then  x/a = ay, and the geometric and arithmetic mean of this number will be the same.

Aha!  I arrive at the same place Dörrie found himself: it is easier to prove a somewhat stronger result.   Instead of untangling (4.1), suppose I say,

If the left side is true, then the right is true.  But the right side says,

• Auxiliary Theorem:  Given two pairs of numbers with the same product, the pair that is closer together will always have the smaller sum.

Is (4) true?  Simplifying the left hand side of (4):
          
Yes.  Skipping to the second property, let the distance x/a − ay go to zero.  This is the maximum value of a.  Are both the geometric and arithmetic mean of this number the same?  Let a = √ x/y.  Then

Yes.
Using the same procedure, the result can be extended to n numbers.  Given a product P = xyz . . . n,  suppose not all the numbers are equal.  Choose any two numbers u ≠ v.  For convenience, I choose the largest and smallest values.   Reduce the difference (u-v) to zero as above,  replacing both u and v with w = √ uv. The product of n numbers is the same; their sum reduced, as above.

Repeat until all the numbers are identical, always choosing the greatest and least numbers.  Each step reduces the sum (and therefore the Arithmetic mean) and eliminates one difference among  the n values.  The product P is constant.  When all the numbers are identical, there is no longer a greatest and least element.  The arithmetic mean of this sequence is
nα/n = α   and the geometric mean is
α ^n/n = α
     Thus, for any sequence of numbers xyz . . . n of constant product P the minimum value of the arithmetic mean is achieved when the numbers are identical.  This number is the geometric mean of the original n numbers.¹
    Since   only when M is minimum, and  held constant, the result follows:

which is Cauchy's mean Theorem.  Now, suppose we replace a, b, c . . . however we wish, as long as their sum remains the same.  The right hand remains constant, and the left varies.  The two will be equal only when a, b, c . . . n are the same number.  Then

• Auxiliary Theorem 2: For n numbers of constant sum, the geometric mean reaches its maximum value when all n numbers are the same. 

 Which, with a bit of noodling to be safe, is the converse of the first Auxiliary Theorem above.  Sneaky, no?  I have now met Dörrie on the other side.

EXPONENTIAL INEQUALITY

Consider a sequence of m numbers, n of which have the positive value a, and the rest equal to one:
          a, a,  . . . , a, 1, 1, . . . , 1           a > 0,  a ≠ 1
There are n a's and (m − n) 1's.  The two means are:
          
Since a ≠ 1, the numbers are not all equal, and therefore
          
Substituting λ for n/m gives


For λ > 1, flip the inequality.

■

...whence, upon having been put into the box and acquiring thereby the passive tense to which the greater part of his misery was owing and, that part outweighing the joyful to such a degree that of the expression relating the two it was conjectured the mathematics were differential with his share of joy -alas!- the quantity to be driven to zero, he began at that precise moment to conjecture it would be his ultimate undoing (for so it was conjectured about him and therefore & c.), and so entraining the drowsy progressive he then began to conjecture that it would not be his ultimate undoing, whereupon the fueling and fanning, oiling and wiping bringing to bear an ever increasing velocity to the transit of time, each passing second closing yet further betwixt gerund  and he the gap, at last the imperfect arrived,
he undid it and freed himself.

It was still two weeks before he would be let out of the box.

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¹There are other, more efficient ways to arrive at the same result.  For example the sequence can be treated as (at most) n/2 different, nested intervals like this:
          
and on the next pass there remain (at most) n/4 distinct intervals... and so on.  Regardless of the procedure, we close in on the same number.  Is there a good way to estimate square roots hiding in here?