GeoTheorem+ 1: UNIZOR.COM - Math+ & Problems - Geometry

Notes to a video lecture on http://www.unizor.com

Geometry+ GeoTheorem 1

Theorem

Prove that non-null vector n(a,b,c) in three-dimensional Cartesian coordinates OXYZ is normal (perpendicular) to a plane α described by an equation
a·x + b·y + c·z + d = 0
where a, b, c, d are real numbers.

Proof

CASE 1 (easy) - Constant d in an equation that describes plane α equals to zero.

The equation for plane α looks in this case as
a·x + b·y + c·z = 0

Then plane α must go through the origin of coordinates O(0,0,0) because this point satisfies the equation for α.

Consider a vector from an origin of coordinate O(0,0,0) to any other point on a plane Q(x,y,z).
Obviously, non-null vector OQ(x,y,z) is lying fully within plane α because both its ends - points O and Q lie within this plane.

We can interpret the equation
a·x + b·y + c·z = 0
that describes plane α in this case as a scalar product of non-null vector n(a,b,c) and non-null vector OQ(x,y,z).

Since this scalar product between vector n(a,b,c) and any vector OQ(x,y,z) lying within plane α is zero, vector n(a,b,c) is perpendicular to plane α.

CASE 2 - Constant d is not equal to zero.

Consider two planes defined by two equations
Plane α: a·x + b·y + c·z + d = 0
Plane β: a·x + b·y + c·z = 0

Since d ≠ 0, any point that satisfies one of these equations will not satisfy another.
Therefore, these planes do not have any common points and, therefore, are parallel.

We have already proven that non-null vector n(a,b,c) is perpendicular to plane β (see CASE 1 above).
Consequently, this vector n is perpendicular to α as well, that is n is normal to α.

End of the proof that n⊥α.

Geometry+ 04: UNIZOR.COM - Math+ & Problems - Geometry

Notes to a video lecture on http://www.unizor.com

Geometry+ 04


This lecture is dedicated to planes, straight lines and their perpendicularity in three-dimensional space with Cartesian coordinates.

Problem A

Given a point P(a,b,c) in three-dimensional Cartesian space not coinciding with the origin of coordinates O(0,0,0).

What will be an equation for coordinates x, y and z that describes a plane going through origin of coordinates O(0,0,0) and is perpendicular to line OP?

Solution A

Let's use Vector Algebra techniques to solve this problem.

Consider a non-null vector OP originated at point O(0,0,0) with endpoint at P(a,b,c) and a plane that goes through point O(0,0,0) perpendicularly to vector OP at this point.
So, vector OP is normal to this plane.

Consider any point Q(x,y,z) on this plane and vector OQ.
This vector is lying completely within a plane because both its ends O and Q belong to this plane.

Therefore, since vector OP is, by assumption, perpendicular to the plane, it's perpendicular to vector OQ lying within this plane.

Two perpendicular vectors have their scalar product equal to zero.
Therefore, OP · OQ = 0


Let's express this in coordinates.
Vector OP in coordinate form is (a,b,c).
Since coordinates of point Q are (x,y,z), vector OQ in coordinate form is (x,y,z).

Now the scalar product of vectors OP and OQ is
OP · OQ =
= a·x + b·y + c·z
which is supposed to be equal to zero since these vectors are perpendicular.

Taking into account that there is one and only one plane perpendicular to a given vector OP at its origin O(0,0,0) and we have chosen any point Q(x,y,z) on this plane, the equation of a plane should be
a·x+b·y+c·z = 0

CONCLUSION 1
If vector (a,b,c) is normal to a plane going through the origin of coordinates (0,0,0) then the equation of this plane is
a·x+b·y+c·z = 0

CONCLUSION 2
The plane described by an equation
a·x+b·y+c·z = 0
is perpendicular to vector (a,b,c).


Problem B

Given a point P(a,b,c) in three-dimensional Cartesian space not coinciding with the origin of coordinates O(0,0,0).

What will be an equation for coordinates x, y and z that describes a plane going through point P and is perpendicular to a line connecting this point with the origin of coordinates O?

Solution B

Let's use Vector Algebra techniques to solve this problem.

Consider a non-null vector OP originated at point O(0,0,0) with endpoint at P(a,b,c) and a plane that is assumed to be perpendicular to OP at point P.
So, vector OP is normal to this plane.

Consider any point Q(x,y,z) on this plane and vector PQ.
This vector is lying completely within a plane because both its ends P and Q belong to this plane.

Therefore, since vector OP is, by assumption, perpendicular to the plane, it's perpendicular to vector PQ lying within this plane.

Two perpendicular vectors have their scalar product equal to zero.
Therefore, OP · PQ = 0


Let's express this in coordinates.
Vector OP in coordinate form is (a,b,c).
Since coordinates of point Q are (x,y,z), vector OQ in coordinate form is (x,y,z).
Vector PQ can be represented as a difference between OQ and OP.
Therefore,
PQ = (x−a, y−b, z−c).

Now the scalar product of vectors OP and PQ is
OP · PQ =
= a·(x−a) + b·(y−b) + c·(z−c) =
= a·x+b·y+c·z − (a²+b²+c²)
which is supposed to be equal to zero since these vectors are perpendicular.

Taking into account that there is one and only one plane perpendicular to a given vector OP at its endpoint P(a,b,c) and we have chosen any point Q(x,y,z) on this plane, the equation of a plane should be
a·x+b·y+c·z − (a²+b²+c²) = 0
or
a·x+b·y+c·z = a²+b²+c²

CONCLUSION 1
If vector (a,b,c) is normal to a plane going through its endpoint (a,b,c) then the equation of this plane is
a·x+b·y+c·z = a²+b²+c²

CONCLUSION 2
The plane described by an equation
a·x+b·y+c·z = a²+b²+c²
is perpendicular to vector (a,b,c) at its endpoint (a,b,c).


Problem C

Given a plane α in three-dimensional Cartesian space going through the origin of coordinates O(0,0,0) and described by an equation
a·x + b·y + c·z = 0

What will be the coordinates p, q and r of an endpoint of a normal (perpendicular to plane α) vector originated at point O in terms of coefficients a,b,c defining the plane α?

Solution C

Obviously, point O(0,0,0) belongs to plane α because substituting x=0, y=0 and z=0 transforms the equation into an identity 0=0.

Let's choose any other point Q with coordinates (x,y,z) satisfying the equation of a plane α
a·x + b·y + c·z = 0

Consider vector OQ=(x,y,z).
Its origin is point O that belongs to our plane α.
Its endpoint Q(x,y,z) also belongs to this plane since plane α is a locus of all points satisfying the above equation.
Therefore, vector OQ lies within our plane α.

Assume, vector OP=(p,q,r) is normal to plane α at point O.
Its origin is the same point O as the origin of vector OQ=(x,y,z) that belongs to our plane.
Its endpoint P(p,q,r) lies somewhere in space.

Since OP ⊥ α, OP is perpendicular to any vector lying within plane α, including vector OQ.

Recall that necessary and sufficient condition for two vectors to be perpendicular to each other is their scalar product is zero.
OP · OQ = 0
Hence,
p·x + q·y + r·z = 0

The above equation is very much alike the equation of the plane
a·x + b·y + c·z = 0
Therefore, we should not look very far for values of p, q and r, but just choose p=a, q=b and r=c.

Now the equation of a plane α can be interpreted as a scalar product of two vectors OP=(a,b,c) and OQ=(x,y,z).
Since this scalar product is equal to zero, vectors OP and OQ are perpendicular to each other.

Since point Q(x,y,z) can be any point on a plane vector OP=(a,b,c) is perpendicular to the whole plane α defined by equation
a·x + b·y + c·z = 0

Answer C
The coordinates of an endpoint of a normal (perpendicular to plane α) vector originated at point O are (a,b,c).


Problem D

Given a plane α in three-dimensional Cartesian space described by a general equation
a·x + b·y + c·z + d = 0

Point O(0,0,0) is an origin of coordinates.
Assume, point P(p,q,r) belongs to plane α and vector OP=(p,q,r) is normal (perpendicular) to this plane.

What will be the coordinates p, q and r of an endpoint of vector OP in terms of coefficients a,b,c,d defining the plane α?

Solution D

We have two conditions for coordinates p, q and r to satisfy:
1. Point P(p,q,r) belongs to plane α, which means
a·p + b·q + c·r + d = 0
2. Vector OP is normal (perpendicular) to plane α.

Consider any point Q(x,y,z) on plane α and vector PQ.
This vector is lying completely within a plane because both its ends P and Q belong to this plane.

Therefore, since vector OP is, by assumption, perpendicular to the plane, it's perpendicular to vector PQ lying within this plane.

Two perpendicular vectors have their scalar product equal to zero.
Therefore, OP · PQ = 0

Let's transform this into coordinates.
Vector PQ is a difference between OQ and OP.
Therefore, the coordinates of PQ are
(x−p, y−q, z−r)

Since OP ⊥ PQ, their scalar product is zero
p·(x−p)+q·(y−q)+r·(z−r) = 0
or
p·x+q·y+r·z−(p²+q²+r²) = 0

From the above equation we have to find p, q and r knowing that x, y and z satisfy the equation of a plane α
a·x + b·y + c·z + d = 0

Simple guess p=a, q=b and r=c is a good idea, but not a solution since
d ≠ −(p²+q²+r²)

To correct this situation, let's scale the vector (p,q,r)=(a,b,c) by some coefficient k to satisfy both conditions for a solution listed above.

First, let's find a scale factor k to make sure that endpoint P(k·a,k·b,k·c) of vector OP belongs to plane α.

Substituting x=k·a, y=k·b and z=k·c into equation that defines plane α, we obtain an equation for k:
a·k·a + b·k·b + c·k·c + d = 0

From this equation
k = −d/(a²+b²+c²)
This value of k is sufficient to put our point P(p,q,r) on the plane α.

Let's check that defined this way vectors OP and PQ are perpendicular to each other for any point Q(x,y,z) that satisfies the equation defining plane α
a·x + b·y + c·z + d = 0

Indeed,
OP · PQ =
= (k·a, k·b, k·c) ·
· (x−k·a, y−k·b, z−k·c) =
= k·(x·a + y·b + z·c) −
− k²·(a² + b² + c²)
where
a·x + b·y + c·z + d = 0 and
k = −d/(a²+b²+c²)

Substitute
a·x + b·y + c·z = −d and
(a²+b²+c²) = −d·k
getting
OP · PQ =
= k·(−d) −k²·(−d/k) = 0

So, since point Q(x,y,z) was chosen arbitrarily on plane α, vector OP is perpendicular to any vector on a plane and, therefore, to plane α itself.

Taking into account that P ∈ α, it proves that the coordinates p, q and r of an endpoint of vector OP are
p = −d·a/(a²+b²+c²)
q = −d·b/(a²+b²+c²)
r = −d·c/(a²+b²+c²)

Geometry+ 03: UNIZOR.COM - Math+ & Problems - Geometry

Notes to a video lecture on http://www.unizor.com

Geometry+ 03


This lecture is dedicated to problems of triangle construction by its certain elements.

The tools of construction are a ruler to draw straight lines and a compass to draw circles.

We will use the following naming rules.
Vertices of triangle will be call by upper case Latin letters A, B and C.
Sides will be called by lower case Latin letters corresponding to opposite vertices: side a is opposite to vertex A etc.
Angles will be called by lower case Greek letters corresponding to names of their vertices: angle α is at vertex A etc.
Medians are named m with a subscript of a side onto which they fall: median m_a is from vertex A to side a etc.
Altitudes are named h with similar subscripts, like h_a etc.
Angle bisectors are named l_a etc.
Radius of a circumscribed circle of a triangle is named R.
Radius of an inscribed circle is named r.

Problem A
Construct a triangle by its three altitudes h_a, h_b and h_c.

Analysis A

As we know, a product of a side by an altitude falling on it is a double area of a triangle.
Therefore,
a·h_a = b·h_b = c·h_c
Hence,
b = a·h_a/h_b
c = a·h_a/h_c

Let's construct a triangle similar to ΔABC defined by sides a, b and c by choosing any segment x and defining two other segments y and z using the equations similar to above.
y ≝ x·h_a/h_b
z ≝ x·h_a/h_c
Segments y and z can be easily constructed from these definitions, knowing x (arbitrarily chosen) and given altitudes.

Let x/a=k be a scaling factor between arbitrarily chosen segment x and side a of triangle ΔABC.
From this follow these relationships:
x = a·k
y = x·h_a/h_b = a·k·h_a/h_b = b·k
z = x·h_a/h_c = a·k·h_a/h_c = c·k
The scaling factor k is the same for x/a=k, y/b=k and z/c=k.

Therefore, triangle ΔXYZ constructed from three segments x, y and z is similar to triangle ΔABC with segments a, b and c we have to construct.

From similarity of triangles follows the congruence of corresponding angles
∠BAC = ∠α = ∠YXZ
∠CBA = ∠β = ∠ZYX
∠ACB = ∠γ = ∠XZY

Therefore, our analysis shows that by constructing ΔXYZ we get all angles of ΔABC.

This is the end of analysis, as the construction of triangle ΔABC, knowing its three angles and altitudes, is straight forward.

Solution A

To enlarge this picture, right click on it and choose "Open image in new tab"

Geometry+ 02: UNIZOR.COM - Math+ & Problems - Geometry

Notes to a video lecture on http://www.unizor.com

Geometry+ 02


This lecture is dedicated to problems of triangle construction by its certain elements.

The tools of construction are a ruler to draw straight lines and a compass to draw circles.

We will use the following naming rules.
Vertices of triangle will be call by upper case Latin letters A, B and C.
Sides will be called by lower case Latin letters corresponding to opposite vertices: side a is opposite to vertex A etc.
Angles will be called by lower case Greek letters corresponding to names of their vertices: angle α is at vertex A etc.
Medians are named m with a subscript of a side onto which they fall: median m_a is from vertex A to side a etc.
Altitudes are named h with similar subscripts, like h_a etc.
Angle bisectors are named l_a etc.
Radius of a circumscribed circle of a triangle is named R.
Radius of an inscribed circle is named r.

Problem A
Construct a triangle by its two sides a, b and a median onto its third side m_c.


Problem B
Construct a triangle by its side a and two medians m_a and m_b.

Hint: Construct ΔBQR where BR=a/2, BQ=2m_b/3 and QR=m_a/3.

Problem C
Construct a triangle by its three medians m_a, m_b and m_c.

Hint: Construct ΔPQC where CQ=2m_a/3, PQ=2m_b/3 and CP=2m_c/3.

Problem D
Construct a triangle by its side a and two altitudes onto other sides h_b and h_c.

Trigonometry+ 01: UNIZOR.COM - Math+ & Problems - Trigonometry

Notes to a video lecture on http://www.unizor.com

Trigonometry+ 01


Problem A

Using the Euler Formula
e^i·x = cos(x)+i·sin(x)
prove the formulas for sine and cosine of a sum of two angles.

Solution A

The Euler formula states that the real part of e^i·x is cos(x) and its imaginary part s i·sin(x)

From i² = −1,
e^i·(α+β) = cos(α+β)+i·sin(α+β),
and a^(p+q) = a^p·a^q
follows:
cos(α+β)+i·sin(α+β) =
= e^i·(α+β) = e^i·α·e^i·β =
= [cos(α)+i·sin(α)] ·
· [cos(β)+i·sin(β)] =
= cos(α)·cos(β)−sin(α)·sin(β) +
+ i·cos(α)·sin(β)+i·sin(α)·cos(β)

The real part of the last expression is
cos(α)·cos(β) − sin(α)·sin(β)
Its imaginary part is
i·[cos(α)·sin(β) + sin(α)·cos(β)]

Therefore,
cos(α+β)
(the real part of e^i·(α+β))
equals to
cos(α)·cos(β) − sin(α)·sin(β),
hence
cos(α+β) =
= cos(α)·cos(β) − sin(α)·sin(β)

Analogously,
i·sin(α+β)
(the imaginary part of e^i·(α+β))
equals to
i·[cos(α)·sin(β) + sin(α)·cos(β)],
hence
sin(α+β) =
= cos(α)·sin(β) + sin(α)·cos(β)


Problem B

Using the Euler Formula
e^i·x = cos(x)+i·sin(x)
prove the formulas for derivatives of sine and cosine functions.

Solution B

Let's differentiate the Euler's formula
e^i·x = cos(x)+i·sin(x)

On the left side the result is
d/dx[e^i·x] = i·e^i·x =
= i·[cos(x) + i·sin(x)] =
= −sin(x) + i·cos(x)

On the right side the result of differentiation is
d/dx[cos(x)] + i·d/dx[sin(x)]

The results of differentiation of left and right sides of the Euler's formula must be equal to each other:
d/dx[cos(x)] + i·d/dx[sin(x)] =
= −sin(x) + i·cos(x)

When two complex numbers are equal to each other, their real and imaginary parts must be equal.
Therefore,
d/dx[cos(x)] = −sin(x)
d/dx[sin(x)] = cos(x)

Arithmetic+ 03: UNIZOR.COM - Math+ & Problems - Arithmetic

Notes to a video lecture on http://www.unizor.com

Arithmetic+ 03


Problem A

I have purchased a share of some company XYZ for price P₀=$100.
After a month the price was lower by ρ=10%.
Another month later the price went up by the same percentage ρ=10% and I sold this share.
Have I lost or gain money, or got even?

Solution A

After the first month the price dropped from P₀=100 by ρ=10%, that is by P₀·0.1 and was
P₁ = P₀ − P₀·ρ =
= 100 − 100·0.1 = 90

After the second month the price went up from P₁=90 by ρ=10%, that is by P₁·0.1 and was
P₂ = P₁ + P₁·ρ =
= 90 + 90·0.1 = 99

Therefore, since I bought a share for P₀=$100 and sold it for P₂=$99, my net result is
ΔP = P₂ − P₀ = −$1
which is a loss.

Notice that percentage of my loss was equal to percentage of my gain, but in dollars it was larger because the same percentage was applied to a larger amount on the way down, than to a smaller amount on the way up.


Problem B

I have purchased a share of some company XYZ for price P₀=$100.
After a month the price was higher by ρ=10%.
Another month later the price went down by the same percentage ρ=10% and I sold this share.
Have I lost or gain money, or got even?

Solution B

After the first month the price went up from P₀=100 by ρ=10%, that is by P₀·0.1 and was
P₁ = P₀ + P₀·ρ =
= 100 + 100·0.1 = 110

After the second month the price went down from P₁=110 by ρ=10%, that is by P₁·0.1 and was
P₂ = P₁ − P₁·ρ =
= 110 − 110·0.1 = 99

Therefore, since I bought a share for P₀=$100 and sold it for P₂=$99, my net result is
ΔP = P2 − P₀ = −$1
which is a loss.

Notice that percentage of my loss was equal to percentage of my gain, but in dollars it was larger because the same percentage was applied to a larger amount on the way down, than to a smaller amount on the way up.

IMPORTANT CONCLUSION:
If market price of a stock regularly goes up and down by the same percentage, the price per share goes down, investors lose money.
To be profitable, the price should go up more frequently or/and by a greater percentage than down.


Problem C

(a) Within the framework of Problems A, what the percentage of gain would neutralize the loss after price moving down by ρ=10%?

(b) Within the framework of Problems B, what the percentage of loss would neutralize the gain after price moving up by ρ=10%?

Solution C

(a) If we lost ρ=10%, the price became
P₁ = P₀ − P₀·ρ =
= 100 − 100·0.1 = 90
To return back to $100 from $90, the growth in price σ should satisfy the equation
100 = 90 + 90·σ
from which
σ = 1/9 ≅ 0.1111
Therefore, if we add approximately 11.11% to P₁=90, the final price will be
P₂ = 90 + 90·0.1111 =
= 99.999 ≅ 100

(b) If we gain ρ=10%, the price became
P₁ = P₀ + P₀·ρ =
= 100 + 100·0.1 = 110
To return back to $100 from $110, the diminishing in price σ should satisfy the equation
100 = 110 − 110·σ
from which
σ = 10/11 ≅ 0.0909
Therefore, if we subtract approximately 9.09% from P₁=110, the final price will be
P₂ = 110 − 110·0.0909 =
= 99.999 ≅ 100

Answer C:

(a) To neutralize a loss of 10% we need a larger (in %) gain of approximately 11.11%.
(b) To neutralize a gain of 10% we need a smaller (in %) loss of approximately 9.09%.


Problem D (suggested by Max)

Concentration of salt dissolved in water is measured in percentage of the mass of salt to the total mass of solution (salt + water).

Initially, a glass contained 100 gram of solution of salt in water with the concentration of salt 1%.
Question 1: how much salt and how much water was in a glass?

This same glass was standing on a table for a week and part of water has evaporated, so the concentration of salt after a week became 2%.
Question 2: how much salt and how much water was in a glass after this process of evaporation?
Question 3: how much water has evaporated in a week?

Solution D

If 100 gram of solution contained 1% of salt, the mass of salt was
m_salt = 100·0.01 = 1 (gram)
Hence, the mass of water was
m_water = 100 − 1 = 99 (gram)

Answer 1 (initially):
m_salt = 1 (gram)
m_water = 99 (gram)

After some water has evaporated, the glass contained the same amount salt, that is 1 gram.
Since we know that it constitutes 2% of a solution, the total mass of solution must be 50 gram because
1/0.02 = 50 (gram)

To make 50 gram of solution with 1 gram of salt, we need 49 gram of pure water.

Answer 2 (after a week):
m_salt = 1 (gram)
m_water = 49 (gram)

Answer 3:
That means, 99−49=50 gram of water has evaporated in a week.

Algebra+ 02: UNIZOR.COM - Math+ & Problems - Algebra

Notes to a video lecture on http://www.unizor.com

Algebra+ 02


Problem

Let {x_i} (where i∈[1,n] and n is any natural number) be a set of n real non-negative numbers.

Their arithmetic average
(or mean) is defined as
A_n = (1/n)·Σ_i∈[1,n] x_i
(symbol Σ_i∈[1,n] is summation for all indices i∈[1,n])

Their geometric average is defined as
G_n = (Π_i∈[1,n] x_i)^1/n
(symbol Π_i∈[1,n] is multiplication for all indices i∈[1,n])

Prove that A_n ≥ G_n.

Solution A
(by Cauchy)

Our plan is multi-step.
First (step 1), we will check our inequality for n=1 and prove it for n=2.
Then (step 2) we will prove that, if it's held for some n, it will be held for 2n, which, if start with n=2 would prove the inequality for n=4, n=8 etc. - for all n=2^k, where k is any natural number.
Finally (step 3), we will prove for all other n by complementing any given n non-negative numbers {x_i} with some new numbers to make the total quantity of numbers equal to 2^k.

Step 1
It's obvious that for n=1 the inequality is held because
A₁ = x₁/1 = x₁
and
G₁ = (x₁)¹ = x₁

For n=2 let's prove it as follows.
Start from the original inequality that needs to be proven, do some reversible invariant transformations to get to an unconditionally true statement (this is the analysis stage) and then use the reversibility of invariant transformations to prove the original inequality by going backward through these transformations, beginning with final unconditionally true statement (the proof proper).

Note that all our participating numbers are real non-negative, it's important when we talk about reversible invariant transformations that involve raising both sides of inequality to the power of 2.

Original inequality to be proven:
(I) (x₁+x₂)/2 ≥ (x₁·x₂)^½
Now do the following series of invariant transformations.
(II) Square both sides (it is an invariant reversible transformation for non-negative numbers).
(x₁+x₂)²/4 ≥ x₁·x₂
(III) Open parenthesis.
x₁²+2·x₁·x₂+x₂² ≥ 4·x₁·x₂
(IV) Bring all members to the left side.
x₁² − 2·x₁·x₂ + x₂² ≥ 0
(V) Notice, we have a full square on the left, let's transform it into an explicit square:
(x₁ − x₂)² ≥ 0
The above is an unconditionally true statement.

Therefore, the above analysis allows us to formulate the proof by starting from this unconditionally true statement (V) and invariantly transform it into (IV), (III), (II) and (I), which proves the original inequality.
Step 1 of the proof is done.

Step 2
Let's prove that the quantity of non-negative numbers participating in arithmetic and geometric averaging can be doubled.
More precisely, we will prove that if the inequality is true for n elements, it's true for 2n elements.

Assume, A_n ≥ G_n for some n.
Let's prove that A_2n ≥ G_2n.
Divide our set of 2n elements into two subsets:
subset #1 {x_i}_i∈[1,n] and
subset #2 {x_i}_i∈[n+1,2n]

Let A_n⁽¹⁾ and G_n⁽¹⁾ be arithmetic and geometric averages of the elements of subset #1 and
A_n⁽²⁾ and G_n⁽²⁾ be arithmetic and geometric averages of the elements of subset #2.
Let A_2n and G_2n be arithmetic and geometric averages of the elements of an entire set of 2n elements.

Each subset contains n elements (non-negative numbers), and we assumed that our original inequality is held for this quantity of elements.
Therefore,
A_n⁽¹⁾ ≥ G_n⁽¹⁾ and
A_n⁽²⁾ ≥ G_n⁽²⁾.

Obviously,
A_2n = [A_n⁽¹⁾+A_n⁽²⁾] /2
G_2n = [G_n⁽¹⁾·G_n⁽²⁾]^½

Using the above assumption about inequality between arithmetic and geometric averages of sets of n elements,
A_2n = [A_n⁽¹⁾+A_n⁽²⁾] /2 ≥
≥ [G_n⁽¹⁾+G_n⁽²⁾] /2
Using the proven above inequality for only two elements G_n⁽¹⁾ and G_n⁽²⁾, we can state
[G_n⁽¹⁾+G_n⁽²⁾] /2 ≥
≥ [G_n⁽¹⁾·G_n⁽²⁾]^½ = G_2n

Combination of the two inequalities obtained above proves that
A_2n ≥ G_2n
Step 2 of the proof is done, and we can state that, since the inequality was proven for n=2, it is true for n=4,8,16..., that is for any n=2^k.

Step 3
Let's prove the inequality for values of n between powers of 2.
Let the value of n be greater than 2^k−1 but less than 2^k.
A_n and G_n are, as before, arithmetic and geometric averages of our n numbers.

Let's complement our set of n numbers {x_i}_i∈[1,n] with m=2^k−n new numbers, each of which is equal to G_n.
Now we have a set of 2^k=m+n numbers, and we can apply the inequality between arithmetic and geometric averages proven to be true for such sets.

Consider A_m+n for our new set that contains n original elements {x_i} and m new elements, each of which equals to G_n.
Obviously,
A_m+n = [n·A_n + m·G_n] /(m+n)

Now calculate G_m+n for our new set that contains n original elements {x_i} and m new elements, each of which equals to G_n.
Obviously,
G_m+n = [G_nⁿ · G_n^m]^1/(m+n) =
= [G_n^m+n]^1/(m+n) = G_n

Since m+n=2^k and our original inequality was proven for this quantity of participating non-negative numbers,
A_m+n ≥ G_m+n
Therefore,
[n·A_n + m·G_n] /(m+n) ≥ G_n

Simple transformations of the last inequality result in the following.
n·A_n + m·G_n ≥ (m+n)·G_n
n·A_n ≥ n·G_n
A_n ≥ G_n
Step 3 of the proof is done, and we can state that it was proven for any set of non-negative numbers that their arithmetic average is greater or equal to their geometric average.

Alternate step 3
Instead of complementing our original set of n numbers with G_n, we can complement it with A_n with similar results.
Then,
A_m+n = [n·A_n+m·A_n] /(m+n) =
= A_n

Now calculate G_m+n for our new set that contains n original elements {x_i} and m new elements, each of which equals to A_n.
G_m+n = [G_nⁿ · A_n^m]^1/(m+n)

Since m+n=2^k and our original inequality was proven for this quantity of participating non-negative numbers,
A_m+n ≥ G_m+n
Therefore,
A_n ≥ [G_nⁿ · A_n^m]^1/(m+n)

Simple transformations of the last inequality result in the following.
A_n^m+n ≥ G_nⁿ·A_n^m
A_n^m·A_nⁿ ≥ G_nⁿ·A_n^m
A_nⁿ ≥ G_nⁿ
A_n ≥ G_n

The equality between arithmetic and geometric averages is obvious if all numbers are equal to each other, that is x_i=x_j for any indices i and j.
Symbolically, it looks like
x_i=x_j ∀ i,j∈[1,n] ⇒ A_n = G_n


Solution B
(by Polya)

From the standard course of Math is known that the function y=e^x at point x=0 has a value y(0)=1 and has a tangential at 45°, which looks like this.

Shifting the graphs by 1 to the right by replacing x with x−1, we will have the following picture

The last picture shows that the following inequality is held for all real values of x
e^x−1 ≥ x

Consider now a set of n non-negative numbers {x_i} (where i∈[1,n] and n is any natural number) with A being their arithmetic average and G being their geometric average (we are omitting a subscript n in averages for brevity).

Using an inequality above for x=x_i /A, we obtain
e^{(x_i /A)−1} ≥ x_i /A for any i or
A·e^{(x_i /A)−1} ≥ x_i

Multiplying the above inequality for all values of index i from 1 to n, we obtain
Aⁿ·e ^{Σ(x_i /A)−n} ≥ Π(x_i )

Interestingly, the exponent in the expression above is zero.
Here is why
Σ(x_i /A) = (1/A)·Σ(x_i) =
= (1/A)·A·n = n,
which makes the exponents in the inequality above equal to zero.

Therefore, considering e⁰=1, the inequality above looks like
Aⁿ ≥ Π(x_i )
A ≥ [Π(x_i )]^1/n = G
End of proof.

Algebra+ 01: UNIZOR.COM - Math+ & Problems - Algebra

Notes to a video lecture on http://www.unizor.com

Algebra+ 01


Problem A

Given an equation of parabola in Cartesian coordinates on a plane
y = x² − A·x + A
where parameter A can be any real number.
Each value of parameter A defines some parabola.

(A1) Find the point on a plane, through which go all the parabolas defined by all real values of parameter A.

(A2) On what curve lie the vertices of all the parabolas defined by all real values of parameter A?

Solution A

A1
Assuming the statement of this problem implies that such a point, where all parabolas are crossing, exists, let's find it by equating any two values of y for different values of parameter A.

Take, for example two parabolas for A=0 and A=1 and find their crosing point, that is such x for which corresponding quadratic forms equal to each other.
A=0 ⇒ y = x²
A=1 ⇒ y = x² − x + 1

Equating both expressions gives
x² = x² − x + 1
⇒ x = 1

Consider now this value x=1.
With this x the equation becomes independent of parameter A:
y = 1² − A·1 + A = 1
for any value of parameter A.

Therefore, any parabola defined by any value of A goes through point with coordinates x=1, y=1 on a plane.


A2
We need a formula for a vertex of a parabola represented by a general quadratic form
y = a·x² + b·x + c
If we do not remember the formula for X-coordinate of its vertex, recall that it is in the middle between the two roots of a quadratic equations
a·x² + b·x + c = 0
The formula for its roots (and that is something to remember) is
x_1,2 = [−b ± √b²−4ac] /2a
Therefore, the X-coordinate of a vertex is in the middle between its roots
x₀ = ½(x₁ + x₂) = −b/2a

Alternatively, using Calculus, we can find x₀ by equating a derivative of a given quadratic form to zero:
y' = 2ax + b = 0
Therefore,
x₀ = −b/2a

The Y-coordinate of a vertex is, therefore,
y₀ = a·x₀² + b·x₀ + c =
= a(b/2a)² + b(−b/2a) + c =
= b²/4a − b²/2a +c =
= −b²/4a + c

In our case of a quadratic form
y = x² − A·x + A
the values of coefficients are
a = 1, b = −A, c = A
Therefore, the coordinates of a parabola's vertex for a specific value of parameter A is
x₀ = A/2
y₀ = −A²/4 + A
As parameter A changes, so do coordinates of a vertex forming some parametrically defined curve.

To convert this parametric definition of a curve into the usual format y₀=f(x₀) we resolve parameter A in terms of x₀ and substitute it into an expression for y₀.
Since
x₀ = A/2
the value of A in terms of x₀ is
A = 2x₀

Substitute it into an expression for y₀:
y₀ = −4x₀²/4 + 2x₀ =
= −x₀² + 2x₀
which is a parabola with coefficients
a = −1
b = 2
c = 0

The standard characteristics of this parabola are as follows.
The "horns" are directed downwards.
Vertex is at
x₀ = −b/2a = −2/−2 = 1
y₀ = −b²/4a + c = −4/−4 + 0 = 1
Roots (X-intercepts) are
x_1,2 = (−2±√4)/−2
⇒ x₁ = 0, x₂ = 2

Geometry+ 01: UNIZOR.COM - Math+ & Problems - Geometry

Notes to a video lecture on http://www.unizor.com

Geometry+ 01

Problem A

Can two sides and all angles of one triangle be equal to two sides and angles of another triangle, but triangles be different (that is, not congruent because the third sides of these triangles have different lengths)?

If yes, are there any requirements on the sides of a triangle to assure the existence of a non-congruent to it another triangle with the same three angles and two sides?

Solution

Here is an approximate picture that seems to illustrate that this is possible.

Both triangles (red and blue) have the same angles α, β and γ and two equal sides a=a and b=b.

Point D lies on the segment AC.
Point E lies on the segment BC.
The length of segment BC=a is greater than the length of segment CE=b.

Sides are parallel: AB || DE
⇒ ∠CAB=∠CDE = α
⇒ ∠ABC=∠DEC = β
Therefore, triangles are similar:
ΔABC ∼ ΔDEC

Congruent sides:
BC ≅ DE = a
AC ≅ CE = b

Now the question is, can we start with any triangle ΔABC and construct ΔDEC that is similar but not congruent to ΔABC.

The answer is negative, we cannot start with any triangle ΔABC, but under certain conditions it is possible, and we will examine specific requirements to make it.

Let's assume that ΔABC is defined by its three sides BC=a, AC=b and AB=c.
Triangle ΔDEC is similar to ΔABC and we know its two sides - DE=a and CE=b.
Assume, the third side of ΔDEC is CD=x.

From similarity of these triangles follows that the ratio of sides lying opposite to equal angles is the same:
CB/CE = CA/CD = AB/DE
Therefore,
a/b = b/x = c/a

From the first equation follows that
x = b²/a
From the second equation follows that
x = a·b/c
Therefore,
b²/a = x = a·b/c
⇒ a²b = b²c
⇒ a² = b·c
The last equation a²=b·c is the necessary condition required to construct triangle ΔDEC.
As we stated above, length a is greater than b. Therefore, to satisfy the last equation, length c must be greater than a, that is, a is somewhere in-between b and c.

The condition a² = b·c with length a being somewhere between unequal b and c is also a sufficient one for the task.

To prove it, assume that we have triangle ΔABC with unequal sides a, b and c, as on the picture above, that satisfy the condition a² = b·c.

Using length b of side AC, we mark point E on the side BC at distance b=b from point C.
Next we draw a line parallel to AB through point E that crosses AC at point D.

Consider triangle ΔDEC.
All its angles are, correspondingly, congruent to angles of triangle ΔABC, so these triangles are similar.

Let's prove that the length of side DE equals to the length of side BC=a. That will conclude the construction (and, therefore, existence) of a triangle with two sides and all angles equal to two sides and all angles of triangle ΔABC, but not congruent to it.

Let the length of side DE be an unknown y.
From similarity of two triangle follows:
a/b = c/y
⇒ y = b·c/a
As we know, b=b and, by assumption, a²=b·c.
Therefore,
y = a²/a = a
which is exactly what we had to prove.

Conclusion
There might exist a triangle with two sides and all angles equal to two sides and angles of a given triangle and not congruent it, but there is certain necessary and sufficient condition for its existence in terms of an equation between the sides a,b,c of a given triangle.
Namely, all sides a,b,c must be different and they must satisfy an equation a²=b·c.

Example
Consider a triangle with sides a=6, b=4 and c=9.
It satisfies the conditions above (all sides are different and 6²=4·9).
Now we know two sides of a triangle we have to construct: a=6 and b=4.
The third side x can be found from the similarity condition
x/b = b/a
⇒ x = b·b/a = 16/6 = 8/3
To prove similarity, let's check the proportionality of all sides.
(8/3)/4 = 2/3
4/6 = 2/3
6/9 = 2/3

Angle Restriction
An interesting restriction can be derived on the value of angle ∠α lying between the longest and the shortest sides of triangle ΔABC across side a.
As we agreed, to make possible the existence of triangle ΔDEC with two (but not all three) sides and all three angles equal to those of ΔABC, the necessary and sufficient condition is
a² = b·c
At the same time, by the law of cosines,
a² = b² + c² −2b·c·cos(α)
Therefore, we have an equality
b·c = b² + c² −2b·c·cos(α)
or
2b·c·cos(α) = b² + c² − b·c =
= (b − c)² + b·c
Since (b−c)² is always positive, we conclude that
2b·c·cos(α) > b·c
or
cos(α) > ½
α < π/3 = 60°
Angle between the longest and the shortest sides of ΔABC should be less than π/3=60°.