Euclidean Geometry

From The Mathematics Student Journal

ACJD, CBGH and BAEF are squares constructed outwardly on the sides of triangle ABC. DE, FG and HJ are drawn. If the sum of the areas of squares  BAEF and CBGH is equal to the area of the rest of the figure, find the measure of angle ABC.

My Solution:

Let [ABC] denotes the area of triangle ABC and angle ABC = α, angle BAC = β, angle BCA = γ. Given that the sum of [BAEF] and[CBGH] is equal to area of the rest of the figure, we have:
(AB)² + (BC)² = [FBG] + [ABC] + (AC)² + [AED] + [HCJ]             (1)

Now [FBG] = [ABC] = [AED] = [HCJ]  because

[FBG] = (1/2)(FB)(BG)sin(180 – α) = (1/2)(AB)(BC)sin α = [ABC] ,

[AED] = (1/2)(AE)(AD)sin(180 – β) = (1/2)(AB)(AC)sin β = [ABC]

and [HCJ] = (1/2)(HC)(CJ)sin(180 – γ) = (1/2)(BC)(AC)sin γ = [ABC] .

Equation (1) becomes

(AB)² + (BC)² = (AC)² + 2(AB)(BC)sin α

Replacing (AC)² using cosine law,

(AB)² + (BC)² = (AB)² + (BC)² – 2(AB)(BC)cos α + 2(AB)(BC)sin α

cos α = sin α

α = angle ABC = 45 degrees.

641

From a friend:

If

prove that p is divisible by 641.

My Solution:

Since 641 is a prime number and it is not present in denominator, hence p is divisible by 641.

Projectile on an Incline

 

My Solution:

Based from the above figure point  can also be denoted as .

From kinematics equations for constant acceleration (at time t):

It follows that

Hence,

sin 18

Compute the exact value of  

 

My Solution:

Since  and if we set 

Then,

Let 

The only applicable value of  is  

Therefore,

Full of Radicals

Simplify:

From Philippine Mathematical Olympiad 2009

Using Binomial Theorem in proving Combinatorial identities

1) Prove:

My Solution:

The above expression can be simplified as

Applying binomial theorem,

setting x=1 and y=8

Thus,

2) Show that

My Solution:

The simplified version of the above expression is

Using binomial theorem,

Letting x=1

Integrating both sides,

setting y=1

The last step is solving for C
If y=0, then

Therefore,

We are not necessary at the center of universe

 

Hubble’s law: Hubble found that distant galaxies are receding with a velocity proportional to their distance (H_O is the Hubble constant) from where we are on Earth. For the ith galaxy: 

with our Milky way galaxy at the origin. Show that this recession of the galaxies from us does not imply that we are at the center of universe.

Source:  Exercise 1.1.9 of Essential Mathematical Methods for Physicists by H.Weber and G.Arfken

My Solution:

By Hubble’s law,

Now by setting the galaxy at r₁ as new origin and using the concept of vectors, we have the following ratios:

Since we have this property of ratio:

Then,

;

Which means that Hubble’s law is still obeyed if another galaxy is set as new origin! (still applicable aside from using the galaxy at r₁)

Therefore, the recession of galaxies from us indicates that we are not necessary at the center of universe…

Determining the molar specific heat at constant pressure from Gibbs Free Energy expression

A system is found to have a Gibbs Free Energy of

Find

Notations:

C_P – molar specific heat at constant pressure
G – Gibbs Free Energy
R – universal gas constant
T – temperature
P – pressure
a – one of the constants in Van der Waals equation
Q – energy transferred (heat)
S – entropy

My Solution:

Method – relate Gibbs free Energy expression to another thermodynamic variable using Maxwell’s relations and then relate this variable to the known equation for molar specific heat at constant pressure.

Using Maxwell’s relations, we find that

Then

Since

We have

By simplifying,

Difference between the isothermal and adiabatic compressibilities

Show that the difference between the isothermal and adiabatic compressibility is:

Notations:

k_T – isothermal compressibility
k_P – adiabatic compressibility
T – Temperature
V – Volume
Beta – Coefficient of volume expansion
C_P – molar specific heat at constant pressure
R – universal gas constant
S – entropy
C_V – molar specific heat at constant volume

From the following definitions:

Coefficient of volume expansion -

Adiabatic compressibility -

Isothermal compressibility -

 

Proof:

For the right side of the equation,

For isothermal compressibility,

For adiabatic compressibility,

Since S is constant it follows that

Then,

Solving first for the derivative of T with respect to P,

We have,

Since

Therefore,

 

Exercise in Algebra (1)

Find a formula for:

My Solution:

Since

We have,

Thus,

Find the Fallacies

Here are some examples of false Mathematical proofs:

Number 1:

Number 2:

Number 3 (the famous one):

let:

Number 4:

(1-1) + (1-1) + … + (1-1) + (1-1) + … = 0

1 + (-1+1) + (-1+1) + … + (-1+1) + … = 0

1 + 0 + 0 +… + 0 + .. = 0

1 = 0

A Simple Exercise in Laplace Transform

Prove that:

Solution:

First,

It follows that:

Then,

Hence,

This 2009!

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The number 5040 – according to Plato

5 040 – Greek philosopher Plato gave this number as the population of an ideal city.

This number has several factors, which implies an efficient division of population for the sake of land distribution, taxes, war etc.

In fact,

5, 040 = 7! = (1)(2)(3)(4)(5)(6)(7)

5, 040 = (2⁴)(3²)(5¹)(7¹)

Thus, 5 040 has (4+1)(2+1)(1+1)(1+1) = 60 factors

A Simple Technique #1

Compute in (less than) 30 seconds without using calculator.

My technique:

Since 2007 is the arithmetic mean of the numbers 2004, 2006, 2008 and 2010, then: Let x = 2007

So, x-3 = 2004, x-1 = 2006, x+1 = 2008, x+3 = 2010

Hence, the expression is equal to

Simplifying,

= x² – 5 = (2007)² – 5 = 4, 028, 044

Value of a in ax+b

When the polynomial x²⁰⁰⁷ + 1 is divided by x² – 3x +2, the remainder can be expressed in the form ax+b. Find the value of a.

My Solution:
Let k be the quotient, then:
x²⁰⁰⁷ + 1 = k(x² – 3x +2) + (ax+b)
Since x² – 3x +2 = (x-2)(x-1),
x²⁰⁰⁷ + 1 = k(x-2)(x-1) + (ax+b)

To omit k and to solve for the value of a easily, set x=2 and x=1:

For x=2
(2)²⁰⁰⁷ + 1= k(2-2)(2-1) + (2a+b)
2²⁰⁰⁷ + 1= 2a+b  (*)

For x=1
(1)²⁰⁰⁷ + 1 = k(1-2)(1-1) + (a+b)
2 = a+b  (**)

To solve for the value of a, perform elimination using (*) and (**).
..Which in turns give: a=2²⁰⁰⁷-1

Number of Perfect Squares

Compute the number of squares between 4⁹ and 9⁴

My short solution: The technique is to express these two numbers into perfect squares so that it’s easy to compute the number of squares between these two numbers.

4⁹ = (2²)⁹ = (2⁹)² = 512² ;

9⁴ = (3²)⁴ = (3⁴)² = 81²

Therefore, the number of squares between 4⁹ and 9⁴ is (512-81)-1 = 430

25th number

Five digit numbers containing all digits from 1-5 are arranged from highest to lowest. What is the 25th number?

(Examples of these numbers are 54 321; 23 451; 41 532)

My Solution:

First, it’s impractical to list these (5)(4)(3)(2)(1)=120 numbers from highest to lowest and then determine the 25th number in the list. So, we can instead do some technique and investigate the distribution of these numbers.

Since it’s from highest to lowest, the first step is to determine the number of these five digit numbers which have a ten thousands digit of 5.

For thousands digit: 4 numbers to choose.

For hundreds digit: 3 numbers to choose.

For tens digit: 2 numbers to choose.

For unit digit: 1 number to choose.

Thus, the number of these 5 digit numbers which have a ten thousands digit of 5 is (4)(3)(2)(1)=24. Then these 24 numbers are the first 24 numbers in the list. Hence, the 25th number is the highest number which has a ten thousands digit of 4. And this 25th number is 45 321.

A Simple Proof # 1

Prove that:    , for any

Solution: Since   , then in order to produce a short

proof, the expression   should be express in the form  

Doing so,

Hence,

Mathematical Beauty

“A physical theory must possess mathematical beauty.”

In the most basic explanation, mathematical beauty is related to simplicity and surprise.
- Paul Dirac, quantum physicist