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.

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.

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,

My Solution:

Since  and if we set 

Then,

Let 

The only applicable value of  is  

Therefore,

From Philippine Mathematical Olympiad 2009

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,

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…

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,

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,

My Solution:

Since

We have,

Thus,