## Monthly Archives: November 2016

### The History of Invention

I had a list of various inventions which were in date order, but the piece of paper on which they were listed was torn in half, one half of which has been lost. Given the list of years in which the inventions were invented and the information below, see if you can match all the inventions with the year in which they were invented.

This is the half I have left, on which is written the years of invention:

$1320, 1607, 1710, 1721, 1763, 1764, 1783, 1815, 1841,$

$1856, 1858, 1876, 1877, 1898, 1899, 1935, 1939, 1941.$

Gunpowder, the telescope, pianoforte and the mercury thermometer were all invented before the spinning jenny, the steam engine and the hot air balloon. Steel, the wireless and the sewing machine were all invented before radar and the tank. The miner’s safety lamp, the telephone, gunpowder, steel, dynamite and the sewing machine were all invented before the phonograph and the wireless, the two of which, and the tank, were invented before radar, the jet engine and polyester.

Gunpowder, the spinning jenny, the steam engine, the mercury thermometer and the hot air balloon were all invented before the miner’s safety lamp and the sewing machine. Polyester was invented after dynamite, radar and the jet engine. The pianoforte was invented after the telescope and gunpowder. Dynamite was invented after steel, the sewing machine and the miner’s safety lamp.

Steel was invented after the sewing machine which in turn was invented after the miner’s safety lamp. The wireless was invented after the phonograph, the jet engine after radar, and the telephone after dynamite. The spinning jenny was invented before the steam engine and the hot air balloon.

Cheers,

Nalin Pithwa

### Non-linear equations for IITJEE Mathematics training

(Non-linear) equations involving three or more variables can only be solved in special cases. We shall here consider some of the most useful methods of solution.

Example 1.

Solve the following system of equations:

$x+y+z=13$ $\ldots \ldots \ldots$ Equation I

$x^{2}+y^{2}+z^{2}=65$ $\ldots \ldots \ldots$ Equation II

$xy=10$ $\ldots \ldots\ldots$ Equation III

Solution I:

From II and III, we get $(x+y)^{2} + z^{2}=85$

Plug in $u=x+y$; then this equation becomes $u^{2}+z^{2}=85$.

Also, from I, we get $u+z=13$

hence, we obtain $u=7 \hspace{0.1in}or \hspace{0.1in}u=6$, $z=6 \hspace{0.1in} or \hspace{0.1in}7$.Thus,we have

$x+y=7$, $xy=10$; and $x+y=6$, $xy=10$. Hence, the solutions are

$x=5, \hspace{0.1in} or \hspace{0.1in} 2$; $y=2, \hspace{0.1in} or \hspace{0.1in} 5$; $z=6$;

or,

$x=3 \pm \sqrt{-1}$; $y=3 \mp \sqrt{-1}$; $z=7$.

Example 2:

Solve the following system of equations:

$(x+y)(x+z)=30$ Equation I

$(y+z)(y+x)=15$ Equation II

$(z+x)(z+y)=18$ Equation III

Solution: 2:

Write u, v, w for $y+z$, $z+x$, $x+y$ respectively; thus

$vw=30$, $uw=15$, $vu=18$…..call this equations A

Multiplying these equations together, we have

$u^{2}v^{2}w^{2}=30 \times 15 \times 18 = 8100$. Hence, $uvw=\pm 90$.

Combining this result with each of the equations in A, we have

$u=3, v=6. w=5$; or, $u=-3, v=-6, w=-5$; therefore, we get two sets of linear equations, as follows:

$y+z=3$

$z+x=6$

$x+y=5$

for which the solution set is: $x=4, y=1, z=2$; and the other set is:

$y+z=-3$

$z+x=-6$

$x+y=-5$

for which the solution set if $x=-4, y=-1, z=-2$.

Example 3:

Solve the following system of equations:

$y^{2}+yz+z^{2}=49$…call this equation (1)

$z^{2}+xz+x^{2}=19$…call this equation (2)

$x^{2}+xy+y^{2}=39$…call this equation (3)

Subtracting (2) from (1), we get the following:

$y^{2}-x^{2}+z(y-x)=30$, that is, $(y-x)(x+y+z)=30$…call this equation (4).

Similarly, from (1) and (3), we get the following:

$(z-x)(x+y+z)=10$…call this equation (5).

Hence, from equations (4) and (5), by division, we get the following:

$\frac{y-x}{z-x}=3$, hence, $y=3x-2z$

Substituting in Equation (3), we obtain $x^{2}-3xz+3z^{2}=13$.

From (2), we get the following: $x^{2}+xz+z^{2}=19$

Solving these homogeneous equations (hint: put $z=mx$, where m is parameter to be found), we get the following:

$x=\pm 2$, $z=\pm 3$; and therefore, $y=\pm 5$

or, $x=\pm \frac{11}{\sqrt{7}}$; $z=\pm \frac{1}{\sqrt{7}}$; and therefore, $y=\mp \frac{19}{\sqrt{7}}$.

Example 4:

Solve the following system of equations:

$x^{2}-yz=a^{2}$

$y^{2}-zx=b^{2}$

$z^{2}-xy=c^{2}$

Solution 4:

Multiply the equations by y, z, and x respectively, and add; then,

$c^{2}x+a^{2}y+b^{2}z=0$…call this equation I

Multiply the equations by z, x and y respectively; and add; then

$b^{2}x+c^{2}y+a^{2}z=0$…call this equation II

From (I) and (II), by cross multiplication,

$\frac{x}{a^{4}-b^{2}c^{2}}=\frac{y}{b^{4}-c^{2}a^{2}}=\frac{x}{c^{4}-a^{2}b^{2}}=k$, suppose.

Substitute in any one of the given equations; then,

$k^{2}(a^{6}+b^{6}+c^{6}-3a^{2}b^{2}c^{2})=1$; hence, we get the following solution:

$\frac{x}{a^{4}-b^{2}c^{2}}=\frac{y}{b^{4}-c^{2}a^{2}}=\frac{z}{c^{4}-a^{2}b^{2}}=\pm \frac{1}{\sqrt{a^{6}+b^{6}+c^{6}-3a^{2}b^{2}c^{2}}}$.

More stuff in the pipeline,

Nalin Pithwa

### Stationary Stationery

The other day, the stationery clerk asked George, Fred and Arthur if they could carry some parcels of stationery to the typist’s office. There were six parcels in all: one large parcel, which weighed the same as the two medium-sized parcels, which in turn weighed the same as the three small parcels. The stationery clerk then said that one person could carry the large parcel, one person could carry the two medium-sized parcels and one person could carry the three small parcels; that way all the three would be carrying the same weight of stationery. George, Fred and Arthur weren’t very happy about having to carry the stationery anyway, and said that they would only carry the stationery to the typists on the following conditions:

George would only carry the two-medium sized parcels if Fred carried the three small parcels. Fred would only carry the three small parcels if George carried the large parcel. George would only carry the large parcel if Arthur carried the two medium sized parcels. Arthur would only carry the two medium sized parcels if Fred carried the large parcel. George would only carry the large small parcels if Arthur carried the large parcel. All this time the stationery remained stationary. Eventually, the stationary clerk came up with a solution to keep all three happy. Who carried what to the typist?

Ref: http://www.amazon.in/Mensa-Book-Logic-Puzzles-Wareham/dp/8122201482/ref=sr_1_1?s=books&ie=UTF8&qid=1479953902&sr=1-1&keywords=A+mensa+book+of+logic+puzzles

I will share more such puzzles later…

Nalin Pithwa

### Fun with Zeta !!!

Mr. and Mrs. Zeta wanted to name their baby Zeta so  that its monogram (first, middle and last initials) will be in alphabetical order with no letters repeated. How many such monograms are possible?

— Nalin Pithwa

(Ref: Prof Titu Andreescu’s literature).

### Who’s who?

Harry and Fred are called Smith and Jones, but I am not sure if it’s Harry Smith and Fred Jones, or Harry Jones and Fred Smith. Given that two of the following statements are false, what is Harry’s surname?

1. Harry’s surname is Jones.
2. Harry’s surname is Smith.
3. Fred’s surname is Smith.

This is called learning to think !! 🙂 🙂 🙂

Nalin Pithwa

### Miscellaneous Example of Algebra

In order to find values of expressions involving a, b, c when these quantities are connected by the equation $a+b+c=0$, we might employ the substitution:

$a = h + k$

$b = {\omega}h + {\omega^{2}}k$

$c = {\omega^{2}}h + {\omega}k$

If, however, the expressions involve a, b, c symmetrically the method exhibited by the following example is preferable:

Example:

If $a+b+c=0$, show that $6(a^{5}+b^{5}+c^{5}) = 5(a^{3}+b^{3}+c^{3})(a^{2}+b^{2}+c^{2})$

Solution:

We have identically

$(1+ax)(1+bx)(1+cx) = 1 + px + qx^{2}+rx^{3}$, where $p = a+b+c$, $q = ab + bc + ca$, $r = abc$.

Hence, using the condition given, $(1+ax)(1+bx)(1+cx) = 1+qx^{2}+rx^{3}$

Taking logarithms and equating the coefficients of $x^{n}$, we have $\frac{(-1)^{n-1}}{n} (a^{n}+b^{n}+c^{n})$, which is equal to the coefficient of $x^{n}$ in the expansion of $\log {(1+qx^{2}+rx^{3})}$, which is equal to the coefficient of $x^{n}$ in the following:

$(qx^{2} + rx^{3}) - \frac{1}{2}(qx^{2}+rx^{3})^{3} + \frac{1}{3}(qx^{2}+rx^{3})^{3} - \ldots$

By putting $n = 2, 3, 5$ we obtain

$q = -\frac{a^{2} + b^{2} + c^{2}}{2}$

$r = \frac{a^{3} + b^{3} + c^{3}}{3}$

$qr = - \frac{a^{5}+b^{5}+c^{5}}{5}$, and hence, we get

$\frac{a^{5}+b^{5}+c^{5}}{5} = \frac{a^{3}+b^{3}+c^{3}}{3} \times \frac{a^{2}+b^{2}+c^{2}}{2}$, and the required result follows at once.

More stuff to follow…

Nalin Pithwa

### Miscellaneous Examples of Algebra: Part 3 for IITJEE Mains

Many identities can be readily established by making use of  the properties of the cube roots of unity; as usual these will be denoted by $1, \omega, \omega^{2}$.

Problem 1:

Show that $(x+y)^{7} - x^{7} - y^{7} = 7xy(x+y)(x^{2}+xy+y^{2})^{2}$

Solution 1:

The expression, E, on the left vanishes when $x=0, y=0, x+y=0$; hence, it must contain $xy(x+y)$ as a factor.

Putting $x = {\omega}y$, we have

$E=((1+\omega)^{7}-\omega^{7}-1)y^{7} = ((-\omega^{2})^{7}-\omega^{7}-1)y^{7}=(-\omega^{2}-\omega-1)y^{7}$

Hence, E contains $x - {\omega}y$ as a factor; and similarly, we may show that it contains $x - \omega^{2}y$ as a factor; that is, E is divisible by

$(x-{\omega}y)(x-\omega^{2}y)$, or $(x^{2}+xy+y^{2})$.

Further, E being of seven, and $xy(x+y)(x^{2}+xy+y^{2})$ of five dimensions, the remaining factor must be of the form $A(x^{2}+y^{2})+Bxy$, thus,

$(x+y)^{7}-x^{7}-y^{7} = xy(x+y)(x^{2}+xy+y^{2})(Ax^{2}+Bxy+Ay^{2})$.

Putting $x=1, y=1$, we have $21 = 2A + B$; putting $x = 2, y = -1$, we have $21= 5A-2B$, and hence, $A=7, B=7$;

$(x+y)^{7} - x^{7} - y^{7} = 7xy(x+y)(x^{2}+xy+y^{2})$.

Problem 2:

Show that the product of $a^{3} + b^{3} + c^{3} -3abc$ and $x^{3} + y^{3} + z^{3} -3xyz$ can be put in the form $A^{3} + B^{3} + C^{3} -3ABC$.

Solution 2:

The product $= (a+b+c)(a+{\omega}b + \omega^{2}c)(a + \omega^{2}b + {\omega}c) \times (x+y+z)(x+{\omega}y+\omega^{2}z)(x+\omega^{2}y + {\omega}z)$

By taking these six factors in the pairs $(a+b+c)(x+y+z)$,

$(a+{\omega}b+\omega^{2}c)(x+\omega^{2}y+{\omega}z)$,

and $(a + \omega^{2}b + {\omega}c)(x + {\omega}y + \omega^{2}z)$,

we obtain the partial products:

$A+B+C$, $A + {\omega}B + \omega^{2}C$, and $A + \omega^{2}B + {\omega}C$

where $A =ax + by + cz$, $B = bx + cy + az$, and $C = cx + ay + bz$

Thus, the product $= (A + B + C)(A + {\omega}B + \omega^{2}C)(A + \omega^{2}B + {\omega}C)$

which, in turn, equals $A^{3} + B^{3} + C^{3} + 3ABC$

More esoteric algebraic miscellany is planned for you!

Nalin Pithwa