Symmetric Functions. Alternating Functions. Algebra for RMO/IITJEE Math

Reference: Higher Algebra by Bernard and Child.

I. Symmetric Functions.

A function which is unaltered by the interchange of any two of the variables which it contains is said to be symmetric with respect to (wrt) these two variables.

Thus, $yz+zx+xy$ and $(x^{2}y+y^{2}z+z^{2}x)(x^{2}z+y^{2}z+z^{2}y)$ are symmetrical w.r.t. x, y, z.

The interchange of any two letters, x, y, z is called the transposition (xy).

Terms of an expression which are such that one can be changed into the other by one or more transpositions are said to be of the same type. Thus, all the terms of $x^{2}y+x^{2}z+y^{2}z+y^{2}x+z^{2}x+z^{2}y$ are of the same type, and the expression is symmetric with respect to x, y, z.

A symmetric function which is the sum of a number of terms of the same type is often written in an abbreviated form thus: Choose any one of the terms and place the letter $\Sigma$ (sigma) before it. For instance:

$x+y+z$ is represented by $\Sigma{x}$ and $xy+yz+zx$ by $\Sigma{xy}$ .

Again, $(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2xy+2yz+2zx=\Sigma{x^{2}}+2\Sigma{xy}$ .

It is obvious that

(i) if a term of some particular type occurs in a symmetric function, then all terms of the same type will also occur.

(ii) The sum, difference, product and quotient of two symmetric functions are also symmetric functions.

(PS: there is no need for a grand proof of the above; just apply the definitions of symmetric functions…check with some examples).

Considerations of symmetry greatly facilitate many algebraical processes as illustrated in the following examples:

Example 1: Expand $(y+z-x)(z+x-y)(x+y-z)$ .

Solution 1:

This expression is symmetric, homogeneous and of the third degree in x, y, z We may therefore assume that

$(y+z-x)(z+x-y)(x+y-z)= a.\Sigma{x^{3}}+b.\Sigma{x^{2}y}+cxyz$ , where a, b, c are independent of x, y, z. In this assumed identity:

(i) put $x=1, y=0, z=0$ , then $-1=a$

(ii) put $x=1, y=1, z=0$ , then $0=2a+2b$ , and so $b=1$ .

(iii) put $x=1, y=1, z=1$ , then $1=3a+6b+c$ and so $c=-2$ .

Hence, the required product is $-x^{2}-y^{2}-z^{2}+y^{2}z+yz^{2}+z^{2}x+zx^{2}+x^{2}y+xy^{2}-2xyz$ .

Example 2: Expand $(a+b+c+d)(ab+ac+ad+bc+bd+cd)$ . Test the result by putting $a=b=c=d=1$ .

Solution 2: the product is the sum of all terms of the product obtained by multiplying any one term of the first expression by any other term of the second expression. Hence, the result will have terms of the type : $a^{2}b, abc$ ,

The coefficient of $a^{2}b$ in the product is 1; because this term is obtained as product of a and ab and in no other way.

The coefficient of abc is 3; because this term is obtained in each of the three ways $a(bc), b(ca), c(ab)$ .

Hence, the required answer is $\Sigma{a^{2}b}+3\Sigma{abc}$

Test: The number of terms of the type $a^{2}b$ is 12 and the number of terms of the type abc is 4; hence, if $a=b=c=d=1$ , then $\Sigma{a}.\Sigma{ab}=4.6=24$ and $\Sigma{a^{2}b}+3.\Sigma{abc}=12+3.4=24$ so that the test is satisfied.

Example 3:

Factorize $(x+y+z)^{5}-x^{5}-y^{5}-z^{5}$ .

Solution 3:

Method I: Brute force is really difficult (I did give it a shot …:-))

Method II; Some of you might try the binomial theorem for positive integral index, but to extract the factors is still ..a little bit like brute force method only.

Method III:

Check whether the given expression is symmetric wrt any two variables (namely, x & y; y & z; z & x; ) and whether it is homogeneous and if so, what is the degree. Also from observations of past solved problems, we need to check how many terms of each type are there:

Observations are as follows: the degree of the expression is five only; and the expression is also homogeneous with each term being of degree five; to check for symmetry, let us proceed as follows:

$E_{1}=(x+y+z)^{5}-x^{5}-y^{5}-z^{5}$ and switching x and y gives us $E_{2}=(y+x+z)^{5}-y^{5}-x^{5}-z^{5}$ . Quite clearly, the expression is symmetric w.r.t. x and y; y and z; and, z and x.

To factorize it, we use fundamental theorem of algebra or factor or remainder theorem. Substitute $x=-y$ so that the expression is equal to $(z)^{5}-x^{5}-(-x)^{5}-z^{5}=z^{5}-x^{5}+x^{5}-z^{5}=0$ so that $(x+y)$ is a factor of the expression. Similarly, the other factors are $(y+z)$ and $(z+x)$ . By the fundamental theorem of algebra, we still need a quadratic factor of x, y and z. This factor should be homogeneous also. Hence, let

$E=(x+y+z)^{5}-x^{5}-y^{5}-z^{5}=(x+y)(y+z)(z+x){A(x^{2}+y^{2}+z^{2})+B(xy+yz+zx)}$ , where A and B are pure numeric coefficients independent of x, y and z.

So, put $x=1,y=1, z=0$ , then $2A+B=15$ and put $x=1, y=1, z=1$ , then $a+b=10$ so that $A=B=5$ .

So, $E=(x+y+z)^{5}-x^{5}-y^{5}-z^{5}=5(x+y)(y+z)(z+x)(x^{2}+y^{2}+z^{2}+xy+yz+zx)$ .

II Alternating Functions:

If a function E of x, y, z …is transformed into -E by the interchange of any two of the set x, y, z, …, then E is called an alternating function of x, y, z…(Note that just as in the case of symmetric functions, we talk of alternating functions w.r.t. a pair of variables at a time.)

((PS: At this juncture, it behooves you to recall the definitions of even and odd functions, and also to recall the fact that every function can be expressed as a sum of an even function and an odd function. Compare all three now: symmetric, alternating and even/odd functions. ))

Such an alternating function is $x^{n}(y-z)+y^{n}(z-x)+z^{n}(x-y)$ ; for, the interchange of any two letters, say x and y, transforms it into

$y^{n}(x--z)+x^{n}(x-y)+z^{n}(y-z)=-E$ .

Observe that the product and the quotient of two alternating functions are symmetric functions. (here, again, it does not require any grand proof…just pore over the definitions in your head…)

Thus, $\frac{x^{3}(y-z)+y^{3}(z-x)+z^{3}(x-y)}{(y-z)(z-x)(x-y)}$ is symmetric w.r.t. x, y, and z. (PS: please do some scribbling and verify this little observation/fact).

Example 1:

Factorize : $x^{3}(y-z)+y^{3}(z-x)+z^{3}(x-y)$ .

Solution 1:

Let $E=x^{3}(y-z)+y^{3}(z-x)+z^{3}(x-y)$

We know that $E=0$ when $x=y$ , $y=z$ and $z=y$ . Hence, the following is a factor of E: $(x-y)(y-z)(z-x)$ . As the given expression E is homogeneous of degree 4, it should have one more homogeneous linear factor. The only such factor possible is $K(x+y+z)$ . So, now,

$E=x^{3}(y-z)+y^{3}(z-x)+z^{3}(x-y)=K(x-y)(y-z)(z-x)(x+y+z)$ . To find K, the numerical coefficient independent of x, y, z, let us equate the coefficient of $x^{3}y$ on each side; then, $K=-1$ . (Alternatively, we could have substituted some numerical values for x, y, z and found K as it is an identity.)

Hence, $E=x^{3}(y-z)+y^{3}(z-x)+z^{3}(x-y)=-(x-y)(y-z)(z-x)(x+y+z)$ .

III. Cyclic Expressions:

An algebraic expression is said to be cyclic with respect to the letters a, b, c, d, …, h, k arranged in this order when it remains the same if we replace a by b, b by c, c by d, …., h by k, and k by a.

This “cycle of interchange of letters” is called the cyclic substitution denoted by $(abcd\ldots hk)$ .

(PS this reminds you of the the right hand unit vectors i, j, k and their cross products).

Thus, the expression $a^{2}b+b^{2}c+c^{2}d+d^{2}a$ is cyclic with respect to a, b, c, and d (in this order only) because the cyclic substitution $(abcd)$ changes the first term to the second term, the second term to the third term and the fourth term to the first term.

It is clear that:

(i) If a term of some particular type occurs in a cyclic expression, then the term which can be derived from this by the cyclic interchange, must also occur; and, the coefficients of these terms must be equal.

(ii) The sum, difference, product and quotient of two cyclic expressions is also cyclic.

In writing a cyclic expression, it is unnecessary to write the whole expression or all the terms explicitly. Thus, instead of writing the full $x^{2}(y-z)+y^{2}(z-x)+z^{2}(x-y)$ it suffices just to abbreviate it as $\Sigma{x^{2}(y-z)}$ . (Please note that the use of $\Sigma$ here has a different meaning than earlier.)

Sometimes, it is also written in short as $x^{2}(y-z)+\ldots+\ldots$ .

We need to be familiar with the following important basic cyclic identities:

$(b-c)+(c-a)+(a-b)=0$
$a(b-c)+b(c-a)+c(a-b)=0$
$a^{2}(b-c)+b^{2}(c-a)+c^{3}(a-b)=-(b-c)(c-a)(a-b)$ .
$bc(b-c)+ca(c-a)+ab(a-b)=-(b-c)(c-a)(a-b)$ .
$a(b^{2}-c^{2})+b(c^{2}-a^{2})+c(a^{2}-b^{2})=-(b-c)(c-a)(a-b)$ .
$a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)=-(b-c)(c-a)(a-b)(a+b+c)$ .
$(a+b+c)(ab+bc+ca)=a(b^{2}+c^{2})+b(c^{2}+a^{2})+c(a^{2}+b^{2})+3abc$
$(b+c)(c+a)(a+b)=a(b^{2}+c^{2})+b(c^{2}+a^{2})+c(a^{2}+b^{2})+2abc$
$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$ .
$(b-c)^{2}+(c-a)^{2}+(a-b)^{2}=2(a^{2}+b^{2}+c^{2}-ab-bc-ca)$
$(a+b+c)(b+c-a)(c+a-b)(a+b-c)=-a^{4}-b^{4}-c^{4}+2b^{2}c^{2}+2c^{2}a^{2}+2a^{2}b^{2}$ .

Note that identity 9 can be proved by at least three non-trivial ways 🙂

PS again :-)) It helps to prove the above identities from LHS to RHS and also from RHS to LHS !!

it will be proved later that

i) Any symmetric function of $\alpha, \beta, \gamma$ can be expressed in terms of $\Sigma{\alpha}$ , $\Sigma{\alpha\beta}$ and $\alpha\beta\gamma$ .

ii) Any symmetric function of $\alpha, \beta, \gamma, \delta$ can be expressed in terms of $\Sigma{\alpha}$ , $\Sigma{\alpha\beta}$ , $\Sigma{\alpha\beta\gamma}$ , and $\alpha\beta\gamma\delta$ .

In the above two cases, the notation $\Sigma$ is used in the sense of a symmetric function.

This mode of expression is extremely useful in factorizing symmetric functions, and in proving identities:

Example 1:

Factorize $a(1-b^{2})(1-c^{2}) + b(1-c^{2})(1-a^{2}) + c(1-a^{2})(1-b^{2})-4abc$ .

Solution 1:

PS: Comment: this is not easy. But, go through it and there is ample scope to improve via exercises in the next blog 🙂

Denoting the given expression by E, we have

$E=a(1-(b^{2}+c^{2})+b^{2}c^{2})+\ldots+\ldots-4abc$

$E=\Sigma{a}-\Sigma{ab^{2}}+abc\Sigma{ab}-4abc$ , but from identity 7 above, we see that $\Sigma{ab^{2}}=\Sigma{a}.\Sigma{ab}-3abc$ ;

Hence, we get $E = \Sigma{a}-\Sigma{a}.\Sigma{ab}-abc+abc\Sigma{ab}$

So, $E=\Sigma{a}.(1-\Sigma{ab})-abc(1-\Sigma{ab})=(1-bc-ca-ab)(a+b+c-abc)$ .

IV. Substitutiions:

We consider processes by which one arrangement (permutation) of a set of elements may be transformed into another:

Taking the permutations $cdba, bdac$ of a, b, c, d, the first is changed into the second by replacing a by c, b by a, c by b and leaving a unaltered. This process is represented by the operator

$\left(\begin{array}{cccc}abcd\\cabd \end{array}\right)$ or $\left(\begin{array}{ccc}abc\\cab\end{array}\right)$

and, we write $\left(\begin{array}{ccc}abc\\cab\end{array}\right)cdba=bdac$ .

Such a process and also the operator which affects it is called a substitution.

As previously stated, the interchange of two elements a, b is called the transposition $(ab)$ .

Also, a substitution such as $\left(\begin{array}{cccc}abcd\\bcda\end{array}\right)$ in which each letter is replaced by the one immediately following it and the last by the first, is called a cyclic substitution or cycle, and is denoted by $(abcd)$ .

If two operators are connected by the sign =, the meaning is that one is equivalent to the other, thus $(abcd)=(bcda)$ .

Two or more substitutioins may be applied successively. This is indicated as follows, the order of operations being from right to left.

Let $S=(ab)$ , $T=(ba)$ , then $STabcd=Sacbd=bcad$ and $TSabcd=Tbacd=cabd$ . Thus, $ST=\left(\begin{array}{cccc}abcd\\bcad\end{array}\right)$ and $TS=\left(\begin{array}{cccc}abcd\\cabd\end{array}\right)$

This process is called multiplication of substitutions, and the resulting substitution is called the product.

Multiplication of this kind is not necessarily commutative, but if the substitutions have no common letter, it is commutative.

The operation indicated by $(ab)(ab)$ , in which $(ab)$ is performed twice, produces no change in the order of the letters, and is called an identical substitution.

Any substituion is cyclic or is the product of two or more cyclic substitutions which have no common element. As an instance, consider the substitution $S = \left(\begin{array}{ccccccccc}abcdefghk\\chfbgaedk\end{array}\right)$

Here, a is changed to c, c to f, f to a, thus completing the cycle $(acf)$ . Also, b is changed to h, h to d, d to b, making the cycle $(bhd)$ . Next, c is changed to g, and g to e, giving the cycle $(eg)$ . The element k is unchanged, and we write

$S=(acf)(bhd)(eg)(k)$ or $S=(acf)(bhd)(eg)$ .

This expression for S in unique, and the order of the factors is indifferent. Moreover, the method applies universally, for in effecting any substitution, we must arrive at a stage when some letter is replaced by the first, thus completing a cycle. The same argument applies to the set of letters not contained in the cycle.

A cyclic substitution of n elements is the product of $(n-1)$ transpositions:

Consider

$(abc)=(ab)(bc)$ , $(abcd)=(abc)(cd)=(ab)(bc)(cd)$ , $(abcde)=(abcd)(de)=(ab)(bc)(cd)(de)$ , and so on.

We also have equalities such as : $(ae)(ad)(ac)(ab)=(abcde)$ and $(ab)(ac)(ad)(ae)=(edcba)$ .

A substitution which deranges n letters and which is the product of r cycles is equivalent to $(n-r)$ transpositions.

This follows at once from our previous work. Thus, if $S = \left(\begin{array}{cccccccc}abcdefgh\\chfbgaed\end{array}\right)$ , then

$S=(acf)(bhd)(eg)=(ac)(cf)(bh)(hd)(eg)$ .

If we introduce the product $(ab)(ab)$ , S is unaltered and the number of transpositions is increased by 2.

Thus, if a given substitutition is equivalent to j transpositions, the number j is not unique. We shall prove that : $j=n-r+2s$ where r is a positive integer or zero.

This is a very important theorem, and to prove it we introduce the notion of “inversions.”

*** Taking the elements a, b, c, d, e choose some arrangement, as $abcde$ , and call it a normal arrangement.

Consider the arrangement $bdeac$ . Here b precedes a, but follows it in the normal arrangement. On this account, we say that the pair ba constitutes an inversion.

Thus, $bdeac$ contains five inversions, namely, $ba, da, dc, ea, ec$ .

Theorem 1:

If i is the number of inversions which are introduced or removed by a substitution which is equivalent to j transpositions, then i and j are both even or both odd.

Proof of theorem 1:

Consider the effect of a single transposition $(fg)$ .

If f, g are consecutive elements, the transposition $(fg)$ does not alter the position of f or g relative to the other elements. It therefore introduces or removes a single inversion due to the interchange of f, g.

If f, g are separated by n elements p, q, r, …, x, then f can be moved to the place occupied by g by $n+1$ interchanges of consecutive elements, and then g can be moved to the place originally occupied by f by n such interchanges.

Thus, the transposition $(fg)$ can be effected by $2n+1$ interchanges of consecutive elements. Therefore, any transposition introduces or removes an odd number of inversions, and the theorem follows. QED.

Again, for a given substitution, i is a fixed number, and therefore whatever value j may have, it must be even or odd, according as i is even or odd. Hence, we get the following:

Theorem 2:

If one arrangement A of a given set of elements is changed into another B by j transpositions, then j is always even or always odd. In other words: the number of transpositions which are equivalent to a given substitution is not unique, but is always even or always odd.

The minimum value of j is $n-r$ .

Thus, substitutions may be divided into two distinct classes. We say that a substitution is even or odd according as it is equivalent to an even or an odd number of transpositions.

Rule:

To determine the class of a substitution S we may express it as the product of cycles, and count the number of cycles with an even number of elements: then S is even or odd according as this number is even or odd.

Or, we can settle the question by counting the number of inversions, but this generally takes longer.

The tutorial exercises follow this blog.

Cheers,

Nalin Pithwa

Number theory: let’s learn it the Nash way !

Posted by Nalin Pithwa

Reference: A Beautiful Mind by Sylvia Nasar.

Comment: This is approach is quite similar to what Prof. Joseph Silverman explains in his text, “A Friendly Introduction to Number Theory.”

Peter Sarnak, a brash thirty-five-year-old number theorist whose primary interest is the Riemann Hypothesis, joined the Princeton faculty in the fall of 1990. He had just given a seminar. The tall, thin, white-haired man who had been sitting in the back asked for a copy of Sarnak’s paper after the crowd had dispersed.

Sarnak, who had been a student of Paul Cohen’s at Stanford, knew Nash by reputation as well as by sight, naturally. Having been told many times Nash was completely mad, he wanted to be kind. He promised to send Nash the paper. A few days later, at tea-time, Nash approached him again. He had a few questions, he said, avoiding looking Sarnak in the face. At first, Sarnak just listened politely. But within a few minutes, Sarnak found himself having to concentrate quite hard. Later, as he turned the conversation over in his mind, he felt rather astonished. Nash had spotted a real problem in one of Sarnak’s arguments. What’s more, he also suggested a way around it. “The way he views things is very different from other people,” Sarnak said later. ‘He comes up with instant insights I don’t know I would ever get to. Very, very outstanding insights. Very unusual insights.”

They talked from time to time. After each conversation, Nash would disappear for a few days and then return with a sheaf of computer printouts. Nash was obviously very, very good with the computer. He would think up some miniature problem, usually very ingeniously, and then play with it. If something worked on a small scale, in his head, Sarnak realized, Nash would go to the computer to try to find out if it was “also true the next few hundred thousand times.”

{What really bowled Sarnak over, though, was that Nash seemed perfectly rational, a far cry from the supposedly demented man he had heard other mathematicians describe. Sarnak was more than a little outraged. Here was this giant and he had been all but forgotten by the mathematics profession. And the justification for the neglect was obviously no longer valid, if it had ever been.}

Cheers,

Nalin Pithwa

PS: For RMO and INMO (of Homi Bhabha Science Foundation/TIFR), it helps a lot to use the following: (it can be used with the above mentioned text of Joseph Silverman also): TI nSpire CAS CX graphing calculator.

https://www.amazon.in/INSTRUMENTS-TI-Nspire-CX-II-CAS/dp/B07XCM6SZ3/ref=sr_1_1?crid=3RNR2QRV1PEPH&keywords=ti+nspire+cx+cas&qid=1585782633&s=electronics&sprefix=TI+n%2Caps%2C253&sr=8-1

A fifth primer: plane geometry tutorial for preRMO and RMO: core stuff

Posted by Nalin Pithwa

Show that three straight lines which join the middle points of the sides of a triangle, divide it into four triangles which are identically equal.
Any straight line drawn from the vertex of a triangle to the base is bisected by the straight line which joins the middle points of the other sides of the triangle.
ABCD is a parallelogram, and X, Y are the middle points of the opposite sides AD, BC: prove that BX and DY trisect the diagonal AC.
If the middle points of adjacent sides of any quadrilateral are joined, the figure thus formed is a parallelogram. Prove this.
Show that the straight lines which join the middle points of opposite sides of a quadrilateral bisect one another.
From two points A and B, and from O the mid-point between them, perpendiculars AP, and BQ, OX are drawn to a straight line CD. If AP, BQ measure respectively 4.2 cm, and 5.8 cm, deduce the length of OX. Prove that OX is one half the sum of AP and BQ. or $\frac{1}{2}(AP-BQ)$ or $\frac{1}{2}(BQ-AP)$ according as A and B are on the same side or on opposite sides of CD.
When three parallels cut off equal intercepts from two transversals, prove that of three parallel lengths between the two transversals the middle one is the Arithmetic Mean of the other two.
The parallel sides of a trapezium are a cm and b cm respectively. Prove that the line joining the middle points of the oblique sides is parallel to the parallel sides, and that its length is $\frac{1}{2}(a+b)$ cm.
OX and OY are two straight lines, and along OX five points 1,2,3,4,5 are marked at equal distances. Through these points parallels are drawn in any direction to meet OY. Measure the lengths of these parallels : take their average and compare it with the lengths of the third parallel. Prove geometrically that the third parallel is the mean of all five.
From the angular points of a parallelogram perpendiculars are drawn to any straight line which is outside the parallelogram : prove that the sum of the perpendiculars drawn from one pair of opposite angles is equal to the sum of those drawn from the other pair. (Draw the diagonals,and from their point of intersection suppose a perpendicular drawn to the given straight line.)
The sum of the perpendiculars drawn from any point in the base of an isosceles triangle to the equal to the equal sides is equal to the perpendicular drawn from either extremity of the base to the opposite side. It follows that the sum of the distances of any point in the base of an isosceles triangle from the equal sides is constant, that is, the same whatever point in the base is taken).
The sum of the perpendiculars drawn from any point within the an equilateral triangle to the three sides is equal to the perpendicular drawn from any one of the angular points to the opposite side, and is therefore, constant. Prove this.
Equal and parallel lines have equal projections on any other straight line. Prove this.

More later,

Cheers,

Nalin Pithwa.

A fourth primer: plane geometry question set including core theorems, preRMO and RMO

Posted by Nalin Pithwa

Hard core definitions of various special quadrilaterals:

A quadrilateral is a plane figure bounded by four straight lines.
A parallelogram is a quadrilateral whose opposite sides are parallel.
A rectangle is a parallelogram which has one of its angles a right angle.
A square is a rectangle which has two adjacent sides equal.
A rhombus is a quadrilateral which has all its sides equal, but its angles are not right angles.
A trapezium if a quadrilateral which has one pair of parallel sides.

Problem 1:

The straight lines which join the extremities of two equal and parallel straight lines towards the same parts are themselves equal and parallel. Prove this.

Problem 2:

The opposite sides and angles of a parallelogram are equal to one another, and each diagonal bisects the parallelogram.

Problem 3:

Corollary 1 of problem 2 above: If one angle of a parallelogram is a right angle, all its angles are right angles.

Corollary 2 of problem 2 above: All the sides of a square are equal; and all its angles are right angles.

Corollary 3 of problem 3 above: The diagonals of a parallelogram bisect one another.

Problem 4:

If the opposite sides of a quadrilateral are equal, then the figure is a parallelogram.

Problem 5:

If the opposite angles of a quadrilateral are equal, then the figure is a parallelogram.

Problem 6:

If the diagonals of a quadrilateral bisect each other, then the figure is a parallelogram.

Problem 7:

The diagonals of a rhombus bisect each other at right angles.

Problem 8:

If the diagonals of a parallelogram are equal, all its angles are right angles.

Problem 9:

In a parallelogram which is not rectangular, the diagonals are not equal.

Problem 10:

Any straight line drawn through the middle point of a diagonal of a parallelogram and terminated by a pair of opposite sides is bisected at that point.

Problem 11:

In a parallelogram, the perpendiculars drawn from one pair of opposite angles to the diagonal which joins the other pair are equal. Prove this.

Problem 12:

If ABCD is a parallelogram, and X, Y respectively the middle points of the sides AD, BC, show that the figure AYCX is a parallelogram.

Problem 13:

ABC and DEF are two triangles such that AB, BC are respectively equal to and parallel to DE, EF; show that AC is equal and parallel to DF.

Problem 14:

ABCD is a quadrilateral in which AB is parallel to DC, and AD equal but not parallel to BC; show that (i) angle A + angle C = 180 degrees = angle B + angle D; (ii) diagonal AC = diagonal BD (iii) the quadrilateral is symmetrical about the straight line joining the middle points of AB and DC.

Problem 15:

AP, BQ are straight rods of equal length, turning at equal rates (both clockwise) about two fixed pivots A and B respectively. If the rods start parallel but pointing in opposite senses, prove that (i) they will always be parallel (ii) the line joining PQ will always pass through a certain fixed point.

Problem 16:

A and B are two fixed points, and two straight lines AP, BQ, unlimited towards P and Q, are pivoted at A and B. AP, starting from the direction AB, turns about A clockwise at the uniform rate of 7.5 degrees a second; and BQ, starting simultaneously from the direction BA, turns about B counter-clockwise at the rate of 3.75 degrees a second. (i) How many seconds will elapse before AP and BQ are parallel? (ii) Find graphically and by calculation the angle between AP and BQ twelve seconds from the start. (iii) At what rate does this angle decrease?

Problem 17 (Intercept theorem or Basic Proportionality Theorem):

If there are three or more parallel straight lines, and the intercepts made by them on any transversal are equal, then the corresponding intercepts on any other transversal are also equal.

Prove the corollary: In a triangle ABC, if a set of lines Pp, Qq, Rr, $\ldots$ , drawn parallel to the base, divide one side AB into equal parts they also divide the other side AC into equal parts.

Definition:

If from the extremities of a straight line AB perpendiculars AX, BY are drawn parallel to a straight line PQ of indefinite length, then XY is said to be the orthogonal projection of AB on PQ.

Problem 18:

Prove: The straight line drawn through the middle point of a side of a triangle parallel to the base bisects the remaining side.

Problem 19:

The straight line which joins the middle points of two sides of a triangle is equal to half the third side.

More later,

Nalin Pithwa.

A third primer: plane geometry question set for preRMO and RMO

Posted by Nalin Pithwa

Problem 1:

Prove: Two right angled triangles which have their hypotenuses equal, and one side of one equal to one side of the other, are equal in all respects. (PS: Please do not use “short-cut or the magic formula — Pythagoras’s theorem; if you do, you are requested to prove Pythagoras’s theorem using plane geometry!) (PS:2: Remember Euclid’s geometry builds up from “scratch”…a small step at a time….axioms, definitions, lemmas, …; you can only use the theorems used and proved so far in “class”).

Problem 2:

If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle included by the two sides of one greater than the angle included by the corresponding sides of the other; then the base of that which has the greater angle is greater than the base of the other.

Problem 3:

Prove converse of the above: If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of one greater than the base of the other, then the angle contained by the sides of that which has the greater base is greater than the angle contained by the corresponding sides of the other.

Problem 4:

(i) The perpendicular is the shortest line that can be drawn to a given straight line from a given point. (ii) Obliques which make equal angles with the perpendicular are equal. (iii) Of two obliques the less is that which makes the smaller angle with the perpendicular. Prove all these.

Problem 5:

If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise the angles opposite to one pair of equal sides equal, then the angles opposite to the other pair of equal sides are either equal or supplementary, and in the former case, the triangles are equal in all respects.

More later,

Cheers,

Nalin Pithwa

A Second primer: plane geometry for preRMO and RMO: basic questions set

Posted by Nalin Pithwa

Problem 1:

If from any point in the bisector of an angle a straight line is drawn parallel to either arm of the angle, the triangle thus formed is isosceles.

Problem 2:

From X, a point in the base BC of an isosceles triangle ABC. a straight line is drawn at right angles to the base, cutting AB in Y, and CA produced in Z, show the triangle AYZ is isosceles.

Problem 3:

If the straight line which bisects an exterior angle of a triangle is parallel to the opposite side, show that the triangle is isosceles.

Problem 4:

The straight line drawn from any point in the bisector of an angle parallel to the arms of the angle, and terminated by them, are equal, and the resulting figure is a quadrilateral having all its sides equal.

Problem 5:

AB and CD are two straight lines intersecting at D, and the adjacent angles so formed are bisected; if through any point X in DC a straight line XYZ is drawn parallel to AB and meeting the bisectors Y and Z, show that XY is equal to XZ.

Problem 6:

Two straight rods PA, QB revolve about pivots at P and Q, PA making 12 complete revolutions a minute, and QB making 10. If they start parallel and pointing the same way, how long will it be before they are again parallel (i) pointing opposite ways (ii) pointing the same way?

Problem 7:

Prove that: if two straight lines are perpendicular to two other straight lines, each to each, the acute angle between the first pair is equal to the acute angle between the second pair.

Problem 8:

Show that the only regular figures which may be fitted together so as to form a plane surface are (i) equilateral triangles (ii) squares (iii) regular hexagons.

Problem 9:

If one side of a regular hexagon is produced, show that the exterior angle is equal to the interior angle of an equilateral triangle.

Problem 10:

If a straight line meets two parallel straight lines, and the two interior angles on the same side are bisected, show that the bisectors meet at right angles.

Problem 11:

If the base of any triangle is produced both ways, show that the sum of the two exterior angles minus the vertical angle is equal to two right angles.

Problem 12:

In the triangle ABC, the base angles at B and C are bisected by BO and CO respectively. Show that the angle BOC is 90 degrees plus A/2.

Problem 13:

In the triangle ABC, the sides AB, AC are produced, and the exterior angles are bisected by BO and CO. Show that the angle BOC is 90 degrees minus A/2.

Problem 14:

Prove: the angle contained by the bisectors of two adjacent angles of a quadrilateral is equal to half the sum of the remaining angles.

Problem 15:

A is the vertex of an isosceles triangle ABC, and BA is produced to D, so that AD is equal to BA; if DC is drawn, show that BCD is a right angle. Prove this.

Problem 16:

Prove: The straight line joining the middle point of the hypotenuse at a right angled triangle to the right angle is equal to half the hypotenuse.

Problem 17:

If two triangles have two angles of one equal to two angles of the other, each to each, and any one side of the first equal to the corresponding side of the other, the triangles are equal in all respects. (ASA congruency test):

Problem 18:

Show that the perpendiculars drawn from the extremities of the base of an isosceles triangle to the opposite sides are equal.

Problem 19:

Prove: Any point on the bisector of an angle is equidistant from the arms of the angle.

Problem 20:

Through O the middle point of a straight line AB, any straight line is drawn, and perpendiculars AX and BY are dropped upon it from A and B: show that AX is equal to BY.

Problem 21:

If the bisector of the vertical angle of a triangle is at right angles to the base, the triangle is isosceles.

Problem 22:

If in a triangle the perpendicular from the vertex on the base bisects the base, then the triangle is isosceles. Prove.

Problem 23:

If the bisector of the vertical angle of a triangle also bisects the base, the triangle is isosceles. Prove this.

Problem 24:

The middle point of any straight line which meets two parallel straight lines, and is terminated by them, is equidistant from the parallels. Prove this.

Problem 25:

A straight line drawn between two parallels and terminated by them is bisected; show that any other straight line passing through the middle point and terminated by the parallels is also bisected at that point.

Problem 26:

If through a point equidistant from two parallel straight lines, two straight lines are drawn cutting the parallels, the portions of the latter thus intercepted are equal. Prove this.

Problem 27:

In a quadrilateral ABCD, if AB=CD, and BC=DC: show that the diagonal AC bisects each of the angles which it joins, and that AC is perpendicular to BD.

Problem 28:

A surveyor wishes to ascertain the breadth of a river which he cannot cross. Standing at a point A, near the bank, he notes an object B immediately opposite on the other bank. He lays down a line AC of any length at right angles to AB, fixing a mark at O the middle point of AC. From C he walks along a line perpendicular to AC until he reaches a point D from which O and B are seen in the same direction. He now measures CD: prove that the result gives him the width of the river.

More later,

Cheers,

Nalin Pithwa.

PS: There is no royal road to geometry —- Plato.

A primer for preRMO and RMO plane geometry with basic exercises

Posted by Nalin Pithwa

Plane geometry is axiomatic deductive logic. I present a quick mention/review of “proofs” which can be “derived” in sequence….building up the elementary theorems …so for example, if there is a question like: prove that the three medians of a triangle are concurrent, please do not use black magic complicated machinery like Ceva’s theorem,etc; or even if say, the question asks you to prove Ceva’s theorem only, you have to prove it using elementary theorems like the ones presented below:

For the present purposes, I am skipping axioms and basic definitions and hypothetical constructions. I am using straight away the reference (v v v old text) : A School Geometry, Metric Edition by Hall and Stevens. (available almost everywhere in India):

Theorem 1:

The adjacent angles which one straight line makes with another straight line on one side of it are together equal to two right angles.

Corollary 1 of Theorem 1:

if two straight lines cut another, the four angles so formed are together equal to four right angles.

Corollary 2 of Theorem 1:

When any number of straight lines meet at a point, the sum of the consecutive angles so formed is equal to four right angles.

Corollary 3 of Theorem 1:

(a) Supplements of the same angle are equal. (ii) Complements of the same angle are equal.

Theorem 2 (converse of theorem 1):

If, at a point in a straight line, two other straight lines, on opposite sides of it, make the adjacent angles together equal to two right angles, then these two straight lines are in one and the same straight line.

Remark: this theorem can be used to prove stuff like three points are in a straight line.

Theorem 3:

If two straight lines cut one another, the vertically opposite angles are equal.

Theorem 4: SAS Test of Congruence of Two Triangles:

If two triangles have two sides of the one equal to two sides of the other, each to each, and the angles included by those sides equal, then the triangles are equal in all respects.

Theorem 5:

The angles at the base of an isosceles triangle are equal.

Corollary 1 of Theorem 5:

If the equal sides AB, AC of an isosceles triangle are produced, the exterior angles EBC, FCB are equal, for they are the supplements of the equal angles at the base.

Corollary 2 of Theorem 5:

If a triangle is equilateral, it is also equiangular.

Theorem 6:

If two angles of a triangle are equal to one another, then the sides which are opposite to the equal angles are equal to one another.

Corollary of Theorem 6:

Hence, if a triangle is equiangular, it is also equilateral.

Theorem 7 (SSS Test of Congruence of Two Triangles):

If two triangles have the three sides of the one equal to the three sides of the others, each to each, they are equal in all respects.

Theorem 8:

If one side of a triangle is produced then the exterior angle is greater than either of the interior opposite angles.

Corollary 1 to Theorem 8:

Any two angles of a triangle are together less than two right angles.

Corollary 2 to Theorem 8:

Every triangle must have at least two acute angles.

Corollary 3 to Theorem 8:

Only one perpendicular can be drawn to a straight line from a given point outside it.

Theorem 9 :

If one side of a triangle is greater than another, then the angle opposite of the greater side is greater than the angle opposite to the less.

Theorem 10:

If one angle of a triangle is greater than another, then the side opposite to the greater angle is greater than the side opposite to the less.

Theorem 11: Triangle Inequality:

Any two sides of a triangle are together greater than the third side.

Theorem 12: Another inequality sort of theorem:

Of all straight lines drawn from a given point to a given straight line the perpendicular is the least.

Corollary 1 to Theorem 12:

Hence, conversely, since there can be only one perpendicular and one shortest line from O to AB: if OC is the shortest straight line from O to AB, then OC is perpendicular to AB.

Corollary 2 to Theorem 12:

Two obliques OP, OQ which cut AB at equal distance from C, the foot of the perpendicular are equal.

Corollary 3 to Theorem 12:

Of two obliques OQ, OR, if OR cuts AB at the greater distance from C. the foot of the perpendicular, then OR is greater than OQ.

Theorem 13 :

If a straight line cuts two other straight lines so as to make: (i) the alternate angles equal or (ii) an exterior angle equal to the interior opposite angle on the same side of the cutting line or (iii) the interior angles on the same side equal to two right angles, then in each case, the two straight lines are parallel.

Theorem 14:

If a straight line cuts two parallel lines, it makes : (i) the alternate angles equal to one another; (ii) the exterior angle equal to the interior opposite angle on the same side of the cutting line (iii) the two interior angles on the same side together equal to two right angles.

Theorem 15:

Straight lines which are parallel to the same straight line are parallel to one another.

Theorem 16:

Sum of three interior angles of a triangle is 180 degrees.

Also, if a side of a triangle is produced the exterior angle is equal to the sum of the two interior opposite angles.

Corollary 1:

All the interior angles of one rectilinear figure, together with four right angles are equal to twice as many right angles as the figure has sides.

Corollary 2:

If the sides of a rectilinear figure, which has no reflex angle, are produced in order, then all the exterior angles so formed are together equal to four right angles.

Theorem 17: AAS test of congruence of two triangles:

If two triangles have two angles of one equal to two angles of the other, each to each, and any side of the first equal to the corresponding side of the other, the triangles are equal in all respects.

Theorem 18:

Two right angled triangles which have their hypotenuses equal, and one side of one equal to one side of the other are equal in all respects.

Theorem 19:

If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle included by the two sides of one greater than the angle included by the two corresponding sides of the other, then the base of that which has the greater angle is greater than the base of the other.

Conversely,

if two triangles have two sides of the one equal to two sides of the other, each to each, but the base of one greater than the base of the other, then the angle contained by the sides of that which has the greater base is greater than the angle contained by the corresponding sides of the other.

Theorem 20:

The straight lines which join the extremities of two equal and parallel straight lines towards the same parts are themselves equal and parallel.

Theorem 21:

The opposite sides and angles of a parallelogram are equal to one another, and each diagonal bisects the parallelogram.

Corollary 1 to Theorem 21:

If one angle of a parallelogram to a right angle, all its angles are equal.

Corollary 2 to Theorem 21:

All the sides of a square are equal and all its angles are right angles.

Corollary 3 to Theorem 21:

The diagonals of a parallelogram bisect each other.

Theorem 22:

If there are three or more parallel straight lines, and the intercepts made by them on any transversal are equal, then the corresponding intercepts on any other transversal are also equal.

Tutorial exercises based on the above:

Problem 1: In the triangle ABC, the angles ABC, ACB are given equal. If the side BC is produced both ways, show that the exterior angles so formed are equal.

Problem 2: In the triangle ABC, the angles ABC, ACB are given equal. If AB and AC are produced beyond the base, show that the exterior angles so formed are equal.

Problem 3: Prove that the bisectors of the adjacent angles which one straight line makes with another contain a right angle. That is to say, the internal and external bisectors of an angle are at right angles to one another.

Problem 4: If from O a point in AB two straight lines OC, OD are drawn on opposite sides of AB so as to make the angle COB equal to the angle AOD, show that OC and OD are in the same straight line.

Problem 5: Two straight lines AB, CD cross at O. If OX is the bisector of the angle BOD, prove that XO produced bisects the angle AOC.

Problem 6: Two straight lines AB, CD cross at O. If the angle BOD is bisected by OX, and AOC by OY, prove that OX, OY are in the same straight line.

Problem 7: Show that the bisector of the vertical angle of an isosceles triangle (i) bisects the base (ii) is perpendicular to the base.

Problem 8: Let O be the middle point of a straight line AB, and let OC be perpendicular to it. Then, if P is any point in OC, prove that PA=PB.

Problem 9: Assuming that the four sides of a square are equal, and that its angles are all right angles, prove that in the square ABCD, the diagonals AC, BD are equal.

Problem 10: Let ABC be an isosceles triangle: from the equal sides AB, AC two equal parts AX, AY are cut off, and BY and CX are joined. Prove that BY=CX.

Problem 11: ABCD is a four-sided figure whose sides are all equal, and the diagonal BD is drawn : show that (i) the angle ABD = the angle ADB (ii) the angle CBD = the angle CDB (iii) the angle ABC = the angle ADC.

Problem 12: ABC, DBC are two isosceles triangles drawn on the same base BC, but on opposite sides of it: prove that the angle ABD = the angle ACD.

Problem 13: ABC, DBC are two isosceles triangles drawn on the same base BC, but on the same side of it: prove that the angle ABD = the angle ACD.

Problem 14: AB, AC are the equal sides of an isosceles triangle ABC, and L, M, N are the middle points of AB, BC and CA respectively; prove that (i) LM = NM (ii) BN = CL (iii) the angle ALM = the angle ANM.

Problem 15: Show that the straight line which joins the vertex of an isosceles triangle to the middle points of the base (i) bisects the vertical angle (ii) is perpendicular to the base.

Problem 16: If ABCD is a rhombus, that is, an equilateral four sided figure, show by drawing the diagonal AC that (i) the angle ABC = the angle ADC (ii) AC bisects each of the angles BAD and BCD.

Problem 17: If in a quadrilateral ABCD the opposite sides are equal, namely, AB = CD and AD=CB, prove that the angle ADC = the angle ABC.

Problem 18: If ABC and DBC are two isosceles triangles drawn on the same base BC, prove that the angle ABD = the angle ACD, taking (i) the case where the triangles are on the same side of BC (ii) the case where they are on the opposite sides of BC.

Problem 19: If ABC, DBC are two isosceles triangles drawn on opposite sides of the same base BC, and if AD be joined, prove that each of the angles BAC, BDC will be divided into two equal parts.

Problem 20: Show that the straight lines which join the extremities of the base of an isosceles triangle to the middle points of the opposite sides are equal to one another.

Problem 21: Two given points in the base of an isosceles triangle are equidistant from the extremities of the base: show that they are also equidistant from the vertex.

Problem 22: Show that the triangle formed by joining the middle points of the sides of an equilateral triangle is also equilateral.

Problem 23: ABC is an isosceles triangle having AB equal to AC, and the angles at B and C are bisected by BC and CO: prove that (i) BO = CO (ii) AO bisects the angle BAC.

Problem 24: Show that the diagonals of a rhombus bisect one another at right angles.

Problem 25: The equal sides BA, CA of an isosceles triangle BAC are produced beyond the vertex A to the points E and F, so that AE is equal to AF and FB, EC are joined: prove that FB is equal to EC.

Problem 26: ABC is a triangle and D any point within it. If BD and CD are joined, the angle BDC is greater than the angle BAC. Prove this (i) by producing BD to meet AC (ii) by joining AD, and producing it towards the base.

Problem 27: If any side of a triangle is produced both ways, the exterior angles so formed are together greater than two right angles.

Problem 28: To a given straight line, there cannot be drawn from a point outside it more than two straight lines of the same given length.

Problem 29: If the equal sides of an isosceles triangle are produced, the exterior angles must be obtuse.

Note: The problems 30 to 43 are based on triangle inequalities:

Problem 30: The hypotenuse is the greatest side of a right angled triangle.

Problem 31: The greatest side of any triangle makes acute angles with each of the other sides.

Problem 32: If from the ends of a side of a triangle, two straight lines are drawn to a point within the triangle, then those straight lines are together less than the other two sides of the triangle.

Problem 33: BC, the base of an isosceles triangle ABC is produced to any point D; prove that AD is greater than either of the equal sides.

Problem 34: If in a quadrilateral the greatest and least sides are opposite to one another, then each of the angles adjacent to the least side is greater than its opposite angle.

Problem 35: In a triangle, in which OB, OC bisect the angles ABC, ACB respectively: prove that if AB is greater than AC, then OB is greater than OC.

Problem 36: The difference of any two sides of a triangle is less than the third side.

Problem 37: The sum of the distances of any point from the three angular points of a triangle is greater than half its perimeter.

Problem 38: The perimeter of a quadrilateral is greater than the sum of its diagonals.

Problem 39: ABC is a triangle, and the vertical angle BAC is bisected by a line which meets BC in X, show that BA is greater than BX, and CA greater than CX. Obtain a proof of the following theorem : Any two sides of a triangle are together greater than the third side.

Problem 40: The sum of the distance of any point within a triangle from its angular points is less than the perimeter of the triangle.

Problem 41: The sum of the diagonals of a quadrilateral is less than the sum of the four straight lines drawn from the angular points to any given point. Prove this, and point out the exceptional case.

Problem 42: In a triangle any two sides are together greater than twice the median which bisects the remaining side.

Problem 43: In any triangle, the sum of the medians is less than the perimeter.

Problem 44: Straight lines which are perpendicular to the same straight line are parallel to one another.

Problem 45: If a straight line meets two or more parallel straight lines, and is perpendicular to one of them, it is also perpendicular to all the others.

Problem 46: Angles of which the arms are parallel each to each are either equal or supplementary.

Problem 47: Two straight lines AB, CD bisect one another at O. Show that the straight line joining AC and BD are parallel.

Problem 48: Any straight line drawn parallel to the base of an isosceles trianlge makes equal angles with the sides.

More later. Get cracking. This perhaps the simplest introduction, step by step, to axiomatic deductive logic…discovered by Euclid about 2500 years before ! Hail Euclid !

Cheers,

Nalin Pithwa