Ratio and proportion tutorial problems: pre RMO or PRMO or IITJEE foundation maths

Example 1:

If (2ma+6mb+3nc+9nd)(2ma-6mb-3nc+9nd)=(2ma-6mb+3nc-9nd)(2ma+6mb-3nc-9nd), prove that a, b, c, and d are proportionals.

Solution 1:

Given that

\frac{2ma+6mb+3nc+9nd}{2ma-6mb+3nc-9nd} = \frac{2mn+6mb-3nc-9nd}{2ma-6mb-3nc+9nd}

We also know that if \frac{x}{y} = \frac{p}{q}, then the property of componendo and dividendo says: \frac{x+y}{x-y} = \frac{p+q}{p-q}. Applying this property to the above “huge” fraction, we get:

\frac{2(2ma+3nc)}{2(6mb+9nd)} = \frac{2(2ma-3nc)}{2(6mb-9nd)}

We know that if \frac{x}{y} = \frac{p}{q}, then \frac{x}{p} = \frac{y}{q}, which is the property called alternendo. Applying this property to the above fraction, we get

\frac{2ma+3nc}{2ma-3nc} = \frac{6mb+9nd}{6mb-9nd},

again, applying componendo and dividendo to the above, we get

\frac{4ma}{6nc} = \frac{12mb}{18nd}

hence, \frac{a}{c} = \frac{b}{d}, that is, a, b, c and d are proportionals. Hence, the proof.

Example 2:

Solve the equation: \frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}} = \frac{4x-1}{2}

Solution 2:

By componendo and dividendo, we get \frac{\sqrt{x+1}}{\sqrt{x-1}} = \frac{4x+1}{4x-3}

Now, squaring both sides of the above equation, we get \frac{x+1}{x-1} = \frac{16x^{2}+8x+1}{16x^{2}-24x+9}.

Again, applying componendo and dividendo,

\frac{2x}{2} = \frac{32x^{2}-16x+10}{32x-8}

x = \frac{16x^{2}-8x+5}{16x-4}

so, we get 16x^{2}-4x=16x^{2}-8x+5

so, we get finally x=\frac{5}{4}

Cheers,

Nalin Pithwa

Find a flaw in this proof: RMO and PRMO tutorial

What ails the following proof that all the elements of a finite set are equal?

The following is the “proof”;

All elements of a set with no elements are equal, so make the induction assumption that any set with n elements has all its elements equal. In a set with n elements, the first and the last n are equal by induction assumption. They overlap at n, so all are equal, completing the induction.

End of “proof:

Regards,

Nalin Pithwa

|log{xx_{1}}| + |log{xx_{2}}| + …+ |log{xx_{n}}| + |log{x/x_{1}}| + |log{x/x_{2}}| + …+|log{x/x_{n}}|= |log{x_{1}}+ log{x_{2}}+ ….+log{x_{n}}|

Solve the following :

Find all positive real numbers x, x_{1}, x_{2}, \ldots, x_{n} such that

|\log{xx_{1}}|+|\log{xx_{2}}| + \ldots + |\log{xx_{n}}| + |\log{\frac{x}{x_{1}}}| + |\log{\frac{x}{x_{2}}}| + \ldots + |\log{\frac{x}{x_{n}}}|= |\log{x_{1}}+ \log{x_{2}}+\log{x_{3}}+ \ldots + \log{x_{n}}|
…let us say this is given equality A

Solution:

Use the following inequality: |a-b| \leq |a| + |b| with equality iff ab \leq 0

So, we observe that : |\log{xx_{1}}|+|\log{\frac{x}{x_{1}}}| \geq |\log{xx_{1}}-\log{\frac{x}{x_{1}}}| = |\log{x_{1}^{2}}|=2 |\log{x_{1}}|,

Hence, LHS of the given equality is greater than or equal to:

2(|\log{x_{1}}|+|\log{x_{2}}|+|\log{x_{3}}|+ \ldots + |\log{x_{n}}|)

Now, let us consider the RHS of the given equality A:

we have to use the following standard result:

|\pm a_{} \pm a_{2} \pm a_{3} \ldots \pm a_{n}| \leq |a_{1}|+|a_{2}| + \ldots + |a_{n}|

So, applying the above to RHS of A:

|\log{x_{1}}+\log{x_{2}}+\ldots + \log{x_{n}}| \leq |\log{x_{1}}|+|\log{x_{2}}|+\ldots + |\log{x_{n}}|.

But, RHS is equal to LHS as given in A:

That is, |\log{xx_{1}}|+|\log{xx_{2}}|+ \ldots + |\log{xx_{n}}| +|\log{\frac{x}{x_{1}}}|+|\log{\frac{x}{x_{2}}}|+ \ldots + |\log{\frac{x}{x_{n}}}| \leq |\log{x_{1}}|+|\log{x_{2}}|+ \ldots + |\log{x_{n}}|

Now, just a few steps before we proved that LHS is also greater than or equal to : That is,

|\log{xx_{1}}|+|\log{xx_{2}}|+\ldots + |\log{xx_{n}}|+ |\log{\frac{x}{x_{1}}}|+|\log{\frac{x}{x_{2}}}| + \ldots + |\log{\frac{x}{x_{n}}}| \geq 2(|\log{x_{1}}|+|\log{x_{2}}|+\ldots + |\log{x_{n}}|)

The above two inequalities are like the following: x \leq y and x \geq 2y; so what is the conclusion? The first inequality means x2y or x=2y; clearly it means the only valid solution is x=2y.

Using the above brief result, we have here:

|\log{x_{1}}|+|\log{x_{2}}|+ \ldots +|\log{x_{n}}| =2(|\log{x_{1}}|+|\log{x_{2}}|+ \ldots + |\log{x_{n}}|)

Hence, we get |\log{x_{1}}|+|\log{x_{2}}|+ \ldots + |\log{x_{n}}|=0, which in turn means that (by applying the definition of absolute value):

|\log{x_{1}}|=|\log{x_{2}}|= \ldots =|\log{x_{n}}|, which implies that x_{1}=x_{2}= \ldots  x_{n}=1.

Substituting these values in the given logarithmic absolute value equation, we get:

n \times |\log{x}|+ n \times |\log{x}|=0, that is 2n \times |\log{x}|=0, and as n \neq 0, this implies that |\log{x}|=0 which in turn means x=1 also.

Every function can be written as a sum of an even and an odd function

Let f(x) be any well-defined function.

We want to express it as a sum of an even function and an odd function.

Let us define two other functions as follows:

F(x) = \frac{f(x)+f(-x)}{2} and G(x)=\frac{f(x)-f(-x)}{2}.

Claim I: F(x) is an even function.

Proof I; Since by definition F(x)= \frac{f(x)+f(-x)}{2}, so F(-x) = \frac{f(-x) +f(-(-x))}{2}=\frac{f(-x)+f(x)}{2} \Longrightarrow F(x) = F(-x) so that F(x) is indeed an even function.

Claim 2: G(x) is an odd function.

Proof 2: Since by definition G(x) = \frac{f(x)-f(-x)}{2}, so G(-x) = \frac{f(-x)-f(-(-x))}{2} = \frac{f(-x)-f(x)}{2} = -\frac{f(x)-f(-x)}{2} = -G(-x) \Longrightarrow G(x) = -G(-x) so that G(x) is indeed an odd function.

Claim 3: f(x)= F(x) + G(x)

Proof 3: F(x)+ G(x) = \frac{f(x)+f(-x)}{2} + \frac{f(x)-f(-x)}{2} = \frac{f(x)}{2} + \frac{f(-x)}{2} + \frac{f(x)}{2} - \frac{f(-x)}{2} = f(x) indeed.

 

Method of undetermined coefficients for PreRMO, PRMO and IITJEE Foundation maths

  1. Find out when the expression x^{3}+px^{2}+qx+r is exactly divisible by x^{2}+ax+b

Solution 1:

Let x^{3}+px^{2}+qx+r=(x^{2}+ax+b)(Ax+B) where A and B are to be determined in terms of p, q, r, a and b. We can assume so because we know from the fundamental theorem of algebra that the if the LHS has to be of degree three in x, the remaining factor in RHS has to be linear in x.

So, expanding out the RHS of above, we get:

x^{3}+px^{2}+qx+r=Ax^{3}+aAx^{2}+bAx+Bx^{2}+Bax+bB

x^{3}+px^{3}+qx+r=Ax^{3}+(aA+B)x^{2}+x(bA+aB)+bB

We are saying that the above is true for all values of x: hence, coefficients of like powers of x on LHS and RHS are same; we equate them and get a system of equations:

A=1

p=aA+B

bA+aB=q

bB=r

Hence, we get p=a+\frac{r}{b} and bp-ba=r or that b(p-a)=r

Also, b+aB=q so that q=b+\frac{ar}{b} which means q-b=\frac{a}{b}r

but \frac{r}{b}=B=p-a and hence, q-b=\frac{a}{b}(p-a)

So, the required conditions are b(p-a)=r and q-b=\frac{a}{b}(p-a).

2) Find the condition that x^{2}+px+q may be a perfect square.

Solution 2:

Let x^{2}+px+q=(Ax+B)^{2} where A and B are to be determined in terms of p and q; finally, we obtain the relationship required between p and q for the above requirement.

x^{2}+px+q=A^{2}x^{2}+B^{2}+2ABx which is true for all real values of x;

Hence, A^{2}=1 so A=1 or A=-1

Also, B^{2}=q and hence, B=\sqrt{q} or B=-\sqrt{q}

Also, 2AB=p so that 2\sqrt{q}=p so q=\frac{p^{2}}{4}, which is the required condition.

3) To prove that x^{4}+px^{3}+qx^{2}+rx+s is a perfect square if (q-\frac{p^{2}}{4})^{2}=4s and r^{2}=p^{2}s.

Proof 3:

Let x^{4}+px^{3}+qx^{2}+rx+s=(Ax^{2}+Bx+C)^{2}

x^{4}+px^{3}+qx^{2}+rx+s=A^{2}x^{4}+B^{2}x^{2}+C^{2}+2ABx^{3}+2BCx+2ACx^{2}

A^{2}=1

2AB=p

q=B^{2}+2AC

2BC=r

C^{2}=s

A=1 or A=-1

2AB=p \longrightarrow 2B=p \longrightarrow B=\frac{p}{2}

q=B^{2}+2AC=\frac{p^{2}}{4}+2\times \sqrt{s} \longrightarrow (q-\frac{p^{2}}{4})^{2}=4s

2 \times \frac{p}{2} \times \sqrt{s}=r \longrightarrow r^{2}=p^{2}s

More later,

Nalin Pithwa.

PS: Note in the method of undetermined coefficients, we create an identity expression which is true for all real values of x.

Miscellaneous Algebra: pRMO, IITJEE foundation maths 2019

For the following tutorial problems, it helps to know/remember/understand/apply the following identities (in addition to all other standard/famous identities you learn in high school maths):

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

By the way, I hope you also know how to derive the above.Let me mention two methods to derive the above :

Method I: Using polynomial division in three variable, divide the dividend a^{3}+b^{3}+c^{3}-3abc by the divisor a+b+c.

Method II: Assume that P(X) is a polynomial with roots a, b and c. So, we know by the fundamental theorem of algebra that P(X)=(X-a)(X-b)(X-c). Now, we also know that a, b and c satisfy P(X). Now, proceed further and complete the proof.

Let us now work on the tutorial problems below:

1) If 2s=a+b+c, prove that \frac{1}{s-a} + \frac{1}{s-b} + \frac{1}{s-c} = \frac{abc}{s(s-a)(s-b)(s-c)}

2) If x^{2}+a^{2}=2(xy+yz+zu-y^{2}-z^{2}), prove that x=y=z=u.

Prove the following identities:

3) b(x^{3}+a^{3})+ax(x^{2}-a^{2})+a^{3}(x+a)=(a+b)(x+a)(x^{2}-ax+a^{2})

4) (ax+by)^{2}+(ay-bx)^{2}+c^{2}x^{2}+c^{2}y^{2}=(x^{2}+y^{2})(a^{2}+b^{2}+c^{2})

5) (x+y)^{3}+ 3(x+y)^{2}z+3(x+y)z^{2}+z^{3}=(x+z)^{3}+3(x+z)^{2}y+3(x+z)y^{2}+y^{3}

6) (a+b+c)(ab+bc+ca)-abc=(a+b)(b+c)(c+a)

7) (a+b+c)^{2}-a(b+c-a)-b(a+c-b)-c(a+b-c)=2(a^{2}+b^{2}+c^{2})

8) (x-y)^{3}+(x+y)^{3}+3(x-y)^{2}(x+y)+3(x+y)^{2}(x-y)=8x^{3}

9) x^{2}(y-z)+y^{2}(z-x)+z^{2}(x-y)+(y-z)(z-x)(z-y)=0

10) a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)=-(b-c)(c-a)(a-b)(a+b+c)

11) Prove that (b-c)^{3}+(c-a)^{3}+(a-b)^{3}=3(b-c)(c-a)(a-b)

12) If3 2s=a+b+c, prove that (s-a)^{2}+(s-b)^{2}+(s-c)^{2}+s^{2}=a^{2}+b^{2}+c^{2}

13) If 2s=a+b+c, prove that (s-a)^{3}+(s-b)^{3}+(s-c)^{3}+3abc=s^{3}

14) If 2s=a+b+c, prove that 16s(s-a)(s-b)(s-c)=2b^{2}c^{2}+2c^{2}a^{2}+2a^{2}b^{2}-a^{4}-b^{4}-c^{4}

15) If   2s=a+b+c, then prove that  2(s-a)(s-b)(s-c)+a(s-b)(s-c)+b(s-c)(s-a)+c(s-a)(s-b)=abc

16) If a+b+c=0, then prove that (2a-b)^{3}+(2b-c)^{3}+(2c-a)^{3}=3(2a-b)(2b-c)(2c-a)

17) If a+b+c=0, then prove that \frac{a^{2}}{2a^{2}+bc} + \frac{b^{2}}{2b^{2}+ca} + \frac{c^{2}}{2c^{2}+ab} =1

18) Prove that (x+y+z)^{3}+(x+y-z)^{3}+(x-y+z)^{3}+(x-y-z)^{3}=4x(x^{2}+3y^{2}+3z^{2})

19) If a+b+c=0 prove that (s+3a)^{3}-(s-3b)^{3}-(s-3c)^{3}-3(s-3a)(s-3b)(s-3c)=0

20) If X=b+c-2a, Y=c+a-2b, Z=a+b-2c, find the value of X^{2}+Y^{2}+Z^{2}-3XYZ

21) Prove that (a-b)^{2}+(b-c)^{2}+(c-a)^{2}=2(c-b)(c-a)+2(b-a)(b-c)+2(a-b)(a-c)

22) Prove that a^{2}(b^{3}-c^{3})+b^{2}(c^{3}-a^{3})+c^{2}(a^{3}-b^{3})=(a-b)(b-c)(c-a)(ab+bc+ca)=a^{2}(b-c)^{3}+b^{2}(c-a)^{3}+c^{2}(a-b)^{3} = -[a^{2}b^{2}(a-b)+b^{2}c^{2}(b-c)+c^{2}a^{2}(c-a)]

23) if (a+b)^{2}+(b+c)^{2}+(c+a)^{2}=4(ab+bc+cd), prove that a=b=c=d.

24) If x=a+d, y=b+d, z=c+d, prove that x^{2}+y^{2}+z^{2}-yz-zx-xy=a^{2}+b^{2}+c^{2}-bc-ca-ab

25) If a+b+c=3, prove that \frac{1}{b^{2}+c^{2}-a^{2}}+ \frac{1}{c^{2}+a^{2}-b^{2}} + \frac{1}{a^{2}+b^{2}-c^{2}}=0

26) If a+b+c=0, simplify: \frac{b+c}{bc}(b^{2}+c^{2}-a^{2}) + \frac{c+a}{ca} (c^{2}+a^{2}-b^{2})+ \frac{a+b}{ab}(a^{2}+b^{2}-c^{2})

27) Prove that the equation (x-a)^{2}+(y-b)^{2}+(a^{2}+b^{2}-1)(x^{2}+y^{2}-1)=0 is equivalent to the equation (ax+by-1)^{2}+(bx-ay)^{2}=0, hence show that the only possible values of x and y are: \frac{a}{a^{2}+b^{2}}, \frac{b}{a^{2}+b^{2}}

28) If 2(x^{2}+a^{2}-ax)(y^{2}+b^{2}-by)=x^{2}y^{2}+a^{2}b^{2}, prove that (x-a)^{2}(y-b)^{2}+(bx-ay)^{2}=0 and therefore that a=x and y=b are the only possible solutions.

Good luck for the PreRMo August 2019 !!

Regards,

Nalin Pithwa

 

Cyclic expressions, fractions: Pre RMO, PRMO, IITJEE foundation 2019

In order to solve the following tutorial sheet, it helps to solve/understand and then apply the following beautiful cyclic relations or identities:

(Note if these look new to you, then you need to check the truth of all them; if all are v v familiar to you, just go ahead and crack the tutorial sheet below):

Core Identities in Cyclic Expressions:
1) (b-c)+(c-a)+(a-b)=0
2) a(b-c)+b(c-a)+c(a-b)=0
3) a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)=-(a-b)(b-c)(c-a)
4) bc(b-c)+ca(c-a)+ab(a-b)=-(a-b)(b-c)(c-a)
5) a(b^{2}-c^{2})+b(c^{2}-a^{2})+c(a^{2}-b^{2})=(a-b)(b-c)(c-a)

Solve or simplify the following:

1) \frac{a}{(a-b)(a-c)} + \frac{b}{(b-c)(b-a)} + \frac{c}{(c-a)(c-b)}
2) \frac{bc}{(a-b)(a-c)} + \frac{ca}{(b-c)(b-a)} + \frac{ab}{(c-a)(c-b)}
3) \frac{a^{2}}{(a-b)(a-c)} + \frac{b^{2}}{(b-c)(b-a)} + \frac{c^{2}}{(c-a)(c-b)}
4) \frac{a^{3}}{(a-b)(a-c)} + \frac{b^{3}}{(b-c)(b-a)} + \frac{c^{3}}{(c-a)(c-b)}
5) \frac{a(b+c)}{(a-b)(c-a)} + \frac{b(a+c)}{(a-b)(b-c)} + \frac{a(a+b)}{(c-a)(b-c)}
6) \frac{1}{a(a-b)(a-c)} + \frac{1}{b(b-c)(b-a)} + \frac{1}{c(c-a)(c-b)}
7) \frac{bc}{a(a^{2}-b^{2})(a^{2}-c^{2})} + \frac{ca}{b(b^{2}-c^{2})(b^{2}-a^{2})} + \frac{ab}{c(c^{2}-a^{2})(c^{2}-b^{2})}
8) \frac{(x-b)(x-c)}{(a-b)(a-c)} + \frac{(x-c)(x-a)}{(b-c)(b-a)} + \frac{(x-a)(x-b)}{(c-a)(c-b)}
9) \frac{bc(a+d)}{(a-b)(a-c)} + \frac{ca(b+d)}{(b-c)(b-a)} + \frac{ab(c+d)}{(c-a)(c-b)}
10) \frac{1}{(a-b)(a-c)(x-a)} + \frac{1}{(b-c)(b-a)(x-b)} + \frac{1}{(c-a)(c-b)(x-c)}
11) \frac{a^{2}}{(a-b)(a-c)(x+a)} + \frac{b^{2}}{(b-c)(b-a)(x+b)} + \frac{c^{2}}{(c-a)(c-b)(x+c)}
12) a^{2}\frac{(a+b)(a+c)}{(a-b)(a-c)} + b^{2}\frac{(b+c)(b+a)}{(b-c)(b-a)} + c^{2}\frac{(c+a)(c+b)}{(c-a)(c-b)}
13) \frac{a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)}{(b-c)^{3}+(c-a)^{3}+(a-b)^{3}}
14) \frac{a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)+2(c-a)(a-b)(b-c)}{(b-c)^{3}+(c-a)^{3}+(a-b)^{3}}
15) \frac{a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)}{a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)}
16) \frac{a^{2}(b-c)^{3}+b^{2}(c-a)^{3}+c^{2}(a-b)^{3}}{(a-b)(b-c)(c-a)}
17) \frac{\frac{b-c}{a} + \frac{c-a}{b} + \frac{a-b}{c}}{\frac{1}{a}(\frac{1}{b^{2}}-\frac{1}{c^{2}})+\frac{1}{b}(\frac{1}{c^{2}}-\frac{1}{a^{2}})+\frac{1}{c}(\frac{1}{a^{2}}-\frac{1}{b^{2}})}^
18) \frac{a^{2}(\frac{1}{a^{2}}-\frac{1}{b^{2}})+b^{2}(\frac{1}{a^{2}}-\frac{1}{c^{2}})+c^{2}(\frac{1}{b^{2}}-\frac{1}{a^{2}})}{\frac{1}{bc}(\frac{1}{c}-\frac{1}{b})+\frac{1}{ca}(\frac{1}{a}-\frac{1}{c})+\frac{1}{ab}(\frac{1}{b}-\frac{1}{c})}
19) \frac{a}{(a-b)(a-c)(x-a)} + \frac{b}{(b-c)(b-a)(x-b)} + \frac{c}{(c-a)(c-b)(x-c)}

More later,
Nalin Pithwa