# A quadratic equation question for pRMO or preRMO

Question:

Find the necessary and sufficient condition that the quadratic equation $ax^{2}+bx+c=0$ where $a \neq 0$ has one root which is the square of the other.

Solution:

Let the two roots of the given quadratic equation $ax^{2}+bx+c=0$, with $a \neq 0$ be $\alpha$ and $\beta$ such that $\beta = \alpha^{2}$.

Then, we know $\alpha+\beta=-\frac{b}{a}$ and $\alpha\beta=\frac{c}{a}$ so that $\alpha+\alpha^{2}=-\frac{b}{a}$ and $\alpha^{3}=\frac{c}{a}$. From the latter relation, we get that $\alpha = (\frac{c}{a})^{\frac{1}{3}}$. Substituting this in the first relation of sum of roots, we get the following necessary and sufficient condition:

$(\frac{c}{a})^{\frac{1}{3}} + (\frac{c}{a})^{\frac{2}{3}} = -\frac{b}{a}$.

The above is the desired solution.

🙂 🙂 🙂

Nalin Pithwa

# A quadratic and trigonometry combo question: RMO and IITJEE maths coaching

Question:

Given that $\tan {A}$ and $\tan {B}$ are the roots of the quadratic equation $x^{2}+px+q=0$, find the value of

$\sin^{2}{(A+B)}+ p \sin{(A+B)}\cos{(A+B)} + q\cos^{2}{(A+B)}$

Solution:

Let $\alpha=\tan{A}$ and $\beta=\tan{B}$ be the two roots of the given quadratic equation: $x^{2}+px+q=0$

By Viete’s relations between roots and coefficients:

$\alpha+\beta=\tan{A}+\tan{B}=-p$ and $\alpha \beta = \tan{A}\tan{B}=q$ but we also know that $\tan{(A+B)}=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}=\frac{-p}{1-q}=\frac{p}{q-1}$

Now, let us call $E=\sin^{2}{(A+B)}+p\sin{(A+B)\cos{(A+B)}}+\cos^{2}{(A+B)}$ which in turn is same as

$\cos^{2}{(A+B)}(\tan^{2}{(A+B)}+p\tan{(A+B)}+q)$

We have already determined $\tan{(A+B)}$ in terms of p and q above.

Now, again note that $\sin^{2}{\theta}+\cos^{2}{\theta}=1$ which in turn gives us that $\tan^{2}{\theta}+1=\sec^{2}{\theta}$ so we get:

$\sec^{2}{(A+B)}=1+\tan^{2}{(A+B)}=1+\frac{p^{2}}{(q-1)^{2}}=\frac{p^{2}+(q-1)^{2}}{(q-1)^{2}}$ so that

$\cos^{2}{(A+B)}=\frac{1}{\sec^{2}{(A+B)}}=\frac{(q-1)^{2}}{p^{2}+(q-1)^{2}}$

Hence, the given expression E becomes:

$(\frac{(q-1)^{2}}{p^{2}+(q-1)^{2}})(\frac{p^{2}}{(q-1)^{2}}+\frac{p^{2}}{q-1}+q)$, which is the desired solution.

🙂 🙂 🙂

Nalin Pithwa.

# preRMO or RMO algebra: my student’s solutions

(These are the solutions from a student of mine, whose identity is private. I will call him, RI, Bengaluru.)

Question I:

Solve as elegantly as possible: $t^{4}=49+20\sqrt{6}$

Solution I (of RI, Bengaluru):

$t^{4}=49+ 2\times 5 \times \sqrt{24}$

$t^{4}=(5+\sqrt{24})^{2}$

$t^{2}=5+\sqrt{24}$

$t^{2}=5 + 2\sqrt{6}$

Hence, $t=\sqrt{2}+\sqrt{3}$.

Question II:

Find the necessary and sufficient conditions on the coefficients p, q, and r of the given cubic equation such that the roots of the cubic are in AP:

$P(t)=t^{3}+pt^{2}+qt+r$

Solution II: (credit to RI, Bengaluru) :

Let the roots of the above cubic be $a-R, a, a+R$.

By Viete’s relations: $(a-R)+a +(a+R)=-p$ and $a(a-R)+(a-R)(a+R)+(a+R)(a)=q$ and $a(a-R)(a+R)=r$ so that we get $3a=p$ so that $a=\frac{p}{3}$ and from the second equation we get $a^{2}-aR+a^{2}-R^{2}+a^{2}+Ra=q$, that is, $3a^{2}-R^{2}=q$ so that $a^{2}-R^{2}=q-2\times a^{2} = q - 2 \times (\frac{p}{3})^{2}=q-\frac{2}{9}\times p^{2}$, and exploiting the third Viete’s relation we get $a \times (a^{2}-R^{2})=r$, that is $(q - (\frac{2}{9})\times p^{2})a=r$, which is the required necessary and sufficient condition, where $a=\frac{p}{3}$.

Method II: For pedagogical purposes. The above solution to question 2 was quick and elegant because of the right choice of three quantities in AP as $a-R, a, a+R$. Do you want to know how ugly and messy it can get if a standard assumption is made:

Let the roots of the cubic be $\alpha, \beta, \gamma$. Let d be the common difference. So that the three roots in AP are $\alpha$, $\beta=\alpha+d$ and $\gamma=\alpha + 2d$

Then, applying Viete’s relations, we get $\alpha + \beta + \gamma = -p$ so that $3\alpha + 2d=-p$ and $\alpha\beta+ \beta\gamma + \gamma\alpha=q$ which changes to $\alpha \times (\alpha + d) + (\alpha+d)(\alpha+2d)+ (\alpha+d)(\alpha+2d) = q$ and the third viete’s relation gives us $\alpha (\alpha+d)(\alpha+2d)=r$.

The second Viete’s relation is a quadratic in $\alpha$ and the third Viete’s relation is a cubic in $\alpha$. This is how messy it can get…at least, you will agree RI’s judicious choice has rendered a clean, quick solution.

Regards,

Nalin Pithwa.

# Ratio and proportion: practice problems: set II: pRMO, preRMO or IITJEE foundation maths

Problem 1:

If $\frac{y+z}{pb+qc} = \frac{z+x}{pc+qa} = \frac{x+y}{pa+qb}$, then show that $\frac{2(x+y+z)}{a+b+c} = \frac{(b+c)x+(c+a)y+(a+b)z}{bc+ca+ab}$

Problem 2:

If $\frac{x}{a} = \frac{y}{b} = \frac{z}{b}$, show that $\frac{x^{3}+a^{3}}{x^{2}+a^{2}} +\frac{y^{3}+b^{3}}{y^{2}+b^{2}} + \frac{z^{3}+c^{3}}{z^{2}+c^{2}} = \frac{(x+y+z)^{3}+(a+b+c)^{3}}{(x+y+z)^{2}+(a+b+c)^{2}}$

Problem 3:

If $\frac{2y+2z-x}{a} = \frac{2z+2x-y}{b} = \frac{2x+2y-z}{c}$, show that $\frac{x}{2b+2c-a} = \frac{y}{2c+2a-b} = \frac{z}{2a+2b-c}$

Problem 4:

If $(a^{2}+b^{2}+c^{2})(x^{2}+y^{2}+z^{2}) = (ax+by+cz)^{2}$, prove that $x:a = y:b = z:c$

Problem 5:

If $l(my+nz-lx) = m(nz+lx-my) = n(lx+my-nz)$, prove that $\frac{y+z-x}{l} = \frac{z+x-y}{m} = \frac{x+y-z}{n}$

Problem 6:

Show that the eliminant of

$ax+cy+bz=0$

$cx+by+az=0$

$bx+ay+cz=0$

is $a^{3}+b^{3}+c^{3}-3abc=0$

Problem 7:

Eliminate x, y, z from the equations:

$ax+hy+gz=0$

$hx+by+fz=0$

$gx+fy+cz=0$.

This has significance in co-ordinate geometry. (related to conics).

Problem 8:

If $x=cy+bz$, $y=az+cx$, $z=bx+cy$, show that $\frac{x^{2}}{1-a^{2}} = \frac{y^{2}}{1-b^{2}} = \frac{z^{2}}{1-c^{2}}$.

Problem 9:

Given that $a(y+z)=x$, $b(z+x)=y$, $c(x+y)=z$, prove that $bc+ab+ca+2abc=1$

Problem 10:

Solve the following system of equations:

$3x-4y+7z=0$

$2x-y-2z=0$

$3x^{3}-y^{3}+z^{3}=18$

Problem 11:

Solve the following system of equations:

$x+y=z$

$3x-2y+17z=0$

$x^{3}+3y^{3}+2z^{3}=167$

Problem 12:

Solve the following system of equations:

$7yz+3zx=4xy$

$21yz-3zx=4xy$

$x+2y+3z=19$

Problem 13:

Solve the following system of equations:

$3x^{2}-2y^{2}+5z^{2}=0$

$7x^{2}-3y^{2}-15z^{2}=0$

$5x-4y+7z=0$

Problem 14:

If $\frac{l}{\sqrt{a}-\sqrt{b}} + \frac{m}{\sqrt{b}-\sqrt{c}} + \frac{n}{\sqrt{c}-\sqrt{a}} =0$,

and $\frac{l}{\sqrt{a}+\sqrt{b}} + \frac{m}{\sqrt{b}+\sqrt{c}} + \frac{n}{\sqrt{c}+\sqrt{c}} = 0$,

prove that $\frac{l}{(a-b)(c-\sqrt{ab})} = \frac{m}{(b-c)(a-\sqrt{ab})} = \frac{n}{(c-a)(b-\sqrt{ac})}$

Problem 15:

Solve the following system of equations:

$ax+by+cz=0$

$bcx+cay+abz=0$

$xyz+abc(a^{3}x+b^{3}y+c^{3}z)=0$

Cheers,

Nalin Pithwa

# Ratio and proportion practice problems: set I: preRMO, pRMO or IITJEE foundation maths

Solve the following equations:

1. $\frac{2x^{3}-3x^{2}+x+1}{2x^{3}-3x^{2}-x-1} = \frac{3x^{3}-x^{2}+5x-13}{3x^{3}-x^{2}-5x+13}$
2. $\frac{3x^{4}+x^{2}-2x-3}{3x^{4}-x^{2}+2x+3} = \frac{5x^{4}+2x^{2}-7x+3}{5x^{4}-2x^{2}+7x-3}$
3. $\frac{(m+n)x-(a-b)}{(m-n)x-(a+b)} = \frac{(m+n)x+a+c}{(m-n)x+a-c}$
4. If a, b, c, d are proportionals, prove that $a+d=b+c+\frac{(a-b)(a-c)}{a}$
5. If a, b, c, d, e are in continued proportion, prove that $(ab+bc+cd+de)^{2}=(a^{2}+b^{2}+c^{2}+d^{2})(b^{2}+c^{2}+d^{2}+e^{2})$
6. If the work done by $(x-1)$ men in $(x+1)$ days is to the work done by $(x+2)$ men in $(x-1)$ days in the ratio of $9:10$, find x.
7. Find four proportionals such that the sum of the extremes is 21, the sum of the means 19, and the sum of the squares of all four numbers is 442.
8. Two casks A and B were filled with two kinds of sherry, mixed in the cask A in the ratio of 2:7, and in the cask B in the ratio of 1:5. What quantity must be taken from each to form a mixture which shall consist of 10 litres of one kind and 45 litres of the other?
9. Nine litres are drawn from a vessel full of wine; it is then filled with water, then 9 litres of the mixture are drawn, and the vessel is again filled with water. If the quantity of wine now in the vessel to the quantity of water in it as 16 to 9, how much does the vessel hold?
10. If four positive quantities, are in continued proportion, show that the difference between the first and last is at least three times as great as the difference between the other two.
11. In a certain country, the consumption of tea is five times the consumption of coffee. If a percent more tea and b percent more coffee were consumed, the aggregate amount consumed would be 7c percent more; but if b percent more tea and a percent more coffee were consumed, the aggregate amount consumed would be 3c percent more; compare a and b.
12. Brass is an alloy of copper and zinc; bronze is an alloy containing 80 percent of copper, 4 percent of zinc, and 16 percent of tin. A fused mass of brass and bronze is found to contain 74 percent of copper, 16 percent of zinc and 10 percent of tin; find the ratio of copper to zinc in the composition of brass.
13. A crew can row a certain course up stream in 84 minutes; they can row the same course down stream in 9 minutes less than they could row it in still water; how long would they take to row down with the stream?

Cheers,

Nalin Pithwa

# 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.