# A fifth degree equation in two variables: a clever solution

Question:

Verify the identity: $(2xy+(x^{2}-2y^{2}))^{5}+(2xy-(x^{2}-2y^{2}))^{5}=(2xy+(x^{2}+2y^{2})i)^{5}+(2xy-(x^{2}+2y^{2})i)^{5}$

let us observe first that each of the fifth degree expression is just a quadratic in two variables x and y. Let us say the above identity to be verified is:

$P_{1}+P_{2}=P_{3}+P_{4}$

Method I:

Use binomial expansion. It is a very longish tedious method.

Method II:

Factorize each of the quadratic expressions $P_{1}, P_{2}, P_{3}, P_{4}$ using quadratic formula method (what is known in India as Sridhar Acharya’s method):

Now fill in the above details.

You will conclude very happily that :

The above identity is transformed to :

$P_{1}=(x+y+\sqrt{3}y)^{5}(x+y-\sqrt{3}y)^{5}$

$P_{2}=(-1)^{5}(x-y-\sqrt{3}y)^{5}(x-y+\sqrt{3}y)^{5}$

$P_{3}=(i^{2}(x-y-\sqrt{3}y)(x-y+\sqrt{3}y))^{5}$

$P_{4}=((-i^{2})(x+y+\sqrt{3}y)(x-y-\sqrt{3}y))^{5}$

You will find that $P_{1}=P_{4}$ and $P_{2}=P_{4}$

Hence, it is verified that the given identity $P_{1}+P_{2}=P_{3}+P_{4}$. QED.

Regards,
Nalin Pithwa.

# A not so easy quadratic equation problem for RMO or IITJEE maths

Question:

Suppose that we are given a monic quadratic polynomial $p(t)$. Prove that for any integer n, there exists an integer k such that $p(n)p(n+1)=p(k)$.

🙂 🙂 🙂

Solution:

Let $p(t)=t^{2}+bt+c$ and which implies $p(t+1)=(t+1)^{2}+b(t+1)+c$. We want $p(n)p(n+1)=p(k)$. By trial and error, we get $k=t(t+1)+bt+c$.

By the way, we could have gotten the same solution by method of undetermined coefficients. But that would also need intelligent guess-works.

Nalin Pithwa

PS: I will post the solution after some time. Meanwhile, please try.

# 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