# Recent content by kshitij

1. ### Challenge Math Challenge - July 2021

I didn't knew there were multiple winning strategies for ##P##! Also, couldn't we generalise that further for when ##P## chooses any value of ##c## in the first move, then the condition for $$f(1)<0$$ would change to $$a+b+c+1<0$$ So if ##Q## chooses a value of ##a## then ##P## would choose...
2. ### Challenge Math Challenge - July 2021

When I first used ##A.M \geq G.M## on$$f(x,y,z)=c(c-2x)(c-2y)(c-2z)$$ the equality would never hold because if $$c=c-2x=c-2y=c-2z$$ then we would get $$x=y=z=0$$ which was not possible
3. ### Challenge Math Challenge - July 2021

So, basically as I understand it, we had $$f(x,y,z)=c(c-2x)(c-2y)(c-2z)$$ And ##c## is a constant, so we use ##A.M \geq G.M## on $$g(x,y,z)=(c-2x)(c-2y)(c-2z)$$ So, we get, \begin{align*} \dfrac{(c-2x)+(c-2y)+(c-2z)}{3} &\geq \left((c-2x)(c-2y)(c-2z)\right)^\frac1 3\\ \dfrac{3c-2c}{3} &\geq...
4. ### Challenge Math Challenge - July 2021

Also I forgot that ##x,y,z \gt 0## so ##x=y=z=0## cannot be correct anyway. I will try again this problem later.
5. ### Challenge Math Challenge - July 2021

But if ##f(x,y,z)## is area of a triangle then ##f(x,y,z)=0## is the least possible value not the maximum value, surely this is wrong!
6. ### Challenge Math Challenge - July 2021

Are you saying that I must prove why the equality holds? or do you want me to show when the equality holds? As I said that for ##a_1,a_2, ... ,a_n##, ##A.M=G.M## when ##a_1=a_2= ... =a_n## So in our case the equality should hold if $$x+y+z=x+y-z=x-y+z=y+z-x$$ But this equality will only hold...
7. ### Challenge Math Challenge - July 2021

I think I have just a strategy but I am not sure because lets say that P plays according to my strategy and chooses a negative value of ##b## say ##b=-1## but then after Q chose a value of ##c## then I don't know what value of ##a## P should choose, I know (more like I feel) that there are some...
8. ### Challenge Math Challenge - July 2021

I just noticed that $$c(c-2x)(c-2y)(c-2z)$$ Looks similar to the formula for area of a triangle! Is that what you meant, is geometrical meaning of ##f(x,y,z)## is that is 16 times the square of the area of a triangle with sides ##x,y,z##! because area of a triangle with sides ##x,y,z## will be...
9. ### Challenge Math Challenge - July 2021

I don't know about why is that maximum, if equality holds then they should be maximum. I think I don't understand what you're trying to say? Also even if ##x,y,z \gt 0##, ##x+y-z## can still be less than zero and we cannot use A.M G.M inequality the way I did, if ##x,y,z## were sides of a...
10. ### Challenge Math Challenge - July 2021

I don't understand what you're saying? How can the equation ##y=x^3+c## have three different real roots for any value of ##c##?
11. ### Challenge Math Challenge - July 2021

I might be wrong, but I thought that when ##a=0## then, if the value of ##b## is chosen to be positive, then the equation ##f(x)=x^3+ax^2+bx+c=0## should have only one solution because we can see that ##f'(x)=3x^2+2ax+b## Now to have two different values of ##x## where ##f'(x)=0## we should have...
12. ### Challenge Math Challenge - July 2021

But I applied ##A.M \geq G.M## on $$(z-x+y)(z+x-y)(x+y-z)(x+y+z)$$ ##(z-x+y),(z+x-y),(x+y-z)## and ##(x+y+z)## aren't necessarily positive right? Also to answer your questions in post #12, I have no idea about the geometric meaning of ##A.M \geq G.M## All I knew was that if we had ##n##...
13. ### Challenge Math Challenge - July 2021

You can ignore this attempt, as the question doesn't say that ##x,y,z\gt 0## so we cannot use ##A.M \geq G.M##
14. ### Challenge Math Challenge - July 2021

Do you want me to derive the equation of director circle? Let ##y=mx+c## be a tangent to the ellipse ##\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1##, so on solving these two we get, \begin{align*} b^2x^2+a^2(mx+c)^2&=a^2b^2\\ (b^2+a^2m^2)x^2+(2a^2mc)x+a^2(c^2-b^2)&=0 \end{align*} For ##y=mx+c## to be a...