# Search results

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...

16. ### Challenge Math Challenge - June 2021

I did realise my mistake, thank you for your help and also for showing how to work backwards from ##\frac{\binom{2n}{n}}{2^{2n}} < \frac{1}{\sqrt{2n+1}} ## to ## \frac{\binom{2n+1}{n}}{2^{2n+1}} < \frac{1}{\sqrt{2n+3}}##
17. ### Challenge Math Challenge - June 2021

How do you consistently come up with an example that I overlook? 😅
18. ### Challenge Math Challenge - June 2021

Now this looks quiet confusing to read as a third person who doesn't know whats going in my head, but to me it makes perfect sense! Still if someone wants to read that, then the case ##e<f## was in the original attempt (see post #21 of this thread) ##\dfrac{\binom {2n} n}{2^{2n}}<...
19. ### Challenge Math Challenge - June 2021

I think I figured out what was wrong (refer to my first attempt of problem 15) I wanted to show that $$b<c$$ I knew that $$e<f$$ is true. I assumed that ##b<c## is also true First I showed that ##a<b## and then I showed ##c<d## so I concluded that $$a<d$$ Now I did the same process with...
20. ### Challenge Math Challenge - June 2021

On rereading, I get this feeling that may be I'm doing something wrong here, but I'm confused as we are dealing with inequalities here, if we had been dealing with equalities then everything you said makes sense to me.
21. ### Challenge Math Challenge - June 2021

I do agree with what you said here, I also agree with your solution which was discussed earlier in this thread, but I think that there is another method of doing this problem which is with the use of orders ##a<b<c<d## which I mentioned above. But this statement is confusing for me. If the...
22. ### Challenge Math Challenge - June 2021

Hey @julian, Thanks for your reply, but I still have the same question, you said that, \begin{align*} \frac{\binom{2n+1}{n}}{2^{2n+1}} < \frac{1}{\sqrt{2n+3}} \qquad (*) \end{align*} is correct But as we have, \begin{align*} \binom{2n}{n} &< \binom{2n+1}{n}\\ \dfrac{\binom {2n}...
23. ### Challenge Math Challenge - June 2021

Problem 12 Let ##\sqrt{n}+\sqrt{n+4}=p## such that ##p\in\mathbb{Q}## So we get, \begin{align} \sqrt{n}+\sqrt{n+4}&=p\nonumber\\ n+n+4+2\sqrt{n}\cdot\sqrt{n+4}&=p^2\nonumber\\ \sqrt{n}\cdot\sqrt{n+4}=\frac{p^2-2n-4}{2}&=q\space\text{(say)}\nonumber \end{align} We can see that...
24. ### Challenge Math Challenge - June 2021

Sorry, I forgot that, For ##n=1## \begin{align*} \frac{\binom{2n}{n}}{2^{2n}} < \frac{1}{\sqrt{2n+1}}\\ \frac{\binom{2}{1}}{2^{2}} < \frac{1}{\sqrt{2+1}}\\ \frac{2}{4}<\frac{1}{\sqrt{2+1}}\\ \frac1 2<\frac{1}{\sqrt3}\\ 2>\sqrt3 \end{align*}
25. ### Challenge Math Challenge - June 2021

I shouldn't write ##c^2=(k-2)(k+2)## I should have directly wrote that from ##c^2+2^2=k^2## the only solution we get is ##c=0## and ##k=2## (as there are no pythagorean triplets with 2), I was thinking something else when I wrote ##c^2=(k-2)(k+2)##. I don't like that solution anyway, maybe I...
26. ### Challenge Math Challenge - June 2021

I though it was like saying if$$a<b$$ And $$b<c$$ Then$$a<c$$

30. ### Challenge Math Challenge - June 2021

Are you talking something like this, \begin{align*} \frac{\binom{2n+1}{n}}{2^{2n+1}} < \frac{1}{\sqrt{2n+3}}\\ \frac{\binom{2n+1}{n+1}}{2^{2n+1}} < \frac{1}{\sqrt{2n+3}}\\ \frac{(2n+1)\cdot \binom{2n}{n}}{(n+1)\cdot 2^{2n+1}} < \frac{1}{\sqrt{2n+3}} \end{align*} But as, \begin{align*} 1<\frac...