Take it to the max
Suppose 2 x 2 + y 2 + z 2 = 6 y − 4 x + 2 x z − 1 3 , where x , y , z ∈ R .
Find the maximum possible value of y + x z .
The answer is 7.00.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
3 + ( − 2 ) ( − 2 ) = 3 + 4 = 7
Log in to reply
Haha. Yeah, apparently I need to get some sleep now. :) I just wanted to get my 100th question posted so I came up with an easy one.
Log in to reply
Congratulations on your 1 0 0 t h Question Sir !!!
Log in to reply
@A Former Brilliant Member – Thanks, Azhaghu. I wonder who has posted the most questions on Brilliant, (other than Calvin Lin, of course)? Sandeep Bhardwaj has posted about 200; that's the most that I'm aware of.
Log in to reply
@Brian Charlesworth – Where's this count? I couldn't find mine on my profile. (On android app)
Log in to reply
@Pranjal Jain – Type in a name in the Search box and what results is a list of all the notes, problems, etc., associated with that person, (either created by them or mentioning their name). Off to the side is a counter for each class of entry, (problems, notes, etc.), from which one can then determine how many problems, etc., a person has created.
Log in to reply
@Brian Charlesworth – Sir, I suggest that you start working on a secret awesome question , The one question that'll be considered as the most difficult one on Brilliant when you post it . One, that even @Jon Haussmann sir won't be able to solve :) I'm sure everyone will like to have such question :)
Log in to reply
@A Former Brilliant Member – How about this: a "LVL 9" problem is not as hard as most people would make it out to be to create. However, here's a challenge: make a problem that requires the mathematical maturity/fundamental concept understanding of Sir Calvin Lin, but requires only the mathematical formula knowledge of a high schooler . Once you are capable of doing this, Imo you have the right to call yourself a problem solving master.
Log in to reply
@Trevor Arashiro – Whoa ! Seriously dude , are you referring to me ? I am definitely not a problem solving master . I am just addicted to solving questions :)
But your idea is good, how about we suggest Calvin Lin sir to think about increasing the levels till , umm 10 ? What do you think ?
Log in to reply
@A Former Brilliant Member – Lol, LVL 10? Like, that's beyond IMO lvl.
Log in to reply
@Trevor Arashiro – Yup! But there are people who can produce masterpieces worthy of L vl 10 , like the @Michael Mendrin sir, @Calvin Lin sir, @brian charlesworth sir and even your and @Finn Hulse 's questions have that feel to it .
Anyway, it was just a thought :)
Log in to reply
@A Former Brilliant Member – Thanks for the compliment, Azhaghu, but I think that a level 10 question would have to be an actual unsolved problem in mathematics. For me the most beautiful "masterpiece" on this site is one that Michael Mendrin composed titled "Gianlino's Disc". It seems like a fairly straightforward problem, and yet the solution is incredibly complicated, (a fact I'm sure Sir Trevor can attest to, as he valiantly attempted to transcribe the solution into LaTeX format). I keep thinking about it from time to time because my instinct is that there must be a simpler solution which remains elusive. (The solution was actually published in a journal by the mysterious Gianlino, I believe.) The only reason it's not a 400 point question is that it is fairly easy to find the solution via Google. As difficult as it is, though, I would "only" put it at a level 7 or 8. :)
Log in to reply
@Brian Charlesworth – Sir Michael made that problem? I thought it was made by Gianlino. And why do you call him the "mysterious Gianlino"?
I'm curious as to who he is since it seems that you and Sir Michael hold him to quite a high degree.
Btw, woohoo, somehow I qualified for the AIME. Lol, it was too hard this year so they accepted 2.5% rather than all scores >120.
Log in to reply
@Trevor Arashiro – Congrats on your selection, @Trevor Arashiro :)
Log in to reply
@A Former Brilliant Member – Thanks Azhaghu :)
@Trevor Arashiro – Fantastic! Congratulations! (I did sorta predict that: just sayin'. :)) So I guess you have a month or so to prepare for the next stage, (and another excuse for skipping school). Good times. :D
Now that I review that question, it was Gianlino that posed the question and both he and Michael formulated solutions. The one that you transcribed was Gianlino's, and I believe it was he that established analytically the exact answer, (and had an article published, I believe). This was quite a few years ago. The funny thing is that I posted the same question myself, (all of this activity being on Yahoo Answers), not knowing its history, and Michael then recalled the work he and Gianlino had already done years prior. I thought it was a perfectly natural question to ask, and was surprised at how difficult it was to solve, (although both Michael and I believe that there just has to be a simpler solution).
As for the "mysterious Gianlino", I don't even know if that is his real name. I believe he lives in Paris, and was likely a professor, given his brilliance, but I have no hard evidence to go on. He could be a bit crusty and cryptic, but there was no doubt he was a genius. He quit Yahoo a few months before I did, and I don't know if he went on to participate on other math sites or not.
Log in to reply
@Brian Charlesworth – Thank you, once again for your words of encouragement. "Good times" lol. The struggle is real though in chemistry.
Wow, that is quite an interesting little story there about Gianlino. As you mentioned, he seems like quite a mysterious person. His solution is so complex, that I can't even latex it because I don't know what it's asking for (that combined with the fact that he uses commas to separate terms in what I believe is a matrix(?)).
Btw, isn't it a little late where u are? Like some time between 2-5 AM? ;P
Log in to reply
@Trevor Arashiro – Ugh, chemistry .... my sympathies. :(
To be honest, I never fully understood Gianlino's solution either. I always felt honored when he took an interest in the questions I posted on YA, and with (sometimes considerable) thought I could always follow his reasoning. But his solution to his own question used some mathematical machinery that I wasn't fluent with, and the typography he used was admittedly confusing.
And yes, it was a bit late when I posted my previous comment; I am a night-owl, to put it mildly. :P
@Brian Charlesworth – I agree with you sir , I was just mentioning a suggestion since most of us have already touched Lvl 5 in most topics . Also have you tried solving any of the unsolved problems sir ?
The Goldbach conjecture has fascinated me quite a lot . In fact I had tried to write a Java Code to try to prove it , but no success .
(BTW I was first introduced to the Goldbach conjecture by @Agnishom Chattopadhyay through this question .
Log in to reply
@A Former Brilliant Member – The only unsolved problem I have played around with is the twin primes conjecture, enough to know just how hard it is to prove. The Goldbach conjecture is fascinating as well, since it seems like it shouldn't be that difficult to prove at first glance, and yet it has defied proof for so long.
If there is to be a change to the level system, I think adding a level 6 would be worthwhile. The level 5 questions vary greatly in their difficulty, which is (often, but not always) reflected in their point ratings, so splitting level 5 in half would probably be a good way of reflecting this variation.
@Brian Charlesworth – I agree with Sir @Trevor Arashiro
@Trevor Arashiro – "Level 9" was a t.v. show about computer hackers and cybercrime. I guess that's why you chose that number rather than, say, 7, huh?
Log in to reply
@Brian Charlesworth – :3, I just chose 9 for no reason. You have quite the knack for over analyzing things :P lol. Although this is a very good skill to have. It allows you to catch things in a problem most people would miss.
Log in to reply
@Trevor Arashiro – Guilty as charged. :) I tend to assume that there is more to everything than first meets the eye, which leads to unwarranted over-analysis on some occasions and interesting discoveries on others. I'm hoping that when I grow up, (yes, haha), I'll get better at determining when to take things at face value and when to "dive in" for a deeper look.
@Brian Charlesworth – What if there is another 'Pranjal Jain'?
Log in to reply
@Pranjal Jain – HAHAHAHHA, try typing in "Agarwal" and see how many results you get :P
Log in to reply
@Trevor Arashiro – Type in "Calvin Lin" and you'll hit the mother lode, as one would expect. :)
@Trevor Arashiro – I wonder why should I try it? And btw FYI, they are two spellings prevalent in India "Agrawal" and "Agarwal" =P
@Pranjal Jain – Oh, right, if there are multiple people with the same name that would be a problem. I guess that this strategy has its limits.
Congrats! 100th question just after a few days from completing an year long streak is awesome.
Log in to reply
@Pranjal Jain – Thanks! My aim is to try and post one new question a day, but it doesn't always work out. I have a bunch of questions I've thought of that I'm still trying to figure out the answers to. :)
Log in to reply
@Brian Charlesworth – Well, for me its easier to know answer first and then framing a question. Credits to @Trevor Arashiro for this tip.
Log in to reply
@Pranjal Jain – That's a good tip; thanks, (to Trevor too; he always posts good questions). I actually took that approach for this question, where I started with the simplified equation and then expanded it to make it look more difficult.
Log in to reply
@Brian Charlesworth – Awww, Thank you. That means a lot to me comming from you :). I haven't posted much recently because I'm storing problems to be released all at once. PS. It's a little competition between me and another member of brilliant.
Btw, congrats on your hundredth amazing problem.
Log in to reply
@Trevor Arashiro – Thanks. I see that you're closing in on 100 as well; the release of your next set will put you over the century mark, I assume. I'm curious who the other member is that you're having a competition with, (I really haven't a clue), and what exactly this competition is, (although I suppose I can guess at that one). I'm sure I'll notice when it happens, anyway. :)
Log in to reply
@Brian Charlesworth – I guess I'll leave most of it as a "surprise" when it does happen. But this little "problem off" won't put me at 100 as we are going for quality over quantity. All but one of the problems in the set are original.
Btw, where do you see the numbers of problems someone has posted? I know how to find mine by typing in "#easymoney" but I don't know how to view other's
@Pranjal Jain – Haha, thanks. I use this trick often, but sometimes i overuse this trick and when I start with the answer and work backwards, I get carried away and end up with a nearly impossible question. But then again, there is no problem that you and Brian can't solve ;P
Log in to reply
@Trevor Arashiro – The Yoda approach to creating problems: Backwards .... Solving..... Problems. :)
My normal approach is to ask "what if?" and then see where that takes me going forwards, but now I feel inspired to take the Yoda approach "to the max" and then blame you if people complain that the problem is tooooo hard, (although no doubt Jon Haussmann will still be able to solve it). :P
Log in to reply
@Brian Charlesworth – I guess you were planning to post one which he can't. Go for it. Good luck.
@Brian Charlesworth – Haha, I think I'll call it the Yoda approach as well from now on. Although the one thing I watch out for on creating problems like that is the "guess and checkability" of a problem. But here you disguised it well so it's not easily guess and checkable.
And also, most of your problems are already hard, so if you make one that's "Yoda hard" I can't imagine what a monster it will be o.O
Log in to reply
@Trevor Arashiro – Well, I'd prefer to create a beautiful monster, but that can be quite a challenge, as I'm sure you know. My recent "Lost in Space" question took about 70 views before anyone got it right, so it probably qualifies as a baby monster. Michael Mendrin has created a few questions I'm still working on, so I would give him the title of Dr. Frankenstein in terms of creating monster problems. :)
Log in to reply
@Brian Charlesworth – Ahh, yes. Sir Michael Mendrin is probably the best problem maker on Brilliant IMO. Some of his problems are so creative and unique it blows my mind how he comes up with them.
@Trevor Arashiro – And I sometimes reach to a place which can be reached by a shortcut in just a moment.
Edit: For example
Yeah, Sir, I also did it this way, but what should we do when sometimes this rearrangement is not so obvious? Thanks!
Log in to reply
Well, if all else fails there is always the method of Lagrange multipliers, although that wouldn't end up working in this case as there is just a single point as a solution. Depending on what is needing to be maximized the AM-GM or CS inequalities might come in handy, but again, in this case those wouldn't be useful either. So I think that sometimes, as with this question, a rearrangement is the only way to go, even if it is not immediately obvious how to rearrange it. Sorry I couldn't give some better advice than this.
Overrated problem. I figured it out after trying a few special cases like x = 0 , z = 0 , x = z … and seeing that most don't fit the condition, so it must be something extreme like rearranging it to sum of squares...
Log in to reply
I don't know how it got to level 5. When I posted it I assigned it as a level 4 problem, thinking that it might eventually drift down to level 3, so I'm surprised that it actually drifted upward. It hasn't been up for that long, though, so over time it may settle to a more appropriate level.
The equation rearranges as ( x + 2 ) 2 + ( y − 3 ) 2 + ( x − z ) 2 = 0 .
Thus the only possible values for x , y , z are − 2 , 3 , − 2 , respectively. The only possible value, and hence also the maximum possible value, for y + x z is therefore 7 .