So it seems that there are some math people around. To follow up on //www.wallstreetoasis.com/forums/harder-brainteaser-than-the-coin-flip here is another one:
Given: a^2+b^2=c^2+d^2 a^3+b^3=c^3+d^3
Show that: a+b=c+d
Requires middle school math, but sure can ruin an interview. As before, SBs for 1st correct answer.
Edit - all numbers are strictly positive. Apologies for confusion, but the problem remains open.
Edit - To provide an update, this has, thus far, been proven in 1 confirmed way by nonTargetChimp 9 (see his text file proof), and also in 1 more way that appears correct by unForseen (or rather by a friend of unForseen that works in HR). Another proof using trigonometry has been proposed and has not been fully evaluated but could be correct, and blastoise tried to prove this using fancy linear algebra but so far that proof does not appear to be correct, since he was able to prove it ignoring the positive constraint, meaning he proved something that is false. The most elegant solution to date, in my opinion, would have to be the HR rep's as presented by unForseen, which is found on page 8.
Background on the problem - came from an 8th grade city level math competition from the late 1960s at a mathematical school in the USSR.
Just divide the 2nd equation by the first it's really simple.
(a^3 + b^3)/(a^2 + b^2) = (c^3 +d^3)/(c^2+b^2) => a+b=c+d
Fail.
Your math is retarded. I guess you're ... a banker ?
I just realized how stupid my answer was.
why can't I just square root the 2nd equation? That's whats throwing me off right now. Are you sure the question doesn't ask you to find what a b c d are?
Take the square root of each variable?
Re-write 2nd equation as:
a^2 . a + b^2 . b = c^2 . c + d^2 . d
then divide it by 1st equation and end up with a+b = c+d ? lol
My guess (disclaimer: I am not mathematically inclined):
Since we have a ^2 and a ^3, it gives us enough information to eliminate any uncertainty around whether a, b, c, or d is negative or positive. Plug in positive and negative numbers, and you should be able to tell that as long as it satisfies the two equations above, a+b will always equal to c+d.
We wouldn't be able to make that assumption if we were only given a^2 + b^2 = c^2 + d^2, because that tells us nothing about the +/- sign of each number, and you can't safely say a + b = c + d.
I fail as well.
Math Proofs....HUMBUG!!!
WHOOPS, edited...
a^2 + b^2 = c^2 + d^2
SO
a^2 + b^2 = a^2 + b^2
AND
a^3 + b^3 = a^3 + b^3
THEREFORE
a^x b^x = a^x b^x
SO
a^1 b^1 = a^1 + b^1
AND
a + b = a + b = c + d
a + b = c + d
Or something like that. Apparently, a 5th grader is smarter than me.
Given: a^2+b^2=c^2+d^2 a^3+b^3=c^3+d^3
just guessing:
FIRST EQUATION, REWRITTEN: (a^2+b^2)=(c^2+d^2) SECOND EQUATION, FACTORED OUT: (a+b)(a^2 + b^2) = (c+d)(c^2 + d^2); Distributive Property
substitute right side of first equation into left side of 2nd equation:
(a+b)(c^2 + d^2) = (c+d)(c^2 + d^2) ; Substitution Method
divide both sides by (c^2 + d^2),
and end up with (a+b) = (c+d)
My GMAT level math tells me that's...kind of not right.
(a+b)(a^2 + b^2) = (c+d)(c^2 + d^2) does not equal a^3 + b^3 = c^3 + d^3.
I also don't get UFO's thing and how that proves a + b = c + d.
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a^3 + b^3 does not equal (a+b)(a^2 + b^2) (a+b)(a^2 + b^2) equals a^3 + 2ab^2 + b^3
no it doesnt.
a+b(a^2 + b^2) = a^3 + ba^2 + ab^2 + b^ 3
The above two corrections are correct (and necessary). Also, all material here is within the scope of the GMAT, so if you prefer, think of it as a data sufficiency question:
Is a+b=c+d? (1) a^2+b^2=c^2+d^2 (2) a^3+b^3=c^3+d^3
yeah my bad. i was thinking about (a+b)^3 lol
Should I really embarrass everyone by doing this right now?
I'll tell you what, I solve this and someone gives me back the 1000 I just lost today on First Solar Calls
I give up do it
haha do it. im curious. i like these math threads. shows how much of an idiot we can all be sometimes :D
You have only had 6 pages of examples (right or wrong) to help you. Please bestow your genius upon me.
I'm with Will I'm sure he did it the way I did it. Many people got burned for the sheer fact you need more than just regular algebra to make a simple proof.
C,
let a = 1 b = -1 c = 1 d = 1 a^2+b^2=c^2+d^2 holds true a+b=c+d is not equal
let a = 1 b = 1 c = 1 d = 1
a^2+b^2=c^2+d^2 holds true a+b=c+d is equal
1 is insufficient
same logic applies to 2
Only when combining can you lock down the sign so C
Well you have proven its not A, B or D but you haven't really proven C.
Lol hoping you overlooked that
That's what I was trying to say in my post with the whole plugging of the numbers thing and picking positive/negative signs.
You guys realise that A^3+B^3 = (A+B )(A^2+B^2 - AB) right? not (A+B)(A^2+B^2) like i saw someone above doing
yeah i realized that too late lol. fail.
somebody go do it the long brute way, if this isn't a data sufficiency question:
first equation becomes (a + b)^2 - 2ab = (c + d)^2 - 2cd then, (a+b)^2 = (c+d)^2 - 2(ab + cd)
and, a^2 + 2ab + b^2 = sqrt((c+d)^2 - 2(ab + cd))
second equation: (a + b) (a^2 - ab + b^2) = (c + d) (c^2 - cd + d^2)
plug all that crap in, solve, then plug in some more and get a+b = c+d but i'm too lazy to do it.
of course there's a shorter way but im an idiot today :D
shhhhh. im trying to make someone go through all that.
Let me give it a shot.
First, a^3 + b^3 = (a+b)(a^2 - ab + b^2) = (a+b)(c^2 - ab + d^2) [since a^2 + b^2 = c^2 + d^2].
Now, we know that a^3 + b^3 = c^3 + d^3, and so
(a+b)(c^2 - ab + d^2) = (c+d)(c^2 - cd + d^2)
For (a+b) = (c+d), we must show that ab = cd (this is a necessary condition).
If a+b = c+d, then
(a+b)^2 = a^2 + 2ab + b^2 (c+d)^2 = c^2 + 2cd + d^2
But we already know that a^2 + b^2 = c^2 + d^2, and so 2ab = 2cd. Hence if a+b = c+d (and our conditions hold) then ab = cd.
I don't know about you guys, but I have a life + work to do so if its wrong, so be it.
.
I'm pretty sure for the next part I have to cross multiply the two equations and show that ab = cd, but man what the hell I don't have time for this shit
Sorry about the redundant proof, I felt bad typing it in
Also, I just showed that if a+b = c+d, I will get this result because
(a+b)^2 = (c+d)^2, and a^2 + b^2 = c^2 + d^2.
I asked a math Phd at Stanford to do it (I can do it but wanted to see how easy it would be for him) and he said FUCK THAT IM NOT FACTORING ALL DAY. LOL
I think he meant factorization. Anyway, I will solve it once I get out of class and have time to type.
Is this doable by direct proof? Or do we have to do RAA or something? When we get to the part where we have to show ab=cd
does it require using imaginary numbers, as in a^2 + b^2 = (a+bi)(a-bi)?
Dr. Joe you're such a tease.
This is probably stupid, but I go to a nontarget so give me abreak.
take the natural log of everything thus:
2lna + 2lnb = 2lnc + 2lnd
divde by 2
lna + lnb = lnc +lnd
raise e^ln. thus
a + b = c + d
8---------------D (|) ratio in this thread is mean 0
a^2+b^2=c^2+d^2 +a^2+b^2=c^2+d^2 = a^2(1+a)+b^2(1+b)=c^2(1+c)+d^2(1+d)
and then somehow you eliminate a^2, b^2, c^2, d^2 through some bullshit because a^2+b^2=c^2+d^2, leaving (1+a)+(1+b)=(1+c)+(1+d), so a+b=c+d eliminating the ones...
Am I on the right track?
Not at all. Sorry.
Apparently the answer is no.
...
I got this on a math test once in high school. Me being the smart ass that I am wrote this.
Since we have 4 undefined variables I can assign any numeric value that I want to each of the variables. Therefore a,b,c,d = 1. Having set the numeric value of the variables. a+b=c+d. I acutally got full credit on that problem because the teacher had not designated that a,b,c,d are not all equal to one another.
I haven't solved the problem (it doesn't look that easy to me), but if this indeed is similar to most complex GMAT problems, you likely need to manipulate one of the formulas (likely the 2nd) in a way that allows you to use the first formula. For example, if you reduced a^3 + b^3 in a way that isolated (a^2+b^2), you could then replace a^2+b^2 with c^2+d^2 and work the formula from there. If that doesn't work, perhaps manipulate BOTH formulas...
I could be wrong, but some food for thought for those that are tackling this.
I've gone through this a few times now and each time I end up proving myself wrong.
Can we set a^2+b^2=c^2+d^2 = 1 and a^3+b^3=c^3+d^3 = 1
such that ac + bd = 0?
IFF this is true we need:
bc-ad = ?
So multiply (by c^2 + d^2) on both sides:
(a^2 + b^2)*(c^2 + d^2) = c^2 + d^2
(ac)^2 + (ad)^2 + (bc)^2 + (bd)^2 = 1
(ac + bd)^2 - 2acbd + (bc-ad)^2 +2abcd = 1
We know that ac + bd= 0
→ 0 -2abcd + (bc-ad)^2 + 2abcd= 1
Now reduce similar terms:
→ (bc-ad)^2 = 1
→ bc - ad = 1
or:
→ bc - ad = -1
accept this approach ignores the second equation.
(I love this stuff even when Im wrong)
I am not sure I follow what you are doing above - and as you admit, without using the 2nd equation, a solution is not possible. Your 1st assumption that everything is equal to 1 is also overly constricting. .
come on monkeys...don't make Fermat shift in his grave.LOL
a²+b²=c²+d² a²-c² = d²-b² (a+c)(a-c) = (d+b)(d-b)
Do the same with a^3+b^3=c^3+d^3
then you get
a+c = d+b and a² + ac + c² = d² + bd + b²
Sqaure both sides of a+c = d+b you get ac = bd
substract a² + ac + c² = d² + bd + b² times 3 you get
a - c = d -b
a + b = c + d
(my math -> english is pretty horrible)
Say, in what domain are we solving this? I'll assume the reals, then the assumptions do not hold. Take the way less general a=1, b=1.
c^2+d^2=2 c^3+d^3=2
So your assumption would imply that c+d=2, but one can find other c and d that solve the above system. Let u=c+d, so from the second:
(c+d)(c^2+d^2-cd)=2 u(2-cd)=2 cd=2-(2/u)
then adding twice to the first equation
c^2+2cd+d^2=2+4-(4/u) u^2=6-(4/u) u^3-6u+4=0
As expected, one solution is u=c+d=1+1=2. But the other roots
u(u-2)(u+2)-2u+4=0 (u-2)(u^2+2u-2)=0 are the roots of the quadratic: u^2+2u-2=0
and those are -1+or-sqrt(3)
So, c+d can be equal to those numbers, and not just to 2=1+1=a+b. For example: c=(a+sqrt(a^2+4a))/2 d=(a+sqrt(a^2+4a))/2
where a=-1+sqrt(3)
one of the roots of the quad.
I (master of fin mathematics), and a team member (PhD. in Mathematics) stand by the above analysis. We want to be proved wrong, however....
what wadtk has shown is that if a=b=1 and assuming c^2+d^2=2 and c^3 +d^3 = 2
then c+d = 2, 1+sqrt(3), 1-sqrt(3)
this proves than c+d is not necessarily equal to a+b and therefore proves that a+b = c+d is not always true even if c^2 + d^2 = a^2 + b^2 and c^3+d^3 = a^3 +b^3 are true
So the problem is not provable as it is not true!
Oh sh*t, it's going to be a showdown. MATH WARRRR!
Too bad Will Hunting hasn't proved his self proclaimed math superiority.
President, you are assuming that xy = ab ==> x= a and y = b. This is not true.
U can proof it by simple induction, assuming positive integers. Look at your solutions of the 1st and 2nd equation. Reformulate them for i and i+1. Get rid of the i-th roots by taking all of it to the power of i. Then set a^i+b^i=c^i+d^i, plug in your solutions and u'll end up with 3a^i+3b^i=3c^i+3d^i and you're done. Also works for i+1...i+n. Should work at first sight, but no clue how to do it GMAT style ;)
...
a^2+b^2 = (a+b)^2 c^2+d^2= (c+d)^2 sqrt(a+b)^2= sqrt(c+d)^2 a+b=c+d
Another high-school flunk out. Next..
the problem does not hold for n^2; n>2 if abcd are all integers.
I actually lol at this and everyone was looking at me "wtf"?
@ Dr. Joe - is this some form of a diophantine equation? Maybe a hint is in order as apparently we are all too retarded to find a proof.
I believe wadtk has shown it to be unprovable, can anyone find a flaw in his/her reasoning?
I didn't do the excercise on paper. Just did some thinking about it, and if u look at my reasoning of above: if you follow it by induction, it should hold true for positive numbers. Looking at the generalized solutions of Eq. (1) or Eq. (2) you will get the roots with a negative sign, where wadtk is definitly right. Prooving it true for negative numbers shouldn't be possible.
This gave me a headache.
My bad folks - all numbers are positive. For what it's worth, the example above was correct in that a possibility is A=1 B=1 C= -0.564579 D=1.296630. But all numbers are strictly positive.
It's ok, you got many people thinking about mathematics which is hall of fame in my book.
President is wrong, you don't get a+c = b+d, all you get is that (a^2+ac+c^2)/(a+c ) = (b^2+bd+d^2)/(b+d)
okay so this isn't really that bad...
But the problem remains open...
a^2+b^2=c^2+d^2
think of c^2+d^2 as u
then a^2+b^2=u
then a, b, and radical u form a right triangle.
so use a triple like 3,4,5
a=3 b=4 and radical u=5
well if a=3 then a^2=9 and if b=4 then b^2=16 and radical u^2=25
radical u^2 is also equal to just "u" which is equal to c^2+d^2
so c^2+d^2=25
which makes sense because 9+16 = 25
no repeat the process with c,d, and 5 as a triangle.
you get c=3 and d=4
hence a+b=c+d
cold pizza, you have no knowledge of formal mathematical proofs, taking specific values is not a fucking proof.
Wtf is this shit, don't make any sense at all. Go back to high school pizza boy.
here's a proof:
http://en.wikipedia.org/wiki/Pythagorean_theorem
[quote=cold pizza 2]here's a proof:
http://en.wikipedia.org/wiki/Pythagorean_theorem[/quote]
Proof that you're retarded
lol
how about this:
a^2+b^2=c^2+d^2
say c^2+d^2 = z^2
so a^2+b^2 = z^2
then a, b, and z form a triangle where z is the hypotenuse
and c, d, and z form the same triangle. because if z=z and it is a right triangle according to pythagorean theorem, then the other sides are the same as well
still no good?
Hey buddy how about you try using the second output a^3+b^3=c^3+d^3 Nevermind it won't make any difference..
Troll ?
Why don't you solve it for us all then, fucking genius?
why do i need to use the second output if i can solve it with one?
We know that: a^3+b^3 = c^3+d^3
Rewrite the equation so that it takes the form: (a+b)(a^2+b^2-ab) = (c+d)(c^2+d^2-cd) ---> call this equation 1
Now, if we can show that (a^2+b^2-ab)=(c^2+d^2-cd) then we will be able to divide them out of the equation, leaving us with our desired result, (a+b)=(c+d). So
Prove: (a^2+b^2-ab) = (c^2+d^2-cd) ----> call this equation 2
We know a^2+b^2=c^2+d^2, from the problem statement. We could subtract this value from both sides to show -ab = -cd and thus (a^2+b^2-ab)=(c^2+d^2-cd), but to show the step more clearly, let us say that:
k = (a^2+b^2) = (c^2+d^2)
plugging k into equation 2 yields: k-ab = k-cd
subtract k from both sides: -ab = -cd
Therefore, since (a^2+b^2) = (c^2+d^2) and -ab = -cd, we have proven that
(a^2+b^2-ab) = (c^2+d^2-cd). We could stop here, but to explicate it further, let j = (a^2+b^2-ab) = (c^2+d^2-cd).
Plug j into equation 1:
(a+b)(a^2+b^2-ab) = (c+d)(c^2+d^2-cd) becomes (a+b)(j) = (c+d)(j).
Divide by j.
(a+b) = (c+d).
QED, that's our proof.
I'll be here all week, feel free to leave tips or better yet, job offers.
I had the same thought process, but I haven't figured out how to prove ab=cd which is the core of this proof. You should check that part of your answer because it looks like an obvious circular argument to me.
yup this just needs a proof that ab=cd and it would work.
^very nice
Nope. Equation 2 is what you need to prove but you use it to prove your conclusion.
Ouch, I just realized what I did. Give me a few minutes; I think I can get it.
Here's my attempt:
a^3+b^3=c^3+d^3 a^2+b^2=c^2+d^2
take the difference of the equations
(a^3-a^2)+(b^3-b^2)=(c^3-c^2)+(d^3-d^2)
for all positive integers, n^3-n^2=k is only true for n. So for that equation to be true, either a=c&b=d or a=d&b=c or a=b=c=d----->therefore a+b=c+d
Not limited to integers
Do we need to show "n^3-n^2=k is only true for n" more rigorously?
I could buy it, but I'm not sure without seeing a proof.
I guess nnn-n*n = k
turns to
n = k/n*n +1
so just one solution.
a^3+b^3 = c^3+d^3
a^3+b^3 = ab ((a^2)/b +(b^2)/a)
c^3+d^3 = cd ((c^2)/d +(d^2)/c)
ab ((a^2)/b +(b^2)/a) = cd ((c^2)/d +(d^2)/c)
a^2+b^2 = c^2+d^2
a^2+b^2 = ab (a/b +b/a)
c^2+d^2 = cd (c/d +d/c)
ab (a/b +b/a) = cd (c/d +d/c)
so, we got:
ab (a/b +b/a) = cd (c/d +d/c)
and
ab ((a^2)/b +(b^2)/a) = cd ((c^2)/d +(d^2)/c)
then
[ab (a/b +b/a) ] / (c/d +d/c) = cd
and
[ab ((a^2)/b +(b^2)/a] / ((c^2)/d +(d^2)/c) = cd
so
[ab (a/b +b/a) ] / (c/d +d/c) = [ab ((a^2)/b +(b^2)/a] / ((c^2)/d +(d^2)/c)
ab cancels out
(a/b +b/a) / (c/d +d/c) = [((a^2)/b +(b^2)/a] / ((c^2)/d +(d^2)/c)
let's say this entire thing = Z
so going back...
[ab ((a^2)/b +(b^2)/a] / ((c^2)/d +(d^2)/c) = cd = abZ
Do the same thing with:
ab (a/b +b/a) = cd (c/d +d/c)
and
ab ((a^2)/b +(b^2)/a) = cd ((c^2)/d +(d^2)/c)
to solve for ab
SO:
ab = [cd (c/d +d/c)] / (a/b +b/a)
ab = [cd ((c^2)/d +(d^2)/c)] / [((a^2)/b +(b^2)/a)]
cd (c/d +d/c)] / (a/b +b/a) = [cd ((c^2)/d +(d^2)/c)] / [((a^2)/b +(b^2)/a)]
cd cancels out
(c/d +d/c)] / (a/b +b/a) = [((c^2)/d +(d^2)/c)] / [((a^2)/b +(b^2)/a)]
let's say this entire thing equals Y
so going back...
ab = cdY
now we have
ab=cdY and abZ=cd
cdY = cd/Z
cdYZ = cd
YZ=1
multiply both equations
ab * cd = cdY * abZ
SO ****************ab = cd ************
that shit is long though lol
ab = cd
going back to:
ab (a/b +b/a) = cd (c/d +d/c) - which came from the first equation
ab and cd cancel out
(a/b +b/a) =(c/d +d/c)
let's take out b and d
so b (a + 1/a) = d (c +1/c)
b (1) = d (1)
b = d
by that logic
a = c
so a+b = c+d
no?
^ wow that was retarded sorry forget that
i still claim the really long one (ab=cd)
just don't know how to use that to prove a+b=c+d
although i can prove a^4+b^4=c^4+d^4
so i'm retarded
but I also stand by the triangle proof. if this was a data sufficiency problem, you only need the first equation.
dude
?
Wow still not solved? I'll really give it ago in a bit. I'm starting to lose my respect for WSO after some of the brainteaser answers I've seen...
Like that guy who didn't know the sum of 1 to 100...
Not knowing how to add 1 to 100, is inexcusable unless you were dropped on your head as a kid. However this problem is legitimately tough, I got a 50 on the Math on the GMAT, but have no clue how to solve this.
MAXIMUM...how about YOU solve this???
U cannot do the triangle proof, as u cannot solve it for i>2. If you can do it for i = 4 u should normally be able to generalize it for i and i +1 and u will be fine - as long as it holds.
a^2+b^2=a^3d^2+b^3+c^2 = (c+d)(c^2-cd+d^2)
Based on Fermats Theory (ad)^e + (bd)^e = (cd)^e
He's like Yoda or something.
No hints lol I wanna see how long this goes on
I got something short and elegant for you.
No homo..
Okay I throw in the towel. A dude with a Math Fin degree and his bud who got a PhD Math came here before so who am I to speak here, just a random non-target chimp..
B = A, C = A, D=A+B-C A=1 B=1 C=1 D=1+1-1=1
D^2 = A^2 + B^2 - C^2 =1^2 + 1^2 - 1^2
D^3 = A^3 + B^3-C^3 = 1^3 = 1^3+1^3-1^3=1^3
B= -A + C +D = 1= -1+1+1
A^2+B^2=C^2+D^2 = (A+B)(A^2-AB+B^2)=(C+D)(C^2-DC+D^2)= 1^2+1^2=1^2+1^2= (1+1)(1^2-1+1^2)=(1+1)(1^2-1+1^2)=2=2
That actually was a lot easier than I originally thought.
That was actually kinda the direction I showed you one page before. It's pretty straight-forward if you do it in a generalized way. This case is i =1 (like Im not very creative showed). If you replace the 1's with i and you can show it another time for i+1 (what you can do) you have a mathematically nice and sound proof (of course not GMAT plug-in style). The trick is really just to calculate the positive solution for D, C, B, and A and then move forward (actually with i, i+1 it's the same and also pretty easy). Dunno, we had to do these induction proofs all the time during high school. I guess there's more exciting stuff at this age.
You're fucking with us right Dr. Joe?
Someone else proved ab=cd so I'll start there. Could be wrong as I haven't taken a math class in almost a decade.
(1) Given: ab = cd So a = cd/b and c = ab/d
(2) a^2 + b^2 = c^2 + d^2 Sub in for a and c from (1) (c^2 * d^2)/b^2 + b^2 = (a^2 * b^2)/d^2 + d^2
(3) multiply each side b^2 * d^2 then rearrange and factor (c^2 * d^4) + (b^4 * d^2) = (a^2 * b^4) + (b^2 * d^4) (c^2 * d^4) - (b^2 * d^4) = (a^2 * b^4) - (b^4 * d^2) d^4 * (c^2 – b^2) = b^4 * (a^2 – d^2)
(4) Rearrange a^2 + b^2 = c^2 + d^2 So c^2 – b^2 = a^2 – d^2 = X
(5) Sub in X in equation (3) d^4 * X = b^4 *X b^4 = d^4 so b=d for all positive numbers
(6) ab=cd and b=d, then a =c so a+b=c+d
That killed some time at work, almost time to start drinking.
Side note - looks like I got someone to join WSO just to solve math problems.
Lol sarcasm doesn't seem to come across very well online. I just picked the most nonsensical thing posted thus far and declared it true.
Edit - I can't believe I just had to clarify that
LOL
You definitely had me until this post. I was like, wtf!?!?!?
Wtf, Okay....
But do you understand why your proof was insufficient?
B=A C=A D=A+B-C A=5 B=5 C=5 D=5+5-5=5
D^2 = A^2 +B^2 - C^2 = 25=25+25-25
D^3 = A^3 +B^3 -C^3 = 125=125+125-125
B= -A + C + D = -5+5+5=5
A^2+B^2=C^2+D^2 = (A+B)(A^2-AB+B^2)=(C+D)(C^2-DC+D^2)= = (5+5)(5^2-25+5^2)=250=(5+5)(5^2-25+5^2)=250 Which means that 5^3+5^3=5^3+5^3 or 125+125=125+125
For n =/=1
C does not always have to equal A...but it will be equal to either A or B...
nonTargetChimp, I came up with the exact same solution as you, but it is wrong because a,b,c, and d don't have to be positive.
Ok Dr. Joe,
I'm going to publicly ask for a hint. Lets be honest, getting to ab=cd is piss easy, its the next part that needs some direction.
I just don't want to spend hours on it on a Friday then find out some smartass has posted the proof. i.e. I'm getting paranoid and continuously refreshing this page.
Edit: Can you PM me the hint? Since now everyone knows. I'll obviously mention it in my proof if I get there.
I'm surprised nobody posted it on a math forum and got it solved in a matter of minutes...
Let a^2+b^2=c^2+d^2 = k
a + b = u, c + d = v
a^3+b^3=c^3+d^3
(a + b)(a^2+b^2 - ab) = (c + d)(c^2+d^2 - cd) (1)
Notice ab = [(a+b)^2 - a^2 - b^2]/2 = (u^2 - k)/2 Same with cd = (v^2 - k)/2
(1) => u(k - (u^2 - k)/2) = v(k - (v^2 - k)/2)
Multiply both sides by 2 u.(2k - u^2 + k) = v(2k - v^2 + k) u.(3k - u^2) = v.(3k - v^2) 3k(u-v) - (u^3 - v^3) = 0
Case 1: u = v. Done
Case 2: 3k = u^2 + v^2 - uv or 3(a^2+b^2) = (a + b)^2 + (c + d)^2 - (a+b)(c+d)
a^2 + b^2 + (a+b)(c+d) - 2ab - 2cd = 0 (2)
If a + b > c + d (2) => (a - b)^2 + (a+b)(c+d) - 2cd = 0
a + b > c + d (a+b)(c+d) > (c + d)^2
=> Left side > (a - b)^2 + (c + d)^2 - 2cd = (a - b)^2 + c^2+d^2 > 0
If a + b c + d
(2) => (c - d)^2 + (a+b)(c+d) - 2ab = 0
c + d > a + b (a+b)(c+d) > (a + b)^2
=> Left side > (c - d)^2 + (a + b)^2 - 2ab = (c - d)^2 + a^2+b^2 > 0
=> Case 2 is impossible
edit:delete (my post)
No I'm not wrong let me clarify further 3(a^2+b^2) = (a + b)^2 + (c + d)^2 - (a+b)(c+d)
3(a^2+b^2) = 2(a^2+b^2) + (c^2 + d^2)
(a + b)^2 + (c + d)^2 - (a+b)(c+d) = a^2+b^2 + 2ab + c^2 + d^2 + 2cd - (a+b)(c+d)
Therefore you can cross a^2+b^2 + c^2 + d^2 from both sides. Please double check it for yourself
u^3 - v^3 = (u-v)(u^2 + v^2 + uv), not (u-v)(u^2 + v^2 - uv),
Oh yeah you're right
P/S: Why can't I edit my previous post anymore ????
i didn't come from a private school or anything, so maybe that's why, but...
none of this shit seems to me like middle school or even high school level math.
what is this "i = 1 and 4" stuff, and from left, and from right?
Can you guys point me in the right direction, so I can try to learn it? what is the name of the topic?
The short answer: slope/intersect form. It appears that non-target is trying to solve it as a linear programming model (although is isnt in y=mx+b form yet), but I do not think this problem requires that degree of complication.
For what its worth, this problem came from ________________
Edit - will cite source after it is solved
Algebra 2 is what? "pre-calc" or something like Linear Algebra
pre-calc
though to be completely honest, to understand/complete the proof (and be able to show that it is true on a deep level), a mathematical proofs course would help
anyways i'll be in the chat room for a bit if anyone wants to discuss this problem
Given: a^2+b^2=c^2+d^2 a^3+b^3=c^3+d^3
Show that: a+b=c+d
a,b,c,d >0 Pf: Assume a+b=c+d
(a+b)^2=(c+d)^2 a^2+2ab+b^2=c^2+2cd+d^2 a^2+b^2-c^2-d^2=2cd-2ab 0=2cd-2ab 0=cd-ab ab=cd
(a+b)^3=(c+d)^3 a^3+3a^2b+3ab^2+b^3=c^3+3c^2d+3cd^2+d^3 a^3+b^3-c^3-d^3=3c^2d+3cd^2-3a^2b-3ab^2 0=3cd(c+d)-3ab(a+b) 3ab(a+b)=3cd(c+d) ab(a+b)=cd(c+d) Using shit above, ab=cd --> (a+b)=(c+d)
Closer?
Here is an idea.... (someone can do the algebra grunt work and get the credit)
Lets get rid of one variable first
rewriting first equation
a = (c^2 + d^2 - b^2) ^ (1/2)
rewriting second equation
a = (c^3 + d^3 - b^3) ^ (1/3)
now we can get rid of "a"
(c^2 + d^2 - b^2) ^ (1/2) = (c^3 + d^3 - b^3) ^ (1/3)
******NOW THIS IS WHERE IT GETS COMPLICATED **********
we can rewrite the equation above isolating "b" so it is in terms of "c" and "d"
i don't know how to factor that shit, but in theory we would have an equation like
b = ( nasty ass equation with "c" and "d" )we could do the same steps up to this point, but instead of getting rid of "a" we get rid of "b"
b = (c^2 + d^2 - a^2) ^ (1/2) b = (c^3 + d^3 - a^3) ^ (1/3)
then
(c^2 + d^2 - a^2) ^ (1/2) = (c^3 + d^3 - a^3) ^ (1/3)
again we isolate "a"
a = (nasty ass equation with "c" and "d")
******** now this is where everything should tie ************
this two equations should be exactly the same
b = ( nasty ass equation with "c" and "d" ) a = (nasty ass equation with "c" and "d")
therefore
a = bthen
c = d (by doing the dame process for "c" and "d" )
by substitution in first equation:
a^2+ a^2 = c^2 + c^2
2 a^2 = 2 c^2
simplified
a = c
and is done.
if we know that a = b , c = d , a = c
then
a + b = c + d
Anyone find anything wrong with this?
How is this a proof? Are you out of your fucking mind? What the hell is a "nasty ass equation"? An equation with gonorrhea? I don't think its possible to get to a=c from the information we have. a+b = c+d is as much as we can infer.
i'm pretty sure a=b and c=d don't have to be true for the equations to work so....
Here is a list of 42 methods of mathematical proofs.......might actually help
Proof by General Agreement: "All in Favor?..."
Proof by Imagination: "Well, we'll pretend its true."
Proof by Convenience: "It would be very nice if it were true, so ..."
Proof by Necessity: "It had better be true or the whole structure of mathematics would crumble to the ground."
Proof by Plausibility: "It sounds good so it must be true."
Proof by Intimidation: "Don't be stupid, of course it's true."
Proof by Lack of Sufficient Time: "Because of the time constraint, I'll leave the proof to you."
Proof by Postponement: "The proof for this is so long and arduous, so it is given in the appendix."
Proof by Accident: "Hey, what have we here?"
Proof by Insignificance: "Who really cares anyway?"
Proof by Mumbo-Jumbo: " For any epsilon> 0 there exists a corresponding delta > 0 s.t. f(x)-L epsilon whenever x-a delta"
Proof by Profanity: (example omitted)
Proof by Definition: "We'll define it to be true."
Proof by Tautology: "It's true because it's true."
Proof by Plagiarism: "As we see on page 238 ..."
Proof by Lost Reference: "I know I saw this somewhere ..."
Proof by Calculus: "This proof requires calculus, so we'll skip it."
Proof by Terror: When intimidation fails ...
Proof by Lack of Interest: "Does anyone really want to see this?"
Proof by Illegibility: " ¥ ª Ð Þ þæ"
Proof by Logic: "If it is on the problem sheet, then it must be true."
Proof by Majority Rule: Only to be used if General Agreement is impossible.
Proof by Clever Variable Choice: "Let A be the number such that this proof works."
Proof by Tessellation: "This proof is just the same as the last."
Proof by Divine Word: "And the Lord said, 'Let it be true,' and it came to pass."
Proof by Stubbornness: "I don't care what you say! It is true!"
Proof by Simplification: "This proof reduces to the statement, 1 + 1 = 2."
Proof by Hasty Generalization: "Well, it works for 17, so it works for all reals."
Proof by Deception: "Now everyone turn their backs ..."
Proof by Supplication: "Oh please, let it be true."
Proof by Poor Analogy: "Well, it's just like ..."
Proof by Avoidance: Limit of Proof by Postponement as t approaches infinity.
Proof by Design: "If it's not true in today's math, invent a new system in which it is."
Proof by Intuition: "I just have this gut feeling ..."
Proof by Authority: "Well, Bill Gates says it's true, so it must be."
Proof by Vigorous Assertion: "And I REALLY MEAN THAT!"
Proof by A.F.K.T. Theorem: "Any Fool Knows That!"
Proof by vigorous handwaving: Works well in a classroom.
Proof by seduction: "Convince yourself that this is true!"
Proof by accumulated evidence: "Long and diligent search has not revealed a counterexample."
B=A. C=B. D=A+B-C
Alternatives:
C=A. D=B. C=B.D=A
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