How to Solve Problem #5 in the 2013 Math Awareness Competition

[Curiousphy]

Homework Statement

http://www.math.ku.edu/mathawareness/2013competition/2013_9-12.pdf

I am trying to figure out how to solve problem #5 in the link above.

Homework Equations

N/A

The Attempt at a Solution

The answer they provided does not make sense to me. Can anyone explain? thanks

 

[skiller]

Your link doesn't work.

 

[haruspex]

Your link doesn't work.

It should be http://www.math.ku.edu/mathawareness/2013competition/2013_9-12.pdf
Curiousphy, you don't say where you got stuck.
Try writing out (x+1)(y+1)(z+1) after substituting for x, y and z. Do the same with (x-1)(y-1)(z-1). You should be able to see that they are actually the same.
Is it clear from there?

 

[Curiousphy]

(x + 1)(y + 1)(z + 1) = (x - 1)(y - 1)(z - 1)
xyz+xy+xz+x+yz+y+z+1 = xyz−xy−xz+x−yz+y+z−1
xy + yz + zx = -1

For xyz+xy+xz+x+yz+y+z+1 to equal xyz−xy−xz+x−yz+y+z−1, you'd need specific combinations of [x, y, z], right? e.g., x = -1, y, z = 1. The left and right hand sides aren't necessarily equal. Am I forgetting a formula, why in the world would anyone set (x + 1)(y + 1)(z + 1) = (x - 1)(y - 1)(z - 1) to get xy + yz + zx = -1? why not set xy + yz + zx = -2 or something else? thanks

 

[verty]

My way for doing this is the following.

Let x = $\frac{a}{b-c} + \frac{b}{c-a} + \frac{c}{a-b}$.
What we are given seems related to x^2 because x^2 = 2 + f(a,b,c). So start by doing this:

#1. Calculate (a + b + c)^2 to get a general formula.
#2. Use this formula to write x^2 in the form 2 + f(a,b,c).
#3. Try to manipulate f(a,b,c) to relate it back to x or x^2.

As you might imagine, one has to find the most likely avenue of progress and follow it.

 

[Mentallic]

(x + 1)(y + 1)(z + 1) = (x - 1)(y - 1)(z - 1)
xyz+xy+xz+x+yz+y+z+1 = xyz−xy−xz+x−yz+y+z−1
xy + yz + zx = -1

For xyz+xy+xz+x+yz+y+z+1 to equal xyz−xy−xz+x−yz+y+z−1, you'd need specific combinations of [x, y, z], right? e.g., x = -1, y, z = 1. The left and right hand sides aren't necessarily equal. Am I forgetting a formula, why in the world would anyone set (x + 1)(y + 1)(z + 1) = (x - 1)(y - 1)(z - 1) to get xy + yz + zx = -1? why not set xy + yz + zx = -2 or something else? thanks

Remember what x,y,z are substitutes for.

Show that

$$(x+1)(y+1)(z+1)\equiv(x-1)(y-1)(z-1)$$

Given that

$$x=\frac{a}{b-c}, y=\frac{b}{c-a}, z=\frac{c}{a-b}$$

 

[Curiousphy]

The exam allows 40 minutes to solve 5 problems. How in the world can anyone expand (x+1)(y+1)(z+1)≡(x−1)(y−1)(z−1) with the a, b, c terms within that time? and I guess my main question is, why would they even do that without working backwards from a known solution? is this supposed to be a known property/formula by high school students? why would I think of or come up with "(x+1)(y+1)(z+1)≡(x−1)(y−1)(z−1)" given what's stated in the problem? nevermind that the left and right hand sides aren't necessarily equal and to prove they are with the given conditions would take forever? thanks

I will follow verty's steps or try to come up with an answer independently, but it just seems unreasonable given the time available for the exam, unless it was a well known property/formula that I have forgotten.

 

[haruspex]

(x + 1)(y + 1)(z + 1) = (x - 1)(y - 1)(z - 1)
xyz+xy+xz+x+yz+y+z+1 = xyz−xy−xz+x−yz+y+z−1
xy + yz + zx = -1

For xyz+xy+xz+x+yz+y+z+1 to equal xyz−xy−xz+x−yz+y+z−1, you'd need specific combinations of [x, y, z], right? e.g., x = -1, y, z = 1. The left and right hand sides aren't necessarily equal. Am I forgetting a formula, why in the world would anyone set (x + 1)(y + 1)(z + 1) = (x - 1)(y - 1)(z - 1) to get xy + yz + zx = -1? why not set xy + yz + zx = -2 or something else? thanks

You overlooked this part of what I wrote:

after substituting for x, y and z

How in the world can anyone expand (x+1)(y+1)(z+1)≡(x−1)(y−1)(z−1) with the a, b, c terms within that time? and I guess my main question is, why would they even do that without working backwards from a known solution?

I don't think it takes that long to realize the expansions lead to the same thing if you consider the symmetry. x+1 = (a+b-c)/(b-c), so in the product (x+1)(y+1)(z+1) the three terms in the numerator must consist of the three forms of 'sum of two of a, b, c minus the third'. x-1 = (a-b+c)/(b-c), leading to the same product in the numerator (and the denominator is obviously the same). Pretty sure I figured that out within 10 seconds.
How you would think of doing that is the much tougher part. I well might not have thought of it at all. But some math whiz somewhere thought of this, and I don't see that knowing that the thing to be proved is true would have helped much. Besides, we had that fact too.
Are you expected to solve all 5 in the 40 minutes? With problems this hard, solving two might be a good result.

 

[Curiousphy]

Thanks Haruspex, their answer is making sense now, but yea, it's nearly impossible to come up with that within the time given, unless the students had dealt regularly with similar problems.

I did try to solve it by arbitrarily assigning values to a and b, solving for c, then plugging the 3 "distinct real numbers" back into the second formula. For example,
a = 10
b = 15

10^2 / (15 - c)^2 + 15^2 / (c - 10)^2 + c^2 / (10 - 15)^2 = 2

Solving for c = -2.9933725394078

10 / (15 + 2.9933725394078) + 15 / (-2.9933725394078 - 10) - 2.9933725394078 / (10 - 15) = -2 x 10^-14 = 0

which I think should be an acceptable answer for the competition, since the answer must hold true for any 3 distinct real numbers for a, b, and c that satisfy the first equation.

thanks all for your help.

 

[awkward]

Here is a quick and dirty solution which would probably not get you full credit and might even get you a serious talking-to from your teacher. We're probably supposed to show that if
$$\frac{a^2}{(b-c)^2} + \frac{b^2}{(c-a)^2} + \frac{c^2}{(a-b)^2} =2$$
then
$$f(a,b,c) = \frac{a}{b-c} + \frac{b}{c-a} + \frac{c}{a-b}$$
is constant and find its value.
But instead, let's apply the meta-theorem "All problems given in math class have a solution" and assume that f(a,b,c) is constant-- so all we have to do is guess its value. Let's try assuming there is a solution with $a=0$. Then we have the simpler problem
$$\frac{b^2}{c^2} + \frac{c^2}{b^2} = 2$$
and
$$f(a,b,c) = \frac{b}{c}-\frac{c}{b}$$
It's easy to see that the equation is satisfied whenever $b=c$, so $$f(a,b,c)=0$$
If we were doing this the right way, we would sweat bullets over having divided by zero in the original equation when we set $a=0$ and $b=c$. But what the heck, we aren't doing it the right way anyhow. (Of course, if this were a multiple choice problem and all you had to do was select the value of f(a,b,c) from a list, they would never know how you got the answer and you would get full credit. Heh heh.)

 

[verty]

My way for doing this is the following.

Let x = $\frac{a}{b-c} + \frac{b}{c-a} + \frac{c}{a-b}$.
What we are given seems related to x^2 because x^2 = 2 + f(a,b,c). So start by doing this:

#1. Calculate (a + b + c)^2 to get a general formula.
#2. Use this formula to write x^2 in the form 2 + f(a,b,c).
#3. Try to manipulate f(a,b,c) to relate it back to x or x^2.

As you might imagine, one has to find the most likely avenue of progress and follow it.

I apologize, my step 3 does not work. One has to simplify f(a,b,c), usually not a great idea in questions like this. This is a very tough question, I think.

The annoying thing about it is, all one has to do is square x and simplify to get the answer. In most competitions, this won't work or will take far too long. It is also strange for the reason that those who plug and chug may get it right while those who look for an angle might not.