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This questions source is the 2010 MMPC Part 1 Competition #33. The answers were released, but no solutions.

I am interested in the solution because I have no idea how to find a good way to do this problem. I will greatly appreciate help as tomorrow is the 2012 MMPC Part 1 Competition and I still cannot solve a problem from 2 years ago!

We let $x$ and $y$ be real numbers that satisfy |$x-1$| + |$y+1$| < 1 . The range of the expression for $\frac{x-1}{y-2}$ is which of the following?

A) $( - \frac{1}{3} , \frac{1}{3} ) $

B) $( - \frac{1}{2} , \frac{1}{2} ) $

C) $( - \frac{1}{2} , \frac{1}{3} ) $

D) $( - \frac{1}{3} , \frac{1}{2} ) $

I know the choices may not be needed but even if it helps slightly, then its worth it. Thank you and if you could, please give a thorough solution. Right now all I am doing is plugging in points in the intervals. On average each question is supposed to be $2$ and a half minutes long so that clearly isn't effective!

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So as not to bury the lead, the answer is (A), namely that the range of $(x-1)/(y-2)$ will be the interval $(-1/3, 1/3)$. The solution essentially consists of two parts.

First part: finding the domain.

We want to find the points $(x, y)$ for which $$ |x-1|+|y+1| < 1 $$ If your geometric intuition is better than mine, you might observe that in three dimensions the region in question is the projection on the $x$-$y$ plane of intersection of a horizontal trough (the $|y+1|$ part) and a vertical trough (the $|x-1|$ part). A less brilliant way is to consider the four possible cases for the absolute values:

  • $x\ge 1, y\ge -1\text{, so }|x-1|+|y+1| = x+y$
  • $x\ge 1, y< -1\text{, so }|x-1|+|y+1| = x-y-2$
  • $x< 1, y\ge -1\text{, so }|x-1|+|y+1| = -x+y+2$
  • $x< 1, y< -1\text{, so }|x-1|+|y+1| = -x-y$

In the first case, the domain will be bounded by $x\ge1,y\ge1,x+y<1$. This is the interior (and the left and bottom edges) of a triangle with vertices $(1,0),(1,-1),(2,-1)$. Each of the other cases also triangular and the union of all four regions is the interior of a square with vertices $(1,0),(2, -1),(1, -2),(0,-1).$

Second part: bounding the function.

We now need to find the upper and lower bounds on the values of $$ f(x, y)=\frac{x-1}{y-2} $$ on the square we found in the first part. For any constant $c\ne0$, the level curves, $f(x,y)=c$, will be the lines $$ y=\frac{1}{c}\;x+\left(\frac{2c-1}{c}\right) $$ and if $c=0$ the level curve will be the vertical line $x=1$. Graph a few of these lines and you'll notice that all the level lines will intersect at the point $(1, 2)$ (which isn't a possible point on $f(x,y)$, but we don't care about that, since it's outside of our square).

Now all you have to do is look at the $c$ values for which these lines intersect the square. It's very easy to see that we're looking at $-1/3<c<1/3$ so we have our answer. Here's a picture (with apologies in advance to the ten percent or so of the color-blind viewers out there).

enter image description here

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