Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
add comment

1 Answer

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.