Stuck on Australian Mathematics Competition Junior Level 2002 Geometry Problem

Question

A rectangle $PQRS$ with $PQ$ = $49$ and $PS=100$ is cut into $4900$ squares of side $1$. $T$ is a point on $QR$ such that $QT=60$. Of these $4900$ squares, how many are cut by the lines $PT$ and $TS$?

Here is the rectangle for the question:

I tried to find the slope of $PT$ and $TS$.

So the slope of $PT$ is equal to $49/60$ and the slope of $TS$ is equal to $49/(100-60)$. I used this approach because with previous experience, especially with the American Mathematics Competitions that finding the slop will be beneficial to finding how many squares the line intersects with.

However, I am now stuck and unable to proceed after this step. I would appreciate it if you were able to solve the problem for me. Thank you very much!

Answer 1

We say that the rectangle $PQRS$ is partitioned into 4900 squares by horizontal and vertical grid lines. There are $48$ horizontal grid lines not including $PQRS$, and $50$ horizontal grid lines including $PQRS$. Likewise there are $99$ vertical grid lines not including $PQRS$ and $101$ vertical grid lines including.

Let's look at how many squares the line segment $L=PT$ from $(0,0)$ passes to $(60,49)$. To this end, for each integer $i=0,1,2,\ldots, 59$, how many squares the line segment $L_i$ of $L$ from $(i,b_i)$ to $(i+1,b_{i+1})$ [whatever $b_i$, $b_{i+1}$ may be] passes through. [So $L=L_0+L_2+\ldots+L_{59}$.]

$1.$ If the line segment $L_i$ does not pass though a horizontal grid line, then the line segment $L_i$ passes only through $1$ square, the square w southwest corner $(i, [b_i])$, where in this answer $[x]$; $x$ a real
number, is defined to be the largest integer no larger than $x$.

$2.$ In general, if the line segment $L_{i+1}$ passes through $j$ horizontal grid lines, then the line segment $L_{i+1}$ passes through $j+1$ squares, unless $L_{i}$ ends at a corner that is.

$3.$ But as $\gcd(49,60)=1$ observe how the line segment $L$ from $(0,0)$ to $(60,49)$ starts and ends at a corner of a square but passes no corners in between. So none of $L_0, L_1,\ldots, L_{58}$ end at a corner.

$4.$ Finally, none of the squares are cut by more than one such $L_i$. Thus from 2. and 3. above the number of squares $L$ passes through is $59$ $+$ $48 +1=108$ [indeed $L_0+L_1+\ldots +L_{58}$ crosses in total $48$ horizontal grid lines none at the corners then finally $L_{59}$ ends at $(60,49)$ and cuts through exactly $1$ square--this we see via direct observation].

By similar reasoning as $1 ,2 ,3 ,4,$ above the number of squares the line segment $M=TS$ from $(60,49)$ to $(0,100)$ passes is $39 + 47 +2=88.$

So $108+88= 196$ total squares cut.

Answer 2

Imagine drawing a grid of horizontal and vertical parallel lines through the rectangle, so that they form the squares mentioned. Hence, there are $48$ horizontal lines interior to the rectangle, and $99$ vertical lines drawn.

For each crossing of the diagonal line across two adjacent vertical lines, a square is cut. Similarly, for each crossing of two adjacent horizontal lines, a square is cut. But if the diagonal line passes through a corner of a square--that is to say, simultaneously crossing a horizontal line and a vertical line, then only one square is cut, not two. For line $PT$, this occurs whenever $49$ and $60$ share a common divisor. But since they are relatively prime, $PT$ never intersects a corner, except at the beginning and end. So there are $60 + 49 - 1 = 108$ squares cut by $PT$, and a similar argument for $TS$ gives another $40 + 49 - 1 = 88$ squares, for a total of $196$ squares cut.

Another way to think about it is like playing a game where you can move on a $100 \times 49$ rectangular array of unit squares. From $P$ to $T$, you may either only move up or to the right. As stated before, because $\gcd(60, 49) = 1$, the path traced by the line does not intersect any square corners interior to the rectangle, so you cannot make any diagonal moves from one square to the next. Since each movement traverses exactly one shared edge between two adjacent squares, we have to make $59$ horizontal moves and $48$ vertical moves, but we must count the starting square, so we intersected $59 + 48 + 1 = 108$ squares from $P$ to $T$. Similarly, from $T$ to $S$, we must make $39$ horizontal and $48$ vertical moves, and again counting the starting square, this makes $88$ from $T$ to $S$.

Answer 3

For these types of problems, it helps to imagine the line during its journey from point $P$ to point $T$. It starts on one of the squares (so it cuts through a square from the beginning) and then, every time it crosses a vertical or horizontal line of the grid it's like it is moving on to cut another square. So the amount of squares it cuts would be the amount of lines it crosses plus one, this is $59+48+1$ for line $PT$ and $39+48+1$ for line $TS$, so they cut through a total of $196$ squares.

One corner case (literally) to watch out for is the line cutting through a corner, but we can easily prove this never happens. Suppose for contradiction that line $PT$ cuts through a corner, that is, it passes through one horizontal and one vertical line simultaneously. Call the point where all $3$ lines intersect $Y$. Also draw points $U$ and $G$ on $PS$ such that segments $YU$ and $TG$ are perpendicular to $PS$. Notice that right triangle $\triangle PTG$ is similar to $\triangle PYU$, and also notice both right triangles have integer size legs. Let $a, b$ be the length of sides $TG$ and $PG$ respectively, also let $c, d$ be the length of sides $YU$ and $PU$ respectively. Then there must exist $x$ such that $0 < x < 1$ and $ax = c$ and $bx = d$. Notice then that $x$ must be rational, we can write is as $x = \frac{p}{q}$ with $p$ and $q$ coprime. So $a\frac{p}{q}$ is an integer so $q$ divides $ap$, but since $q$ and $p$ are coprimes, $q$ divides $a$. Similarly $q$ divides $b$. But remember $a = 49$ and $b = 60$, this means and integer greater than $1$ divides both $49$ and $60$ and this contradicts the fact that $49$ and $60$ are coprime. Since we found a contradiction, our assumption that line $PT$ goes through a corner must be false. Same for line $TS$.