Line through a Square

Put $A(0,0); B(1,0); C(1,1); D(0,1)$ . Assume $P$ is on $AB$ such that $\frac{BP}{BA}=k$ , where $k \in [0,1]$ . Then $C$ and $A$ are on the same side of $PQ$ if and only if Q is contained in the triangle $BPC$ , which occurs with probability $p=\frac{[BPC]}{[ABCD]}=\frac{k}{2}$ . This is the same as the uniform distribution on $[0,1]$ , so the probability is the same as the expected value of $\frac{[BPC]}{[ABCD]}$ . The expected value of $[BPC]$ is $\frac{1}{2}[BAC]$ so the probability is $\frac{[BAC]}{2[ABCD]}=\frac{1}{4}$ . We follow the same deduction for $P$ on $BC$ , $CD$ , $DA$ and have the same reesult.

Label $A=(0,0), B=(0,1), C=(1,1), D=(0,1)$ . The situation of $Q$ on the four sides are symmetrical with respect to the diagonal $AC$ , therefore we can assume $Q$ to be randomly chosen just from one side, e.g. the segment $AB$ .

When $Q=(x,0)$ , the points $P$ such that $A$ and $C$ are on the same side of the line $PQ$ are those points that are in the triangle $QBC$ , with area $x/2$ . So the desired probability is

$\int_0^1 \frac{[QBC]}{[ABCD]}dx = \int_0^1 \frac{x}{2}dx = \frac{1}{4}.$

For simplicity let us denote the corners of the square as: $O=(0,0$ ), $N=(0,1)$ , $M=(1,1)$ , and $L=(1,0)$ .

We are concerned with the probability of event $A$ where $A$ is the event that $O$ and $M$ are on the same side of the line $L_1$ determined by $P$ and $Q$ .

We note that $P$ is equally likely to be on each side, i.e, if $P_{RS}$ is the event that $P$ is on side $RS$ , then $P(P_{RS})=\frac{1}{4}$ for $RS\in\Omega=\{LM,MN,NO,OL\}$ . The $P_{RS}$ 's also form a partition of our probability space.

Let us denote by $A_{RS}$ the probability that $O$ and $M$ are on the same side of $L_1$ given that $P$ is on $RS\in\Omega$ $\Rightarrow P(A_{RS})=P(A|P_{RS})=\displaystyle\frac{P(A\cap P_{RS})}{P(P_{RS})}$ $\Rightarrow P(A\cap P_{RS})=P(A_{RS})\times P(P_{RS})$ . It should be noted that $P(A)=\displaystyle\sum_{RS\in\Omega}{P(A\cap P_{RS})}$

Let us consider that $P$ is on $MN$ . Then $P(x,1)$ for $x\in[0,1]$ . Now, we place $Q$ . The event $A_{MN}$ is then equivalent to the event that $Q$ is inside the region $\Delta NOP$ . The area of $\Delta NOP$ is $\frac{x}{2}$ . Then, $P(A_{MN})=\displaystyle\int_0^1{\frac{x}{2}dx}$ since the area of the square is $1$ .

Thus, $P(A_{MN})=\frac{1}{4}\Rightarrow P(A\cap P_{MN})=\frac{1}{16}$ . In a similar fashion, by considering $\Delta LOP$ , $P(A_{LM})=\frac{1}{4}\Rightarrow P(A\cap P_{LM})=\frac{1}{16}$ .

Likewise, let us consider that $P$ is on $ON$ with $P(0,y)$ for $y\in[0,1]$ . The event $A_{ON}$ is then equivalent to the event that $Q$ is inside the region $\Delta PMN$ with area $\frac{1-y}{2}$ $\Rightarrow P(A_{ON})=\displaystyle\int_0^1{\left(\frac{1}{2}-\frac{y}{2}\right)dy}=\frac{1}{4}$ $\Rightarrow P(A\cap P_{ON})=\frac{1}{16}$ . Again, we can obtain $P(A\cap P_{OL})=\frac{1}{16}$ in a similar way.

Therefore, $P(A)=4\times\frac{1}{16}=\frac{1}{4}$ and the answer to the question, $5$ , follows.

Name the points (0,0) as A, (1,0) as B, (1,1) as C and (0,1) as D. So for (1,1) and (0,0) to lie on 1 side of $L_{1}$ , there has to 2 cases:

• P lies on AB or BC and Q lies in $\bigtriangleup$ ABC
• Q lies on AD or CD and Q lies in $\bigtriangleup$ ADC

We know that that AB=BC=CD=AD=1 and area $\bigtriangleup$ ABC=area $\bigtriangleup$ ADC =1/2.

Favourable cases for $\bigtriangleup$ ABC:

(P lying on AB $\times$ Q lying in $\bigtriangleup$ ABC) + (P lying on BC $\times$ Q lying in $\bigtriangleup$ ABC) =

(length of AB $\times$ area of $\bigtriangleup$ ABC) + (length of BC $\times$ area of $\bigtriangleup$ ABC) =

[(1 $\times$ $\frac {1}{2}$ ) + (1 $\times$ $\frac {1}{2}$ )] / 2 = $\frac {1}{2}$

It is to be noted that the equation is further divided by 2, since it includes each case of $L_{1}$ twice. As $L_{1}$ intersects both AB and BC, both the intersecting points can be taken as P, therefore the cases are halved to eliminate redundancy when P is on AB or BC.

Similarly favourable cases for $\bigtriangleup$ ADC):

(P lying on AD $\times$ Q lying in $\bigtriangleup$ ADC) + (P lying on CD $\times$ Q lying in $\bigtriangleup$ ADC) =

(length of AD $\times$ area of $\bigtriangleup$ ADC) + (length of CD $\times$ area of $\bigtriangleup$ ADC) =

[(1 $\times$ $\frac {1}{2}$ ) + (1 $\times$ $\frac {1}{2}$ )] / 2 = $\frac {1}{2}$

Therefore total favourable cases by adding we have $\frac {1}{2}$ + $\frac {1}{2}$ = 1.

For total cases in square ABCD: (P lying on AB $\times$ Q lying in ABCD) + (P lying on BC $\times$ Q lying in ABCD) + (P lying on CD $\times$ Q lying in ABCD) + (P lying on AD $\times$ Q lying in ABCD) =

(length of AB $\times$ area of ABCD) + (length of BC $\times$ area of ABCD) + (length of CD $\times$ area of ABCD) + (length of AD $\times$ area of ABCD) =

(1 $\times$ 1) + (1 $\times$ 1) + (1 $\times$ 1) + (1 $\times$ 1) =

1+1+1+1=4 = Total cases.

Probability = $\frac {Favourable Cases}{Total Cases}$ = $\frac {1}{4}$ .

Since 1 & 4 are coprime, their sum, i.e ans = 1+4 =5

Since both halves of the square divided by the line (0,0) (1,1) are the same, the probabilities are the same. So, we assume that that P is on the point x on line (1,0) to (0,0). It doesn't matter since (1,0) to (0,0) is the same length as (1,0) to (1,1). Then, we see that the maximum "boundary" Q can be on is point (1,1) itself. thus the summation of the area that point Q can be drawn is \frac {1}{2} \times 1 \times x which is \frac {x}{2}. thus we integrate \frac {x}{2} from 1 to zero and we get \frac {1}{4} and 1 + 4 =5

We may ignore the case when $(0,0)$ or $(1,1)$ is on the line $L_1$ , since this happens with probability 0.

By symmetry, we may assume that $P$ lies on the $x$ -axis. Let $p$ be the $y$ - coordinate of $P$ . If we draw the line $L_2$ from $(1,1)$ to $P$ , we will have $(0,0)$ and $(1,1)$ on the same side of $L_1$ if and only if $Q$ is to the right of line $L_2$ . The area of this triangle is $\frac{1-p}{2}$ .

We can construct a new unit square where the $x$ -axis represents the value of the point $p$ and the $y$ -coordinate represents the probability that $Q$ will make $L_1$ have the desired property. The square in the figure below represents this, and the green region represents when the line $L_1$ will have the desired property. This area is $\frac{1}{4}$ the area of the square, so the probability is $\frac{1}{4}$ , and $a + b = 1 + 4 = 5$ .

After we draw the square with given points on the coordinate plane,it is clear that the points (0,0) & (1,1) are both on the same side of L1 only when both points P & Q are on the same side of the diagonal joining (0,0)& (1,1).It means point P must lie on two adjacent sides,whose probability= 2/4 = 1/2.Now point Q must lie on either side of the diagonal joining (0,0) & (1,1),whose probability =1/2.Hence net probability = (1/2) *(1/2) =1/4 which implies that a+b=1+4=5

We are asked to find the probability that the points (0,0) and (1,1) are both on the same side of L1, so, the point P and Q should be at the same side of the diagonal joining (0,0) and (1,1) The point P must lie on either on the side joining (0,1) and (1,1) or on the side joining (0,0) and (1,0), so we have 2 ways for this or its probability is 2/4 or 1/2 Now, the point Q must either lie on the upper part of the diagonal or on the lower part of the diagonal, so its probability is 1/2

Thus, the required probability is 1/2*1/2=1/4 so, a+b=1+4=5

THE POINTS (0,0) AND (1,1) WILL BE ON THE SAME SIDE OF THE LINE (L 1) WHEN THE POINT P AND Q BOTH ARE PRESENT ON THE SAME SIDE OF THE DIAGONAL JOINING (0,0),(1,1).THEN THE POINT P MUST LIE ON THE TWO ADJACENT SIDES i.e 2/4 = 1/2

now q must lie on either one half of the diagonal so as the diagonal divides the square in 2 equal halves so probability is 1/2.

now net probability = (1/2) *(1/2)