An intersection of geometry and combinatorics

If the terminal point of the blue segment is $(u, v)$ (while the other point is $(1, 0)$ ), then the probability that the two segments won't intersect is the the probability that the terminal red point is in either above the line $y=\frac { v }{ u } x$ , or below both of the line $y=\frac { v }{ u } x$ and the blue segment.

Thus, it remains to integrate this area for all the possible terminal blue points $(u, v)$ in the given square; there are two cases to compute this integral.

The first case, if $u \le v$ , then the area will be $\frac {1}{2} (v+\frac { u }{ v } )$ and the probability is $\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ \int _{ 0 }^{ v }{ (v+ \frac{u}{v})\quad du} dv } =\frac { 7 }{ 24 }$

The second case, if $u > v$ , then the area will be $\frac { 1 }{ 2 } (1+(1-\frac { v }{ u } )+v)$ and the probability is $\frac {1}{2}\int _{ 0 }^{ 1 }{ \int _{ 0 }^{ u }{ (1+(1-\frac { v }{ u } ) +v)\quad dv} du} =\frac { 11 }{ 24 }$

Therefore the probability that both lines intersect is $1-(\frac { 11 }{ 24 } +\frac { 7 }{ 24 } ) = \frac { 1 }{4}$

A slightly more intuitive answer:

We order all points within the square first by y-coordinate, and secondly by x-coordinate. We'll describe all red and blue segments by the cross product of all unique ordered pairs $(P_1, P_2)$ of points in the square, with the assignments $P_1$ red, $P_2$ blue, and vice versa.

Fix $P_1$ , and let $y_1$ be its y-coordinate. $P_2$ can thus be any point in an area of size $y_1$ . (Note that $P_2$ might have the same y-coordinate as $P_1$ , but these cases measure as nothing compared to the remaining area.) Note, as well, that the triangle with points $(0, 0)$ , $P_1$ , and $(1, 0)$ has area $\frac{y_1}{2}$ .

$P_2$ has probability $\frac{1}{2}$ of falling inside the triangle, in which case, neither red-blue assignment will yield an intersection. If it falls outside the triangle, one red-blue assignment will yield an intersection, and the other will not. Thus, for arbitrarily fixed $P_1$ , the probability that a uniformly chosen $(P_2, \text{red-blue-choice})$ pair will result in an intersection is exactly $\frac{1}{4}$ .

Trivially integrate this result across all possible $P_1$ , and we get our final answer, $\frac{1}{4}$ .

Mostafa has provided an excellent solution. Here's another approach.

Instead of calculating the probability of intersection directly, I'll first calculate the probability that the line segments do not intersect. We label the points $O(0,0), P, Q, R(1,0)$ , where $P, Q$ are the first and second random points chosen, respectively. If these $4$ points are not in convex position, then either $P$ lies inside triangle $OQR$ or $Q$ lies inside triangle $OPR$ , and in both of these cases the line segments $OP$ and $QR$ will not meet. Each of these cases occurs with probability $\int_{0}^{1} \frac{1}{2} y dy = \frac{1}{4}$ , for a combined probability of $2 * \frac{1}{4} = \frac{1}{2}$ . (The integral simply evaluates the expected fraction of the square occupied by, in turn, triangle $OPR$ and triangle $OQR$ .)

The probability that the $4$ points are in convex position is then $1 - \frac{1}{2} = \frac{1}{2}$ . Now if they are in convex position, then it is equally likely that $OP$ and $QR$ will form either the sides of the convex hull or the diagonals. In the former case the line segments will not intersect, and in the latter they will. Thus the probability that the points are in convex position and do not intersect is $\frac{1}{2} * \frac{1}{2} = \frac{1}{4}$ .

Combining these two results gives us a probability of $\frac{3}{4}$ that the line segments will not intersect, and thus a probability of $\frac{1}{4}$ that they will. This makes $a = 1, b = 4$ and so $a + b = \boxed{5}$ .

Comment: By "in convex position" I mean that the $4$ points are the vertices of their convex hull, which in turn is the smallest convex quadrilateral that contains the $4$ points.

Consider the first point $(a,b)$ . Notice that the area of the region that the second line crosses the first is a triangle where the points $(0,0)$ and $(0,1)$ create the base, and the height is simply $a$ . Thus the area of the region is $a/2$ , and dividing this over the total, 1, we get $a/2$ . Since $a$ varies equally from $0$ to $1$ , we can take the average, $a=\frac{1}{2}$ , and our answer is $\frac{1}{4}$ . Thus $\boxed {5}$

i think its a simple solution, but im struggling to explain this properly, so please bear with me. consider any point pair of points P{(a,b),(c,d)} in the square. from P, it is possible to draw two pairs of lines to the vertices of the lowermost edge. identify L1(from (0,0) to (a,b)) L2(from (1,0) to (c,d)) as one pair; and M1(from 0,0 to c,d) and M2(from 1,0 to a,b) as another. in a similar manner, pairs of lines can be identified for EACH edge of the square. it can be proved by geometry that for every P, 2 pairs of such lines MUST intersect, and these two pairs will in fact belong to opposite sides of the square.

But we observe, that for all of the pairs of lines that have been drawn on lines other than the lowermost edge, an identical situation is obtained by rotating the "innards" of the square (without changing the coordinates), to obtain a different pair of points P', with each point in an different quadrant of the square, but having lines drawn from the base edge.

Summarizing, for every pair P, 2 lines out of 8 drawn to the sides of the square intersect. By rotating the square, we can make the other edges coincide with the lowermost one. Thus, for every pair selected there exist four other pairs corresponding to each of the other four quadrants such that 2 OUT OF 8 OF THE PAIRS OF LINES, DRAWN TO THE BASE, INTERSECT. Each point belongs to a unique set of four such pairs of points. therefore the probability is 2/8 for each such set of four pairs, and owing to the mutually exclusive and exhaustive nature of such pairs, so is the probability of intersection.

If the first point is (x1, y1) and the second is (x2, y2), then once the first point is picked, x2 has to be less than x1...in a uniform distribution, the probability is 1/2...then, y2 has to be greater than y1, probability = 1/2...so 1/4 is the probability, and the answer to the question is 5..

as we can see wherever i select the first random point, i get the same answer(according to question) .so jst select the first random point in the midpoint of square. then, area where i can get the second random P(lines will intersect)=point to cut the line(think geometrically) / total area =1/4 /1 =1/4 =a/b that gives ,a+b =5 this is a particular solution not a general, however u can use it jst to get the correct answer in minimum time but its not for knowledge concern.

red point=terminal of red line segment

blue point=terminal of blue line segment

There'll be two cases;

$\boxed{CASE(i)}$ Suppose I choose red point s.t. x+y>1 where x,y are the coordinates of red point, then for the 2 segments to intersect blue point must be chosen in the region bounded by

->red line segment ->x=0 ->y=1 and

-> line segment passing through (1,0) and red point (CONVINCE YOURSELF)

Note: area of this fig.= ar(trepezium) - ar(triangle)

probability of choosing red point is (dxdy)/1 and blue point is (area of the above mentioned figure)/1

so for this case prob. comes out to be..

$\int\limits_0^1 \int\limits_{1-x}^1 (1+x+((x-1)(1-y)/y))/2 - y/2) dxdy = 5/24$

here x goes from 0 to 1 and y goes from 1-x to 1

$\boxed{CASE(ii)}$ Now I choose red point s.t. x+y<1, then segments would intersect iff blue point is chosen in the region bounded by

->x=0 ->red line segment -> -> line segment passing through (1,0) and red point

In this case its a triangle. Using similar arguments as above probability comes out to be

$\int\limits_0^1 \int\limits_{1-x}^1 ((y/2 + xy/1-x)x) dxdy = 1/24$

therefore summing the prob. ans = 5/24 + 1/24 =1/4 $\Rightarrow$ 4+1 = 5