Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Twenty distinct points are marked on a circle and labeled $1$ through $20$ in clockwise order. A line segment is drawn between every pair of points whose labels differ by a prime number. Find the number of triangles formed whose vertices are among the original $20$ points.

Solution: As above, we must deduce that the sum of two primes must be equal to the third prime. Then, we can finish the solution using casework.
If the primes are $2,3,5$ , then the smallest number can range between $1$ and $15$ .
If the primes are $2,5,7$ , then the smallest number can range between $1$ and $13$ .
If the primes are $2,11,13$ , then the smallest number can range between $1$ and $7$ . 
If the primes are $2,17,19$ , then the smallest number can only be $1$
Adding all cases gets $15+13+7+1=36$ . However, due to the commutative property, we must multiply this by 2. For example, in the $2,17,19$ case the numbers can be $1,3,20$ or $1,18,20$ . Therefore the answer is $36\cdot2=\boxed{072}$

Answer:  072
Problem: Let $x_1\leq x_2\leq \cdots\leq x_{100}$ be real numbers such that $|x_1| + |x_2| + \cdots + |x_{100}| = 1$ and $x_1 + x_2 + \cdots + x_{100} = 0$ . Among all such $100$ -tuples of numbers, the greatest value that $x_{76} - x_{16}$ can achieve is $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: To find the greatest value of $x_{76} - x_{16}$ $x_{76}$ must be as large as possible, and $x_{16}$ must be as small as possible. If $x_{76}$ is as large as possible, $x_{76} = x_{77} = x_{78} = \dots = x_{100} > 0$ . If $x_{16}$ is as small as possible, $x_{16} = x_{15} = x_{14} = \dots = x_{1} < 0$ . The other numbers between $x_{16}$ and $x_{76}$ equal to $0$ . Let $a = x_{76}$ $b = x_{16}$ . Substituting $a$ and $b$ into $|x_1| + |x_2| + \cdots + |x_{100}| = 1$ and $x_1 + x_2 + \cdots + x_{100} = 0$ we get: \[25a - 16b = 1\] \[25a + 16b = 0\] $a = \frac{1}{50}$ $b = -\frac{1}{32}$
$x_{76} - x_{16} = a - b = \frac{1}{50} + \frac{1}{32} = \frac{41}{800}$ $m+n = \boxed{841}$

Answer:  841
Problem: Let $x_1\leq x_2\leq \cdots\leq x_{100}$ be real numbers such that $|x_1| + |x_2| + \cdots + |x_{100}| = 1$ and $x_1 + x_2 + \cdots + x_{100} = 0$ . Among all such $100$ -tuples of numbers, the greatest value that $x_{76} - x_{16}$ can achieve is $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Define $s_N$ to be the sum of all the negatives, and $s_P$ to be the sum of all the positives.
Since the sum of the absolute values of all the numbers is $1$ $|s_N|+|s_P|=1$
Since the sum of all the numbers is $0$ $s_N=-s_P\implies |s_N|=|s_P|$
Therefore, $|s_N|=|s_P|=\frac 12$ , so $s_N=-\frac 12$ and $s_P=\frac 12$ since $s_N$ is negative and $s_P$ is positive.
To maximize $x_{76}-x_{16}$ , we need to make $x_{16}$ as small of a negative as possible, and $x_{76}$ as large of a positive as possible.
Note that $x_{76}+x_{77}+\cdots+x_{100}=\frac 12$ is greater than or equal to $25x_{76}$ because the numbers are in increasing order.
Similarly, $x_{1}+x_{2}+\cdots+x_{16}=-\frac 12$ is less than or equal to $16x_{16}$
So we now know that $\frac 1{50}$ is the best we can do for $x_{76}$ , and $-\frac 1{32}$ is the least we can do for $x_{16}$
Finally, the maximum value of $x_{76}-x_{16}=\frac 1{50}+\frac 1{32}=\frac{41}{800}$ , so the answer is $\boxed{841}$

Answer:  841
Problem: A circle with radius $6$ is externally tangent to a circle with radius $24$ . Find the area of the triangular region bounded by the three common tangent lines of these two circles.

Solution: [asy] //Created by isabelchen  size(12cm, 12cm);  draw(circle((0,0),24)); draw(circle((30,0),6)); draw((72/5, 96/5) -- (40,0)); draw((72/5, -96/5) -- (40,0)); draw((24, 12) -- (24, -12)); draw((0, 0) -- (40, 0)); draw((72/5, 96/5) -- (0,0)); draw((168/5, 24/5) -- (30,0)); draw((54/5, 72/5) -- (30,0));  dot((72/5, 96/5)); label("$A$",(72/5, 96/5),NE); dot((168/5, 24/5)); label("$B$",(168/5, 24/5),NE); dot((24,0)); label("$C$",(24,0),NW); dot((40, 0)); label("$D$",(40, 0),NE); dot((24, 12)); label("$E$",(24, 12),NE); dot((24, -12)); label("$F$",(24, -12),SE); dot((54/5, 72/5)); label("$G$",(54/5, 72/5),NW); dot((0, 0)); label("$O_1$",(0, 0),S); dot((30, 0)); label("$O_2$",(30, 0),S);  [/asy]
$r_1 = O_1A = 24$ $r_2 = O_2B = 6$ $AG = BO_2 = r_2 = 6$ $O_1G = r_1 - r_2 = 24 - 6 = 18$ $O_1O_2 = r_1 + r_2 = 30$
$\triangle O_2BD \sim \triangle O_1GO_2$ $\frac{O_2D}{O_1O_2} = \frac{BO_2}{GO_1}$ $\frac{O_2D}{30} = \frac{6}{18}$ $O_2D = 10$
$CD = O_2D + r_2 = 10 + 6 = 16$
$EF = 2EC = EA + EB = AB = GO_2 = \sqrt{(O_1O_2)^2-(O_1G)^2} = \sqrt{30^2-18^2} = 24$
$DEF = \frac12 \cdot EF \cdot CD = \frac12 \cdot 24 \cdot 16 = \boxed{192}$

Answer:  192
Problem: A circle with radius $6$ is externally tangent to a circle with radius $24$ . Find the area of the triangular region bounded by the three common tangent lines of these two circles.

Solution: Let the center of the circle with radius $6$ be labeled $A$ and the center of the circle with radius $24$ be labeled $B$ . Drop perpendiculars on the same side of line $AB$ from $A$ and $B$ to each of the tangents at points $C$ and $D$ , respectively. Then, let line $AB$ intersect the two diagonal tangents at point $P$ . Since $\triangle{APC} \sim \triangle{BPD}$ , we have \[\frac{AP}{AP+30}=\frac14 \implies AP=10.\] Next, throw everything on a coordinate plane with $A=(0, 0)$ and $B = (30, 0)$ . Then, $P = (-10, 0)$ , and if $C = (x, y)$ , we have \[(x+10)^2+y^2=64,\] \[x^2+y^2=36.\] Combining these and solving, we get $(x, y)=\left(-\frac{18}5, \frac{24}5\right)$ . Notice now that $P$ $C$ , and the intersections of the lines $x=6$ (the vertical tangent) with the tangent containing these points are collinear, and thus every slope between a pair of points will have the same slope, which in this case is $\frac{-\frac{18}5+10}{\frac{24}5}=\frac34$ . Thus, the other two vertices of the desired triangle are $(6, 12)$ and $(6, -12)$ . By the Shoelace Formula, the area of a triangle with coordinates $(-10, 0)$ $(6, 12)$ , and $(6, -12)$ is \[\frac12|-120-0-72-72+0-120|=\boxed{192}.\]

Answer:  192
Problem: A circle with radius $6$ is externally tangent to a circle with radius $24$ . Find the area of the triangular region bounded by the three common tangent lines of these two circles.

Solution: [asy] //Created by isabelchen and edited by afly  size(12cm, 12cm);  draw(circle((0,0),24)); draw(circle((30,0),6)); draw((72/5, 96/5) -- (40,0)); draw((72/5, -96/5) -- (40,0)); draw((24, 12) -- (24, -12)); draw((0, 0) -- (40, 0)); draw((72/5, 96/5) -- (0,0)); draw((168/5, 24/5) -- (30,0)); draw((54/5, 72/5) -- (30,0)); draw((30,0)--(30,15/2));  dot((72/5, 96/5)); label("$A$",(72/5, 96/5),NE); dot((168/5, 24/5)); label("$B$",(168/5, 24/5),NE); dot((24,0)); label("$C$",(24,0),NW); dot((40, 0)); label("$D$",(40, 0),NE); dot((24, 12)); label("$E$",(24, 12),NE); dot((24, -12)); label("$F$",(24, -12),SE); dot((54/5, 72/5)); label("$G$",(54/5, 72/5),NW); dot((0, 0)); label("$O_1$",(0, 0),S); dot((30, 0)); label("$O_2$",(30, 0),S); dot((30,15/2)); label("$H$",(30,15/2),N);  label("$x$",(30,0)--(40,0),N); [/asy]
First, we want to find $O_2D$ . We know that $\angle O_1AD = \angle O_2BD = 90^{\circ}$ , so by AA similarity, $\triangle O_1AD \sim \triangle O_2BD$ . We want to find the length of $x$ , and using the similar triangles, we write an equation: $\frac{30 + x}{4} = x$ . Solving, we get $x=10$ . Therefore, $CD = 10 + 6 = 16$ . Next, we find that using AA similarity, $\triangle O_2BD \sim \triangle HO_2D \sim \triangle ECD$ and they are 3-4-5 triangles. We can quickly compute $EF = 2EC = 2 \cdot \left( \frac{3}{4} \cdot 16 \right) = 2 \cdot 12 = 24$ . Therefore, the area is $\frac{1}{2} \cdot 16 \cdot 24 = \boxed{192}$

Answer:  192
Problem: Find the number of positive integers $n \le 600$ whose value can be uniquely determined when the values of $\left\lfloor \frac n4\right\rfloor$ $\left\lfloor\frac n5\right\rfloor$ , and $\left\lfloor\frac n6\right\rfloor$ are given, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to the real number $x$

Solution: We need to find all numbers between $1$ and $600$ inclusive that are multiples of $4$ $5$ , and/or $6$ which are also multiples of $4$ $5$ , and/or $6$ when $1$ is added to them.
We begin by noting that the LCM of $4$ $5$ , and $6$ is $60$ .  We can therefore simplify the problem by finding all such numbers described above between $1$ and $60$ and multiplying the quantity of such numbers by $10$ $600$ $60$ $10$ ).
After making a simple list of the numbers between $1$ and $60$ and going through it, we see that the numbers meeting this condition are $4$ $5$ $15$ $24$ $35$ $44$ $54$ , and $55$ .  This gives us $8$ numbers. $8$ $10$ $\boxed{080}$

Answer:  080
Problem: Find the number of positive integers $n \le 600$ whose value can be uniquely determined when the values of $\left\lfloor \frac n4\right\rfloor$ $\left\lfloor\frac n5\right\rfloor$ , and $\left\lfloor\frac n6\right\rfloor$ are given, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to the real number $x$

Solution: This is Solution 1 with a slick element included.  Solution 1 uses the concept that $60k+l$ is a solution for $n$ if $60k+l$ is a multiple of $3$ $4$ , and/or $5$ and $60k+l+1$ is a multiple of $3$ $4$ , and/or $5$ for positive integer values of $l$ and essentially any integer value of $k$ .  But keeping the same conditions in mind for $k$ and $l$ , we can also say that if $60k+l$ is a solution, then $60k-l-1$ is a solution!  Therefore, one doesn't have to go as far as determining the number of values between $1$ and $60$ and then multiplying by $10$ .  One only has to determine the number of values between $1$ and $30$ and then multiply by $20$ .  The values of $n$ that work between $1$ and $30$ are $4$ $5$ $15$ , and $24$ .  This gives us $4$ numbers. $4$ $20$ $\boxed{080}$

Answer:  080
Problem: Find the number of positive integers $n \le 600$ whose value can be uniquely determined when the values of $\left\lfloor \frac n4\right\rfloor$ $\left\lfloor\frac n5\right\rfloor$ , and $\left\lfloor\frac n6\right\rfloor$ are given, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to the real number $x$

Solution: 1. For $n$ to be uniquely determined, $n$ AND $n + 1$ both need to be a multiple of $4, 5,$ or $6.$ Since either $n$ or $n + 1$ is odd, we know that either $n$ or $n + 1$ has to be a multiple of $5.$ We can state the following cases:
1. $n$ is a multiple of $4$ and $n+1$ is a multiple of $5$
2. $n$ is a multiple of $6$ and $n+1$ is a multiple of $5$
3. $n$ is a multiple of $5$ and $n+1$ is a multiple of $4$
4. $n$ is a multiple of $5$ and $n+1$ is a multiple of $6$
Solving for each case, we see that there are $30$ possibilities for cases 1 and 3 each, and $20$ possibilities for cases 2 and 4 each. However, we over-counted the cases where
1. $n$ is a multiple of $24$ and $n+1$ is a multiple of $5$
2. $n$ is a multiple of $5$ and $n+1$ is a multiple of $24$
Each case has $10$ possibilities.
Adding all the cases and correcting for over-counting, we get $30 + 20 + 30 + 20 - 10 - 10 = \boxed{080}.$

Answer:  080
Problem: Find the number of positive integers $n \le 600$ whose value can be uniquely determined when the values of $\left\lfloor \frac n4\right\rfloor$ $\left\lfloor\frac n5\right\rfloor$ , and $\left\lfloor\frac n6\right\rfloor$ are given, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to the real number $x$

Solution: Here is a detailed solution for Solution 2.
Over-counted cases:
$30 + 20 + 30 + 20 - 10 - 10 = \boxed{080}$

Answer:  080
Problem: Find the number of positive integers $n \le 600$ whose value can be uniquely determined when the values of $\left\lfloor \frac n4\right\rfloor$ $\left\lfloor\frac n5\right\rfloor$ , and $\left\lfloor\frac n6\right\rfloor$ are given, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to the real number $x$

Solution: The problem is the same as asking how many unique sets of values of $\lfloor\frac{n}{4}\rfloor$ $\lfloor\frac{n}{5}\rfloor$ , and $\lfloor\frac{n}{6}\rfloor$ can be produced by one and only one value of $n$ for positive integers $n$ less than or equal to 600.
Seeing that we are dealing with the unique values of the floor function, we ought to examine when it is about to change values, for instance, when $n$ is close to a multiple of 4 in $\lfloor\frac{n}{4}\rfloor$
For a particular value of $n$ , let $a$ $b$ , and $c$ be the original values of $\lfloor\frac{n}{4}\rfloor$ $\lfloor\frac{n}{5}\rfloor$ , and $\lfloor\frac{n}{6}\rfloor$ , respectively.
Notice when $n$ $\equiv0\mod4$ and $n$ $\equiv4\mod5$ , the value of $\lfloor\frac{n-1}{4}\rfloor$ will be 1 less than the original $a$ . The value of $\lfloor\frac{n+1}{5}\rfloor$ will be 1 greater than the original value of $b$
More importantly, this means that no other value less than or greater than $n$ will be able to produce the set of original values of $a$ $b$ , and $c$ , since they make either $a$ or $b$ differ by at least 1.
Generalizing, we find that $n$ must satisfy:
Where $j$ and $k$ are pairs of distinct values of 4, 5, and 6.
Plugging in the values of $j$ and $k$ , finding the solutions to the 6 systems of linear congruences, and correcting for the repeated values, we find that there are $\boxed{080}$ solutions of $n$

Answer:  080
Problem: Find the number of positive integers $n \le 600$ whose value can be uniquely determined when the values of $\left\lfloor \frac n4\right\rfloor$ $\left\lfloor\frac n5\right\rfloor$ , and $\left\lfloor\frac n6\right\rfloor$ are given, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to the real number $x$

Solution: By Chinese Remainder Theorem , the general solution of systems of $2$ linear congruences is:
$lcm[4, 5, 6] = 60$ , we solve the number of values for $n \le 60$ , then multiply by $10$ to get the number of values for $n \le 600$ . We are going to solve the following $6$ systems of linear congruences:
$n \in \{ 4, 5, 15, 24, 35, 44, 54, 55 \}$ , there are $8$ values for $n \le 60$ . For $n \le 600$ , the answer is $8 \cdot 10 = \boxed{080}$

Answer:  080
Problem: Let $\ell_A$ and $\ell_B$ be two distinct parallel lines. For positive integers $m$ and $n$ , distinct points $A_1, A_2, \allowbreak A_3, \allowbreak \ldots, \allowbreak A_m$ lie on $\ell_A$ , and distinct points $B_1, B_2, B_3, \ldots, B_n$ lie on $\ell_B$ . Additionally, when segments $\overline{A_iB_j}$ are drawn for all $i=1,2,3,\ldots, m$ and $j=1,\allowbreak 2,\allowbreak 3, \ldots, \allowbreak n$ , no point strictly between $\ell_A$ and $\ell_B$ lies on more than 1 of the segments. Find the number of bounded regions into which this figure divides the plane when $m=7$ and $n=5$ . The figure shows that there are 8 regions when $m=3$ and $n=2$ [asy] import geometry; size(10cm); draw((-2,0)--(13,0)); draw((0,4)--(10,4)); label("$\ell_A$",(-2,0),W); label("$\ell_B$",(0,4),W); point A1=(0,0),A2=(5,0),A3=(11,0),B1=(2,4),B2=(8,4),I1=extension(B1,A2,A1,B2),I2=extension(B1,A3,A1,B2),I3=extension(B1,A3,A2,B2); draw(B1--A1--B2); draw(B1--A2--B2); draw(B1--A3--B2); label("$A_1$",A1,S); label("$A_2$",A2,S); label("$A_3$",A3,S); label("$B_1$",B1,N); label("$B_2$",B2,N); label("1",centroid(A1,B1,I1)); label("2",centroid(B1,I1,I3)); label("3",centroid(B1,B2,I3)); label("4",centroid(A1,A2,I1)); label("5",(A2+I1+I2+I3)/4); label("6",centroid(B2,I2,I3)); label("7",centroid(A2,A3,I2)); label("8",centroid(A3,B2,I2)); dot(A1); dot(A2); dot(A3); dot(B1); dot(B2); [/asy]

Solution: We can use recursion to solve this problem:
1. Fix 7 points on $\ell_A$ , then put one point $B_1$ on $\ell_B$ . Now, introduce a function $f(x)$ that indicates the number of regions created, where x is the number of points on $\ell_B$ . For example, $f(1) = 6$ because there are 6 regions.
2. Now, put the second point $B_2$ on $\ell_B$ . Join $A_1~A_7$ and $B_2$ will create $7$ new regions (and we are not going to count them again), and split the existing regions. Let's focus on the spliting process: line segment formed between $B_2$ and $A_1$ intersect lines $\overline{B_1A_2}$ $\overline{B_1A_3}$ , ..., $\overline{B_1A_7}$ at $6$ points $\Longrightarrow$ creating $6$ regions (we already count one region at first), then $5$ points $\Longrightarrow$ creating $5$ regions (we already count one region at first), 4 points, etc. So, we have: \[f(2) = f(1) + 7 + (6+5+...+1) = 34.\]
3. If you still need one step to understand this: $A_1~A_7$ and $B_3$ will still create $7$ new regions. Intersecting \[\overline{A_2B_1}, \overline{A_2B_2};\] \[\overline{A_3B_1}, \overline{A_3B_2};\] \[...\] \[\overline{A_7B_1}, \overline{A_7B_2}\] at $12$ points, creating $12$ regions, etc. Thus, we have: \[f(3) = f(2)+7+(12+10+8+...+2)=34+7+6\cdot 7=83.\]
Yes, you might already notice that: \[f(n+1) = f(n)+7+(6+5+...+1)\cdot n = f(n) + 7 + 21n.\]
5. Finally, we have $f(4) = 153$ , and $f(5)=244$ . Therefore, the answer is $\boxed{244}$

Answer:  244
Problem: Let $\ell_A$ and $\ell_B$ be two distinct parallel lines. For positive integers $m$ and $n$ , distinct points $A_1, A_2, \allowbreak A_3, \allowbreak \ldots, \allowbreak A_m$ lie on $\ell_A$ , and distinct points $B_1, B_2, B_3, \ldots, B_n$ lie on $\ell_B$ . Additionally, when segments $\overline{A_iB_j}$ are drawn for all $i=1,2,3,\ldots, m$ and $j=1,\allowbreak 2,\allowbreak 3, \ldots, \allowbreak n$ , no point strictly between $\ell_A$ and $\ell_B$ lies on more than 1 of the segments. Find the number of bounded regions into which this figure divides the plane when $m=7$ and $n=5$ . The figure shows that there are 8 regions when $m=3$ and $n=2$ [asy] import geometry; size(10cm); draw((-2,0)--(13,0)); draw((0,4)--(10,4)); label("$\ell_A$",(-2,0),W); label("$\ell_B$",(0,4),W); point A1=(0,0),A2=(5,0),A3=(11,0),B1=(2,4),B2=(8,4),I1=extension(B1,A2,A1,B2),I2=extension(B1,A3,A1,B2),I3=extension(B1,A3,A2,B2); draw(B1--A1--B2); draw(B1--A2--B2); draw(B1--A3--B2); label("$A_1$",A1,S); label("$A_2$",A2,S); label("$A_3$",A3,S); label("$B_1$",B1,N); label("$B_2$",B2,N); label("1",centroid(A1,B1,I1)); label("2",centroid(B1,I1,I3)); label("3",centroid(B1,B2,I3)); label("4",centroid(A1,A2,I1)); label("5",(A2+I1+I2+I3)/4); label("6",centroid(B2,I2,I3)); label("7",centroid(A2,A3,I2)); label("8",centroid(A3,B2,I2)); dot(A1); dot(A2); dot(A3); dot(B1); dot(B2); [/asy]

Solution: We want to derive a general function $f(m,n)$ that indicates the number of bounded regions. Observing symmetry, we know this is a symmetric function about $m$ and $n$ . Now let's focus on $f(m+1, n)-f(m, n)$ , which is the difference caused by adding one point to the existing $m$ points of line $\ell_A$ . This new point, call it #m, when connected to point #1 on $\ell_B$ , crosses $m*(n-1)$ lines, thus making additional $m*(n-1)+1$ bounded regions; when connected to point #2 on $\ell_B$ , it crosses $m*(n-2)$ lines, thus making additional $m*(n-2)+1$ bounded regions; etc. By simple algebra/recursion methods, we see
$f(m+1, n)-f(m, n)=m*\frac{n(n-1)}{2} +n$
Notice $f(1,n)=n-1$ . Not very difficult to figure out:
$f(m, n)=\frac{m(m-1)n(n-1)}{4} +mn-1$
The fact that $f(3,2)=8$ makes us more confident about the formula. Now plug in $m=5, n=7$ , we get the final answer of $\boxed{244}$

Answer:  244
Problem: Let $\ell_A$ and $\ell_B$ be two distinct parallel lines. For positive integers $m$ and $n$ , distinct points $A_1, A_2, \allowbreak A_3, \allowbreak \ldots, \allowbreak A_m$ lie on $\ell_A$ , and distinct points $B_1, B_2, B_3, \ldots, B_n$ lie on $\ell_B$ . Additionally, when segments $\overline{A_iB_j}$ are drawn for all $i=1,2,3,\ldots, m$ and $j=1,\allowbreak 2,\allowbreak 3, \ldots, \allowbreak n$ , no point strictly between $\ell_A$ and $\ell_B$ lies on more than 1 of the segments. Find the number of bounded regions into which this figure divides the plane when $m=7$ and $n=5$ . The figure shows that there are 8 regions when $m=3$ and $n=2$ [asy] import geometry; size(10cm); draw((-2,0)--(13,0)); draw((0,4)--(10,4)); label("$\ell_A$",(-2,0),W); label("$\ell_B$",(0,4),W); point A1=(0,0),A2=(5,0),A3=(11,0),B1=(2,4),B2=(8,4),I1=extension(B1,A2,A1,B2),I2=extension(B1,A3,A1,B2),I3=extension(B1,A3,A2,B2); draw(B1--A1--B2); draw(B1--A2--B2); draw(B1--A3--B2); label("$A_1$",A1,S); label("$A_2$",A2,S); label("$A_3$",A3,S); label("$B_1$",B1,N); label("$B_2$",B2,N); label("1",centroid(A1,B1,I1)); label("2",centroid(B1,I1,I3)); label("3",centroid(B1,B2,I3)); label("4",centroid(A1,A2,I1)); label("5",(A2+I1+I2+I3)/4); label("6",centroid(B2,I2,I3)); label("7",centroid(A2,A3,I2)); label("8",centroid(A3,B2,I2)); dot(A1); dot(A2); dot(A3); dot(B1); dot(B2); [/asy]

Solution: Let some number of segments be constructed. We construct a new segment. We start from the straight line $l_B.$ WLOG from point $B_3.$ Segment will cross several existing segments (points $A,B,C,...$ ) and enter one of the points of the line $l_A (A_1).$
Each of these points adds exactly 1 new bounded region (yellow bounded regions).
The exception is the only first segment $(A_1 B_1),$ which does not create any bounded region. 
Thus, the number of bounded regions is $1$ less than the number of points of intersection of the segments plus the number of points of arrival of the segments to $l_A.$
Each point of intersection of two segments is determined uniquely by the choice of pairs of points on each line.
The number of such pairs is $\dbinom{n}{2} \cdot \dbinom{m}{2}.$
Exactly one segment comes to each of the $n$ points of the line $l_A$ from each of the $m$ points of the line $l_B.$ The total number of arrivals is equal to $mn.$ Hence, the total number of bounded regions is $N = \dbinom{n}{2} \cdot \dbinom{m}{2} + mn – 1.$
We plug in $m=5, n=7$ , we get the final answer of $\boxed{244}$

Answer:  244