Finding number of diagonals of polygon knowing number of points of intersection In a polygon, no three diagonals are concurrent. If the total number of points of intersection of the diagonals be 70, then the number of diagonals in the polygon is:
My attempt:
Let the number of diagonals in the polygon be $n$. Then, the number of points of intersection given that no three are concurrent will be ${ n \choose 2 } = 70$. However, this has no solutions. I understand this may be wrong since some of the diagonals may be parallel to each other, hence having no point of intersection.
How do I approach this problem? Hints or solutions are appreciated.
EDIT: @user3733558's answer is correct and matches what is given in my book, and I understand the approach used as well.
However, I'm still uncertain as to why the method I was using is incorrect, so could someone help me understand where my thinking is flawed?
 A: Every intersection maps uniquely to a set of $4$ distinct points. Therefore ${n \choose 4} = 70$, which implies that there are $n=8$ vertices, and consequently, there are ${n \choose 2} = 28$ diagonals in the polygon.
A: 
However, I'm still uncertain as to why the method I was using is incorrect, so could someone help me understand where my thinking is flawed?

The problem is that not all pairs of diagonals intersect. For example, if you have a convex hexagon ABCDEF diagonals AC and DF do not intersect
A: This doesn't seem possible as stated. Suppose that the polygon has $m$ vertices.  Each vertex lies on $m-3$ diagonals, one joining it to each vertex other than itself and its two neighbors.  That is, $m-3$ diagonals concur at each vertex.  If $m-3<3$ then $m\leq5$.  A pentagon has $5$ diagonals, and for $\binom n2\geq70$ we need $n\geq13$.
Perhaps the problem should say, "No three are concurrent, except at the vertices."  In that case, suppose the polygon has $m$ vertices, and that no diagonals are parallel.  Then the polygon has $$d=\frac{m(m-3)}2$$ diagonals.  There are $\binom d2$ pairs of diagonals but this would count each vertex $\binom{m-3}2$ times, once for each pair of diagonals passing through it.  Therefore the number of points of intersection is $$\binom d2-m\left(\binom{m-3}2-1\right)$$  Setting $m=7$ gives $56$ and setting $m=8$ gives $118$, so there's no solution without parallel diagonals.  In that case, it's hard to know how to find a solution, let alone show that it's unique.
I originally did the problem incorrectly.  I said, "Each vertex is counted $m-3$ times, once for each diagonal that passes through it, so that the number of points of intersection is $\binom d2-m(m-4)$."  In this case, setting $m=7$ gives $70$.
I suspect that this is the solution intended, and that whoever composed the problem made the same mistake that I did.
