Ok so I stumbled across a problem(I found a solution just to be clear) and it got me thinking.The problem is a classic,it was challenge to prove that line can intersect circle in at most 2 points.
So I found a solution with quadratics,and parametric line and all that but what I thought was better solution is following.
If we prove that circle has a corresponding colinear point for each point that is part of the circle(aka that it is made of pairs of colinear points,which really is true).If we could prove that then we could easily prove that there is no line that intersects the circle at more than two points.
My idea is to prove it by saying that if we take one axis of symmetry for circle then for each point on one side,there is point colinear to it on the other side of the axis,but how can we prove that the number of these colinear points does not exceed two,or does such thing even need to be proven?
Best Answer
Circle is just a set of points that are equidistant to its centre. Say there is a circle and a line that intersects at more than two different points, and let any distinct three of these intersections be $P$, $Q$ and $R$. Without loss of generality, let $Q$ is between $P$ and $R$ on the straight line. Let the centre of the circle be $O$. $$\begin{align*} \angle PQO + \angle OQR =& 180^\circ&\text{(angles on a straight line)}\\ \angle PQO =& \angle OPQ&\text{(base angle, }OP=OQ\text{)}\\ \angle OQR =& \angle QRO&\text{(base angle, }OQ=OR\text{)}\\ (\angle OPQ + \angle PQO) + (\angle OQR + \angle QRO) =& 360^\circ\\ (180^\circ-\angle POQ)+(180^\circ - \angle QOR) =&360^\circ &\text{(Internal angles of triangle)}\\ \angle POQ + \angle QOR =&0^\circ \end{align*}$$
This means either: both angles $\angle POQ$ and $\angle QOR$ are $0^\circ$, or one of these angles is positive and the other triangle has their bases overlapped. In any case, some of the intersection points are shown to be the same.