# How/when do we use circle inversions to solve problems?

Given an angle AOB and a point M inside it, construct a segment PQ such that

• M is the midpoint of PQ
• P is on side OA
• Q is on side OB

So i've been thinking about this problem and of course the best and easiest case i can construct for, is when M lies on the angle bisector of AOB but of course there's a construction that includes for ANY arbitrary point of M.

The only idea i have in my head, is that this line segment PMB has its average located at M. So in general, M is the line's average length.

Question: How should i go about solving this construction problem? We are given the tools of circle inversions, homothety, isometries constructing trivial things such as perpendiculars, angle bisectors etc...

Reflect $O$ in $M$ to give point $O'$. Draw a line $l$ through $O'$ parallel to $OB$. Then set $P$ as the intersection of $l$ and $OA$, and $Q$ as the reflection of $P$ in $M$.

This works, since $l$ is the reflection of $OB$ in $M$ so the reflection $Q$ of $P$ must land on $OB$.

• Thank you for the help, i was going over this construction of yours in geoalgebra and i'd like to ask, what was your thought process whilst solving this problem? i love you. did you automatically think of the diagonals of a parrallelogram? – cavell May 11 '18 at 5:53
• A while ago I came across the problem: Given three parallel lines, find an equilateral triangle with a vertex on each line. The solution was to fix a point on the first line and rotate the second line around it by $60^\circ$. The intersection with the third line solves the problem. I had this problem (and solution) in the back of my mind, and so I tried to transform line $OB$ around $M$ to solve this: it's not hard to see that reflecting through $M$ is the appropriate transformation. – B. Mehta May 11 '18 at 12:08
• Drop a perpendicular from $M$ to $OB$, hitting $OB$ at $S$.
• Let $T$ be the reflection of $S$ about $M$.
• Draw a line $l$ through $T$, perpendicular to $MT$.
• Let $P$ be the point where $l$ hits $OA$.
• Let $Q$ be the point where $PM$ extended hits $OB$.

To prove $PM=QM$, show that right triangles $PTM$ and $QSM$ are congruent.

$1)$ Draw a line parallel to $OB$ passing through $M$. Let's call the point $C$ where it cuts $OA$.

$2)$ Take a point $P$ on the line $OA$ such that $OC=CP$

$3)$ Draw a line passing through $P$ and $M$. Let's call the point $Q$ where it cuts the line $OB$. $PQ$ is the required line.