1
$\begingroup$

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

image of the problem

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...

$\endgroup$
0
$\begingroup$

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 $M$.

enter image description here

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

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – cavell May 11 '18 at 5:53
  • $\begingroup$ 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. $\endgroup$ – B. Mehta May 11 '18 at 12:08
0
$\begingroup$
  • 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.

$\endgroup$
0
$\begingroup$

$1)$ 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.