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Consider four points $P,Q,R,S$ on a line such that $PQ = m, QR= n , RS= o$ , $m \neq n \neq o $. Let $X$ be a point not on the line such that lines $L_1,L_2,L_3,L_4$ are constructed as shown in figure : enter image description here

  • $1)$ Show that there exist many lines $l$ which cuts the other branch of pencil of lines from X in equal portion (i.e. $P'Q': Q'R' :R'S'= 1:1:1$) while varying X to any position . If given that there exists a line $l_°$ in which there exists $P'_°Q'_° :Q'_°R'_°:R'_°S'_°= 1:1:1$ for some specific $X$ position ( that is just one existence shows thats its always possible for other $X$ positions )

    Now consider making the lines $A,B,C,D$ pependicular to one of possible line $l$ and passing through each point of intersection : $P',Q'R',S'$ . Let $Z$ be point of feet of perpendicular from $X$ to line $l$ , let $ZP$ line intersects line $A$ at $a$, simiarily others like $ZQ$ intersecting line $B$ at $b$ ,

  • $2)$ Show that the points $a,b,c,d$ are collinear .

Note: i was able to realize this from a physics problem, this all came into use there, but was not able to prove it mathematically so asked this. My progress: i tried using some similar triangles to make a possible line but fails :(

ProblemDestroyer
  • 967
  • Just for part 1), I wonder if it is always possible to have such line $l$ with $4$ intersections and $3$ equal segments? I see this part as: given $3$ angles $\angle P'XQ'$, $\angle Q'XR'$ and $\angle R'XS'$, and then ignore the right side about the line $PQRS$. I asked and solved the simpler version with only $3$ intersections, $2$ angles and $2$ equal segments: Arithmetic sequence of tangent values, and the $4$th intersection shouldn't always be on the new line regardless of what the $3$rd angle is. – peterwhy Sep 17 '22 at 00:57
  • Sorry i forgot giving one major condition for the problem now hope so its fine @peterwhy – ProblemDestroyer Sep 17 '22 at 03:17

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