Welcome to our community

Be a part of something great, join today!

Locus

sbhatnagar

Active member
Jan 27, 2012
95
Challenge Problem: A variable line $L$ passing through the point $B(2,5)$ intersects the lines $2x^2-5xy+2y^2=0$ at $P$ and $Q$. Find the locus of the point $R$ on $L$ such that distances $BP$, $BR$ and $BQ$ are in harmonic progression.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Challenge Problem: A variable line $L$ passing through the point $B(2,5)$ intersects the lines $2x^2-5xy+2y^2=0$ at $P$ and $Q$. Find the locus of the point $R$ on $L$ such that distances $BP$, $BR$ and $BQ$ are in harmonic progression.
Hi sbhatnagar, :)

Let \(P\equiv (x_{p},2x_{p})\), \(Q\equiv (x_{q},\frac{x_{q}}{2})\) and \(R\equiv (x_{r},y_{r})\). Since $BP$, $BR$ and $BQ$ are in harmonic progression we can write,

\[\frac{1}{BR}-\frac{1}{BP}=\frac{1}{BQ}-\frac{1}{BR}\]

\[\frac{1}{\sqrt{(x_{r}-5)^2+(x_r-2)^2}}-\frac{1}{\sqrt{(2x_{p}-5)^2+(x_p-2)^2}}=\frac{1}{\sqrt{(\frac{x_{q}}{2}-5)^2+(x_q-2)^2}}-\frac{1}{\sqrt{(x_{r}-5)^2+(x_r-2)^2}}\]

Since, \(\displaystyle\frac{2x_p -5}{x_p -2}=\frac{\frac{1}{2}x_q -5}{x_q-2}\Rightarrow x_{q}=-\frac{2x_p}{3x_p-8}\) we get,

\[\frac{2}{\sqrt{(x_{r}-5)^2+(x_r-2)^2}}-\frac{1}{\sqrt{(2x_{p}-5)^2+(x_p-2)^2}}=\frac{1}{\sqrt{\left(-\frac{x_p}{3x_p -8}-5\right)^2+\left(-\frac{2x_p}{3x_p -8}-2\right)^2}}\]

\[\Rightarrow \frac{2}{\sqrt{(x_{r}-5)^2+(x_r-2)^2}}=\frac{|3x_p-8|+8}{8\sqrt{(2x_{p}-5)^2+(x_p-2)^2}}~~~~~~(1)\]

Considering the line \(L\) we can write,

\[\frac{2x_p -5}{x_p -2}=\frac{y_r -5}{x_r -2}~~~~~~~~(2)\]

By (1) and (2) we get,

\[\frac{2}{x_r -2}=\frac{|3x_p-8|+8}{8(x_p -2)}=\frac{\pm(3x_p-8)+8}{8(x_p -2)}\]

It is clear from, \(\displaystyle\frac{1}{BR}-\frac{1}{BP}=\frac{1}{BQ}-\frac{1}{BR}\) that if \(x_p=0\) then, \(x_r=0\). This happens only when we take the negative sign in the above equation. That is,

\[\frac{2}{x_r -2}=\frac{-3x_p+16}{8(x_p -2)}\]

\[\Rightarrow x_p=\frac{16x_r}{3x_r+10}\]

Substituting for \(x_p\) in (2) we get,

\[y_r=\frac{17x_r}{10}\mbox{ for }x_{r}\neq 2\]

When \(x_p=x_q=x_r=2\), \(P\equiv (2,4),\,Q\equiv (2,1)\mbox{ and }R\equiv (2,y_r)\). Then, \(BP=1,\,BQ=4\mbox{ and }BR=5-y_r \).

\[\frac{1}{BR}-\frac{1}{BP}=\frac{1}{BQ}-\frac{1}{BR}\]

\[\Rightarrow \frac{1}{5-y_r}-1=\frac{1}{4}-\frac{1}{5-y_r}\]

\[\Rightarrow \frac{1}{5-y_r}=\frac{5}{8}\]

\[\therefore y_r=5-\frac{8}{5}=\frac{17}{5}\]

Therefore,

\[y_r=\begin{cases}\frac{17x_r}{10}&\mbox{ when }&x_{r}\neq 2\\\\\frac{17}{5}&\mbox{ when }&x_r=2\end{cases}\]

\[\Rightarrow y_r=\frac{17x_r}{10}\]

Kind Regards,
Sudharaka.
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
Challenge Problem: A variable line $L$ passing through the point $B(2,5)$ intersects the lines $2x^2-5xy+2y^2=0$ at $P$ and $Q$. Find the locus of the point $R$ on $L$ such that distances $BP$, $BR$ and $BQ$ are in harmonic progression.
If four collinear points $B,P,R,Q$ satisfy the condition that the distances $BP$, $BR$ and $BQ$ are in harmonic progression, then $$\frac2{BR} = \frac1{BP} + \frac1{BQ} = \frac{BP+BQ}{BP.BQ},$$ $$2BP.BQ = BR(BP+BQ),$$ $$BP(BQ-BR) + (BP-BR)BQ = 0,$$ $$BP.QR + BQ.{P}R = 0.$$
Thus the four points have cross-ratio $-1$ and form a harmonic range. If the points are all connected by lines to a point $O$ not on the line $BPQR$, then those four lines form a harmonic pencil. It follows that if the lines $OB$, $OP$ and $OQ$ are given, then the locus of $R$ is the fourth line of the harmonic pencil determined by them.

For the problem in this thread, $2x^2-5xy+2y^2=(x-2y)(2x-y)$, so the two lines $OP$ and $OQ$ have equations $y=2x$ and $y=\frac12x$. The point $B$ is $(2,5)$, so the line $OB$ has equation $y=\frac52x.$ The harmonic conjugate line $OR$ is the line through $O$ with gradient $h$ such that the gradients $\frac52,\,2,\,h,\,\frac12$ are in harmonic progression, in other words the difference $\frac52-h$ should be the harmonic mean of $\frac52-2$ and $\frac52-\frac12.$ Thus $$\frac2{\frac52-h} = \frac1{\frac52-2} + \frac1{\frac52-\frac12}.$$ You can check that the solution to that equation is $h=\frac{17}{10}.$

Finally, the locus of $R$ is the line $y=\frac{17}{10}x$.

I learned about harmonic ranges and pencils in high school 55 years ago and had forgotten about them until this thread brought it all back. That sort of projective geometry seems to have fallen out of fashion, but it is a beautiful theory and I was glad to be reminded of it.