Constant Ratio on Circle

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11 March 2016, Created with GeoGebra

Problem

Given a circle and on it four points: $U,V,W,P\;$ such that $V$ is the midpoint of an arc $UW\;$ while $P\;$ is arbitrary in the other arc $UW.$

Constant Ratio on Circle - Problem and Solution

Then the ratio $\displaystyle\frac{PU+PW}{PV}\;$ does not depend on the position of $P.$

Proof

Assume we deal with the unit circle and let $V=(\cos s, \sin s),\;$ $U=(\cos (s-d),\sin (s-d)),\;$ $W=(\cos (s+d),\sin (s+d)),\;$ and $P=(\cos t, \sin t).\;$ Let $M=(\cos m, \sin m)\;$ be a random point on the circle. Then

$\displaystyle\begin{align}MP^2=|M-P|^2&=((\cos m-\cos t)^2+(\sin m-\sin t)^2\\ &=(\cos^2 m+\sin^2m)+(\cos^2t+\sin^2t)-2(\cos m\cos t+\sin m\sin t)\\ &=2-2\cos (m-t)\\ &=4\sin^2\frac{m-t}{2}. \end{align}$

We may assume that $\displaystyle 0\lt\frac{m-v}{2}\lt 180^{\circ},\;$ such that $\displaystyle MP=2\sin\frac{m-t}{2}.$

Using that

$\displaystyle\begin{align} \frac{PU+PW}{PV} &= \frac{\displaystyle 2\sin\frac{s-d-t}{2}+2\sin\frac{s+d-t}{2}}{\displaystyle 2\sin\frac{s-t}{2}}\\ &=\frac{\displaystyle 2\sin\frac{s-t}{2}\cos\frac{d}{2}}{\displaystyle\sin\frac{s-t}{2}}\\ &=2\cos\frac{d}{2}. \end{align}$

Variant

Given a circle and on it six points: $U_2,U_1,V,W_1,W_2,P\;$ such that $V$ is the midpoint of both arcs $U_1W_1\;$ and $U_2W_2\;$ while $P\;$ is arbitrary in the other arc $U_2W_2.$

Then the ratio $\displaystyle\frac{PU_1+PW_1}{PU_2+PW_2}\;$ is independent of position of $P.$

For a proof consider

$\displaystyle\frac{PU_1+PW_1}{PU_2+PW_2}=\frac{PU_1+PW_1}{PV}\cdot\frac{PV}{PU_2+PW_2}$

and apply the previous result.

Acknowledgment

There were two earlier posts: one by Dorin Marghidanu, concerning a constant ratio in a square and another by Dao Thanh Oai, with a comment to the latter by Tim Robinson that suggested a more general, unifying property, as above.

 

Trigonometry

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