Points in Pairs
Move the point P to see how P' moves. Then use your insights to calculate a missing length.
Age
14 to 16
Challenge level
Problem
Two points, one inside a circle and the other outside, are related in the following way :
A line starting at the centre of the circle and passing through the first point ($P$) goes on to pass through the second point ($P'$)
Positions along the line are such that the ratio of $OP$ to the radius of the circle matches the ratio of the radius of the circle to $OP'$
For example if $OP$ happened to be $2/3$ of the radius then $OP'$ would be $3/2$ of the radius.
You can use the interactivity below to help you explore how the positions of a pair of points relate to each other.
Once you have a feel for how the points $P$ and $P'$ relate to each other, use the checkbox to add a point $Q$ and its corresponding point $Q'$.
What do you notice about triangles $OPQ$ and $OQ'P'$, where $O$ is the origin?
Final challenge
In this diagram, the radius of the circle is 10 units, $OP$ is 8 units and $OQ$ is 6 units.
Image
If the distance $PQ$ is 5 units what is the distance $P'Q'$?
Getting Started
What is the ratio $OP : OQ$ ?
Compare it with the ratio $OQ' : OP'$.
Student Solutions
We had a couple of correct solutions (500/48 or 10.4 to 3sf) using the Cosine Rule, and one solution which used the much simpler method of similar triangles:
Image
I noticed that, since $OP'=100/OP$ and $OQ'=100/OQ$, the ratio $\frac{OP}{OQ}$ is $\frac{OQ'}{OP'}$ and so the triangles $OPQ$ and $OQ'P'$ are similar.
So, in this question $OP'=100/8$ and, since the ratio between lengths of sides is the same by similar triangles, $$ \frac{P'Q'}{OP'}=\frac{PQ}{OQ}.$$ Therefore, $$P'Q'=OP'\times\frac{PQ}{OQ}=\frac{100}{8}\times\frac{5}{6}=\frac{500}{48}.$$.