# Circle Geometry Question 4

O, Q and R are fixed points on a circle. A tangent is drawn to O. K and M lies of OQ and OR respectively and KM is parallel to the tangent. The perpendicular bisectors of QK and RM intersect at T. Prove that the locus of T is a straight line.

I am guessing you need to prove that $\angle TOB$ is a constant angle or OT bisects $\angle QOR$. Initially I thought that T lies on the diameter since when KM approach O, they bisectors meet at the center. So I am guessing I need to prove that OT is perpendicular to AB or something like that. But I can't seem to prove it. Can someone give me a clue to solve this problem?

Any help is greatly appreciated.

## 2 Answers

Note that points $Q,K,M,R$ lie on a common circle. This is because $$\angle KQR = \angle BPR = \angle KMP = 180^\circ - \angle RMK.$$

So perpendicular bisectors of $QK, MR$ meet at the center of this circle. In particular it lies on perpendicular bisector of $QR$.

Here is a treatment by coordinate geometry (hopefully, this can be converted in pure geometry reasoning).

Let us take $P$ as origin, let us take PD as $x$-axis, the orthogonal line to the $x$ axis in $P$ will be the $y$-axis.

Let $Q(2a,2b)$, $R(2c,2d)$. Let $Q_1(a,b)$ (resp. $R_1(c,d)$) be the midpoint of $PQ$ (resp. $PR$). Let $t$ be a parameter (time...) such that $K(a+ta,b+tb)$, resp $M(c+tkc,d+tkd)$ where $K$ has unit speed and $M$ as a speed that is in a constant ratio $k$ with the former (this is a central point), due to Thales property.

As we can take as "guiding" vectors of the perpendicular bissectors the normal vectors to $PQ$ (resp. $PR$), which are $\binom{-b}{a}$ (resp. $\binom{-d}{c}$), the parametric equations of these bissectors are:

$$\tag{1}(P_1) \cases{x=a(t+1)-bu\\y=b(t+1)+au} \ \ \ (P_2)\cases{x=c(kt+1)-dv\\y=d(kt+1)+cv}$$

for certain parameters $u$ and $v$.

Finding the point of intersection of $(P_1)$ and $(P_2)$ amounts to find $u$ and $v$ such that we have in (1) the same $x$ and $y$, said otherwise such that ($t$ remaining fixed):

$$\tag{2}(i)\cases{a(t+1)-bu=c(kt+1)-dv\\b(t+1)+au=d(kt+1)+cv} \ \ \Leftrightarrow \ \ (ii)\cases{-bu+dv=c(kt+1)-a(t+1)\\au-cv=d(kt+1)-b(t+1)}$$

System (ii) will clearly give rise to a solution where $u$ and $v$ will be expressed in the form of first degree expressions:

$$\tag{3}\cases{u=et+f\\v=gt+h}$$

Plugging the first relationship (3) in (1)$(P_1)$ gives again first degree equations for $x$ and for $y$, proving that the intersection point moves along a straight line (more exactly on a line segment).

• A comment that has nothing to do with the proof: the figure seems zerox-copied from an XVIII-th century book. Is it the case?... – Jean Marie Oct 16 '16 at 14:31
• No, it is copied from a online maths past paper – Nanoputian Oct 16 '16 at 20:41