Wikipedia:Reference desk/Archives/Mathematics/2020 November 22

= November 22 =

Vector proofs of circle geometry
The mathematics syllabi for high school were recently changed in the state where I live. Vectors have been introduced into the second highest and highest level courses. The second highest level course includes dot products but not cross products for vectors in 2D, and 3D vectors are in the highest level course. At the same time, almost all circle geometry has been removed. In fact, all geometry has been significantly de-emphasised, though much is placed as assumed knowledge that can be examined in the context of a broader question on content within the syllabus. I am looking at past exam papers to adapt for students and there are geometry questions that are obviously provable by vector methods, but the problems that involve circle geometry are posing challenges for me. One such question requires a proof of the tangent-secant theorem as a precursor to an interesting question that remains in the syllabus.

Tangent-Secant Theorem: If P, Q, and R are points on a circle and the tangent at P meets QR projected (beyond R) at T, then PT2 = QT.RT

Now, I know this is easy to prove using the fact that triangles TPR and TQP are similar as a consequence of the alternate segment theorem, but the alternate segment theorem is not part of the syllabus. They would know that, if we call the centre of the circle O, then OP and PT are perpendicular and also that OP = OQ = OR (equal radii). Even if I state the result from the alternate segment theorem, that the angles PQR and RPT are equal, so as to use dot products of vectors, I still am unable to find a way to derive the required result without using the similar triangles (which negate the point of using vectors). Can anyone offer a vector proof for this theorem?

More generally, can anyone point to resources that I can use for vector proofs of circle geometry results? The highest level high school mathematics course also covers more difficult examples of 2D vectors.

Many Thanks,

EdChem (talk) 00:28, 22 November 2020 (UTC)


 * It is fairly straightforward (using the Pythagorean theorem and the fact that the diameter through the midpoint of a chord is perpendicular to it) to show that both quantities are equal to $$PO^2 - r^2$$ (see power of a point) but after fiddling around for a while I did not come up with a direct proof that avoids the intermediate term. --JBL (talk) 17:29, 23 November 2020 (UTC)


 * It's not the same theorem, but perhaps Circles of Apollonius has the kind of proof you're looking for. Note that, even applying some tricks that no one would be expected to some up with right away, the algebra gets complicated and hard to follow, but when you restrict yourself to analytic proofs that's often the case. --RDBury (talk) 19:04, 23 November 2020 (UTC)


 * PS. The assumed fact that the diameter through the midpoint of the chord is perpendicular to it is perhaps easier to prove with vectors than without. We're given
 * $$|\overrightarrow{\mathrm{OA}}| = |\overrightarrow{\mathrm{OB}}|$$
 * and want to prove
 * $$\overrightarrow{\mathrm{OM}} \cdot \overrightarrow{\mathrm{AB}} =0$$
 * where M is the midpoint of A and B. We have
 * $$\overrightarrow{\mathrm{OM}} = \frac{1}{2} (\overrightarrow{\mathrm{OB}} + \overrightarrow{\mathrm{OA}})$$
 * so it's enough to show
 * $$(\overrightarrow{\mathrm{OB}} + \overrightarrow{\mathrm{OA}}) \cdot \overrightarrow{\mathrm{AB}} = 0.$$
 * But this is
 * $$(\overrightarrow{\mathrm{OB}} + \overrightarrow{\mathrm{OA}}) \cdot (\overrightarrow{\mathrm{OB}} - \overrightarrow{\mathrm{OA}}) $$
 * $$= \overrightarrow{\mathrm{OB}} \cdot \overrightarrow{\mathrm{OB}} - \overrightarrow{\mathrm{OA}} \cdot \overrightarrow{\mathrm{OA}} $$
 * $$= |\overrightarrow{\mathrm{OB}}|^2 - |\overrightarrow{\mathrm{OA}}|^2 = 0. $$
 * --RDBury (talk) 20:18, 24 November 2020 (UTC)


 * Perhaps, although I feel that this is one of those facts where putting a formal proof on it is a bad idea from a pedagogical point of view: hardly any facts in any part of mathematics are as easy to represent convincingly by a single picture. (Not that I have anything against your proof -- it's a nice, elegant proof!) --JBL (talk) 21:48, 24 November 2020 (UTC)