Problem: 1 (unit), a and b are given plotted on a plane. Using only an unmarked ruler and compass, construct lengths (a+b), (a-b), (ab), (a/b), a² and √(ab). Also, hence one can draw lengths 1/a and √a.

Solution:

To draw a±b is trivially easy. I shall draw a*b. a/b can be similarly be drawn. Draw AP = a and PB = b such that A-P-B. Through P, draw a line l different from AB. Mark a point C on l so that PC = 1. Draw the (unique: prove!) circumcircle of ∆ABC. Let it cut l at Q (Q≠P) Then PQ = ab. (cf diagram)

Taking AP = a, PB = 1 and PC = b gives PQ = a/b.

Note that you cannot draw length √(ab), or for that matter, √a. This is because, you have to know three of the four quantities in ab=cd.

Method for √(ab):

Draw AP = a, PB = b. Draw circle with AB as diameter and C as centre. (cf diagram)

AP . PB = (PQ)²

and so, (PQ) = √(ab)…… But wait..what is CD? (the radius perpendicular to AB) Indeed, it is (a+b)/2, because our diameter was a+b. And it is obvious from the diagram that CD > PQ, which implies that AM > GM for two numbers!!

Proof:

The two triangles APC and BPQ (upper diagram) are similar, their angles being inscribed in the same arc. Hence their sides are in proportion. Hence proved. (I shall call this theorem as TWO CHORD THEOREM)

Inadequacies in the proof:

The proof obviously doesn’t derieve everything from Euclid’s postulates but it is left to the reader to prove it! An insight as to what is to be proved is as shown:

* If two triangles have two angles congruent, then their sides are in proportion.

* Angles inscribed in the same arc are all congruent (Inscribed Angle theorem) (does this gives an alternate definition of a circle?)

* Circumcentre of a triangle exists and is unique.

* Basic proportionality theorem for line parallel to one side of a triangle.

Further insight:

* AB is diameter of circle with centre C and radius r. PQ is a chord at distance a from C and perpendicular to AB. Let point of intersection of AB and PQ be D. Let DP = DQ = b. Then the two chord theorem tells that AD.DB = PD.DQ i.e. (r-a).(r+a) = b.b i.e. r² = a² + b² i.e. Pythagoras theorem.

*Given lengths s (sum) and p (product) , can I find lengths a and b satisfying (a+b)=s and ab = p? Note that it is equivalent to solving a quadratic in geometry. When will the discriminant be zero?

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