RECTANGLE

A Note on the Circum Circles Associated With the Rectangle

A Note on the Circum Circles Associated With the Rectangle

The following two results were motivated by a problem that appeared in an AMTI competition in the late 70's. The problem was to find tan 75º geometrically. The following are the two results motivated by a solution to the above problem.

Result I

Let ABCD be a square. Let E be a point inside ABCD such that (EAB = (EBA = 15º as shown in the figure (i) below. Then CED is an equilateral(.

Figure (ii)

Result I is an exercise in the book 'Geometry revisited' by Coxeter and Greitter published by the Mathematical association of America. The authors hint at an indirect proof at the back of the book. What is new here is that we give a simple direct trigonometric proof of Result I using the value of tan 75º and tan 15º.

Result II

Let PQRS be a rectangle such that PS = 2PQ

Let T be a point inside the rectangle such that (TPQ = (TQP = 15º as in figure (ii), then ('s STP and RTQ are right triangles and the circum circle, of ('s PTQ, TQR, STR and PTS are equal (equal in radius).

We first give a solution to the problem that appeared in the AMTI competition. Although it may have appeared in an earlier issue of the mathematics teacher we give it here for completeness.

Let ABC be a right angled (, right angled at B. let (C = 75º. Cut off (BCD = 60º (refer to figure (iii) below). Let BC = a then CD = 2a.

Let BD = (3a. Also since (ACD = (DAC = 15º.

Figure III

tan75º = = = 2 + (3

tan15 º= = = = 2-(3

Solution to Result I

Figure IV

We refer to figure (iv). Draw EF ( to AB and produce GE to meet DC in G.

Let AB = AD = 2a

Then AF = FD = DG = GC = a

Since EF = AF tan 15º

we have EF = a (2 - (3)

And EG = FG - EF = 2a - (2a - (3 a) = (3 a

( In right triangle DGE, DG = a

GE = (3 a ( (GDE = 60º

(GCE = 60º. ( DEC is equilateral.

The above solution although uses trigonometry, can be considered geometrical since the value of tan 15º used in the proof was obtained geometrically, not sacrificing the spirit of 'Geometry revisited'.

Solution to Result II

We deduce result II from result I.

Let U and V be the mid points of PS and QR join UV. (Refer to fig.(v))

Figure IV

Now since PQVU is a square.

We have from result I UTV is an equilateral(. ( UT = UV = TV = PU

In (PTS U is the midpoint of PS

And UT = PU = US. ( (PTS is right angled and (STP = 90º

(RTQ is right angled and (RTQ = 90º.

( These circum circles are equal.

Also the common chord TQ of circum circles of ('s PTQ and TQR ...