Texas Mathematics Teachers' Bulletin . Eeverting now to the isosceles triangle having each angle atthe base double of the third angle, let us drop one of the condi-tions and find the locus of the vertex of a triangle which has agiven base and one of the base angles double of the verticalangle. Let ABC (Fig. 4) be a triangle having BC fixed inmagnitude and position and B^2A, it is required to find thelocus of the vertex A. Denote BA by r, BC by a: r sin BCA Law of Sines: —= a sin A sin ACD sin (B+A)sin A sin A 10 University of Texas Bulletin3B sin B •, since A=i/^B sm B 3 sin B/2—4 sin^ —2 sin
Texas Mathematics Teachers' Bulletin . Eeverting now to the isosceles triangle having each angle atthe base double of the third angle, let us drop one of the condi-tions and find the locus of the vertex of a triangle which has agiven base and one of the base angles double of the verticalangle. Let ABC (Fig. 4) be a triangle having BC fixed inmagnitude and position and B^2A, it is required to find thelocus of the vertex A. Denote BA by r, BC by a: r sin BCA Law of Sines: —= a sin A sin ACD sin (B+A)sin A sin A 10 University of Texas Bulletin3B sin B •, since A=i/^B sm B 3 sin B/2—4 sin^ —2 sin 3/2B =3—4 sin^ —2 B = 3—2 (2 sin= —)2 =3—2 (1—cos B) =1+2 cos B hence r=a (1+2 cos B). This is the equation of the locus of A in polar co-ordinates,B being the pole and BD being the polar Plotting the locus for values of B from 0° to 180° we have thecurve DNOP (Fig. 5). For values of B from 180^ to 360° the re-sulting curve will be sjTnmetrical to DNOP with respect to trisect an angle make at the point P in the axis OD an anglesay NPD equal the given angle. Join N with the pole 0 thenthe angle ONP equals one-third of the given angle. Mathematics Teachers Bulletin 11 The equation of the curve in rectangular co-ordinates reduces to x-+y-—2ax=a Ax-+y- This, of course, is a higher plane curve. In plane elementarygeometry the only curves used are the straight line and circle soour problem is outside the domain of the geometry of the point,straight line and circle. PRACTICAL SOLUTION The following is an easy practical solution of the problem oftrisecting an angle. Let ABC (Fig. 6) be the angle to be trisected, ^ith B as
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