Let the quadrant _OAB_ (in Fig. 146) be divided into two parts of equal area, by the circular arc _MP_. It is required to determine (1) the position of _P_ upon the arc of the quadrant, that is to say the angle _BOP_; (2) the position of the point _M_ on the side _OA_; and (3) the length of the arc _MP_ in terms of a radius of the quadrant. (1) Draw _OP_; also _PC_ a tangent, meeting _OA_ in _C_; and _PN_, perpendicular to _OA_. Let us call _a_ a radius; and θ the angle at _C_, which is obviously equal to _OPN_, or _POB_. Then _CP_ = _a_ cot θ; _PN_ = _a_ cos θ; _NC_ = _CP_ cos θ = _a_ ⋅ (cos^2 θ)/(sin θ). The area of the portion _PMN_ = ½_C_(_P_^2)θ − ½_PN_ ⋅ _NC_ = ½_a_^2(cot^2 θ) − ½_a_(cos θ) ⋅ _a_(cos^2 θ)/(sin θ) = ½_a_^2(cot^2 θ − (cos^3 θ)/(sin θ)). {361} And the area of the portion _PNA_ = ½_a_^2(π/2 − θ) − ½_ON_ ⋅ _NP_ = ½_a_^2(π/2 − θ) − ½_a_(sin θ) ⋅ _a_(cos θ) = ½_a_^2(π/2 − θ − sin θ ⋅ cos θ). Therefore the area of the whole portion _PMA_ = (_a_^2)/2 (π/2 − θ + θ cot^2 θ − (cos^3 θ)/sin θ − sin θ ⋅ cos θ) = (_a_^2)/2 (π/2 − θ + θ cot^2 θ − cot θ), and also, by hypothesis, = ½ ⋅ area of the quadrant, = (π_a_^2)/8. [Illustration: Fig. 146.] Hence θ is defined by the equation _a_^2/2 (π/2 − θ + θ cot^2 θ − cot θ) = (π_a_^2)/8, or π/4 − θ + θ cot^2 θ − cot θ = 0. We may solve this equation by constructing a table (of which the following is a small portion) for various values of θ. θ π/4 − θ − cot θ + θ cot^2 θ = _x_ 34° 34′ ·7854 − ·6033 − 1·4514  + 1·2709 = ·0016 35′ ·7854 ·6036 1·4505 1·2700 ·0013 36′ ·7854 ·6039 1·4496 1·2690 ·0009 37′ ·7854 ·6042 1·4487 1·2680 ·0005 38′ ·7854 ·6045 1·4478 1·2671 ·0002 39′ ·7854 ·6048 1·4469 1·2661 −·0002 40′ ·7854 ·6051 1·4460 1·2652 −·0005 {362} We see accordingly that the equation is solved (as accurately as need be) when θ is an angle somewhat over 34° 38′, or say 34° 38½′. That is to say, a quadrant of a circle is bisected by a circular arc cutting the side and the periphery of the quadrant at right angles, when the arc is such as to include (90° − 34° 38′), i.e. 55° 22′ of the quadrantal arc. This determination of ours is practically identical with that which Berthold arrived at by a rough and ready method, without the use of mathematics. He simply tried various ways of dividing a quadrant of paper by means of a circular arc, and went on doing so till he got the weights of his two pieces of paper approximately equal. The angle, as he thus determined it, was 34·6°, or say 34° 36′. (2) The position of _M_ on the side of the quadrant _OA_ is given by the equation _OM_ = _a_ cosec θ − _a_ cot θ; the value of which expression, for the angle which we have just discovered, is ·3028. That is to say, the radius (or side) of the quadrant will be divided by the new partition into two parts, in the proportions of nearly three to seven.