110 



P O R I S M S. 



Porisms 



determinate number of values ; such values of x satis- 

 fy the conditions of the problem ; but in order to dis- 

 cover whether any porismatic case exists, we must exa- 

 mine whether the data of the problem admit of such a 

 relation amongst themselves, that we may have at the 

 same time 



A = B = C = . . . = N = 0. 



If this is the case, the equation () is verified inde- 

 pendently of any particular value ofx ; and, instead of a 

 limited, we have an indefinite number of solutions. 

 This principle may be stated more generally thus, if 

 a, b, c, . . are given quantities, and x, y, z, . . . those 

 which are to be found ; then the solution of the pro- 

 blem leads to several equations of the form 



And since these two quantities must be equal by the Porism*. 

 conditions of the problem, we have the following equa v -y~w 

 tion for determining cos. 6, 



F (x, y, . 



.)=o. 



Fig. 4. 



Jf any relation which can be established amongst the 

 constants a, b, c, . . . shall cause the equation which re- 

 sults from the elimination of the unknown quantities 

 from these equations to be independent of any of them, 

 then by supposing that relation to exist we have a 

 porism. 



PLATE As an example, let us take the following problem. 



CCCCLXVII. Suppose a circle (Fig, 4.) whose radius is r, and a 

 point C in its diameter, such that OC r= v, also a 

 straight line, FL, perpendicular to this diameter ; let it 

 be required to find the angle which the chord, PQ, 

 must make with the diameter ; so that if another chord, 

 P/Q/> be drawn at right angles to it, and from the ex- 

 tremities of these chords perpendiculars be drawn to 

 the given line, the rectangle under those let fall from 

 one chord on this line shall be equal to that under the 

 perpendiculars let fall on the same line from the other 

 chord. Let the required angle PCF =: 6, then 



CE rr v cos. 6 and OE = v sin. 6 



The solutipn of this equation gives 



1 

 cos. 6 = rr . 



x 

 or t = == - , which evidently satisfies the condition ; 



but there exists a possible relation amongst the quan- 

 tities a, v, and r, which will fulfil the equation (1) 

 without determining the value of cos. 6, for that equa- 

 tion is the same as 



? ( ' _,= COS. f=0; 



and by assuming, 



Hence CP =r tj r 1 ^ sin. 02 _ . v C os. 6 



V COS. $. 



CQ = 4/ r 2 v 2 sin. 0* 

 a v 

 cos. 

 o v 



Also CG 



Therefore PG = ^^ + v cos. == y' 7 * u &in> fl ' 

 cos. 



ft . fli 



QG = 



_. - . 



a v a 2 + r * = 0, or esrffidfc.y'a* r 2 ; 



this equation is verified independently of the value of 

 cos. 6. Here then is an indeterminate or porismatic 

 case, and the porism so discovered may be thus enun- 

 ciated. 



A circle and a straight line being given in position, a 

 point may be found within the circle, such, that if any two 

 chords are drawn through that point at right angles to 

 each other, the rectangle under the perpendiculars, let fall 

 from the extremities of the first chord to the line given in 

 position, shall always be equal to the rectangle under the 

 perpendiculars let fall from the extremities of the other 

 chord on the same line. 



If the problem had been to find the angle which 

 the first chord should make with the diameter, that the 

 sum of the squares of the perpendiculars from the first 

 chord should be equal to the sum of the squares of the 

 perpendiculars from the extremities of the second 

 chord, we should have found 



cos. 



cos. 

 a v 

 cos. t 



cos. 



v cos. 6 -3=. \/ r" 1 



sin. 



These multiplied by cos. 6 produce respectively 



PL = a v -j- v cos. t 1 rt cos. \/ r y v 2 cos. 2 

 QM = v -j- i) cos. 2 zt: cos. */ r 2 w* cos. 1 



The rectangle under these two lines is 



(a v + M cos. 2 ) 2 cos. 2 (r 1 v 1 sin. 2 



And this by proper reductions becomes 



PL.QM = (a- v y+( 2av 2 



7T 



and putting - + * f r I* anc ^ making the proper re- 

 SE 



ductions P y L / 2 + Q / M / 2 =2(a I + r 2 ) 2( s w + u 2 -f r 2 ) 



cos. 6 *-}-4<v' t cos. ft ; 

 and the equation determining cos. t, is, 



or 



2 (a v*a* r 1 ) 



cos. 0= 



like the former, gives ^zzziz:-: but here also 



In order to find the rectangle of the perpendiculars 

 let fall from the extremities of another chord at right 



angles to this, we have only to change into -J- 6 



in this expression, and since cos.( + 0)= sin. we 



must substitute 1 cos. s , instead of cos. 2 , .which 

 gives 



- PX. Q,M, = t _ r * - f I2a V i: L, 1 cos. 



4 



an indeterminate case exists, and 

 by making the coefficients vanish. 



a v~ a 2 r*Q, gives v =r a rt= \/ 



may be found 

 The equation 



V*4-r' 2 ; and this 



value of v verifies the equation independently of the 

 magnitude of cos. 6. This porism may be enunciated 

 as follows. 



A circle and a right line being given in position, a 

 point may be found within the circle, such that drawing 

 through it any two chords at right angles to each other, 

 the sums of the squares of the perpendiculars from the ex- 

 tremities of the first chord to the given line, may be equal 



