383] 
PROBLEMS AND SOLUTIONS. 
565 
3. Solution by Professor Cayley. 
Since the theorem (7) follows at once from (/3), and (a) is included in (/3), 
it is only necessary to prove (/3). Consider three given circles, and let it be proposed 
to construct a triangle the sides whereof touch the given circles, and which is similar 
to a given triangle; the direction of one side may be assumed at pleasure, and then 
the triangle is determined. Impose now on the triangle the condition that the area 
is equal to a given quantity; we obtain for the given area an expression involving 
the angle 0 which fixes the direction of one of the sides, and we have thus an 
equation for the determination of the angle 6. But, for a properly determined 
relation between the data of the problem, the expression for the area becomes 
independent of the angle 0, that is, every triangle, the sides whereof touch the three 
circles, and which is similar to a given triangle, is of the same area, or say, the area 
of every such triangle is equal to a given quantity A; and, this being so, it is 
clear that, if we construct a triangle similar to a given triangle and of the given 
area A (that is, a triangle equal to a given triangle), in such manner that two of 
the sides touch two of the given circles, then the envelope of the remaining side will 
be the remaining given circle; which is in fact the theorem (/3). 
It only remains therefore to show that the foregoing porismatic case of the problem 
exists. 
For the first circle, let the coordinates of the centre be a, b, and the radius 
be c ; and suppose in like manner that we have a', b', and c' for the second circle, 
and a", b", and c" for the third circle. Let A, A', A" be the inclinations to the axis 
of x of the perpendiculars on the sides which touch these circles respectively; then 
the equations of the three sides respectively are 
(x — a) cos \ + (y — b) sin A — c = 0, (x — a') cos A' + (y — b’) sin A' — c' — 0, 
(x — a") cos A" + (y — b") sin A" — c" = 0. 
If the triangle be similar to a given triangle, then the differences of the angles 
X, X', X" will be given angles, or, what is the same thing, we may write 
\=0 + %, \' = 0' + £, A" = 0" + £, 
where 0, 0', 0" are given angles, and £ is a variable angle. Let A be the area of 
the triangle, then (disregarding a merely numerical factor) we have 
VA = sin (A' — A") (a cos A + b sin A + c) 
+ sin (A" — A) (a' cos A' + b' sin A' -1- c') + sin (A — A') (a" cos A" + b" sin A" + c"); 
or, what is the same thing, 
VA = sin (0' — 0’’) {a cos (0 + |) + b sin {0 + £) + c } 
+ sin (0" — 0 ){a' cos (0' + f) + b' sin (0' + f) + c'} 
+ sin (0 -0') [a" cos (0" + |) + b" sin (0" + f) + c"}.