TO GEOMETRICAL PROBLEMS» 
271 
PROBLEM XVIII. 
The area, the perimeter, and one of the angles of any 
plane, triangle ABC, being given ; to determine the tri 
angle. 
Suppose a circle to be inscribed in the triangle, touch 
ing the sides thereof in the points D, E, and F ; also 
from the centre O, suppose OA, OD, OC, OF, OB, 
and OE to be drawn: 
and upon BC let fall the 
perpendicular AG; put 
ting AB -4- BC + AC zz 
b, the given area zz a 2 , 
the sine of the angle ACB 
(the radius being l) — m, 
the co-tangent of half that 
angle (or the tangent of 
DOC) zz n, and AC zz a. 
Therefore, since the area 
of the triangle is equal to 
¿AB x OE + ¿BC x OF 
p ¿AC x OD, that is, equal to a rectangle under 
half the perimeter and the radius of the inscribed circle, 
we have — x OE 
2 
a a ; and therefore OE zz 
2 aa 
But 
AD being zz AE, and BF zz BE; it is manifest that the 
sum of the sides, C A 4- CB, exceeds the base AB, by 
the sum of the two equal segments CD and CF ; and so 
is greater than half the perimeter by one of those equal 
segments CD; that is, CA 4- CB zz ¿6 4- CD: but 
[by trigonometry) as 1 (radius) : n (the tangent ot 
DOC)::~ (OD) : DC zz whence CA + 
CB (zz lb p CD) \b 4- —; which, taken from [b) 
Yl 0?“ 
the whole perimeter, leaves \b zz the base AB. 
Make now ¿6 4- —zz c; then will BC zz c — ;r: also 
2 0 
[by trigonometry) it will be, as 1 (radius) : m (the sine