In number theory, the optic equation is an equation that requires the sum of the reciprocals of two positive integers a and b to equal the reciprocal of a third positive integer c:
1 a + 1 b = 1 c . {\displaystyle {\frac {1}{a}}+{\frac {1}{b}}={\frac {1}{c}}.}
Multiplying both sides by abc shows that the optic equation is equivalent to a Diophantine equation (a polynomial equation in multiple integer variables).
Solution All solutions in integers a, b, c are given in terms of positive integer parameters m, n, k by
a = k m ( m + n ) , b = k n ( m + n ) , c = k m n , {\displaystyle {\begin{aligned}a&=km(m+n),\\b&=kn(m+n),\\c&=kmn,\end{aligned}}}
where m, n are coprime.
Appearances in geometry
The optic equation, permitting but not requiring integer solutions, appears in several contexts in geometry. In a bicentric quadrilateral, the inradius r, the circumradius R, and the distance x between the incenter and the circumcenter are related by Fuss' theorem according to
1 ( R − x ) 2 + 1 ( R + x ) 2 = 1 r 2 , {\displaystyle {\frac {1}{(R-x)^{2}}}+{\frac {1}{(R+x)^{2}}}={\frac {1}{r^{2}}},}
and the distances of the incenter I from the vertices A, B, C, D are related to the inradius according to
1 I A 2 + 1 I C 2 = 1 I B 2 + 1 I D 2 = 1 r 2 . {\displaystyle {\frac {1}{IA^{2}}}+{\frac {1}{IC^{2}}}={\frac {1}{IB^{2}}}+{\frac {1}{ID^{2}}}={\frac {1}{r^{2}}}.}
In the crossed ladders problem, two ladders braced at the bottoms of vertical walls cross at the height h and lean against the opposite walls at heights of A and B. We have 1 h = 1 A + 1 B . {\displaystyle {\tfrac {1}{h}}={\tfrac {1}{A}}+{\tfrac {1}{B}}.} Moreover, the formula continues to hold if the walls are slanted and all three measurements are made parallel to the walls. Let P be a point on the circumcircle of an equilateral triangle △ABC, on the minor arc AB. Let a be the distance from P to A and b be the distance from P to B. On a line passing through P and the far vertex C, let c be the distance from P to the triangle side AB. Then 1 a + 1 b = 1 c . {\displaystyle {\tfrac {1}{a}}+{\tfrac {1}{b}}={\tfrac {1}{c}}.}
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