Putnam 2003, Problem B-5

Let \( A, B, \) and \( C \) be equidistant points on the circumference of a circle of unit radius centered at \( O \), and let \( P \) be any point in the circle’s interior. Let \( a, b, c \) be the distance from \( P \) to \( A, B, C \), respectively. Show that there is a triangle with side lengths \( a, b, c \), and that the area of this triangle depends only on the distance from \( P \) to \( O \).

Solution:
### First solution:  
Place the unit circle on the complex plane so that \( A,B,C \) correspond to the complex numbers \( 1, \omega, \omega^2 \), where \( \omega = e^{2\pi i/3} \), and let \( P \) correspond to the complex number \( x \). The distances \( a,b,c \) are then \( |x - 1|, |x - \omega|, |x - \omega^2| \). Now the identity  
\[
(x - 1) + \omega(x - \omega) + \omega^2(x - \omega^2) = 0
\]  
implies that there is a triangle whose sides, as vectors, correspond to the complex numbers \( x - 1, \omega(x - \omega), \omega^2(x - \omega^2) \); this triangle has sides of length \( a,b,c \).  

To calculate the area of this triangle, we first note a more general formula. If a triangle in the plane has vertices at \( 0, v_1 = s_1 + it_1, v_2 = s_2 + it_2 \), then it is well known that the area of the triangle is \( |s_1 t_2 - s_2 t_1|/2 = |v_1 \overline{v_2} - v_2 \overline{v_1}|/4 \). In our case, we have \( v_1 = x - 1 \) and \( v_2 = \omega(x - \omega) \); then  
\[
v_1 \overline{v_2} - v_2 \overline{v_1} = (\omega^2 - \omega)(x \overline{x} - 1) = i\sqrt{3}(|x|^2 - 1).
\]  
Hence the area of the triangle is \( \sqrt{3}(1 - |x|^2)/4 \), which depends only on the distance \( |x| \) from \( P \) to \( O \).  


### Second solution: (by Florian Herzig)  
Let \( A', B', C' \) be the points obtained by intersecting the lines \( AP, BP, CP \) with the unit circle. Let \( d \) denote \( OP \). Then \( A'P = (1 - d^2)/a \), etc., by using the power of the point \( P \). As triangles \( A'B'P \) and \( BAP \) are similar, we get that \( A'B' = AB \cdot A'P/b = \sqrt{3}(1 - d^2)/(ab) \). It follows that triangle \( A'B'C' \) has sides proportional to \( a, b, c \), by a factor of \( \sqrt{3}(1 - d^2)/(abc) \). In particular, there is a triangle with sides \( a, b, c \), and it has circumradius \( R = (abc)/(\sqrt{3}(1 - d^2)) \). Its area is \( abc/(4R) = \sqrt{3}(1 - d^2)/4 \).  


### Third solution: (by Samuel Li)  
Consider the rotation by the angle \( \pi/3 \) around \( A \) carrying \( B \) to \( C \), and let \( P_A \) be the image of \( P \); define \( P_B, P_C \) similarly. Let \( A' \) be the intersection of the tangents to the circle at \( B,C \); define \( B',C' \), similarly. Put \( \ell = AB = BC = CA \); we then have  
\[
AB' = AC' = BC' = BA' = CA' = CB' = \ell, \quad PA = PPA = PAA = PBC = PCA' = a, \quad PB = PPB = PBB = PCA' = PAB' = b, \quad PC = PPC = PCC = PAB' = PBA' = c.
\]