HMMT-Feb 2016 alg, Problem 1

Let $z$ be a complex number such that $|z|=1$ and $|z-1.45|=1.05$. Compute the real part of $z$.

Solution: From the problem, let $A$ denote the point $z$ on the unit circle, $B$ denote the point 1.45 on the real axis, and $O$ the origin. Let $A H$ be the height of the triangle $O A H$ and $H$ lies on the segment $O B$. The real part of $z$ is $O H$. Now we have $O A=1, O B=1.45$, and $A B=1.05$. Thus \[ O H=O A \cos \angle A O B=\cos \angle A O B=\frac{1^{2}+1.45^{2}-1.05^{2}}{2 \cdot 1 \cdot 1.45}=\frac{20}{29} \] $\fbox{\frac{20}{29}}$.

Answer: \frac{20}{29}