A problem on geometry and circle Let $A,B, C,D$ and $E$ be five points marked, in clockwise order, on the
unit circle in the plane (with centre at origin). Let $α$ and $β$ be real numbers
and set $f (P ) = αx + βy$ where $P$ is a point whose coordinates are ($x, y$).
Assume that $f (A) = 10$, $f (B) = 5$, $f (C ) = 4$ and $f (D) = 10$. Which of the
following are impossible?
(a) $f (E) = 2$
(b) $f (E) = 4$
(c) $f (E) = 5$   
totally stuck on it. how can I able to solve this problem 
 A: Hint: For a constant $k$, the set of points where $f(P) = k$ is a straight line.
A: I claim that none of the three values presented are possible. As Calvin Lin notes, the locus of points for which the equation takes on a particular value $c$ is a line. More specifically, if we have
$$c = \alpha x + \beta y$$
then we have
$$y = -\frac{\alpha}{\beta}x + \frac{c}{\beta}$$
Notice that different values of the function are obtained by translating a line of slope $-\frac{\alpha}{\beta}$ up and down. 
It follows that $A$ and $D$ lie on such a line with intercept $\frac{10}{\beta}$. Moreover, the line $\overline{AD}$ acts as a divider: all lines with $c>10$ lies on one side of the line and all lines with $c<10$ lies on the other side. 
Since $B$ and $C$ lies within arc $\widehat{AD}$, it follows that they lie on the same side of $\overline{AD}$ and the side containing $B$ and $C$ is the side for which $c<10$. Now $E$ is necessarily contained on the other side of $\overline{AD}$ and therefore we must have $f(E) > 10$.
A: Take the plane $z = \alpha x + \beta y$ and examine the $z$ values as they vary over the unit circle. The values change direction twice, and since there is a direction change between $B$ and $D$, the value of $f(E)$ must be greater than $10$.
