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2007 Alabama ARML TST Problems/Problem 2

Problem

The points $A, B, C ,$ and $D$ lie in that order on a line. Point $E$ lies in a plane with $A, B, C ,$ and $D$ such that $\angle BEC = 78^{\circ}$ . Given that $\angle EBC > \angle ECB$ , $\angle ABE = 4x + y$ , and $\angle ECB = x + y$ , compute the number of positive integer values that $y$ can take on.

Solution

$\angle ABE$ is external to $\triangle BEC$ at $\angle B$ . Therefore it is equal to the sum: $\angle E + \angle C$

Then, according to the problem statement:

$\angle ABE = 4x + y = 78 + x + y$

$x = 26$

As y cancels, its value is not bounded by this algebraic relation.

However we note that by the problem statement $\angle ECB$ cannot be greater than $\angle EBC$ .

The two angles sum to $102^\circ$ , thus $m\angle ECB < 51^\circ$

Noting that $m\angle ECB = 26 + y$ , it becomes clear that $1 \le m\angle ECB \le 24$ $\longrightarrow \boxed {24}$

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