Difference between revisions of "2007 Alabama ARML TST Problems/Problem 2"

Revision as of 18:46, 17 July 2011

$\angle ABE$ 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 $1 \le m\angle ECB \le 24$ $\longrightarrow \boxed {24}$

@@ Line 14: / Line 14: @@
 However we note that by the problem statement <math>\angle ECB</math> cannot be greater than <math>\angle EBC</math>.
-The two angles sum to <math>102^\circ</math>, thus <math>m\angle ECB < 56^\circ</math>
+The two angles sum to <math>102^\circ</math>, thus <math>m\angle ECB < 51^\circ</math>
-Noting that <math>m\angle ECB = 26 + y</math>, it becomes clear that <math>1 \le m\angle ECB \le 29</math> <math>\longrightarrow \boxed {29}</math>
+Noting that <math>m\angle ECB = 26 + y</math>, it becomes clear that <math>1 \le m\angle ECB \le 24</math> <math>\longrightarrow \boxed {24}</math>