2007 Alabama ARML TST Problems/Problem 2: Difference between revisions

Revision as of 18:27, 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 $m\angle ECB < 56^\circ$

Noting that $m\angle ECB = 26 + y$ , it becomes clear that $1 \le m\angle ECB \le 29$ $\longrightarrow \boxed {29}$

Revision as of 18:26, 17 July 2011 view source Gravel (talk \| contribs) 22 edits Created page with "===== Solution 1 ===== <math>\angle ABE</math> is external to <math>\triangle BEC</math> at <math>\angle B</math>. Therefore it is equal to the sum: <math>\angle E + \angle C</..."		Revision as of 18:27, 17 July 2011 view source Gravel (talk \| contribs) 22 edits →Solution 1 Newer edit →
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	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 < 56^\circ</math>

	Noting that <math>m\angle ECB = 26 + y</math>, it becomes clear that <math>1 \le m\angle ECB \le 29</math> \longrightarrow \boxed {29}		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>