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

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===== Solution 1 =====
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== Problem ==
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The points <math>A, B, C ,</math> and <math>D</math> lie in that order on a line. Point <math>E</math> lies in a plane with <math>A, B, C ,</math> and <math>D</math> such that <math>\angle BEC = 78^{\circ}</math> . Given that <math>\angle EBC > \angle ECB</math>, <math>\angle ABE = 4x + y</math>, and <math>\angle ECB = x + y</math>, compute the number of positive integer values that <math>y</math> can take on.
  
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== Solution ==
  
 
<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</math>
 
<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</math>

Latest revision as of 00:54, 21 December 2021

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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