BTGmoderatorDC wrote:The lengths of the sides of triangle ABC are such that AB = x + 2, BC = x - 3, and AC = y - 1. Which angle in triangle ABC is the largest?
(1) x - y = 1
(2) y = 8
Source: Magoosh
In any triangle, a greater (length of) side is always "facing" a greater (measure of) angle, and vice-versa. Hence:
$$?\,\,\,\,:\,\,\,{\rm{greater}}\,\,{\rm{angle}}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,?\,\,\,\,:\,\,\,{\rm{greater}}\,\,{\rm{side}}$$
$$x + 2\,\,\mathop > \limits^{{\rm{always}}} \,\,\,x - 3\,\,\,\,\,\, \Rightarrow \,\,\,\,\,AB > BC\,\,\,\,\left( * \right)$$
$$\left( * \right)\,\,\,\, \Rightarrow \,\,\,\,?\,\,\,:\,\,\,\max \left\{ {AB,AC} \right\} = \,\,\boxed{\,\,\max \left\{ {x + 2\,\,,\,\,y - 1} \right\}\,\,}$$
$$\left( 1 \right)\,\,x - y = 1\,\,\,\,\mathop \Rightarrow \limits^{{\text{FOCUS}}\,\,!} \,\,\,\,\,x + 2 = 1 + y + 2\,\,\,\,\mathop > \limits^{{\text{always}}} \,\,\,\,y - 1\,\,\,\,\, \Rightarrow \,\,\,\,{\text{SUFF}}.$$
$$\left( 2 \right)\,\,\,?\,\,:\,\,\,\max \left\{ {x + 2\,\,,\,\,8 - 1} \right\}\,\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,x = 6\,\,\,\, \Rightarrow \,\,\,\,\Delta = \left( {8,3,7} \right)\,\,{\rm{exists}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,? = AB\,\,\,\, \hfill \cr
\,{\rm{Take}}\,\,x = 4.5\,\,\,\, \Rightarrow \,\,\,\,\Delta = \left( {6.5,1.5,7} \right)\,\,{\rm{exists}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,? = AC\,\, \hfill \cr} \right.$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
