fskilnik wrote:The square ACEG shown below has an area of 36 units squared. What is the value of x that maximizes the area of the polygon ABDFG?



(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

Answer: [spoiler]__(C)____[/spoiler]
Difficulty Level: 650 - 700
Source: www.GMATH.net

Almost no math is needed here.

Answer choice A implies the following:
BCD is a 45-45-90 triangle in which each leg has a length of 1.
DEF is a 45-45-90 triangle in which each leg has a length of 5.
Answer choice E implies the following:
BCD is a 45-45-90 triangle in which each leg has a length of 5.
DEF is a 45-45-90 triangle in which each leg has a length of 1.
In each case, one of the 45-45-90 triangles has a leg of 1, while the other 45-45-90 triangle has a leg of 5.
Implication:
The two answer choices will each yield the same total area for the two triangles and thus the same area for polygon ABDFG.
Since A and E cannot both be correct, eliminate A and E.

Answer choice B implies the following:
BCD is a 45-45-90 triangle in which each leg has a length of 2.
DEF is a 45-45-90 triangle in which each leg has a length of 4.
Answer choice D implies the following:
BCD is a 45-45-90 triangle in which each leg has a length of 4.
DEF is a 45-45-90 triangle in which each leg has a length of 2.
In each case, one of the 45-45-90 triangles has a leg of 2, while the other 45-45-90 triangle has a leg of 4.
Implication:
The two answer choices will each yield the same total area for the two triangles and thus the same area for polygon ABDFG.
Since B and D cannot both be correct, eliminate B and D.

The correct answer is C.

fskilnik wrote:The square ACEG shown below has an area of 36 units squared. What is the value of x that maximizes the area of the polygon ABDFG?



(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Difficulty Level: 650 - 700
Source: www.GMATH.net

\[?\,\,:\,\,x\,\,{\text{to}}\,\,\max \,\,S\left( {ABDFG} \right)\,\,\,\, \Leftrightarrow \,\,\,x\,\,{\text{to}}\,\,\min \,\,\,\left[ {S\left( {\Delta {\text{BCD}}} \right) + S\left( {\Delta DEF} \right)} \right]\]
\[S\left( {\Delta {\text{BCD}}} \right) + S\left( {\Delta DEF} \right) = \frac{{{x^2}}}{2} + \frac{{{{\left( {6 - x} \right)}^2}}}{2} = \frac{{2{x^2} - 12x + 36}}{2} = {x^2} - 6x + 18\]
\[?\,\,\,:\,\,\,x\,\,{\text{to}}\,\,\min \,\,{x^2} - 6x + 18\,\,\,\,\, \Leftrightarrow \,\,\,\,\,? = x = {x_{vert}} = - \frac{b}{{2a}} = - \frac{{ - 6}}{2} = 3\]

Almost no lines, almost no arguments, almost no effort were needed here. Just the old and powerful good math.

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.