Difference between revisions of "2005 Alabama ARML TST Problems/Problem 11"

(See also)
 
Line 3: Line 3:
  
 
==Solution==
 
==Solution==
{{solution}}
+
Join four such hexagons and let the points be labeled as shown below:
 +
 
 +
[[Image:Fourhexsquare.JPG]]
 +
 
 +
Since the sum of the angles in a hexagon is <math>(6-2)180^\circ = 720^\circ</math>, <math>\angle E = 720^\circ - (3\cdot 90^\circ + 80^\circ + 100^\circ) = 270^\circ</math>.
 +
 
 +
Since <math>\angle D_iE_iF_i = 360^\circ - \angle E = 360^\circ - 270^\circ = 90^\circ</math> and <math>\angle E_iF_iE_{i+1} = \angle F + \angle D = 80^\circ + 100^\circ = 180^\circ</math>, <math>E_1E_2E_3E_4</math> is a square with a side length of <math>EF+DE = 12</math>.
 +
 
 +
Also, Since <math>\angle A_iB_iC_i = \angle B = 90^\circ</math> and <math>\angle B_iA_iB_{i+1} = \angle A + \angle C = 90^\circ + 90^\circ = 180^\circ</math>, <math>B_1B_2B_3B_4</math> is a square with a side length of <math>AB+BC = 7+9 = 16</math>.
 +
 
 +
Therefore, <math>4[ABCDEF] + [E_1E_2E_3E_4] = [B_1B_2B_3B_4] \Longrightarrow</math> <math>[ABCDEF] =  \frac{[B_1B_2B_3B_4] - [E_1E_2E_3E_4]}{4} = \frac{16^2-12^2}{4} = \boxed{28}</math>.
  
 
==See also==
 
==See also==
 
{{ARML box|year=2005|state=Alabama|num-b=10|num-a=12}}
 
{{ARML box|year=2005|state=Alabama|num-b=10|num-a=12}}

Latest revision as of 23:31, 16 June 2008

Problem

In concave hexagon $ABCDEF$, $\angle  A = \angle B = \angle C = 90^\circ$, $\angle D = 100^\circ$, and $\angle F = 80^\circ$. Also, $CD=FA$, $AB=7$, $BC=9$, and $EF+DE=12$. Compute the area of the hexagon.

Solution

Join four such hexagons and let the points be labeled as shown below:

Fourhexsquare.JPG

Since the sum of the angles in a hexagon is $(6-2)180^\circ = 720^\circ$, $\angle E = 720^\circ - (3\cdot 90^\circ + 80^\circ + 100^\circ) = 270^\circ$.

Since $\angle D_iE_iF_i = 360^\circ - \angle E = 360^\circ - 270^\circ = 90^\circ$ and $\angle E_iF_iE_{i+1} = \angle F + \angle D = 80^\circ + 100^\circ = 180^\circ$, $E_1E_2E_3E_4$ is a square with a side length of $EF+DE = 12$.

Also, Since $\angle A_iB_iC_i = \angle B = 90^\circ$ and $\angle B_iA_iB_{i+1} = \angle A + \angle C = 90^\circ + 90^\circ = 180^\circ$, $B_1B_2B_3B_4$ is a square with a side length of $AB+BC = 7+9 = 16$.

Therefore, $4[ABCDEF] + [E_1E_2E_3E_4] = [B_1B_2B_3B_4] \Longrightarrow$ $[ABCDEF] =  \frac{[B_1B_2B_3B_4] - [E_1E_2E_3E_4]}{4} = \frac{16^2-12^2}{4} = \boxed{28}$.

See also

2005 Alabama ARML TST (Problems)
Preceded by:
Problem 10
Followed by:
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Invalid username
Login to AoPS