1.A,B,C,D, and E are points on a circle of radius 2 in counterclockwise order. We know AB=BC=DE=2 and CD = EA. Find [ABCDE]

Enter your answer in the form x + y sqrt(z) in simplest radical form.

2. ABCDEF is a regular hexagon with area 1. The intersection of ACE and BDF is a smaller hexagon. What is the area of the smaller hexagon?

abhinav Jan 28, 2018

#1**+2 **

Here's the first one:

See the way I've set it up

angle ABC = 120° ......so the area of triangle ABC = (1/2)*2^2 sin (120) = √3 units^2

And ACDE forms an isosceles trapezoid

Let the bottom base be DE = 2

A = (0,2) and C = (-√ (3) , -1) E = (2, 0) D = ( 1, -√3)

And AC forms the top base = √ [ 3 + 9 } = √12 units

Call G the mid-point of AC = (-√3/2 , 1/2 )

And F is the midpoint of DE = ( 3/2, -√3/2 )

And FG is the height of the trapezoid = √ [ ( 3 + √3)^2 + (1 + √3)^2 ] / 2 =

√ [ 4 ( 4 + 2√3) ]/ 2 = √ (4 + 2√3) = √[ 1 + √3 ]^2 = 1 + √3 units

So.....the area of the trapezoid is

(1/2) (height) (sum of the bases) =

(1/2) ( 1 + √3) [ 2 + √12 ] =

(1/2) (1 + √3) [ 2 + 2√3 ] =

( 1 + √3)^2 = [ 4 + 2√3 ] units^2

So.....the total area is

[ √3 + 4 + 2√3 ] = 4 + 3√3 units^2

CPhill Jan 28, 2018

#2**+2 **

2)

Here's a pic of the way I set this up :

The radius of the larger hexagon can be found as follows :

Area = 1 ....so....

3 r^2 sin (60) = 1

3r^2 * √3/2 = 1

r^2 = 2 / √27

r = √ [ 2 / √27 ]

The y coordinate of B is r * sin (60°) = √ [ 2 / √27 ] * [ √3 / 2 ]

And the y coordinate of F is r * sin (-60°) - √ [ 2 / √27 ] * [ √3 / 2 ]

And the distance between them, BF, is just 2 * √ [ 2 / √27 ] * [ √3 / 2 ] =

√ [ 2 / √27 ] * √3 =

√ [ 6 / √27]

Note that because they span equal arcs, chords AC and BF are equal

And by SSS, triangle CBA is congruent to triangle BAF and both are isosceles

And angle ACB = angle BAC = angle BAF = angle FBA

So angle FBA = angle BAC....so BH = HA

And angle CBA = angle BAE = 90°

So...subtracting equal angles , then angle HBI = angle HAM

And angle BHI = angle AHM

So...by SAS, triangle BHI is congruent to triangle AHM

But since HI is a transversal that cuts parallels DB and AE....then angle HAM = angle HIB

But angle HAM = angle HBI......so angle HBI = angle HIB

And angle HIB = angle HMA.....so angle HAM = angle HMA

So HM = AH and HI = HB

So.....HM = HI = AH = HB

So .....since triangles BHI and AHM are congruent.....then HM = HB

And by similar logic, we can prove that HM = MF

So..... HB = HM = HF

So....side HM of hexagon HIJKLM is 1/3 of BF

So....the area of the hexagon is

3 * [ 6 / √27] / 9 * √3 / 2 =

√3 / √27 =

√3 / [ 3 √3] =

1/3 units^2

CPhill Jan 28, 2018