# **Hexagon Help**

• Apr 30th 2006, 11:20 PM
Chuck_3000
**Hexagon Help**
I am stuck with this question, can anyone help?

Hexagon ABCDEF has the following properties:
i. diagonals AC, CE and EA are all the same length
ii. angles ABC and CDE are both 90 degrees
iii. all the sides of the hexagon have lengths which are different integers

a) What is the minimum perimeter of ABCDEF if AC = √85 (sqr root of 85)
b) What is the smallest length of AC for which ABCDEF has all these properties? What is the minimum perimeter in this case?

Any help would be much appreciated

thanks
• May 1st 2006, 12:43 AM
CaptainBlack
Quote:

Originally Posted by Chuck_3000
I am stuck with this question, can anyone help?

Hexagon ABCDEF has the following properties:
i. diagonals AC, CE and EA are all the same length
ii. angles ABC and CDE are both 90 degrees
iii. all the sides of the hexagon have lengths which are different integers

a) What is the minimum perimeter of ABCDEF if AC = √85 (sqr root of 85)
b) What is the smallest length of AC for which ABCDEF has all these properties? What is the minimum perimeter in this case?

Any help would be much appreciated

thanks

First the diagram: see attachment.

Now Part (a):

Now $\triangle\ ABC$ and $\triangle\ CDE$ are right triangles with hypotenuses of length
$\sqrt{85}$, so for $\triangle ABC$:

$
AB^2+BC^2=85
$

with $AB$ and $BC$ integers, and trail and error shows that the pair
of lengths $(AB,BC)$ is one of $(2,9),\ (6,7),\ (9,2),\ (7,6)$.

The same argument applies to $\triangle CDE$.

So for part (a) we have the minimum of $AB+BC+CD+DE=24$
(each pairs $(AB,BC),\ (CD,DE)$ is one of $(2,9),\ (6,7),\ (9,2),\ (7,6)$
but with none of the sides equal, so these sides is one of $2,6,7,9$ in some order, without repetition)

Also $EF+FA$ cannot be $10$ or less (trial and error shows this can't be
$10$, and it must be greater than $9$ as it must be greater than $\sqrt{85}$).
But $EF+FA$ can be $11$.

So the minimum perimeter is $24+11=35$.

RonL
• May 1st 2006, 01:27 AM
earboth
Quote:

Originally Posted by Chuck_3000
I am stuck with this question, can anyone help?

Hexagon ABCDEF has the following properties:
i. diagonals AC, CE and EA are all the same length
ii. angles ABC and CDE are both 90 degrees
iii. all the sides of the hexagon have lengths which are different integers

a) What is the minimum perimeter of ABCDEF if AC = √85 (sqr root of 85)
b) What is the smallest length of AC for which ABCDEF has all these properties? What is the minimum perimeter in this case?

Any help would be much appreciated

thanks

Hello,

to a)
You have to do with right triangles. The squares of the legs must be 85, that means, you have to split 85 into 2 squares. Thus:
AB = 2
BC = 9

DC = 6
ED = 7

Now you are looking for integers, which are not present in the list of legs above(1, 3, 4, 5, 8). The sum of these integers must exceed √(85) approximately 9.22. Therefore you have as possible combination:

EF = 3, AF = 8
EF = 4, AF = 8
EF = 5, AF = 8

I'm awfully sorry, but I can't offer you any help for part b).

Greetings

EB
• May 1st 2006, 01:54 AM
Chuck_3000
thank you so much guys!

CaptainBlack, how do you get those images on there?
• May 1st 2006, 03:07 AM
CaptainBlack
Quote:

Originally Posted by Chuck_3000
I am stuck with this question, can anyone help?

Hexagon ABCDEF has the following properties:
i. diagonals AC, CE and EA are all the same length
ii. angles ABC and CDE are both 90 degrees
iii. all the sides of the hexagon have lengths which are different integers

a) What is the minimum perimeter of ABCDEF if AC = √85 (sqr root of 85)
b) What is the smallest length of AC for which ABCDEF has all these properties? What is the minimum perimeter in this case?

Any help would be much appreciated

thanks

part (b)

Just an idea:

The smallest length for AC is the square root of the smallest number
that can be written as the sum of two squares in two different ways
with no common square among the four.

I will think about this later after we have finished playing HeroScape here :eek:

RonL
• May 1st 2006, 03:36 AM
CaptainBlack
Quote:

Originally Posted by Chuck_3000
thank you so much guys!

CaptainBlack, how do you get those images on there?

I use PowerPoint as a drawing package, then capture the image as
a bitmap into Paint Shop Pro (or any other photo editor) then
resize and save as a JPEG.

Then when composing a post I upload the image from the manage
attachments dialog.

RonL
• May 1st 2006, 06:12 AM
earboth
Quote:

Originally Posted by Chuck_3000
I am stuck with this question, can anyone help?

Hexagon ABCDEF has the following properties:
i. diagonals AC, CE and EA are all the same length
ii. angles ABC and CDE are both 90 degrees
iii. all the sides of the hexagon have lengths which are different integers

a) What is the minimum perimeter of ABCDEF if AC = √85 (sqr root of 85)
b) What is the smallest length of AC for which ABCDEF has all these properties? What is the minimum perimeter in this case?...

Hello,

to b) How about
$\sqrt{65}=1^2 + 8^2 = 4^2 + 7^2$

Greetings

EB
• May 2nd 2006, 04:09 AM
topsquark
An almost entirely unrelated question.

Doesn't "hexagon" imply equal sides? So wouldn't the question be asking about a "hexalateral" or some such?

Just curious.

-Dan
• May 2nd 2006, 04:58 AM
CaptainBlack
Quote:

Originally Posted by topsquark
An almost entirely unrelated question.

Doesn't "hexagon" imply equal sides? So wouldn't the question be asking about a "hexalateral" or some such?

Just curious.

-Dan

No hexagon=six sided planar figure, with all sides equal we are talking about
a regular hexagon.

RonL
• May 2nd 2006, 09:20 AM
topsquark
Quote:

Originally Posted by CaptainBlack
No hexagon=six sided planar figure, with all sides equal we are talking about
a regular hexagon.

RonL

:o Now I just feel silly. I knew that. (Ahem!)

-Dan
• May 2nd 2006, 09:23 AM
CaptainBlack
Quote:

Originally Posted by CaptainBlack
No hexagon=six sided planar figure, with all sides equal we are talking about
a regular hexagon.

RonL

I can make that a large hexagon for an extra 50 cents,
and would you like fries with that? :D

RonL
• May 2nd 2006, 09:26 AM
topsquark
Quote:

Originally Posted by CaptainBlack
I can make that a large hexagon for an extra 50 cents,
and would you like fries with that? :D

RonL

SuperSize me!

Wait, don't. I'm big enough already! :D

-Dan