BrainDen.com - Brain Teasers
• 0

Star area

Go to solution Solved by Logophobic,

Question

Suppose we have two congruent equilateral triangles with side length of 8ft.  They are on top of each other in the same orientation.  Now  One triangle is rotated by 180 degrees and is laying on top the other triangle forming a star. The triangles are positioned relative to each other such that the distance from the base of one triangle is 6ft to the base of the other triangle. What is the area of the star?

Recommended Posts

• 0
• Solution

I believe you are mistaken.

Move the top triangle far enough and it will not overlap the bottom triangle, so the area of overlap is clearly not constant.

The area of overlap is a hexagon. It appears that you have only accounted for a rectangular area of overlap.

The area of the star, by my calculation, is 50 * sqrt(3) - 48 square feet.

Share on other sites
• 0

If I am not mistaken ...

Spoiler

about the area remaining the same while we move the top triangle left and right keeping it parallel at 6ft to the base of the bottom triangle, then 56*sqrt(3) - 48 square ft.

Share on other sites
• 0

@LogophobicWhat I meant was:

Spoiler

That you can move the up-side down triangle parallel to the base at 6 ft any way left to right and still create a star. The "star" doesn't have to be symmetrical from the OP as far as I read it.

So IF the area of the star does change when you move it, THEN the OP is obviously ambiguous, one could get a lot of stars with different areas by moving it around a bit (but stopping when the edges of the upside down triangle touch the edges of the base of the upright triangle).

Hence, it makes sense that the reverse hypothesis is true: the area of the "star" does not change when you move the upside down triangle left-right. I'm not talking about the area of the overlap (be it a hexagon or a parallelogram), the area of the "star" composed of the super-imposed image. It felt easier to compute it in a corner scenario such as this:

I could have messed up the calculations, but if both mine and yours check out, the area is anywhere between those two numbers (and my hypothesis about the orange area being constant is false).

Edit: my calculations are 6*8 (middle parallelogram with base 8 and height 6ft) + 2*area of the little triangle (up and down in left figure). For the little triangle i get (8-4sqrt(3))^2*sqrt(3)/4 which yields an orange area of 48+56sqrt(3)-96.

Edited by araver
Share on other sites
• 0
Spoiler

I got 32sqrt(3)-48+54/sqrt(3)

Share on other sites
• 0

54/sqrt(3) = 18*sqrt(3)

Share on other sites
• 0

The area of the star is the combined area of the triangles minus the are of overlap. Since the area of overlap is not constant when moving the triangle, we can only conclude that symmetry of the star is implied.

Share on other sites
• 0
14 hours ago, Logophobic said:
Hide contents

The area of the star is the combined area of the triangles minus the are of overlap. Since the area of overlap is not constant when moving the triangle, we can only conclude that symmetry of the star is implied.

my mistake.  I did mean to state symmetry.  well done!

Share on other sites
• 0

The definition of ambiguity is when two different people undestand different things. I do admit that my answer is not correct for what the OP *intended* but have you thought what the answer would have been if no symmetry was involved?

Share on other sites
• 0
29 minutes ago, araver said:

The definition of ambiguity is when two different people undestand different things. I do admit that my answer is not correct for what the OP *intended* but have you thought what the answer would have been if no symmetry was involved?

Not until you have raised the question.  Clearly their must be an upper bound and lower bound between which the area could exist.

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

×   Pasted as rich text.   Paste as plain text instead

Only 75 emoji are allowed.

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.