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Star Stack (Posted on 2004-04-12) Difficulty: 3 of 5
Three pentagram-shaped stars (the stars formed from the diagonals of a regular pentagon) are stacked up so that the bottom two ends of the tips touch the middle ends of the tips of the star below. (See diagram.)

The distance from the top of the stack to the floor (where the bottom star's "feet" rest) is 4 feet.

What is the distance between the bottom two ends of the tips of the stack that touch the floor?

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Solution Really Long Explanation | Comment 2 of 19 |
The answer I got for the distance between the two “legs” of the bottom pentagram is 0.613625.  Here is my solution:
I numbered the pentagrams 1-3, starting with the bottom one.  I also drew in the 3 pentagons that enclosed each pentagram (basically I just connected the points of the stars).
I labeled certain segments of each pentagram/pentagon as follows: S = the length of one side of the pentagon (the distance between two adjacent vertices of the pentagon), W = the width of the pentagon (the distance between two non-adjacent vertices of the pentagon), H = the height of the pentagon (the perpendicular distance between one vertex and the opposite side of the pentagon), E = the edge of one of the pentagram points (the distance between a convex vertex of the pentagram and an adjacent concave vertex of the pentagram), and X = the perpendicular distance between a concave vertex of the pentagram and the neighboring side of the pentagon.
So if I were talking about the lengths of the sides of all three pentagons, I’d refer to them as S1 for the bottom, S2 for the middle, and S3 for the top.  But if I were talking about general relationships in a pentagon I might drop the 1, 2, or 3.
Ok, hopefully my terminology is clear and I can start working on some relationships.  Eventually what I want is an equation that just contains S1 and some known stuff, which equals 4 (the distance from the floor to the tip of Pentagram 3).  Moving on…
Well, the total number of degrees inside an n-sided polygon is (n-2)*180.  Here we are dealing with pentagons, so (5-2)*180 = 540.  Since these are regular pentagons, the angle at each vertex is 540/5=108.  Let’s look at one of the concave vertices of the pentagram.  Notice that the large angle on the inside is part of a smaller pentagon.  So this angle is also 108.  Since this vertex is the intersection of two straight lines, the other large angle at this vertex is also 108 degrees.  And the smaller angle is 180-108 = 72 degrees.
Now let’s look at a particular triangle by picking one vertex on the pentagon, and the two adjacent vertices on the pentagon (so the sides would be S, S, and W, so it’s isosceles).  Well, we’ve already determined that the large angle is 108 degrees.  This means that the remaining angles are each (180-108)/2 = 36 degrees.  Notice that this triangle is similar to the triangle with vertices at a concave vertex of the pentagram and the two adjacent convex vertices (so the sides would be E, E, and S, so it’s isosceles).  Remember that in the previous paragraph, we determined that the large angle is 108, so again, the other two angles must each be 36 degrees.  Since these are similar, we know that S/E = W/S, or E = S^2/W.
Ok, so using the law of sines, we can see that W/sin(108) = S/sin(36), or W/S = sin(108)/sin(36).  I’m going to say this factor of sin(108)/sin(36) = F for simplicity because we’ll be running into this factor a bit, and this will keep the equations cleaner.
Alrighty!  Now I feel like we have a little bit of a base of information.  Let’s see where we want to go with this.  We’re told the height of this star-stack, so let’s see what we can do with that.  I can see two ways of breaking up this stack so that we have little heights we can figure out and add up.  The first way is 4 = H1 + H3 + “the inner pentagon inside Pentagram 2.”  Another way is 4 = H1 + (H2 – X2) + (H3 – X3).  I’m going to use this second way, but there’s not reason you couldn’t use the previous and following information to find the relationships needed for the first way.
Well, to calculate the height of a pentagon, let’s consider the triangle formed by the top vertex and the bottom two vertices of a pentagon.  The bottom edge is S, and the two sides are W.  We can find the height by using the Pythagorean theorem on the triangle with W as the hypotenuse, and S/2 & H as the two legs.  So W^2 = H^2 + (S/2)^2.
H^2 = W^2 – (S/2)^2 = (F*S)^2 – (S^2)/4 = [F^2 –1/4]*S^2.
So H = sqrt(F^2 – 1/4)*S.
To calculate the distance X of a pentagon, let’s revisit the triangle with vertices at a concave vertex of the pentagram and the two adjacent convex vertices (so the sides would be E, E, and S).  We can find X by using the Pythagorean theorem on the triangle with E as the hypotenuse, and S/2 & X as the two legs.  So E^2 = X^2 + (S/2)^2.  X^2 = E^2 – (S/2)^2.  Remember that E = S^2/W = S^2/(FS) = S/F.
So X^2 = (S/F)^2 – (S/2)^2 = [1/(F^2) – 1/4]*S^2.
So X = sqrt(1/(F^2) – 1/4)*S.
I think we’re ready for pentagram to pentagram relationships.  Notice that for Pentagram 1 and Pentagram 2, S1 = E2, and W1 = S2.  Similarly for Pentagram 2 and Pentagram 3, S2 = E3, and W2 = S3.  So…
S2 = W1 = F*S1.
S3 = W2 = F*S2 = F*(W1) = F*(F*S1) = F^2*S1.
So we have:
H1 = sqrt(F^2 – 1/4)*S1
H2 = sqrt(F^2 – 1/4)*S2 = sqrt(F^2 – 1/4)*(F*S1)
H3 = sqrt(F^2 – 1/4)*S3 = sqrt(F^2 – 1/4)*(F^2*S1)
X1 = sqrt(1/(F^2) – 1/4)*S1
X2 = sqrt(1/(F^2) – 1/4)*S2 = sqrt(1/(F^2) – 1/4)*(F*S1)
X3 = sqrt(1/(F^2) – 1/4)*S3 = sqrt(1/(F^2) – 1/4)*(F^2*S1)
So 4 = H1 + (H2 – X2) + (H3 – X3)
 4 =  sqrt(F^2 - 1/4)*S1 +
       [sqrt(F^2 - 1/4)*(F*S1) - sqrt(1/(F^2) - 1/4)*(F*S1)] +
       [sqrt(F^2 - 1/4)*(F^2*S1) - sqrt(1/(F^2) - 1/4)*(F^2*S1)]
    =S1*[sqrt(F^2 - 1/4) +
             sqrt(F^2 - 1/4)*F - sqrt(1/(F^2) - 1/4)*F +
             sqrt(F^2 - 1/4)*F^2 - sqrt(1/(F^2) - 1/4)*F^2]
    =S1*[sqrt(F^2-1/4)*(1+F+F^2) - sqrt(1/(F^2)-1/4)*(F+F^2)]
S1 = 4 /[sqrt(F^2-1/4)*(1+F+F^2) - sqrt(1/(F^2)-1/4)*(F+F^2)]
     = 0.613625
Well, you made it to the end!  I hope you enjoyed it!  Later!

  Posted by nikki on 2004-04-13 19:32:55
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