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La Ora Stelo

Jacques Ferroul is a prolific puzzle designer and creator of utterly beautiful tilings. Kadon has produced several of his designs, including Poly-Spidrons, Tetrapentos, and Stelo.

During the 2007 Christmas holidays, Jacques was vacationing in the Alps. On December 26, he learned of the death of a friend, Jean-Pierre Puisais-Hée. On the 27th, Jacques was in Marseille for his friend’s funeral. At the far end of the church, where the ceremony was taking place, Jacques noticed a large gold-colored star, perhaps part of the church’s Christmas decorations.

Because the death of his friend weighed much on his heart and mind, he turned his thoughts to the star and there, in a few seconds, he visualized it cut into several pieces. On the 28th he returned to the Alps, and by the 29th, La Ora Stelo (Esperanto for The Golden Star) was ready, with many solutions for pentagons, diamonds, kites and more...

La Ora Stelo is herewith dedicated to Jean-Pierre Puisais-Hée.


The tiles of La Ora Stelo are an original polyform set, the polyores, based on the two golden triangles shown here. At left is the "wide" isosceles triangle, with angles of 36-108-36; at right is the "tall" isosceles triangle, with angles of 72-36-72. The base of the wide triangle matches the sides of the tall, and the base of the tall matches the sides of the wide:

The two golden triangles

These tiles come from dissecting a pentagon, and they share the pentagon's golden-ratio and non-periodic characteristics. The lengths of their sides are related by the golden ratio (an endless decimal beginning with 1.6180). They also occur within a tiling in a frequency that matches the golden ratio, with about 5/8 tall triangles and 3/8 the wide triangle.

Note the difference between La Ora Stelo's building blocks and those in our Collidescape sets, wherein the sides of the two triangles are the matching length. These occur in the reverse ratio of 5/8 wide and 3/8 tall because their wide triangle is actually the size of two unit polyore triangles joined (the order-2 polyores triangle shown in the first row below).

Jacques derived the polyores by combining the unit triangles in length-matched groupings from one to twos and threes. Here are all 32 shapes of polyores orders 1 through 3:

The 32  La Ora Stelo tiles

This complete set contains 42 each of the two unit triangles. In solving the wealth of designs provided, you will not always need to use the full set. Each design will demand its own inventory. A large collection of beautiful shapes will use just 25 pieces.

This puzzle is made for Jacques Ferroul and for you by special order. We laser-cut it from transparent tinted acrylic in jewel colors. The Lucite cut edges glow and make it easy to see the divisions between the individual pieces. Color arrangements may vary. If you have a preference for all pieces in one solid color—we offer topaz (transparent yellow) and aqua (transparent green)—please indicate your choice in the special instructions space of the order form. Otherwise we will ship the 8-color ensemble.

The tray has 9 rings on its floor, and you can use them to form ascending sizes of pentagons, rhombs, trapezoids and embedded stars. See Jacques' webpage for the entire and growing collection of themes and challenges. Recommended for ages 10 to adult.   $59

Click the green logo to enter on Order Form


Prize Competition!

Jacques Ferroul is offering cash prizes for the best solutions to a two-fold TRAPEZOID challenge to run during 2009. Details below.

Prize Challenges for La Ora Stelo

Prizes

  1. 150 dollars for the first challenge—
    be first to submit longest series in 2009 with 37 pieces.

  2. 150 Euros for the second challenge—
    be first to submit longest series in 2009 with the 32 polyores only.

 
Details

Lengths of sides of the trapezoids are counted in units of the sides of the golden triangles. The longer dimension (the sides of the tall triangle and base of the wide triangle) is called phi (phi); the shorter dimension (the sides of the wide triangle and base of the tall triangle) is simply "1".

Jacques enumerates 148 different sizes of trapezoids, where x is the length of the top horizontal side. For the special case where x=0, it's the tip of a triangle. The oblique sides of each trapezoid in the series are always the same length:   phi+1. The bottom horizontal sides have the length of x+phi. So the height of the trapezoids stays constant while they progressively stretch sideways like an accordion.

We list here for you the 148 sizes. It is not known whether they all have solutions. You're not required to solve all 148, only the longest continuous sequence within the series. If you skip one, that ends your streak. Here's a sample trapezoid of size phi:

Sample trapezoid of size phi, the length of top edge

The dimensions for x:

  • 0
  • 1
  • phi
  • 2
  • phi+1
  • 3
  • 2phi
  • phi+2
  • 4
  • 2phi+1
  • phi+3
  • 3phi
  • 5
  • 2phi+2
  • phi+4
  • 3phi+1
  • 6
  • 2phi+3
  • 4phi
  • phi+5
  • 3phi+2
  • 7
  • 2phi+4
  • 4phi+1
  • phi+6
  • 3phi+3
  • 8
  • 5phi
  • 2phi+5
  • 4phi+2
  • phi+7
  • 3phi+4
  • 9
  • 5phi+1
  • 2phi+6
  • 4phi+3
  • phi+8
  • 6phi
  • 3phi+5
  • 10
  • 5phi+2
  • 2phi+7
  • 4phi+4
  • phi+9
  • 6phi+1
  • 3phi+6
  • 11
  • 5phi+3
  • 2phi+8
  • 7phi
  • 4phi+5
  • phi+10
  • 6phi+2
  • 3phi+7
  • 12
  • 5phi+4
  • 2phi+9
  • 7phi+1
  • 4phi+6
  • phi+11
  • 6phi+3
  • 3phi+8
  • 8phi
  • 13
  • 5phi+5
  • 2phi+10
  • 7phi+2
  • 4phi+7
  • phi+12
  • 6phi+4
  • 3phi+9
  • 8phi+1
  • 14
  • 5phi+6
  • 2phi+11
  • 7phi+3
  • 4phi+8
  • 9phi
  • phi+13
  • 6phi+5
  • 3phi+10
  • 8phi+2
  • 15
  • 5phi+7
  • 2phi+12
  • 7phi+4
  • 4phi+9
  • 9phi+1
  • phi+14
  • 6phi+6
  • 3phi+11
  • 8phi+3
  • 16
  • 5phi+8
  • 10phi
  • 2phi+13
  • 7phi+5
  • 4phi+10
  • 9phi+2
  • phi+15
  • 6phi+7
  • 3phi+12
  • 8phi+4
  • 17
  • 5phi+9
  • 10phi+1
  • 2phi+14
  • 7phi+6
  • 4phi+11
  • 9phi+3
  • phi+16
  • 6phi+8
  • 3phi+13
  • 8phi+5
  • 18
  • 5phi+10
  • 10phi+2
  • 2phi+15
  • 7phi+7
  • 4phi+12
  • 9phi+4
  • phi+17
  • 6phi+9
  • 3phi+14
  • 8phi+6
  • 19
  • 5phi+11
  • 10phi+3
  • 2phi+16
  • 7phi+8
  • 4phi+13
  • 9phi+5
  • phi+18
  • 6phi+10
  • 3phi+15
  • 8phi+7
  • 20
  • 5phi+12
  • 10phi+4
  • 2phi+17
  • 7phi+9
  • 4phi+14
  • 9phi+6
  • phi+19
  • 6phi+11
  • 3phi+16
  • 8phi+8
  • 21
  • Each solution will have a different combination or "inventory" of the two unit triangles. For example, the sample size shown above contains 6 tall and 3 wide triangles. Jacques' website shows the inventory for a few other sizes. The rest are up to you to figure out. See more illustrations and explanations on Jacques' Trapezium Series webpage.

     
    Send us your solutions

    We wish you successful solving and await your solutions! Will you be the one to win the prize? You may enter more than once, if you improve your results. We will only consider your best entry. In case of ties, the best solution received the earliest wins. Anytime during the year 2009, email your solutions as electronic image files to Jacques Ferroul or to Kadon Enterprises, Inc., or mail drawings on paper to:

    Kadon Enterprises, Inc.
    1227 Lorene Drive, Suite 16
    Pasadena, MD 21122
    USA

    See also Jacques Ferroul's TetrapentosTM and Poly-SpidronsTM and SteloTM

    Visit Jacques' blog Kaprompiloj (in French and Esperanto)
    for many other great polyform explorations



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