Fractals and Visualising Solids | FIO

Question 2

Find the number of holes, and the triangles that remain at each step of the shape sequence that leads to the Sierpinski Triangle.

Question diagram 1
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Solution

The Sierpinski Triangle is a special shape called a fractal. We make it by starting with a triangle and repeatedly removing its middle part.

Step 1 — Counting the holes

Let us count the number of empty, uncolored triangular spaces, which we call holes, at each step.

At Step 0, we start with one solid triangle. There are no empty spaces inside it. Number of holes at Step 0=0\text{Number of holes at Step 0} = \mathbf{0}

Diagram 6

At Step 1, we remove the middle inverted triangle from the original triangle. This creates one hole. Number of holes at Step 1=1\text{Number of holes at Step 1} = \mathbf{1}

Diagram 2

At Step 2, we take each of the three solid triangles from Step 1. From each of these three triangles, its own middle inverted triangle is removed. This adds 3 new holes. We already had 1 hole from Step 1. Number of holes at Step 2=(holes from Step 1)+(new holes added) \text{Number of holes at Step 2} = (\text{holes from Step 1}) + (\text{new holes added}) =1+3= 1 + 3

4 holes\boxed{\mathbf{4} \text{ holes}}

Diagram 3

At Step 3, we continue the process. There are 9 solid triangles remaining from Step 2. Each of these 9 triangles will have its middle part removed. This adds 9 new holes. We already had 4 holes from Step 2. Number of holes at Step 3=(holes from Step 2)+(new holes added) \text{Number of holes at Step 3} = (\text{holes from Step 2}) + (\text{new holes added}) =4+9= 4 + 9

13 holes\boxed{\mathbf{13} \text{ holes}}

Diagram 4

At Step 4, we repeat the process again. There are 27 solid triangles remaining from Step 3. Each of these 27 triangles will have its middle part removed. This adds 27 new holes. We already had 13 holes from Step 3. Number of holes at Step 4=(holes from Step 3)+(new holes added) \text{Number of holes at Step 4} = (\text{holes from Step 3}) + (\text{new holes added}) =13+27= 13 + 27

40 holes\boxed{\mathbf{40} \text{ holes}}

Diagram 5

We can see a pattern for the number of new holes added at each step. At Step 1, we added 30=13^0 = 1 hole. At Step 2, we added 31=33^1 = 3 holes. At Step 3, we added 32=93^2 = 9 holes. At Step 4, we added 33=273^3 = 27 holes. So, at Step nn (for n1n \ge 1), we add 3n13^{n-1} new holes. The total number of holes at Step nn is the sum of all new holes added up to that step. Number of holes at Step n=30+31+32++3n1\text{Number of holes at Step n} = 3^0 + 3^1 + 3^2 + \dots + 3^{n-1}

Number of holes at Step n=30+31+32++3n1\boxed{\text{Number of holes at Step n} = 3^0 + 3^1 + 3^2 + \dots + 3^{n-1}}

Step 2 — Counting the remaining triangles

Let us count the number of solid, colored triangles that remain at each step.

At Step 0, we start with one large solid triangle. Number of remaining triangles at Step 0=1\text{Number of remaining triangles at Step 0} = \mathbf{1}

<DIAGRAM: A single solid purple triangle labeled "Step 0".>

At Step 1, the middle part is removed. This leaves three smaller solid triangles at the corners. Number of remaining triangles at Step 1=3\text{Number of remaining triangles at Step 1} = \mathbf{3}

Diagram 7

At Step 2, each of the three solid triangles from Step 1 is further divided. Each of them becomes three smaller solid triangles. So, we multiply the previous number of triangles by 3. Number of remaining triangles at Step 2=3×(triangles from Step 1) \text{Number of remaining triangles at Step 2} = 3 \times (\text{triangles from Step 1}) =3×3= 3 \times 3

9 triangles\boxed{\mathbf{9} \text{ triangles}}

Diagram 8

At Step 3, each of the nine solid triangles from Step 2 is further divided into three smaller solid triangles. Number of remaining triangles at Step 3=3×(triangles from Step 2) \text{Number of remaining triangles at Step 3} = 3 \times (\text{triangles from Step 2}) =3×9= 3 \times 9

27 triangles\boxed{\mathbf{27} \text{ triangles}}

Diagram 9

At Step 4, each of the twenty-seven solid triangles from Step 3 is further divided into three smaller solid triangles. Number of remaining triangles at Step 4=3×(triangles from Step 3) \text{Number of remaining triangles at Step 4} = 3 \times (\text{triangles from Step 3}) =3×27= 3 \times 27

81 triangles\boxed{\mathbf{81} \text{ triangles}}

Diagram 10

We can see a pattern for the number of remaining triangles. At Step 0, we have 30=13^0 = 1 triangle. At Step 1, we have 31=33^1 = 3 triangles. At Step 2, we have 32=93^2 = 9 triangles. At Step 3, we have 33=273^3 = 27 triangles. At Step 4, we have 34=813^4 = 81 triangles. So, at Step nn, the number of remaining triangles is 3n3^n.

Number of remaining triangles at Step n=3n\boxed{\text{Number of remaining triangles at Step n} = 3^n}

Answer

(i) Number of holes: Step 0: 0 hole Step 1: 1 hole Step 2: 4 holes Step 3: 13 holes Step 4: 40 holes Step n: 30+31+32++3n13^0 + 3^1 + 3^2 + \dots + 3^{n-1} holes

(ii) Number of triangles that remain: Step 0: 1 triangle Step 1: 3 triangles Step 2: 9 triangles Step 3: 27 triangles Step 4: 81 triangles Step n: 3n3^n triangles

More questions in FIO

Q1

Draw the initial few steps (at least till Step 2) of the shape sequence that leads to the Sierpinski Triangle.

Q2

Find the number of holes, and the triangles that remain at each step of the shape sequence that leads to the Sierpinski Triangle.

Q3

Find the area of the region remaining at the nnth step in each of the shape sequences that lead to the Sierpinski fractals. Take the area of the starting square/triangle to be 1 sq. unit.

Q4

Draw the initial few steps (at least till Step 2) of the shape sequence that leads to the Koch Snowflake.

Q5

Find the number of sides in the nth step of the shape sequence that leads to the Koch Snowflake.

Q6

Find the perimeter of the shape at the nth step of the sequence. Take the starting equilateral triangle to have a sidelength of 1 unit.

Q7

Figure it Out

  1. Which of the following are the nets of a cube? First, try to answer by visualisation. Then, you may use cutouts and try.
Q8

A cube has 11 possible net structures in total. In this count, two nets are considered the same if one can be obtained from the other by a rotation or a flip. For example, the following nets are all considered the same —

Find all the 11 nets of a cube.

Q9

Draw a net of a cuboid having sidelengths:

(i) 5 cm, 3 cm, and 1 cm

(ii) 6 cm, 3 cm, and 2 cm

Q10

Observe the front view, top view and side view of the different lines in Fig. 4.6. Is there any relation between their lengths?

Q11

Find the front view, top view and side view of each of the following solids, fixing its orientation with respect to the vertical, horizontal and side planes: cube, cuboid, parallelepiped, cylinder, cone, prism, and pyramid. If needed, see the next problem for clues.

Q12

Match each of the following objects with its projections.

Q13

Draw the top view, front view and the side view of each of the following combinations of identical cubes.

Q14

Imagine eight identical cubes, glued together along faces to form the letter 'C'.

Q15

Which solid corresponds to the given top view, front view, and side view?

Solids to choose from:

Q16

Using identical cubes, make a solid that gives the following projections:

Q17

Find the number of cubes in this stack of identical cubes.

Q18

What are the different shapes the projection of a cube can make under different orientations?

Q19

In addition to the 5 ways shown in Fig. 4.8, are there any additional ways of gluing four cubes together along faces? Can you visualise and draw these as well?

Q20

Draw the following figures on the isometric grid.

[Hint: It may be useful to determine whether the edge to be currently drawn — say, along the height — goes from down to up or up to down. Accordingly, draw the line segment on the grid either in the direction of the height axis or opposite to it.]

Q21

Is there anything strange about the path of this ball? Recreate it on the isometric grid.

[Hint: Consider a portion of this figure that is physically realisable and identify the 3 primary directions.]

Q22

Observe this triangle.

(i) Would it be possible to build a model out of actual cubes? What are the front, top, and side profiles of this impossible triangle? (ii) Recreate this on an isometric grid. (iii) Why does the illusion work?

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