The segment of a polygon

 

From the table above, we can spot clearly that the more ‘n’ (from 1:n) in the ratio, the more length increases in the sides and the area of the inner (the small) triangle. However, the ratio between the area of the outer and the inner triangle’s gap became smaller (1:7 (inner: outer) can be written as 1: 0.143, 4: 13; 1: 0.308 and 3: 7; 0.429). This shows that the area of the inner triangle increases when the ratio sides increases while the outer triangle stays the same. There is also a pattern shown in the ratio of the area of the outer and inner triangle. The difference of the bigger and ratio is increased by 2 (6 and 8) while the pattern of the inner triangle is the square of the number before ‘n’ ((2-1)²=1, (3-1)²=4, (4-1)²=9)

Moreover, it should be noted that the ratios of the sides are the square root of the ratio between the areas. The square of 0.378 for example is 0.1429, while 1 divide by 7 is also 0.1429. This also happen in the 1:3. 0.55² is 0.30, same with 4 divide by 7. For 1:4, 0.655² is 0.429, the same with 3 divide by 7. So, basically, like other shapes, if the two shapes are similar, the ratio of the area is the square of the ratio between the sides.

 Using the data above, we can obtained a conjecture of;

Outer triangle: inner triangle=

This conjecture can be used to obtain the ratio result between the area of the outer and the inner triangle. ‘n’ represent the number of ratio of sides. We can prove the conjecture by trialing the ratio of sides.

Ratio of 1:2

Ratio of 1:3

Ratio of 1:4

From the proofs and trials above, the conjecture is supposedly true. We can also prove that the answer is true by using manual calculation. To prove that the conjecture is true, the answer of the manual calculation should match the geometry tool. This is to prevent the conjecture from the possible error on the geometry tool.

In this drawing of triangle 1:2, let the sides be ‘S’.

The area of ∆CAB can be determined as below

AR=PB=QC

∆PBU = ∆ATR = ∆QCV let the smallest triangular be ‘t’

=

=

=

Using the sine rule we continue and get

From the Pythagoras theory (tan= ) I got

 =

from this point, we know that sinΘ = (sinΘ=)

while cosΘ = (cosΘ=)

To find AT we use the sine rule.

 

The ratio of the bigger equilateral triangle to the smaller equilateral triangle is acquired from

So, the ratio (after crossing the  is 4:28 or simplified to be 1:7. The calculation of the “Geometer’s sketchpad” and the formula is true.

Of course, checking the other ratios (of the triangle shaped polygons) are a bit different from the one calculated above. Every one of the triangle have only 3 parts overlaying another part. When it is 1:2, the portion of the section is , which makes the whole part of the triangle covered if not minus the overlaying parts. In other ratios though, an additional calculation is needed.

An example is for the 1:3

In this drawing of triangle 1:3, let the sides be ‘S’.

Exactly the same as the 1:2, the area of ∆CAB can be determined as below

AR=PB=QC

∆PBU = ∆ATR = ∆QCV let the smallest triangular be ‘t’

=

=

= 

There will be an additional  to get the area of the smaller triangle.

For 1:4 it is basically

=  +3t

Since

=

=

= 

Therefore, to find the area of the smaller triangle, we are able to construct a formula of  t.  is the formula of the smaller triangle (S for the length of the sides) and t is for the area of the smallest triangle 9which usually are on the one being overlaid in the edge of the object. This conjecture is to find the area of the inner triangle.

Of course the area of t is not always the same. Because of that, we should follow the procedure mentioned above for the case of 1:2 (by obtaining the triangle over layering areas and adding it with the percentage mentioned from the formula above.

        

The conjecture of  though, will only work for an equilateral triangle. It will not work for non equilateral triangle because the sides and the angle are not the same. We therefore cannot calculate the degree and the sides properly. Even though the sine rule can still be use, the equation would not have the same conjecture like the one that I have made. The formula of the smaller triangle in the middle ( t. ) would not work for a non equilateral triangle because of two reasons. One is because the area of the triangle will not be the same (since the degrees are different) and two is because the area ratio of the smaller triangle and the bigger triangle wouldn’t be the same for every side.

For a similar construction of inner segments in a square which is divided in the ratio of 1:2, the result will be as show in the diagram below.

From the result above, I was able to make a table to sum it all up.

Similar to the triangle shaped, the length of the inner square’s sides increases as the ‘n’ of 1: n increases. The area of the inner square also increases, however, the gaps between the ratios of the area become smaller. 5:2 can be written as 1: 2.5 (inner: outer), 17:9; 1: 1.88 and 13.8; 1: 525. From the ratio of 1: 2.5, to 1: 1.88 and to 1: 525, we are able to know that the areas of the inner square indeed become bigger (since the area of the bigger square stays the same, the ratio decreases). Like what I have mentioned above, the ratios of the length of sides are the square root of the ratio between the area. There is also a pattern shown in the ratio of the area of the outer and inner square. Similar to the triangle pattern, the difference of the bigger and ratio is increased by 2 but starts with an odd number (7 and 9) while the pattern of the inner square is the square of the number ‘n’ (2²=4, 3²=9, 4²=16)

The conjecture that I formulated from seeing the results above from the relationship of 1: n and the ratio between the area of the outer square and inner square is;

Outer square: inner square =

Proving the conjecture;

The ratio of 1:2

The ratio of 1:3

The ratio of 1:4

This proves that my conjecture is correct and that it works for squares in the same state as these ones.

I believe that f segments were constructed in a similar manner in other regular (same length of sides and same angles) polygons such as pentagons, hexagon, octagon, etc, a similar relationship will exist. Even though the conjecture will be different, since the calculation of the triangles and the manner of the overlaid triangles are the same, a similar relationship will exist. Even with a different length of sides, the ratio will stays the same that’s why it doesn’t matter how long the examples of the sides is for the formula to work out.

To investigate the relationship of the polygons, I have made a table such as below to clearly see the patters of the area ratio.

From the table above, I have found out a pattern that states that the smallest segments possible to be constructed in this manner is triangle and the starting point of the ratio on the smaller ratio is the square of 1 for triangle, the square for 2 for square, and I believe that the one with 5 sides will be the square of 3, and so on. While for the starting point of the bigger ratio starts with 7 for triangle, 13 for square and I believe it will keep increasing like the manner of the smaller ratio. So if polygon were constructed the ratio bigger ratio will be 14 and the smaller ratio will be 9 (14:9) and if a hexagon were constructed the ratio will be 19:16 (outer: inner).

For the ratio of 1: n, we can know that the difference of bigger triangle is 6 (13-7) and 8 (21-13) while for the smaller ones are the square of 1 (1), 2 (4), and 3 (9). While for the ratio of 1: n of square, the difference of the bigger square and the smaller square is 7 (17-10) and 9 (26-17), whereas the difference of triangle and square was 3 (for 1:2 segment side ratio), 4 for 1:3 and 5 for 1:4.

I believe that the pattern of 1: n for other polygons will be similar. So if the sides are 5 (pentagon), the difference should be 8 and 10 so after the first side segment ratio of 1:2 is 14:9, the next ratio would be 22:16 and for the side segment ratio of 1:3 is 32:25. While for hexagon, the difference will be 9 and 11. So for the segment ratio of 1:2 is 19:16, 1:3 is 28:25 and for 1:4 side segment ratios is 39:36.

So from the predictions of relationship above, I made this table below to summarize my findings and my predictions.

The ratio of the outer shape addition should increase since the inner shape also increases (the square of the next number). If we look at the outer ratio of the triangle and square, the pattern value increases every time, from 4 (17-13) for the 1:2, 5 (22-17) for 1:3. That is why I predict the pattern of the other polygons would be similar to that. Even though I cannot verify whether the patterns of the other kinds of polygon are exactly like my prediction, I know that the segments were constructed in a similar manner for other regular polygons since to find the triangles inside the shapes; we basically use the same formula of trigonometry (sine rule, etc).