Equiangular Non-regular and Cyclic Non-regular Polygons Resulting From Images Formed When an Object Is Placed Between Two Plane Mirrors That Are Not Parallel

This Instructables:

• is an extension of an earlier Instructables that looked at images of an object in two plane mirrors that were placed so the angle between them was 180/n degrees, where n is an integer; that Instructables restricted the placement of the object to positions that were equidistant from the two mirrors; in this Instructables, the object can be placed at any position between the two mirrors.
• provides an example that can be used by teachers to introduce the concepts of equiangular non-regular polygons and cyclic non-regular polygons that have a real-world physical application; students studying geometric shapes are usually introduced to the properties of regular polygons, but rarely look at the properties of polygons that are non-regular and equiangular.

As in the previous Instructables, the ideas presented here were first obtained using a ruler, protractor and set square to find the reflections of an object placed between the two mirrors for values of n equal to 3, 4, 5 and 6. Rather than starting this Instructables with looking at each of these individual polygons in turn, Steps 1 to 8 of this Instructables analyse a more general situation where some of the geometrical consequences regarding the position of images obtained when an object is placed between two plane mirrors under conditions that:

• the two mirrors are oriented at any angle with respect to each other;
• the object is placed at any position between the two mirrors.

Steps 9 and 10 use the results from Steps 1 to 8 to look at two special cases:

1. equiangular non-regular polygons that describe the situation when the angle between the two mirrors is 360/n where n is an even number;
2. the situation when the angle between the two mirrors is 360/n where n is an odd number.

These results are compared with those in the previous Instructables.

Figure 1 above shows the first 12 images obtained, using the laws of reflection, when an object, labelled O, is placed between two mirrors, labelled M1 and M2, represented by solid red lines (the dashed red lines are extensions of the solid red lines). The two mirrors are oriented so that the angle between them is s degrees, with the mirrors meeting at a point labelled X. The object is placed so that it is not equidistant from the two mirrors. The images are labelled A, B, C, …, with the blue lines showing the various reflections in both mirrors after the object is first reflected in Mirror M1, and, the orange lines showing the path taken by the reflections in both mirrors after the object is first reflected in Mirror M2. These lines are obtained by drawing lines perpendicular to each mirror so that:

• the distance from the object (or an image acting as an object) to a mirror equals the distance from the image of the object to the same mirror;
• the line joining the object (or an image acting as an object) and its image lie along the same straight line.

Figure 2a shows Figure 1 with angle symbols shown as short black arcs at the various acute angles that result as images are formed in both mirrors. These angles are also equal to s degrees. This follows since lines drawn at right angles to another pair of lines oriented so the angle between them is s degrees also subtend an angle of s degrees or its supplementary angle (180 – s) degrees.

Since the lines OA, BD, CE, FH, GI, and JL are all perpendicular to mirror M1, they are parallel and since the lines OB, AC, DF, EG, HJ and IK are all perpendicular to mirror M2, they are also parallel.

Another property not shown in Figure 2 regards answering what happens to lines drawn from images K and L if one continues to follow the method used in the previous step to draw the lines connecting the various further images? The lines connecting various images will eventually either:

• meet at a single point as shown by the examples in Step 9;
• start to proceed in the opposite direction, moving towards the position of the object rather than away from the object as shown in Figure 2a.

In Figure 2b, where the angle between the two mirrors is larger than that shown in Figure 2a, the lines emanating from images C and F acting as objects are moving towards the object O. Of course, this cannot happen as images C and F acting as objects cannot be reflected as the lines drawn from C and F are moving in a direction towards the back surface of mirror M1, which, we assume. does not reflect. Step 10 will provide examples of this situation.

Figure 3 shows Figure 2a with:

• the short black arcs at the various acute angles removed;
• a line joining the point of intersection of the two mirrors to the object;
• the point of intersection of mirror M1 with the line OA labelled V;
• the point of intersection of mirror M2 with the line OB labelled W;
• black lines connecting each image to its nearest neighbouring image as well as connecting the object to its nearest neighbouring images (the primary images A and B); this results in a number of trapezia whose parallel sides are parallel to either OA or OB.

Since mirror M1 bisects and is perpendicular to the parallel sides of the trapezia parallel to OA, all these trapezia are isosceles trapezia and:

• OB = AD;
• BC = DE;
• CF = EH;
• FG = HI;
• GJ = IL.

Since mirror M2 bisects and is perpendicular to the parallel sides of the trapezia parallel to OB, all these trapezia are isosceles trapezia and:

• OA = BC;
• AD = CF;
• DE = FG;
• EH = GJ;
• HI = JK.

In some of the following steps various pairs of trapezia are considered with one member of each pair consisting of a trapezium with parallel sides parallel to OA and the other member with parallel sides parallel to OB. The five pairs of trapezia in Figure 3 are given numbers as listed below (those on the left are parallel to OA and those on the right are parallel to OB):

• Trapezia No. 1: OADB, OBCA
• Trapezia No. 2: BDEC, ACFD
• Trapezia No. 3: CEHF, DFGE
• Trapezia No. 4: FHIG, EGJH
• Trapezia No. 5: GILJ, HJKI

Note that for each of the above pairs of trapezia, one non-parallel side of one member of the pair is shorter than the other non-parallel side of the other member of the pair:

• OB = AD < OA = BC;
• AD = CF < BC = DE;
• CF = EH < DE = FG;
• EH = GJ < FG = HI;
• GJ = IL < HI = JK.

We now define the position of the object O relative to the two mirrors (see Figure 3 in the previous step). We let the position of the object be such that line joining X to O subtends an angle a from mirror M1 and b from mirror M2. If the angle between the two mirrors is s, then, s = a + b. It follows that:

∠OXV = ∠AXV = a degrees;

∠OXW = ∠BXW = b degrees;

∠OXA = 2a degrees;

∠OXB = 2b degrees;.

∠XOA = ∠XAO = (90 - a) degrees;

∠XOB = ∠XBO = (90 - b) degrees;

triangles XOA and XOB are isosceles triangles;

XO = XA = XB.

Figure 4 shows Figure 2a with black lines connecting each image to its nearest neighbouring image as well as connecting the object to its nearest neighbouring images (the primary images A and B).

This step is broken up into a number of sub-steps to aid in following the analysis below.

Sub-step 1

Starting with the first pair of isosceles trapezia OADB and OBCA we note the following:

For the trapezium parallel to OA (OADB)

• the angle between the upper parallel side (BD) and the left non-parallel side (∠DBO) equals the angle between the upper parallel side and the right non-parallel side (∠ADB) which equals s degrees (see ∠DBO in Figure 2; ∠ADB is marked by a short red arc in Figure 4);.
• the angle between the lower parallel side (OA) and the left non-parallel side (∠BOA) equals the angle between the lower parallel side and the right non-parallel side (∠OAD) which equals (180 – s) degrees.

The latter statement follows from the facts that:

• the sum of the angles of a trapezoid is 360 degrees;
• the sum of the angle between the lower parallel side (OA) and the left non-parallel side (∠BOA) plus the angle between the lower parallel side and the right non-parallel side thus equals (360 – 2s) degrees;
• each of these latter angles are equal (∠BOA = ∠OAD);
• each of them thus equals (180 – s) degrees).

For the trapezium parallel to OB (OBCA)

• the angle between the upper parallel side (AC) and the right non-parallel side (∠CAO) equals the angle between the upper parallel side and the left non-parallel side (∠BCA) which equals s degrees (angle BCA is also marked by a red short arc in Figure 4);
• the angle between the lower parallel side (OB) and the right non-parallel side (∠AOB) equals the angle between the lower parallel side and the left non-parallel side (∠OBC) which equals (180 – s) degrees.
• Thus ∠OAD = ∠BOA = ∠OBC = (180 - s) degrees.

Sub-step 2

Considering the next pair of trapezia BDEC and ACFD and using similar arguments as above, it follows that:

For the trapezium parallel to OA (BDEC)

• ∠ECB = ∠DEC = 2s degrees (since ∠ECB = ∠ECA + ∠ACB; ∠DEC is labelled as a red angle in Figure 4) .
• ∠BDE = ∠CBD = (180 – 2s) degrees.

For the trapezium parallel to OB (ACFD)

• ∠FDA = ∠CFD = 2s degrees (since ∠FDA = ∠FDB + ∠BDA; ∠CFD is labelled as a red angle in Figure 4) .
• ∠DAC = ∠ACF = (180 – 2s) degrees.

Sub- step 3

Combining the results of Sub-steps 1 and 2

• ∠BDE = ∠CBD = (180 – 2s) degrees;
• ∠ADB = ∠DBO = s degrees;
• ∠ADE = ∠BDE + ∠ADB = (180 – s) degrees;
• ∠OBC = ∠CBD + ∠DBO = (180 – s) degrees.

Thus ∠ADE = ∠OBC = (180 - s) degrees.

Similarly,

• ∠DAC = ∠ACF = (180 – 2s)
• ∠CAO = ∠BCA = s degrees
• ∠OAD = ∠DAC + ∠CAO = (180 - s) degrees;
• ∠BCF = ∠ACF + ∠BCA = (180 – s) degrees.

Thus ∠OAD = ∠BCF = (180 - s) degrees.

Sub-step 4

Conclusion regarding black lines in Figure 4

Continuing in this way it can be shown that all angles between adjacent black lines in Figure 4 are equal to (180 – s) degrees (angles labelled in red in Figure 4 should help with carrying out the above analysis).

Sub-step 5

Generalization of the above results when there are more images than those shown in Figure 1

The angle between the adjacent non-parallel sides of the (i - 1) th and i th trapezia parallel to OA is equal to the sum of:

• the angles between the lower parallel side of the i th trapezium and its adjacent upper non-parallel sides both equal (180 – is) degrees;
• the angles between the upper parallel side of the (i -1) th trapezium and its adjacent lower non-parallel sides both equal (i - 1)s degrees.

The sum of these angles equals (180 – is) + (i - 1)s = (180 -s) degrees.

A similar argument applies for the i th and (i - 1) th trapezia parallel to OB.

Sub-step 6

Conclusion

Thus the angle between the lines joining nearest neighbouring images as well as the angle between the lines joining the object to its nearest neighbouring images are equal.

Comment

In Step 8 it will be shown that the lengths of the non-parallel sides of the smaller trapezia in each of the pairs of trapezia listed in Step 3 are equal and that the lengths of the non-parallel sides of the larger trapezia in each of the pairs of trapezia listed in Step 3 are equal. Thus, adjacent black sides form an alternating sequence of a long side followed by a short side or a short side followed by a long side.

This step is involved with finding what are the angles between the point of intersection of the two mirrors (X) and adjacent pairs of images. We will make use of the results in Step 4 and wish to find a general formula that describes these angles for any of the pairs of trapezia listed in Step 3.

As in Step 5, this step is broken up into a number of sub-steps to aid in following the analysis below.

Sub -step 1

Figure 5a repeats Figure 3 with green lines joining X to the five vertices of the first pair of trapezia listed in Step 3.

Considering the first trapezium parallel to OA, namely, OADB, we want to work out the angles between X and its four vertices.

For the lower parallel side of trapezium OADB, namely OA (see Step 4):

• ∠OXV = ∠AXV = a. ...............................................................................................(1)

For the upper parallel side of trapezium OADB, namely BD (note that triangle BXD is an isosceles triangle since Mirror M1 is perpendicular to and bisects DB):

• ∠BXV = ∠DXV = ∠BXW + ∠WXV = b + s = a + 2b.....................................(2)

Thus, ∠AXD is found by subtraction Equation (2) from Equation (1):

• ∠AXD = ∠DXV - ∠AXV = a + 2b - a = 2b......................................................(3)

Next, consider first trapezium parallel to OB, namely, OBCA and work out the angles between X and its four vertices.

For the lower parallel side of trapezium OBCA, namely OB (see Step 4):

• ∠OXW = ∠BXW = b. ..........................................................................................(4)

For the upper parallel side of trapezium OBCA, namely CA, using the expression for ∠AXV in Equation (1) (note that triangle AXC is an isosceles triangle since Mirror M2 is perpendicular to and bisects CA):

• ∠AXW = ∠CXW = ∠AXV + ∠WXV = a + s = 2a + b. ...............................(5)

Thus, ∠BXC is found by subtracting Equation (5) from Equation (4):

∠BXC = ∠CXW - ∠BXW = 2a + b - b = 2a. ......................................................(6)

Therefore:

• the angle between XO and XA = the angle between XB and XC = 2a;
• the angle between XO and XB = the angle between XA and XD = 2b.

Sub-step 2

Figure 5b repeats Figure 3 with green lines joining X to the six vertices of the second pair of trapezia listed in Step 3.

We now look at the second trapezium parallel to OA, namely, BDEC.

For the lower parallel side of trapezium BCED, namely BD, the angle ∠DXV is given by Equation (2).

For the upper parallel side of trapezium BDEC, namely CE, using Equation (5) for ∠CXW, we have (note that triangle CXE is an isosceles triangle since Mirror M1 is perpendicular to and bisects CE):

• ∠CXV = ∠EXV = ∠CXW + ∠WXV = 2a + b +s = 3a + 2b. ........................(7)

Thus, subtracting Equation (2) from Equation (7):

• ∠DXE = ∠EXV - ∠DXV = 3a + 2b - (a +2b) = 2a. ........................................(8)

Next we consider the second trapezium parallel to OB, namely ACFD.

For the lower parallel side of trapezium ACFD, namely CA, the angle ∠CXW is given by Equation (5).

For the upper parallel side of trapezium ACFD, namely FD, using Equation (2) for ∠DXV, we have (note that triangle DXF is an isosceles triangle since Mirror M2 is perpendicular to and bisects DF):

• ∠DXW = ∠FXW = ∠DXV + ∠WXV = a + 2b +s = 2a + 3b. ....................(9)

Thus, ∠CXF is found by subtracting Equation (5) from Equation (9):.

• ∠CXF = ∠FXW - ∠CXW = 2a + 3b -(2a + b) = 2b. ....................................(10)

Therefore:

• the angle between XO and XA = the angle between XB and XC = the angle between XD and XF = 2a;
• the angle between XO and XB = the angle between XA and XD = the angle between XC and XF = 2b.

Continuing on with this approach, one can show that:

• ∠OXA = ∠BXC = ∠ DXE = ∠FXG = ∠HXI = ∠JXK = 2a;
• ∠OXB = ∠AXD = ∠ CXF = ∠EXH = ∠GXJ = ∠IXL = 2b.

Sub-step 3

Generalization of the above results

For the i th trapezium parallel to OA if:

i is an odd number:

• the angle between X and the points on the lower parallel side of the trapezium = ia + (i – 1) b;
• the angle the angle between X and the points on the upper parallel side of the trapezium = ia + (i + 1) b;
• difference between the latter pair of angles = 2b.

i is an even number:

• the angle between X and the points on the lower parallel side of the trapezium = (i - 1) a + ib;
• the angle between X and the points on the upper parallel side of the trapezium = (i + 1) a + ib;
• difference between the latter pair of angles = 2a.

For the ith trapezium parallel to OB if:

i is an odd number:

• the angle between X and the points on the lower parallel side of the trapezium = (i - 1) a + ib;
• the angle between X and the points on the upper parallel side of the trapezium = (i + 1) a + ib;
• difference between the latter pair of angles = 2a.

i is an even number:

• the angle between X and the points on the lower parallel side of the trapezium = ia + (i – 1)b;
• the angle between X and the points on the upper parallel side of the trapezium = ia + (i + 1)b;
• difference between the latter pair of angles = 2b.

Sub-step 4

Conclusion

Consider adjacent lines joining nearest neighbouring images as well as the lines joining the object to its primary images being given sequential numbers as follows:

• start with a line joining the object to one of its primary images being labelled with the number 1;
• label adjacent lines to the right of the line numbered 1 with the numbers 2, 3, 4, ..., that is, numbering in an anticlockwise direction;
• label adjacent lines to the left of the line numbered 1 with the numbers 2, 3, 4, ..., that is, numbering in a clockwise direction.

Then the results of this step show that:

• the angles subtended by the two mirrors and the odd numbered lines are all equal;
• the angles subtended by the two mirrors and the even numbered lines are all equal.

The angles subtended by the odd and even numbered lines are equal if, as in the earlier Instructables, the object is placed equidistant from the two mirrors (angle a = angle b).

Figure 3 is redrawn in Figures 6 with the labels V and W removed and green lines added to show triangles having a common vertex X (the point of intersection of the two mirrors), which we refer to as the apices of the triangles. The bases of these triangles (the sides opposite the apices) are the adjacent lines between adjacent images as well as the lines joining the object to its primary images A and B.

Sub-step 1

For this step we look at the triangles starting with triangle XBO which we refer to as triangle 1 and refer to the triangles going in an anticlockwise direction from this triangle as triangle 2 (triangle XOA), triangle 3 (triangle XAD), triangle 4 (triangle XDE), etc. (see Figure 6b).

For Triangles 1, 2 and 3

Using the results of Steps 4, 5 and 6, it follows that:

1. the lengths of the left legs of triangles 1 and 3 are equal (XB = XA);
2. the angle between the right leg and base of triangle 2 (∠ XAO) is (90 – a) degrees;
3. the angle between the left leg and base of triangle 1 (∠ XBO) is (90 – b) degrees;
4. the angles at the apices of triangle 1 (∠BXO) and triangle 3 (∠AXD) are 2b degrees;
5. the angle between the bases of triangles 2 and 3 (∠OAD) is (180 – s) degrees.

It follows that the angle between the left leg (XA) of triangle 3 and its base (AD) (that is ∠XAD) is equal to the difference between:

1. the angle between the bases of triangles 2 and 3 (∠OAD);
2. the angle between the right leg of triangle 2 and its base (∠XAO).

This difference equals (180 - s) - (90 - a) = (90 - b) degrees.

Thus we have the following:

1. the lengths of the left legs of triangles 1 and 3 are equal (XB = XA);
2. the angles at the apices of triangle 1 (∠BXO) and triangle 3 (∠AXD) are 2b degrees;
3. the angles between the left leg and the base of triangle 1 (∠XBO) equals the angle between the left leg and the base of triangle 3 (∠XAD). This angle equals (90 – b) degrees.

Thus triangles 1 and 3 are congruent. Like triangle 1, triangle 3 is an isosceles triangle with the lengths of its left and right legs equal. Also, the bases of triangles 1 and 3 are equal.

As the distances from the apices to the bases of triangles 1, 2 and 3 are equal, the vertices of these bases lie on a circle whose center is X and whose radius is XO.

Sub-step 1a

Generalising the above for triangles i, i + 1 and i + 2, when i is odd.

The following statements hold (when i = 1, this restates Sub-step 1)

1. the lengths of the left legs of triangles i and i + 2 are equal;
2. the angle between the right leg and base of triangle i + 1 is (90 – a) degrees;
3. the angle between the left leg and base of triangle i is (90 – b) degrees;
4. the angles at the apices of triangle i and triangle i + 2 are 2b degrees;
5. the angle between the bases of triangles i + 1 and i + 2 is (180 – s) degrees.

It follows that the angle between the left leg of triangle i + 2 and its base is equal to the difference between:

1. the angle between the bases of triangles i + 1 and i + 2;
2. the angle between the right leg of triangle i + 1 and its base.

This difference equals (180 - s) - (90 - a) = (90 - b).

Thus we have the following:

1. the lengths of the left legs of triangles i and i + 2 are equal;
2. the angles at the apices of triangle i and triangle i + 2 are 2b degrees;
3. the angles between the left leg and the base of triangle i equals the angle between the left leg and the base of triangle i+ 2. This angle equals (90 – b) degrees.

Thus triangles i and i + 2 are congruent. Like triangle i, triangle i + 2 is an isosceles triangle with the lengths of its left and right legs equal. Also, the bases of triangles i nd i + 2 are equal.

As the distances from the apices to the bases of triangles i, i + 1 and i + 2 are equal, the vertices of these bases lie on a circle whose center is X and whose radius is XO.

Sub-step 1b
Generalising the above for triangles i, i + 1 and i + 2, when i is even.

All the statements and conclusions made for Sub-step 1a hold except that wherever angle a is mentioned, it is replaced by angle b and vice versa.

Sub-step 2

We now look at the triangles starting with triangle XAO which we now refer to as triangle 1 and refer to the triangles going in an clockwise direction from this triangle as triangle 2 (triangle XOB), triangle 3 (triangle XBC), triangle 4 (triangle XCF), etc. (see Figure 6c).

For Triangles 1, 2 and 3

Again, using the results of Steps 4, 5 and 6, it follows that:

1. the lengths of the right legs of triangles 1 and 3 are equal (XA = XB);
2. the angle between the left leg and base of triangle 2 (∠ XBO) is (90 – b) degrees;
3. the angle between the right leg and base of triangle 1 (∠ XAO) is (90 – a) degrees;
4. the angles at the apices of triangle 1 (∠AXO) and triangle 3 (∠BXC) are 2a degrees;
5. the angle between the bases of triangles 2 and 3 (∠OBC) is (180 – s) degrees.

It follows that the angle between the left leg (XB) of triangle 3 and its base (BC) (that is ∠XBC) is equal to the difference between:

1. the angle between the bases of triangles 2 and 3 (∠OBC);
2. the angle between the left leg of triangle 2 and its base (∠XBO).

This difference equals (180 - s) - (90 - b) = (90 - a) degrees.

Thus we have the following:

1. the lengths of the right legs of triangles 1 and 3 are equal (XA = XB);
2. the angles at the apices of triangle 1 (∠AXO) and triangle 3 (∠BXC) are 2a degrees;
3. the angle between the right leg and base of triangle 1 (∠XAO) equals the angle between the right leg and the base of triangle 3 (∠XBC). This angle equals (90 – a) degrees.

Thus triangles 1 and 3 are congruent. Like triangle 1, triangle 3 is an isosceles triangle with the lengths of its left and right legs equal. Also, the bases of triangles 1 and 3 are equal.

As the distances from the apices to the bases of triangles 1, 2 and 3 are equal, the vertices of these bases lie on a circle whose center is X and whose radius is XO.

Sub-step 2a
Generalising the above for triangles i, i + 1 and i + 2, when i is odd. The following statements hold (when i = 1, this restates Sub-step 2)

1. the lengths of the right legs of triangles i and i + 2 are equal;
2. the angle between the left leg and base of triangle i + 1 is (90 – b) degrees;
3. the angle between the right leg and base of triangle i is (90 – a) degrees;
4. the angles at the apices of triangle i and triangle i + 2 are 2a degrees;
5. the angle between the bases of triangles i + 1 and i + 2 is (180 – s) degrees.

It follows that the angle between the left leg of triangle i + 2 and its base is equal to the difference between:

1. the angle between the bases of triangles i + 1 and i + 2;
2. the angle between the left leg of triangle i + 1 and its base.

This difference equals (180 - s) - (90 - b) = (90 - a) degrees.

Thus we have the following:

1. the lengths of the left legs of triangles i and i + 2 are equal;
2. the angles at the apices of triangle i and triangle i + 2 are 2a degrees;
3. the angle between the right leg and base of triangle i equals the angle between the right leg and the base of triangle i + 2. This angle equals (90 – a) degrees.

Thus triangles i and i + 2 are congruent. Like triangle i, triangle i + 2 is an isosceles triangle with the lengths of its left and right legs equal. Also, the bases of triangles i and i + 2 are equal.

As the distances from the apices to the bases of triangles i, i + 1 and i + 2 are equal, the vertices of these bases lie on a circle whose center is X and whose radius is XO.

Sub-step 2b

Generalising the above for triangles i, i + 1 and i + 2, when i is even.

All the statements and conclusions made for Sub-step 2a hold except that wherever angle a is mentioned, it is replaced by angle b and vice versa.

Sub-step 3

Conclusions

From the results presented in Sub-steps 1 and 2 and using Step 6 it follows that:

• triangles XOA, XBC, XDE, XFG, XHI, and XJK are all congruent;
• triangles XOB, XAD, XCF, XEH, XGJ and XIL are all congruent.

Therefore:

• OA = BC = DE = FG = HI =JK;
• OB = AD = CF = EH =GJ = IL;
• XO = XA = XB = XC =XD = XE = XF = XG = XH = XI =XJ = XK = XL;

Thus the lengths of all the shorter non-parallel sides of the five pairs of trapezia listed in Step 3 are equal and the lengths of the longer non-parallel sides of the five pairs of trapezia listed in Step 3 are equal.

Therefore adjacent black sides in Figures 3, 4a and 4b form an alternating sequence of a short side followed by a long side or a long side followed by a short side.

It can also be concluded that the lengths of all lines drawn from X in Figure 3 to the object O and to all images of the object (A, B, C, …) are equal and therefore the object and all its images lie on a circle whose center is O and whose radius is equal to the distance between the point of intersection of the two mirrors and the object (XO).

Given values for the angles a and b, as well as the radius of the circle OX, the lengths of OA and OB are:
OA = 2 XO sin a;

OB = 2 XO sin b.

Thus, the positions of the images can readily be found by drawing a circle of radius OX and marking off alternating chord lengths along the circumference of the circle equal to (2 XO sin a) and (2 XO sin b) starting from the position where the object O is located. If the sequence of lengths starts from O with the length (2 XO sin a) going in an anticlockwise direction, then the sequence of lengths starting from O that go in a clockwise direction should start with a length equal to (2 XO sin b) and vice versa.

Sub-step 1

If angle s (the angle between the two mirrors M1 andM2) is chosen so that points C and D in Figure 3 coincide, the resulting figure is a rectangle (a non-regular polygon). For this to happen s = (360/4) = 90 degrees (see Figure 7a above).

If we can regard a rectangle as a special case of a trapezium, then the rectangle in Figure 7a is a single trapezium whose parallel sides are parallel to OA and OB.

Sub-step 2

If angle s is chosen so that points E and F in Figure 3 coincide, the resulting hexagon is an equiangular non-regular hexagon (see Figure 7b above). For this to happen s = (360/6) = 60 degrees (see Figure 7b above). This equiangular hexagon can be regarded as being made up of four trapezia, with two of the trapezia having parallel sides parallel to OA and the other two trapezia having parallel sides parallel to OB.

Sub-step 3

If angle s is chosen so that points G and H in Figure 3 coincide, the resulting figure is an equiangular non-regular octagon (see Figure 7c above). For this to happen, s = (360/8)= 45 degrees. This equiangular octagon can be regarded as being made up of six trapezia, with three of the trapezia having parallel sides parallel to OA and the other three trapezia having parallel sides parallel to OB.

Sub-step 4

Conclusion

Continuing in this way, it follows that if the angle between the two plane mirrors is 360/n with n an even integer greater than 2, the resulting polygons are equiangular non-regular polygons.

It should also be noted that, as well being equiangular, these polygons are also cyclic polygons since all the vertices of the polygon (the points representing the images as well as the object) lie on a circle as shown in Step 7.

The total number of images obtained when the two plane mirrors are placed at an angle of 360/n degrees, (where n is an even integer greater than 2) relative to each other with the object not equidistant from the two mirrors is (n - 1), which is identical to the results found in the previous Instructables, where the object was placed so that it was equidistant form both mirrors.

Sub-step 1

Figure 8a shows that three images are obtained when the angle between two mirrors is (360/3) = 120 degrees. When the object was equidistant from the two mirrors, Images A and B lay on the extensions of Mirrors M2 and M1, respectively, and only two images existed. When the object is not equidistant from the two mirrors, Image B does not lie on the extension of Mirror M1 and can act as an object in Mirror M1 giving rise to Image D. Note that Image A cannot undergo reflection in Mirror 2 as it lies to the back side of this mirror.

If a straight line joined Images A and D in Figure 8a (these images on the circumference of the circle are not joined by a straight line in Figure 8a), then a cyclic non-regular quadrilateral would be formed, which in this case results in an isosceles trapezium whose parallel sides are parallel to OA.

Sub-step 2

Figure 8b shows that 5 images are obtained when the angle between the two mirrors is (360/5) = 72 degrees. When the object was equidistant from the two mirrors, Images C and D lay on the extensions of Mirrors M1 and M2, respectively, and only four images existed. When the object is not equidistant from the two mirrors, Image D acts as an object in Mirror M2 giving rise to image F. Note again, that image C cannot undergo reflection in Mirror M1 as it lies to the back side of this mirror.

If straight lines joined the pairs of Images A and D, B and C, and C and F in Figure 8b (these pairs of images on the circumference of the circle are not joined by a straight lines in Figure 8b), then a cyclic non-regular pentagon would be formed, which in this case would result in two isosceles trapezia whose parallel sides are parallel to OB.

Sub-step 3

Figure 8c shows that 7 images are obtained when the angle between the two mirrors is (360/7) = 51 and 3/7 degrees. When the object was equidistant from the two mirrors, Images E and F lay on the extensions of Mirrors M2 and M1, respectively, and only six images existed. When the object is not equidistant from the two mirrors, Image F acts as an object in Mirror M1 giving rise to image H. Note that image E cannot undergo reflection in Mirror M2 as it lies to the back side of this mirror.

If straight lines joined the pairs of Images B and C, D and E, and E and H in Figure 8c (these pairs of images on the circumference of the circle are not joined by a straight lines in Figure 8c), then a cyclic non-regular heptagon would be formed, which in this case would result in three isosceles trapezia parallel to OA.

Sub-step 4

Conclusion

Continuing in this way, it follows that if the angle between the two plane mirrors is 360/n with n an odd integer greater than 2, the resulting polygons are cyclic non-regular polygons.

The total number of images obtained when the two plane mirrors are placed at an angle of 360/n degrees (where n is an odd integer greater than 2) relative to each other with the object not equidistant from the two mirrors is n. This result is not the same as the results found in the previous Instructables, where the object was placed so that it was equidistant form both mirrors. When the object is not equidistant from the two mirrors the number of images is one greater than that found when the object is equidistant from the two mirrors. When the object is equidistant from the two mirrors the number if images is (n - 1).