Mathematics with Cinderella: Various Curves: Cissoid
We will draw a cissoid using a drawing instrument as described in the "Encyclopedia of Curves" (Masami Isoda et al., Kyoritsu Shuppan, 2009). On the web, photos and explanations of the drawing instrument can be found on the following page.
Definition of a Cissoid
Let line segments AB and CD be diameters of a circle that are perpendicular to each other. Points E and F are points on arcs BD and BC, respectively, taken such that the lengths of arcs BE and BF are equal. Draw line segments EG and FH perpendicular to line segment DC, and let P be the intersection of line CE and line FH. The locus of point P as point E moves is called a cissoid.
In the days of Newton and Suardi, there were no computers, so they built drawing instruments to draw them. First, let's try drawing it in Cinderella using this definition as it is.
First, use the magnet icon tool to display the coordinate axes and grid, and set it to snap mode.
1. Aligning with the background grid and axes, draw line segments AB on the y-axis and CD on the x-axis, positioned symmetrically left-right and top-bottom.
2. Draw a circle centered at the origin with AB as the diameter, and take a point on the upper-left circumference. (F)
3. Select the reflection tool, choose line segment AB as the mirror, and point F as the object to reflect. Point H is created on the opposite side.
4. Using the perpendicular line tool, drop perpendiculars from F and H to the x-axis.
5. Connect F and C with a line segment and find the intersection point. (K)
6. Select the locus tool, choose F as the point to move and K as the point to draw the locus; the curve drawn is the cissoid.
Note that the names of the points may change slightly depending on the drawing procedure.
As you will see when drawing it, the perpendicular from F is not actually necessary. Also, instead of the reflection tool, you could take points F and H by drawing a circle centered at point B and finding the intersections, and there are likely other drawing procedures as well.
By the way, this definition requires a slight supplement. Draw a tangent to the circle at D, and let L be the intersection with line CK. At this time, CK = FL holds. The same applies if CF = KL. (The triangle formed by drawing a line through F parallel to the x-axis and the triangle at C are congruent.)
The same holds true even if the moving point F moves to the left of the y-axis.
Therefore, you can consider that if this holds, a cissoid can be drawn. The subsequent explanations of curves using Newton's and Suardi's drawing instruments use this fact.
Note that although it is not in the "Encyclopedia of Curves," there is also something called the Cissoid of Diocles.
Newton's Drawing Instrument
A right-angled triangle QRK with a right angle at K is given, and vertex Q moves along line p. Also, for a line q perpendicular to line p, let O be the intersection of the two lines, and take point H on line q such that HO = QK. As vertex Q moves, the side RK of the right-angled triangle QRK moves such that it always passes through point H. At this time, the midpoint M of line segment QK draws a cissoid related to the circle with center O and diameter QK. Also, point K draws a strophoid with line HO as the axis of symmetry.
Unlike other drawing instruments, the explanation of the figure on the left does not seem to match the drawing instrument.Looking at the web version,it says "a transparent plate." There is a rod inside a wooden frame, and a transparent plate is attached to it. The point located just below the origin is the moving point. Therefore, the figure when point Q is brought below O in the figure on the right is the photograph of the drawing instrument.
For now, let's try drawing it in Cinderella according to the figure on the right. You must read the explanation of the drawing instrument above and decipher what does not change when point Q is moved. It starts by mentioning triangle QRK, but this triangle does not change. Therefore, point H, where HO = QK, is also a fixed point that does not change. Furthermore, reading the subsequent explanation (which I will not quote), point R is not related to the locus of M. It is simply taken to match the drawing instrument. Therefore, draw it using the following procedure.
1. Set the coordinate axes and origin... Draw the axes to match the background using the line tool.
2. Take moving point Q on the y-axis.
3. Take point H.
4. Draw a circle with QH as the diameter... Take the midpoint of QH and draw the circle.
5. Take a point on the circumference of the circle in step 4 such that HO = QK... Use the compass tool.
6. Take the midpoint of QK.
Note that depending on the drawing order, the names of each point will not be as above. You can continue as is, or change the names in the inspector each time you take a point. It is more efficient to change them all at once later. The following figure shows the state after completing step 6.
In this state, select the locus tool, specify F in the figure as the point to move, and L in the figure as the point to draw the locus.
A cissoid has been drawn. As mentioned above, the position of point R in the "right-angled triangle QRK" is not related to the drawing of the locus.
After that, draw each line segment, draw a line passing through K and G in the figure, draw a circle of an appropriate size centered at Q, and take the point corresponding to R as the intersection. Finally, tidy up the appearance by changing point names or deleting auxiliary lines.
I added various lines to this and created a model of the drawing instrument. You can also draw a strophoid.
Suardi's Drawing Instrument
As shown in the figure, for an L-shaped square ROZ with a right angle at vertex O, vertex O is fixed to the plane, and vertex R moves on the circumference of circle $${\gamma}$$ with center M and diameter OP passing through point O. For another L-shaped square RVK with a right angle at vertex V, vertex V moves on a line s passing through point O and perpendicular to PO, and shares vertex R. When point R moves on circle $${\gamma}$$, the intersection Z of the two triangles draws a cissoid.
This drawing instrument is straightforward, so it should be easy to draw. Let's draw it using the following steps.
1. Set the coordinate axes and origin: Use the line tool to draw the axes according to the background.
2. Draw a circle centered at M passing through the origin. M is at an appropriate position on the x-axis.
3. Take points on the circumference of the circle in step 2. One is point P on the opposite side of the origin, and the other is a moving point R on the circumference.
4. Take the intersection of the line passing through P and R with the y-axis. Also, draw the ray OR.
5. Use the perpendicular line tool to take point Z.
Note that depending on the drawing order, the names of each point will not be as described above. You can continue as is, or change the names using the inspector each time you take a point. It is more efficient to change them all at once later. The following figure shows the state after completing step 6.
In this state, select the locus tool, and try specifying H in the figure as the moving point and L in the figure as the point to draw the locus.
Once successful, let's tidy up the appearance. To make the lines intersecting at a right angle look like an L-shaped square, take a point on the line and set it to a line segment using 'Display Method' and 'Endpoint Handling' in the inspector.
Note that the L-shaped squares of the drawing instrument each have a groove dug in the center so they can be moved. Since the groove corresponding to the y-axis only exists below the origin, the drawing instrument can only draw the lower part. A photo of the drawing instrument is on the next page, and you can enlarge it to see it.
Two rolling parabolas
Parabola P with focus F and directrix d, and parabola P' with focus F' and directrix d' are congruent curves, and they are tangent to each other such that each has the other's focus on its own directrix.
At this time, the line perpendicular to the axis of parabola P' and passing through focus F' is tangent to the circle centered at focus F and tangent to directrix d. Furthermore, as parabola P' moves, focus F' becomes line-symmetric with respect to the tangent line, and the vertex V' of parabola P' traces a cissoid.
There are explanatory texts, photos of the drawing instrument, and diagrams, but they might be a bit difficult to understand. The explanatory text contains conditions and properties, and it is difficult to decipher which are the 'conditions'. The photos and diagrams are upside down.
The photo on the web is here.
The explanatory text consists of two paragraphs and is divided by 'at this time', so it is normal to consider the first half as conditions and the second half as properties. Let's pick out the conditions or properties.
1. Parabolas P and P' are congruent.
2. They have each other's focus on their own directrix.
3. Parabolas P and P' are tangent.
4. The line perpendicular to the axis of parabola P' and passing through focus F' is tangent to the circle centered at focus F and tangent to directrix d.
5. Focus F' is line-symmetric with respect to the tangent line.
This 'tangent line' is not a tangent to the circle, but a tangent to the parabola. And F and F' are line-symmetric. This part is difficult to understand.
It is quite difficult to draw all of these as conditions. Couldn't 2 and 5 be derived as properties from the conditions in 1 and 3? The following figure was drawn based on that. First, draw a parabola with the y-axis as its axis, take a point, and draw a tangent line using the 'polar line' tool. Then, I used the reflection tool to symmetrically move the parabola, focus, and vertex.
I have not drawn the circle in 4. However, looking at the photo of the drawing instrument, there is a circle centered at focus F and tangent to directrix d, and it seems that a rod is rotating so as to be tangent to this. The rod is attached to a transparent plate in the shape of a parabola. So, let's draw it using these steps.
(1) Draw a circle centered at the focus and tangent to the directrix (x-axis).
(2) Take a point on the circumference and draw a tangent line using the polar line tool.
(3) Take the intersection with the x-axis. This becomes the focus of P'.
(4) Following 2, draw a line passing through the focus of the original parabola and parallel to the tangent of the circle.
(5) Draw a parabola using the parabola tool with (4) as the directrix.
(6) Draw a line passing through the focus and perpendicular to the tangent of the circle, and find the intersection with the parabola. This is the vertex, and draw the locus of the vertex as point E moves on the circumference.
This is the figure corresponding to the drawing instrument. Let's tidy up the appearance to look like the drawing instrument.
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