Editing Cycloid (section)

==Involute==
[[File:Evolute generation.png|thumb|Generation of the involute of the cycloid unwrapping a tense wire placed on half cycloid arc (red marked)]]
The [[involute]] of the cycloid has exactly the [[Congruence (geometry)|same shape]] as the cycloid it originates from. This can be visualized as the path traced by the tip of a wire initially lying on a half arch of the cycloid: as it unrolls while remaining tangent to the original cycloid, it describes a new cycloid (see also [[Cycloid#Cycloidal pendulum|cycloidal pendulum]] and [[Cycloid#Arc length|arc length]]).

===Demonstration===
[[File:Evolute demo.png|thumb|Demonstration of the properties of the involute of a cycloid]]
This demonstration uses the rolling-wheel definition of cycloid, as well as the instantaneous velocity vector of a moving point, tangent to its trajectory. In the adjacent picture, <math>P_1</math> and <math>P_2</math> are two points belonging to two rolling circles, with the base of the first just above the top of the second. Initially,  <math>P_1</math> and <math>P_2</math> coincide at the intersection point of the two circles. When the circles roll horizontally with the same speed, <math>P_1</math> and <math>P_2</math> traverse two cycloid curves. Considering the red line connecting <math>P_1</math> and <math>P_2</math> at a given time, one proves ''the line is always'' ''tangent to the lower arc at <math>P_2</math> and orthogonal to the upper arc at <math>P_1</math>''. Let <math>Q</math> be the point in common between the upper and lower circles at the given time. Then:
*<math>P_1,Q,P_2</math> are colinear: indeed the equal rolling speed gives equal angles <math>\widehat{P_1O_1Q}=\widehat{P_2O_2Q}</math>, and thus <math>\widehat{O_1 Q P_1} = \widehat{O_2QP_2}</math> . The point <math>Q</math> lies on the line <math>O_1O_2</math> therefore <math>\widehat{P_1 Q O_1} + \widehat{P_1QO_2}=\pi</math> and analogously <math>\widehat{P_2QO_2}+\widehat{P_2QO_1}=\pi</math>. From the equality of <math>\widehat{O_1QP_1}</math> and <math>\widehat{O_2QP_2}</math> one has that also  <math>\widehat{P_1QO_2}=\widehat{P_2QO_1}</math>. It follows <math>\widehat{P_1QO_1}+\widehat{P_2QO_1}=\pi</math> .
*If <math>A</math> is the meeting point between the perpendicular from <math>P_1</math> to the line segment <math>O_1O_2</math> and the tangent to the circle at <math>P_2</math> , then the triangle <math>P_1AP_2</math> is isosceles, as is easily seen from the construction: <math>\widehat{QP_2A}=\tfrac{1}{2}\widehat{P_2O_2Q}</math> and <math>\widehat{QP_1A} = \tfrac{1}{2}\widehat{QO_1R}=</math><math>\tfrac{1}{2}\widehat{QO_1P_1}</math> . For the previous noted equality between <math>\widehat{P_1O_1Q}</math> and <math>\widehat{QO_2P_2}</math> then <math>\widehat{QP_1A}=\widehat{QP_2A}</math> and <math>P_1AP_2</math> is isosceles.
*Drawing from <math>P_2</math> the orthogonal segment to <math>O_1O_2</math>, from <math>P_1</math> the straight line tangent to the upper circle, and calling <math>B</math> the meeting point, one sees that <math>P_1AP_2B</math> is a [[rhombus]] using the theorems on angles between parallel lines
*Now consider the velocity <math>V_2</math> of <math>P_2</math> . It can be seen as the sum of two components, the rolling velocity <math>V_a</math> and the drifting velocity <math>V_d</math>, which are equal in modulus because the circles roll without skidding. <math>V_d</math> is parallel to <math>P_1A</math>, while <math>V_a</math> is tangent to the lower circle at <math>P_2</math> and therefore is parallel to <math>P_2A</math>. The rhombus constituted from the components <math>V_d</math> and <math>V_a</math> is therefore similar (same angles) to the rhombus <math>BP_1AP_2</math> because they have parallel sides. Then <math>V_2</math>, the total velocity of <math>P_2</math>, is parallel to <math>P_2P_1</math> because both are diagonals of two rhombuses with parallel sides and has in common with <math>P_1P_2</math> the contact point <math>P_2</math>. Thus the velocity vector <math>V_2</math> lies on the prolongation of <math>P_1P_2</math> . Because <math>V_2</math> is tangent to the cycloid at <math>P_2</math>, it follows that also <math>P_1P_2</math> coincides with the tangent to the lower cycloid at <math>P_2</math>.
*Analogously, it can be easily demonstrated that <math>P_1P_2</math> is orthogonal to <math>V_1</math> (the other diagonal of the rhombus).
*This proves that the tip of a wire initially stretched on a half arch of the lower cycloid and fixed to the upper circle at <math>P_1</math> will follow the point along its path ''without changing its length'' because the speed of the tip is at each moment orthogonal to the wire (no stretching or compression). The wire will be at the same time tangent at <math>P_2</math> to the lower arc because of the tension and the facts demonstrated above. (If it were not tangent there would be a discontinuity at <math>P_2</math> and consequently unbalanced tension forces.)