Once more again: Amazing News about the Cardioid
by Klaus-Anton, Jun 24, 2024, 9:21 PM
We have so often seen that the Carioid is very surprizing. We know that the circumcircle has diameter 
. Now you move the center from this circle to the right hand so that in walks through the maximum and through the minimum of the cardioid. What is the intersection area?
It is said:
(APEX Calculus CalculusIII_Version2.pdf, page 521 following).
![[asy]
// Figure 1
size(0,150);
real GR=(sqrt(5)+1)/2;
real gr=(sqrt(5)-1)/2;
import graph;
import geometry;
line l1=line((-.25,-sqrt(3)/4),(0,0));
line l2=line((-.25,sqrt(3)/4),(0,0));
line l3=line((-1,0),(2,0));
real m(real t) {return 1+cos(t);}
path h=polargraph(m,0,2pi,operator ..)--cycle
,cardioid=h;
pair p0= (-.25,sqrt(3)/4);
fill(Circle((.5,0),1.5),red+white);
fill(cardioid,yellow+red);
fill(Circle((1,0),1),rgb(205, 080, 080) );
fill(Circle((.5,0),.5),green);
fill(Circle((1.5,0),.5),yellow);
real[] myarray1={-1.5,-1,-.5,0,.5,1,1.5,2,2.5,3,3.5};
for(real k : myarray1) { draw(shift(k,0)*l1,grey);
draw(shift(k,0)*l2,grey); }
real[] myarray2={0
,sqrt(3)/4
,2*sqrt(3)/4
,3*sqrt(3)/4
,-1*sqrt(3)/4
,-2*sqrt(3)/4
,-3*sqrt(3)/4
};
for(real k : myarray2) draw(shift(0,k)*l3,grey);
draw(Circle((1,0),1));
draw(Circle((.5,0),.5));
draw(Circle((1.5,0),.5));
draw(Circle((.5,0),1.5),linewidth(gr));
draw((0,0)--(1,sqrt(GR)),yellow,PenMargin);
draw((0,0)--(1,0)--(1,sqrt(GR)),yellow,PenMargin);
pair p1=relpoint((0,0)--(1,sqrt(GR)),gr^2);
draw(p1--(1,0),yellow);
pair p2=relpoint(p1--(1,0),GR^-GR);
pair p3=(1.25,sqrt(3)/4)
,p4=(1.25,3*sqrt(3)/4)
,p5=relpoint(p3--p4,gr)
,p6=(p3.x-.5,p3.y)
,p7=(p4.x-.5,p4.y)
,p8=(p5.x-.5,p5.y)
,p9=(p3.x-1,p3.y)
,p10=(p4.x-1,p4.y)
// ,p11=relpoint(p9--p10,gr^exp(1))
,p12=(.5,0)
,p13=relpoint(p12--p0,gr);
draw(p3--p5
^^p6--p8
^^p12--p13
,blue+linewidth(2));
draw(p4--p5^^p7--p8^^p0--p13,red+linewidth(2));
dot(p3^^p4^^p6^^p7^^p12^^p0
//^^p9^^p10
);draw(p3^^p4^^p6^^p7^^p12^^p0,white+linewidth(gr));
dot(p5^^p8^^p13
//^^p11
,yellow);draw(p5^^p8^^p13,linewidth(gr));
draw(circle(origin+1.5,1.5));
draw(cardioid);
draw(p0^^p3^^p4^^p6^^p7^^p12, white+linewidth(gr));
shipout(scale(sqrt(2))*bbox(gr*1cm,invisible,Fill(rgb(064,224,208))));
[/asy]](//latex.artofproblemsolving.com/e/7/e/e7e7b892354716444d832cf94cbb9cf0e7e062dd.png)
![[asy]
// Figure 2a
size(0,150);
real GR=(sqrt(5)+1)/2;
real gr=(sqrt(5)-1)/2;
import graph;
import geometry;
line l1=line((-.25,-sqrt(3)/4),(0,0));
line l2=line((-.25,sqrt(3)/4),(0,0));
line l3=line((-1,0),(2,0));
real f(real t) {return 1+cos(t);}
path cardioid=polargraph(f,0,2pi,operator ..)--cycle;
real tick=1/3*.25;
path p1=polargraph(f,-1/3 *pi,1/3 *pi,operator ..);
path p2=arc((1.5,0),1.5,120,120*2);
filldraw(circle((1.5,0),1.5),lightblue+white,blue+linewidth(1.25));
fill(cardioid,white);
fill (p1..p2..cycle,lightred+white);
draw(cardioid,red+linewidth(1.25));
draw(arc((1.5,0),1.5,120,240),blue+linewidth(1.25));
draw(
(-tick,1)--(tick,1)
^^(-tick,-1)--(tick,-1)
^^(1,-tick)--(1,tick)
^^(2,-tick)--(2,tick)
^^(3,-tick)--(3,tick)
);
draw((-.25,0)--(3.375,0),Arrow(TeXHead,size=.6*mm));
draw((0,-1.5)--(0,1.5),Arrow(TeXHead,size=.6*mm));
label("$x$",(3.375,0),1.5*S);
label("$y$",(0,1.5),1.5*W);
label(scale(GR^.5)*"$\mathbf\pi$",(.8+1.5,.5*GR^.5),.8blue);
label(scale(GR^.5)*"$\mathbf\pi$",(.8+1.5,-.5*GR^.5),.8blue);
label("$-1$",(-tick,-1),W);
label("$1$",(-tick,1),W);
label("$1$",(1,-tick),S);
label("$2$",(2,-tick),SE);
label("$3$",(3,-tick),SE);
shipout(bbox(2mm, FillDraw(rgb(064,224,208),rgb(064,224,208))));
[/asy]](//latex.artofproblemsolving.com/5/1/4/5145ec9617eb3afb8c0b728609e052a72dcb9657.png)
I have put
with lower limit 
and upper limit 
into Integralrechner. It gave me the
and confirmed the result:
.
![[asy]
// Figure 2b
size(0,150);
real GR=(sqrt(5)+1)/2;
real gr=(sqrt(5)-1)/2;
import graph;
import geometry;
line l1=line((-.25,-sqrt(3)/4),(0,0));
line l2=line((-.25,sqrt(3)/4),(0,0));
line l3=line((-1,0),(2,0));
real f(real t) {return 1+cos(t);}
path cardioid=polargraph(f,0,2pi,operator ..)--cycle;
real tick=1/3*.25;
path p1=polargraph(f,-1/3 *pi,1/3 *pi,operator ..);
path p2=arc((1.5,0),1.5,120,120*2);
filldraw(circle((1.5,0),1.5),lightblue+white,blue+linewidth(1.25));
fill(cardioid,white);
fill (p1..p2..cycle,lightred+white);
filldraw(circle((1,0),1),palered+white,red);
draw(cardioid,red+linewidth(1.25));
draw(arc((1.5,0),1.5,120,240),blue+linewidth(1.25));
draw(
(-tick,1)--(tick,1)
^^(-tick,-1)--(tick,-1)
^^(1,-tick)--(1,tick)
^^(2,-tick)--(2,tick)
^^(3,-tick)--(3,tick)
);
draw((-.25,0)--(3.375,0),Arrow(TeXHead,size=.6*mm));
draw((0,-1.5)--(0,1.5),Arrow(TeXHead,size=.6*mm));
label("$x$",(3.375,0),1.5*S);
label("$y$",(0,1.5),1.5*W);
label(scale(GR^.5)*"$\mathbf\pi?$",(.8+1.5,.5*GR^.5),.8blue);
label(scale(GR^.5)*"$\mathbf\pi?$",(.8+1.5,-.5*GR^.5),.8blue);
real gr=GR^(-1);
label(scale(gr^.5)*"$\mathbf.5\pi$",(1,sqrt(2)-1),.8red);
label(scale(gr^.5)*"$\mathbf.5\pi$",(1,-1.375*(sqrt(2)-1)),.8red);
label("$-1$",(-tick,-1),W);
label("$1$",(-tick,1),W);
label("$1$",(1,-tick),S);
label("$2$",(2,-tick),SE);
label("$3$",(3,-tick),SE);
shipout(bbox(2mm, FillDraw(rgb(064,224,208),rgb(064,224,208))));
[/asy]](//latex.artofproblemsolving.com/6/3/8/638de21023eef5951cdf9f07c6adfbe61d179df1.png)
The area of the cardioid is
(Lubov Vassilevskaya schwerpunkt-2.pdf, page 14). It follows that the togeteher of the both cardioid crescents (white + lightred) must have an area of 
. So one single cardioid crescents has as area 
.
![[asy]
// Figure 2c
size(0,150);
real GR=(sqrt(5)+1)/2;
real gr=(sqrt(5)-1)/2;
import graph;
import geometry;
line l1=line((-.25,-sqrt(3)/4),(0,0));
line l2=line((-.25,sqrt(3)/4),(0,0));
line l3=line((-1,0),(2,0));
real f(real t) {return 1+cos(t);}
path cardioid=polargraph(f,0,2pi,operator ..)--cycle;
real tick=1/3*.25;
path p1=polargraph(f,-1/3 *pi,1/3 *pi,operator ..);
path p2=arc((1.5,0),1.5,120,120*2);
filldraw(circle((1.5,0),1.5),lightblue+white,blue+linewidth(1.25));
fill(cardioid,white);
fill (p1..p2..cycle,lightred+white);
filldraw(circle((1,0),1),palered+white,red);
draw(cardioid,red+linewidth(1.25));
draw(arc((1.5,0),1.5,120,240),blue+linewidth(1.25));
draw(
(-tick,1)--(tick,1)
^^(-tick,-1)--(tick,-1)
^^(1,-tick)--(1,tick)
^^(2,-tick)--(2,tick)
^^(3,-tick)--(3,tick)
);
draw((-.25,0)--(3.375,0),Arrow(TeXHead,size=.6*mm));
draw((0,-1.5)--(0,1.5),Arrow(TeXHead,size=.6*mm));
label("$x$",(3.375,0),1.5*S);
label("$y$",(0,1.5),1.5*W);
//Complete Area blue Circle: 1-5^2*pi=2.25pi
// There from substracting 2 pi equals .25 pi. 2.25pi-2pi=.25pi
// But the upper palered halfcircle already has Area .5pi
// Because the palered circle has radius 1. Area_palered_circle=1^2*pi=1pi
label(scale(GR^.5)*"$\mathbf.5\pi$",(.8+1.5,.5*GR^.5),.8blue);
label(scale(GR^.5)*"$\mathbf.5\pi$",(.8+1.5,-.5*GR^.5),.8blue);
real gr=GR^(-1);
label(scale(gr^.5)*"$\mathbf.5\pi$",(1,sqrt(2)-1),.8red);
label(scale(gr^.5)*"$\mathbf.5\pi$",(1,-1.375*(sqrt(2)-1)),.8red);
label("$-1$",(-tick,-1),W);
label("$1$",(-tick,1),W);
label("$1$",(1,-tick),S);
label("$2$",(2,-tick),SE);
label("$3$",(3,-tick),SE);
shipout(bbox(2mm, FillDraw(rgb(064,224,208),rgb(064,224,208))));
[/asy]](//latex.artofproblemsolving.com/6/5/9/659971868df2c21f2338a9e3dea00ae4bbd5c824.png)
The radius of the blue circle is 1.5. So the area is:
. Wouldt you substract from this value you would come to only as the common cover region of cardioid and blue circle. And this - this is obviously not true, because the unitcircle area already has the area of 
.
![[asy]
// Figure 3a
size(0,150);
real GR=(sqrt(5)+1)/2;
real gr=(sqrt(5)-1)/2;
import graph;
import geometry;
line l1=line((-.25,-sqrt(3)/4),(0,0));
line l2=line((-.25,sqrt(3)/4),(0,0));
line l3=line((-1,0),(2,0));
real f(real t) {return 1+cos(t);}
path cardioid=polargraph(f,0,2pi,operator ..)--cycle;
path p1=polargraph(f,-1/3 *pi,1/3 *pi,operator ..);
path p2=arc((1.5,0),1.5,120,120*2);
real tick=1/3*.25;
filldraw(circle((1.5,0),1.5),lightblue+white,blue+linewidth(1.25));
filldraw(circle((0,0),1.5),lightblue+white,blue+linewidth(1.25));
fill(cardioid, white);
fill(p1..p2..cycle,lightred+white);
draw(p2,blue+linewidth(1.25));
draw(cardioid,red+linewidth(1.25));
draw(
(-tick,1)--(tick,1)
^^(-tick,-1)--(tick,-1)
^^(1,-tick)--(1,tick)
^^(2,-tick)--(2,tick)
^^(3,-tick)--(3,tick)
);
draw((-.25,0)--(3.375,0),Arrow(TeXHead,size=.6*mm));
draw((0,-1.5)--(0,1.75),Arrow(TeXHead,size=.6*mm));
label("$x$",(3.375,0),1.5*S);
label("$y$",(0,1.75),1.5*W);
label(scale(GR^.5)*"$\mathbf.5\pi$",(.8+1.5,.5*GR^.5),.8blue);
label(scale(GR^.5)*"$\mathbf.5\pi$",(.8+1.5,-.5*GR^.5),.8blue);
label("$-1$",(-tick,-1),W);
label("$1$",(-tick,1),W);
label("$1$",(1,-tick),S);
label("$2$",(2,-tick),SE);
label("$3$",(3,-tick),SE);
shipout(bbox(2mm, FillDraw(rgb(064,224,208),rgb(064,224,208))));
[/asy]](//latex.artofproblemsolving.com/0/b/6/0b6b48f2d876f6e337f50dbb440f64920f23cf21.png)
On the left hand you see a blue moon with nose. And as parts of the cardioid there are two white crescents.
And on the right hand there is a big blue crescent as rest part of the right circle. On the inner side of this crescent there is the outline of the cardioid untill the minimum and maximum of the cardioid.
![[asy]
size(0,150);
real GR=(sqrt(5)+1)/2;
real gr=(sqrt(5)-1)/2;
import graph;
import geometry;
line l1=line((-.25,-sqrt(3)/4),(0,0));
line l2=line((-.25,sqrt(3)/4),(0,0));
line l3=line((-1,0),(2,0));
real f(real t) {return 1+cos(t);}
path cardioid=polargraph(f,0,2pi,operator ..)--cycle;
path p1=polargraph(f,-1/3 *pi,1/3 *pi,operator ..);
path p2=arc((1.5,0),1.5,120,120*2);
real tick=1/3*.25;
filldraw(circle((1.5,0),1.5),lightblue+white,blue+linewidth(1.25));
filldraw(circle((0,0),1.5),lightblue+white,blue+linewidth(1.25));
fill(cardioid, white);
fill(p1..p2..cycle,lightred+white);
draw(p2,blue+linewidth(1.25));
draw(arc(origin,1.5*dir(-60),1.5*dir(60)),dotted+.8blue+linewidth(1.25));
draw(cardioid,red+linewidth(1.25));
draw(
(-tick,1)--(tick,1)
^^(-tick,-1)--(tick,-1)
^^(1,-tick)--(1,tick)
^^(2,-tick)--(2,tick)
^^(3,-tick)--(3,tick)
);
draw((-.25,0)--(3.375,0),Arrow(TeXHead,size=.6*mm));
draw((0,-1.5)--(0,1.75),Arrow(TeXHead,size=.6*mm));
label("$x$",(3.375,0),1.5*S);
label("$y$",(0,1.75),1.5*W);
label(scale(GR^.5)*"$\mathbf.5\pi$",(.8+1.5,.5*GR^.5),.8blue);
label(scale(GR^.5)*"$\mathbf.5\pi$",(.8+1.5,-.5*GR^.5),.8blue);
label("$-1$",(-tick,-1),W);
label("$1$",(-tick,1),W);
label("$1$",(1,-tick),S);
label("$2$",(2,-tick),SE);
label("$3$",(3,-tick),SE);
dot(origin);
shipout(bbox(2mm, FillDraw(rgb(064,224,208),rgb(064,224,208))));
[/asy]](//latex.artofproblemsolving.com/9/d/1/9d1fec539d158d660b5dc1bd2c0b0fbb1f76cafc.png)
![[asy]
// Figure 4
size(0,150);
real GR=(sqrt(5)+1)/2;
import geometry;
real f(real t) {return 1+cos(t);}
path cardioid=polargraph(f,0,2pi,operator ..)--cycle;
path cardioid_subpath=polargraph(f,pi/3,2pi-pi/3,operator ..)--cycle;
fill(cardioid,lightred+white);
fill(circle((0,0),1.5),lightblue+white);
draw(cardioid,red);
fill(cardioid_subpath,red+white);
draw(cardioid_subpath,.8red+linewidth(GR^2));
/*
Gerade ausgerechnet: Der Kreis mit Mittelpunkt (xm,ym) und Radius rk hat
in Polarkoordinaten die Gleichung
r(φ)=x_m cosφ+y_m sinφ + sqrt( r_k^2 - ( x_m sinφ−y_mcosφ )^2 )
https://www.mikrocontroller.net/topic/132819
*/
//draw(circle((.5,0),1.5),dotted);//Umkreis
// also: xm=.5, ym=0
// Jetzt läuft es. Warum vorher nicht?!
real xm=.5, ym=0, rk=1.5;// rk=Radius Kreis
// rk^2 wird zu 2.25, wenn rk=1.5
real c6(real t) {return xm*cos(t) + //ym*sin(t)// wird ja Null
//+ sqrt( rk^2 - ( xm*sin(t)
+ sqrt( 2.25 - ( xm*sin(t)
//− ym*cos(t)// wird ja Null
)^2 );}
// invalid token + sqrt( rk^2 - ( xm*sin(t) − ym*cos(t) )^2 );
//soll: erzeugt Kreis um (.5,0), Radius 1.5 (Umkreis der Kardiode)
path circle6=polargraph(c6,0,2pi,operator ..)--cycle;
//draw(circle6,dashed+red);// Umkreis
//real xm=0, ym=0,rk=1.5;
/*
real c0(real t) {return xm*cos(t) + //ym*sin(t)// wird ja Null
//+ sqrt( rk^2 - ( xm*sin(t)
+ sqrt( 2.25 - ( xm*sin(t)
//− ym*cos(t)// wird ja Null
)^2 );}
*/
// gekürzt:
real xm=0, ym=0,rk=1.5;
real c0(real t) {return xm*cos(t) + sqrt(2.25 - (xm*sin(t))^2);}
path circle0=polargraph(c0,0,2pi,operator ..)--cycle;
path circle0_subpath=polargraph(c0,-pi/3,pi/3,operator ..)--cycle;
draw(circle0,blue);
dot((xm,ym),blue+linewidth(.85),UnFill);
fill(circle0_subpath,blue+white);
draw(circle0_subpath,.8blue+linewidth(GR^2));
pair cardioid_max=(.75,3*sqrt(3)/4);
// sqrt(.75^2+(3*sqrt(3)/4)^2) = 3/2 = 1.5
dot(cardioid_max^^conj(cardioid_max),red+linewidth(.85),UnFill);
//dot(1.5*dir(60)^^1.5*dir(-60),blue,UnFill);// the same
shipout(bbox(2mm, FillDraw(white,white)));
[/asy]](//latex.artofproblemsolving.com/d/4/9/d4992a19c9d44a14ad1483ad03a928386f5b8a70.png)
You see again (in lightblue) the moon with nose as in Figure 3. And in blue the circle segment we are too trying to trace it out in Intersection Area of two same large Circles. And around this circle segment in lightred there appears a crescent, which is formed on the right outside not of a circle but as the remaining right part of the cardioid.
. Now you move the center from this circle to the right hand so that in walks through the maximum and through the minimum of the cardioid. What is the intersection area?It is said:
APEX Calculus CalculusIII_Version2.pdf, page 521 following wrote:
Amazingly enough, the area between these curves has a “nice” value.
(APEX Calculus CalculusIII_Version2.pdf, page 521 following).
![[asy]
// Figure 1
size(0,150);
real GR=(sqrt(5)+1)/2;
real gr=(sqrt(5)-1)/2;
import graph;
import geometry;
line l1=line((-.25,-sqrt(3)/4),(0,0));
line l2=line((-.25,sqrt(3)/4),(0,0));
line l3=line((-1,0),(2,0));
real m(real t) {return 1+cos(t);}
path h=polargraph(m,0,2pi,operator ..)--cycle
,cardioid=h;
pair p0= (-.25,sqrt(3)/4);
fill(Circle((.5,0),1.5),red+white);
fill(cardioid,yellow+red);
fill(Circle((1,0),1),rgb(205, 080, 080) );
fill(Circle((.5,0),.5),green);
fill(Circle((1.5,0),.5),yellow);
real[] myarray1={-1.5,-1,-.5,0,.5,1,1.5,2,2.5,3,3.5};
for(real k : myarray1) { draw(shift(k,0)*l1,grey);
draw(shift(k,0)*l2,grey); }
real[] myarray2={0
,sqrt(3)/4
,2*sqrt(3)/4
,3*sqrt(3)/4
,-1*sqrt(3)/4
,-2*sqrt(3)/4
,-3*sqrt(3)/4
};
for(real k : myarray2) draw(shift(0,k)*l3,grey);
draw(Circle((1,0),1));
draw(Circle((.5,0),.5));
draw(Circle((1.5,0),.5));
draw(Circle((.5,0),1.5),linewidth(gr));
draw((0,0)--(1,sqrt(GR)),yellow,PenMargin);
draw((0,0)--(1,0)--(1,sqrt(GR)),yellow,PenMargin);
pair p1=relpoint((0,0)--(1,sqrt(GR)),gr^2);
draw(p1--(1,0),yellow);
pair p2=relpoint(p1--(1,0),GR^-GR);
pair p3=(1.25,sqrt(3)/4)
,p4=(1.25,3*sqrt(3)/4)
,p5=relpoint(p3--p4,gr)
,p6=(p3.x-.5,p3.y)
,p7=(p4.x-.5,p4.y)
,p8=(p5.x-.5,p5.y)
,p9=(p3.x-1,p3.y)
,p10=(p4.x-1,p4.y)
// ,p11=relpoint(p9--p10,gr^exp(1))
,p12=(.5,0)
,p13=relpoint(p12--p0,gr);
draw(p3--p5
^^p6--p8
^^p12--p13
,blue+linewidth(2));
draw(p4--p5^^p7--p8^^p0--p13,red+linewidth(2));
dot(p3^^p4^^p6^^p7^^p12^^p0
//^^p9^^p10
);draw(p3^^p4^^p6^^p7^^p12^^p0,white+linewidth(gr));
dot(p5^^p8^^p13
//^^p11
,yellow);draw(p5^^p8^^p13,linewidth(gr));
draw(circle(origin+1.5,1.5));
draw(cardioid);
draw(p0^^p3^^p4^^p6^^p7^^p12, white+linewidth(gr));
shipout(scale(sqrt(2))*bbox(gr*1cm,invisible,Fill(rgb(064,224,208))));
[/asy]](http://latex.artofproblemsolving.com/e/7/e/e7e7b892354716444d832cf94cbb9cf0e7e062dd.png)
Figure 1: Cardioid with to the right Hand moved Circumcircle
![[asy]
// Figure 2a
size(0,150);
real GR=(sqrt(5)+1)/2;
real gr=(sqrt(5)-1)/2;
import graph;
import geometry;
line l1=line((-.25,-sqrt(3)/4),(0,0));
line l2=line((-.25,sqrt(3)/4),(0,0));
line l3=line((-1,0),(2,0));
real f(real t) {return 1+cos(t);}
path cardioid=polargraph(f,0,2pi,operator ..)--cycle;
real tick=1/3*.25;
path p1=polargraph(f,-1/3 *pi,1/3 *pi,operator ..);
path p2=arc((1.5,0),1.5,120,120*2);
filldraw(circle((1.5,0),1.5),lightblue+white,blue+linewidth(1.25));
fill(cardioid,white);
fill (p1..p2..cycle,lightred+white);
draw(cardioid,red+linewidth(1.25));
draw(arc((1.5,0),1.5,120,240),blue+linewidth(1.25));
draw(
(-tick,1)--(tick,1)
^^(-tick,-1)--(tick,-1)
^^(1,-tick)--(1,tick)
^^(2,-tick)--(2,tick)
^^(3,-tick)--(3,tick)
);
draw((-.25,0)--(3.375,0),Arrow(TeXHead,size=.6*mm));
draw((0,-1.5)--(0,1.5),Arrow(TeXHead,size=.6*mm));
label("$x$",(3.375,0),1.5*S);
label("$y$",(0,1.5),1.5*W);
label(scale(GR^.5)*"$\mathbf\pi$",(.8+1.5,.5*GR^.5),.8blue);
label(scale(GR^.5)*"$\mathbf\pi$",(.8+1.5,-.5*GR^.5),.8blue);
label("$-1$",(-tick,-1),W);
label("$1$",(-tick,1),W);
label("$1$",(1,-tick),S);
label("$2$",(2,-tick),SE);
label("$3$",(3,-tick),SE);
shipout(bbox(2mm, FillDraw(rgb(064,224,208),rgb(064,224,208))));
[/asy]](http://latex.artofproblemsolving.com/5/1/4/5145ec9617eb3afb8c0b728609e052a72dcb9657.png)
Figure 2a: Cardioid with to the right Hand moved Circumcircle
Intersection Area (lightblue) 
I have put
with lower limit 
and upper limit 
into Integralrechner. It gave me the
and confirmed the result:
.![[asy]
// Figure 2b
size(0,150);
real GR=(sqrt(5)+1)/2;
real gr=(sqrt(5)-1)/2;
import graph;
import geometry;
line l1=line((-.25,-sqrt(3)/4),(0,0));
line l2=line((-.25,sqrt(3)/4),(0,0));
line l3=line((-1,0),(2,0));
real f(real t) {return 1+cos(t);}
path cardioid=polargraph(f,0,2pi,operator ..)--cycle;
real tick=1/3*.25;
path p1=polargraph(f,-1/3 *pi,1/3 *pi,operator ..);
path p2=arc((1.5,0),1.5,120,120*2);
filldraw(circle((1.5,0),1.5),lightblue+white,blue+linewidth(1.25));
fill(cardioid,white);
fill (p1..p2..cycle,lightred+white);
filldraw(circle((1,0),1),palered+white,red);
draw(cardioid,red+linewidth(1.25));
draw(arc((1.5,0),1.5,120,240),blue+linewidth(1.25));
draw(
(-tick,1)--(tick,1)
^^(-tick,-1)--(tick,-1)
^^(1,-tick)--(1,tick)
^^(2,-tick)--(2,tick)
^^(3,-tick)--(3,tick)
);
draw((-.25,0)--(3.375,0),Arrow(TeXHead,size=.6*mm));
draw((0,-1.5)--(0,1.5),Arrow(TeXHead,size=.6*mm));
label("$x$",(3.375,0),1.5*S);
label("$y$",(0,1.5),1.5*W);
label(scale(GR^.5)*"$\mathbf\pi?$",(.8+1.5,.5*GR^.5),.8blue);
label(scale(GR^.5)*"$\mathbf\pi?$",(.8+1.5,-.5*GR^.5),.8blue);
real gr=GR^(-1);
label(scale(gr^.5)*"$\mathbf.5\pi$",(1,sqrt(2)-1),.8red);
label(scale(gr^.5)*"$\mathbf.5\pi$",(1,-1.375*(sqrt(2)-1)),.8red);
label("$-1$",(-tick,-1),W);
label("$1$",(-tick,1),W);
label("$1$",(1,-tick),S);
label("$2$",(2,-tick),SE);
label("$3$",(3,-tick),SE);
shipout(bbox(2mm, FillDraw(rgb(064,224,208),rgb(064,224,208))));
[/asy]](http://latex.artofproblemsolving.com/6/3/8/638de21023eef5951cdf9f07c6adfbe61d179df1.png)
Figure 2b: Cardioid with Incircle, Area
Incircle Area (palered): 
The area of the cardioid is
(Lubov Vassilevskaya schwerpunkt-2.pdf, page 14). It follows that the togeteher of the both cardioid crescents (white + lightred) must have an area of 
. So one single cardioid crescents has as area 
.![[asy]
// Figure 2c
size(0,150);
real GR=(sqrt(5)+1)/2;
real gr=(sqrt(5)-1)/2;
import graph;
import geometry;
line l1=line((-.25,-sqrt(3)/4),(0,0));
line l2=line((-.25,sqrt(3)/4),(0,0));
line l3=line((-1,0),(2,0));
real f(real t) {return 1+cos(t);}
path cardioid=polargraph(f,0,2pi,operator ..)--cycle;
real tick=1/3*.25;
path p1=polargraph(f,-1/3 *pi,1/3 *pi,operator ..);
path p2=arc((1.5,0),1.5,120,120*2);
filldraw(circle((1.5,0),1.5),lightblue+white,blue+linewidth(1.25));
fill(cardioid,white);
fill (p1..p2..cycle,lightred+white);
filldraw(circle((1,0),1),palered+white,red);
draw(cardioid,red+linewidth(1.25));
draw(arc((1.5,0),1.5,120,240),blue+linewidth(1.25));
draw(
(-tick,1)--(tick,1)
^^(-tick,-1)--(tick,-1)
^^(1,-tick)--(1,tick)
^^(2,-tick)--(2,tick)
^^(3,-tick)--(3,tick)
);
draw((-.25,0)--(3.375,0),Arrow(TeXHead,size=.6*mm));
draw((0,-1.5)--(0,1.5),Arrow(TeXHead,size=.6*mm));
label("$x$",(3.375,0),1.5*S);
label("$y$",(0,1.5),1.5*W);
//Complete Area blue Circle: 1-5^2*pi=2.25pi
// There from substracting 2 pi equals .25 pi. 2.25pi-2pi=.25pi
// But the upper palered halfcircle already has Area .5pi
// Because the palered circle has radius 1. Area_palered_circle=1^2*pi=1pi
label(scale(GR^.5)*"$\mathbf.5\pi$",(.8+1.5,.5*GR^.5),.8blue);
label(scale(GR^.5)*"$\mathbf.5\pi$",(.8+1.5,-.5*GR^.5),.8blue);
real gr=GR^(-1);
label(scale(gr^.5)*"$\mathbf.5\pi$",(1,sqrt(2)-1),.8red);
label(scale(gr^.5)*"$\mathbf.5\pi$",(1,-1.375*(sqrt(2)-1)),.8red);
label("$-1$",(-tick,-1),W);
label("$1$",(-tick,1),W);
label("$1$",(1,-tick),S);
label("$2$",(2,-tick),SE);
label("$3$",(3,-tick),SE);
shipout(bbox(2mm, FillDraw(rgb(064,224,208),rgb(064,224,208))));
[/asy]](http://latex.artofproblemsolving.com/6/5/9/659971868df2c21f2338a9e3dea00ae4bbd5c824.png)
Figure 2c: My opinion. - Lightblue Crescent Area: 
The radius of the blue circle is 1.5. So the area is:
. Wouldt you substract from this value you would come to only as the common cover region of cardioid and blue circle. And this - this is obviously not true, because the unitcircle area already has the area of 
.![[asy]
// Figure 3a
size(0,150);
real GR=(sqrt(5)+1)/2;
real gr=(sqrt(5)-1)/2;
import graph;
import geometry;
line l1=line((-.25,-sqrt(3)/4),(0,0));
line l2=line((-.25,sqrt(3)/4),(0,0));
line l3=line((-1,0),(2,0));
real f(real t) {return 1+cos(t);}
path cardioid=polargraph(f,0,2pi,operator ..)--cycle;
path p1=polargraph(f,-1/3 *pi,1/3 *pi,operator ..);
path p2=arc((1.5,0),1.5,120,120*2);
real tick=1/3*.25;
filldraw(circle((1.5,0),1.5),lightblue+white,blue+linewidth(1.25));
filldraw(circle((0,0),1.5),lightblue+white,blue+linewidth(1.25));
fill(cardioid, white);
fill(p1..p2..cycle,lightred+white);
draw(p2,blue+linewidth(1.25));
draw(cardioid,red+linewidth(1.25));
draw(
(-tick,1)--(tick,1)
^^(-tick,-1)--(tick,-1)
^^(1,-tick)--(1,tick)
^^(2,-tick)--(2,tick)
^^(3,-tick)--(3,tick)
);
draw((-.25,0)--(3.375,0),Arrow(TeXHead,size=.6*mm));
draw((0,-1.5)--(0,1.75),Arrow(TeXHead,size=.6*mm));
label("$x$",(3.375,0),1.5*S);
label("$y$",(0,1.75),1.5*W);
label(scale(GR^.5)*"$\mathbf.5\pi$",(.8+1.5,.5*GR^.5),.8blue);
label(scale(GR^.5)*"$\mathbf.5\pi$",(.8+1.5,-.5*GR^.5),.8blue);
label("$-1$",(-tick,-1),W);
label("$1$",(-tick,1),W);
label("$1$",(1,-tick),S);
label("$2$",(2,-tick),SE);
label("$3$",(3,-tick),SE);
shipout(bbox(2mm, FillDraw(rgb(064,224,208),rgb(064,224,208))));
[/asy]](http://latex.artofproblemsolving.com/0/b/6/0b6b48f2d876f6e337f50dbb440f64920f23cf21.png)
Figure 3a: Cardioid with two shifted Circumcircles
Mirror Situation
On the left hand you see a blue moon with nose. And as parts of the cardioid there are two white crescents.
And on the right hand there is a big blue crescent as rest part of the right circle. On the inner side of this crescent there is the outline of the cardioid untill the minimum and maximum of the cardioid.
![[asy]
size(0,150);
real GR=(sqrt(5)+1)/2;
real gr=(sqrt(5)-1)/2;
import graph;
import geometry;
line l1=line((-.25,-sqrt(3)/4),(0,0));
line l2=line((-.25,sqrt(3)/4),(0,0));
line l3=line((-1,0),(2,0));
real f(real t) {return 1+cos(t);}
path cardioid=polargraph(f,0,2pi,operator ..)--cycle;
path p1=polargraph(f,-1/3 *pi,1/3 *pi,operator ..);
path p2=arc((1.5,0),1.5,120,120*2);
real tick=1/3*.25;
filldraw(circle((1.5,0),1.5),lightblue+white,blue+linewidth(1.25));
filldraw(circle((0,0),1.5),lightblue+white,blue+linewidth(1.25));
fill(cardioid, white);
fill(p1..p2..cycle,lightred+white);
draw(p2,blue+linewidth(1.25));
draw(arc(origin,1.5*dir(-60),1.5*dir(60)),dotted+.8blue+linewidth(1.25));
draw(cardioid,red+linewidth(1.25));
draw(
(-tick,1)--(tick,1)
^^(-tick,-1)--(tick,-1)
^^(1,-tick)--(1,tick)
^^(2,-tick)--(2,tick)
^^(3,-tick)--(3,tick)
);
draw((-.25,0)--(3.375,0),Arrow(TeXHead,size=.6*mm));
draw((0,-1.5)--(0,1.75),Arrow(TeXHead,size=.6*mm));
label("$x$",(3.375,0),1.5*S);
label("$y$",(0,1.75),1.5*W);
label(scale(GR^.5)*"$\mathbf.5\pi$",(.8+1.5,.5*GR^.5),.8blue);
label(scale(GR^.5)*"$\mathbf.5\pi$",(.8+1.5,-.5*GR^.5),.8blue);
label("$-1$",(-tick,-1),W);
label("$1$",(-tick,1),W);
label("$1$",(1,-tick),S);
label("$2$",(2,-tick),SE);
label("$3$",(3,-tick),SE);
dot(origin);
shipout(bbox(2mm, FillDraw(rgb(064,224,208),rgb(064,224,208))));
[/asy]](http://latex.artofproblemsolving.com/9/d/1/9d1fec539d158d660b5dc1bd2c0b0fbb1f76cafc.png)
Figure 3b: Cardioid with two shifted Circumcircles
(in dotted and blue the hidden Arc of the left Circle)
![[asy]
// Figure 4
size(0,150);
real GR=(sqrt(5)+1)/2;
import geometry;
real f(real t) {return 1+cos(t);}
path cardioid=polargraph(f,0,2pi,operator ..)--cycle;
path cardioid_subpath=polargraph(f,pi/3,2pi-pi/3,operator ..)--cycle;
fill(cardioid,lightred+white);
fill(circle((0,0),1.5),lightblue+white);
draw(cardioid,red);
fill(cardioid_subpath,red+white);
draw(cardioid_subpath,.8red+linewidth(GR^2));
/*
Gerade ausgerechnet: Der Kreis mit Mittelpunkt (xm,ym) und Radius rk hat
in Polarkoordinaten die Gleichung
r(φ)=x_m cosφ+y_m sinφ + sqrt( r_k^2 - ( x_m sinφ−y_mcosφ )^2 )
https://www.mikrocontroller.net/topic/132819
*/
//draw(circle((.5,0),1.5),dotted);//Umkreis
// also: xm=.5, ym=0
// Jetzt läuft es. Warum vorher nicht?!
real xm=.5, ym=0, rk=1.5;// rk=Radius Kreis
// rk^2 wird zu 2.25, wenn rk=1.5
real c6(real t) {return xm*cos(t) + //ym*sin(t)// wird ja Null
//+ sqrt( rk^2 - ( xm*sin(t)
+ sqrt( 2.25 - ( xm*sin(t)
//− ym*cos(t)// wird ja Null
)^2 );}
// invalid token + sqrt( rk^2 - ( xm*sin(t) − ym*cos(t) )^2 );
//soll: erzeugt Kreis um (.5,0), Radius 1.5 (Umkreis der Kardiode)
path circle6=polargraph(c6,0,2pi,operator ..)--cycle;
//draw(circle6,dashed+red);// Umkreis
//real xm=0, ym=0,rk=1.5;
/*
real c0(real t) {return xm*cos(t) + //ym*sin(t)// wird ja Null
//+ sqrt( rk^2 - ( xm*sin(t)
+ sqrt( 2.25 - ( xm*sin(t)
//− ym*cos(t)// wird ja Null
)^2 );}
*/
// gekürzt:
real xm=0, ym=0,rk=1.5;
real c0(real t) {return xm*cos(t) + sqrt(2.25 - (xm*sin(t))^2);}
path circle0=polargraph(c0,0,2pi,operator ..)--cycle;
path circle0_subpath=polargraph(c0,-pi/3,pi/3,operator ..)--cycle;
draw(circle0,blue);
dot((xm,ym),blue+linewidth(.85),UnFill);
fill(circle0_subpath,blue+white);
draw(circle0_subpath,.8blue+linewidth(GR^2));
pair cardioid_max=(.75,3*sqrt(3)/4);
// sqrt(.75^2+(3*sqrt(3)/4)^2) = 3/2 = 1.5
dot(cardioid_max^^conj(cardioid_max),red+linewidth(.85),UnFill);
//dot(1.5*dir(60)^^1.5*dir(-60),blue,UnFill);// the same
shipout(bbox(2mm, FillDraw(white,white)));
[/asy]](http://latex.artofproblemsolving.com/d/4/9/d4992a19c9d44a14ad1483ad03a928386f5b8a70.png)
Figure 4: Cardioid with one to the left Hand moved Circumcircle
Circle through the origin and through Minimum and Maximum of the Cardioid
You see again (in lightblue) the moon with nose as in Figure 3. And in blue the circle segment we are too trying to trace it out in Intersection Area of two same large Circles. And around this circle segment in lightred there appears a crescent, which is formed on the right outside not of a circle but as the remaining right part of the cardioid.
This post has been edited 28 times. Last edited by Klaus-Anton, Jul 16, 2024, 2:30 PM
June 2025
May 2025