Circle inscribed in a semicircleinternally and externally tangent circlesCenter of Soddy CircleDeriving formula for externally tangent circle to internally tangent circleWhat will be the value of $r$?Find the radius of the third circle given three circlesArea of bounded region questionWhat is the radius of this circle?If $S, S_1, S_2$ be the circles of radii 5,3 and 2 respectively. If $S_1$ and $S_2$ touch externally and they touch internally with $S$.Kind of hard(?) geometry proofCongruent triangles in 3 tangent circle configuration
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Circle inscribed in a semicircle
internally and externally tangent circlesCenter of Soddy CircleDeriving formula for externally tangent circle to internally tangent circleWhat will be the value of $r$?Find the radius of the third circle given three circlesArea of bounded region questionWhat is the radius of this circle?If $S, S_1, S_2$ be the circles of radii 5,3 and 2 respectively. If $S_1$ and $S_2$ touch externally and they touch internally with $S$.Kind of hard(?) geometry proofCongruent triangles in 3 tangent circle configuration
$begingroup$
There is black semicircle, which radius is $ R $. The red circle is tangentially inward to the semicircle and to the diameter in its center. The yellow one is tangent externally to the red circle, internally to the semicircle and tangent to the diameter of the semicircle.
My question is :
What is the relationship between $R$ and the radius $r$ of yellow one?
My attempts :
I tried to use similarity of triangles, but always I had the third unknown number and two equations. I am sure that the radius of the red one is $ 0.5R$
geometry circles
asked Mar 28 at 20:10
KukozKukoz
529
$endgroup$
add a comment |
$begingroup$
There is black semicircle, which radius is $ R $. The red circle is tangentially inward to the semicircle and to the diameter in its center. The yellow one is tangent externally to the red circle, internally to the semicircle and tangent to the diameter of the semicircle.
My question is :
What is the relationship between $R$ and the radius $r$ of yellow one?
My attempts :
I tried to use similarity of triangles, but always I had the third unknown number and two equations. I am sure that the radius of the red one is $ 0.5R$
geometry circles
asked Mar 28 at 20:10
KukozKukoz
529
$endgroup$
$begingroup$
Draw a line passing through the centers of the black and of the yellow circle and a line connecting the centers of the yellow and red circles and then use Pythagorean Theorem. Does that work?
$endgroup$
– Matteo
Mar 28 at 20:41
add a comment |
$begingroup$
There is black semicircle, which radius is $ R $. The red circle is tangentially inward to the semicircle and to the diameter in its center. The yellow one is tangent externally to the red circle, internally to the semicircle and tangent to the diameter of the semicircle.
My question is :
What is the relationship between $R$ and the radius $r$ of yellow one?
My attempts :
I tried to use similarity of triangles, but always I had the third unknown number and two equations. I am sure that the radius of the red one is $ 0.5R$
geometry circles
asked Mar 28 at 20:10
KukozKukoz
529
$endgroup$
There is black semicircle, which radius is $ R $. The red circle is tangentially inward to the semicircle and to the diameter in its center. The yellow one is tangent externally to the red circle, internally to the semicircle and tangent to the diameter of the semicircle.
My question is :
What is the relationship between $R$ and the radius $r$ of yellow one?
My attempts :
I tried to use similarity of triangles, but always I had the third unknown number and two equations. I am sure that the radius of the red one is $ 0.5R$
geometry circles
geometry circles
asked Mar 28 at 20:10
KukozKukoz
529
asked Mar 28 at 20:10
KukozKukoz
529
asked Mar 28 at 20:10
KukozKukoz
529
asked Mar 28 at 20:10
KukozKukoz
529
asked Mar 28 at 20:10
KukozKukoz
529
529
$begingroup$
Draw a line passing through the centers of the black and of the yellow circle and a line connecting the centers of the yellow and red circles and then use Pythagorean Theorem. Does that work?
$endgroup$
– Matteo
Mar 28 at 20:41
add a comment |
$begingroup$
Draw a line passing through the centers of the black and of the yellow circle and a line connecting the centers of the yellow and red circles and then use Pythagorean Theorem. Does that work?
$endgroup$
– Matteo
Mar 28 at 20:41
$begingroup$
Draw a line passing through the centers of the black and of the yellow circle and a line connecting the centers of the yellow and red circles and then use Pythagorean Theorem. Does that work?
$endgroup$
– Matteo
Mar 28 at 20:41
$begingroup$
Draw a line passing through the centers of the black and of the yellow circle and a line connecting the centers of the yellow and red circles and then use Pythagorean Theorem. Does that work?
$endgroup$
– Matteo
Mar 28 at 20:41
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let the radius of the "black" semicircle be $[CK]=2R$. Thus, the radius of the "red" circumference will be $[DC]=R$. In virtue of the Pythagorean Theorem in the triangles $triangle DGE$ and $triangle ECF$ respectively
beginalign* GE^2=CF^2&=DE^2-DG^2=(R+r)^2-(R-r)^2=4Rr\ &=CE^2-EF^2=(2R-r)^2-r^2=4R^2-4Rrendalign*
Hence
$$4Rr=4R^2-4Rriff colorblue2r=R$$
The radius of the yellow circumference is, consequently, one-fourth of the radius of the black circle.
answered Mar 28 at 21:08
Dr. MathvaDr. Mathva
3,190529
$endgroup$
$begingroup$
:) already in my comment and in my hint below
$endgroup$
– Matteo
Mar 28 at 21:09
$begingroup$
Oh, see... We had the same idea...
$endgroup$
– Dr. Mathva
Mar 28 at 21:10
add a comment |
$begingroup$
HINT
Consider the Figure below and note that
$overlineO_2H = fracR2-r$;
$overlineO_2O_3 = fracR2 + r$;
$overlineHO_1 = r$;
$overlineO_1O_3 = R-r$;
Use Pythagorean theoerm on $triangle O_2HO_3$ and on $triangle O_1HO_3$ to determine $overlineHO_3$ in two ways and thus get an equation, which, once solved, will give you the desider result of $r$ as a function of $R$.
answered Mar 28 at 21:03
MatteoMatteo
1,302313
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let the radius of the "black" semicircle be $[CK]=2R$. Thus, the radius of the "red" circumference will be $[DC]=R$. In virtue of the Pythagorean Theorem in the triangles $triangle DGE$ and $triangle ECF$ respectively
beginalign* GE^2=CF^2&=DE^2-DG^2=(R+r)^2-(R-r)^2=4Rr\ &=CE^2-EF^2=(2R-r)^2-r^2=4R^2-4Rrendalign*
Hence
$$4Rr=4R^2-4Rriff colorblue2r=R$$
The radius of the yellow circumference is, consequently, one-fourth of the radius of the black circle.
answered Mar 28 at 21:08
Dr. MathvaDr. Mathva
3,190529
$endgroup$
$begingroup$
:) already in my comment and in my hint below
$endgroup$
– Matteo
Mar 28 at 21:09
$begingroup$
Oh, see... We had the same idea...
$endgroup$
– Dr. Mathva
Mar 28 at 21:10
add a comment |
$begingroup$
Let the radius of the "black" semicircle be $[CK]=2R$. Thus, the radius of the "red" circumference will be $[DC]=R$. In virtue of the Pythagorean Theorem in the triangles $triangle DGE$ and $triangle ECF$ respectively
beginalign* GE^2=CF^2&=DE^2-DG^2=(R+r)^2-(R-r)^2=4Rr\ &=CE^2-EF^2=(2R-r)^2-r^2=4R^2-4Rrendalign*
Hence
$$4Rr=4R^2-4Rriff colorblue2r=R$$
The radius of the yellow circumference is, consequently, one-fourth of the radius of the black circle.
answered Mar 28 at 21:08
Dr. MathvaDr. Mathva
3,190529
$endgroup$
$begingroup$
:) already in my comment and in my hint below
$endgroup$
– Matteo
Mar 28 at 21:09
$begingroup$
Oh, see... We had the same idea...
$endgroup$
– Dr. Mathva
Mar 28 at 21:10
add a comment |
$begingroup$
Let the radius of the "black" semicircle be $[CK]=2R$. Thus, the radius of the "red" circumference will be $[DC]=R$. In virtue of the Pythagorean Theorem in the triangles $triangle DGE$ and $triangle ECF$ respectively
beginalign* GE^2=CF^2&=DE^2-DG^2=(R+r)^2-(R-r)^2=4Rr\ &=CE^2-EF^2=(2R-r)^2-r^2=4R^2-4Rrendalign*
Hence
$$4Rr=4R^2-4Rriff colorblue2r=R$$
The radius of the yellow circumference is, consequently, one-fourth of the radius of the black circle.
answered Mar 28 at 21:08
Dr. MathvaDr. Mathva
3,190529
$endgroup$
Let the radius of the "black" semicircle be $[CK]=2R$. Thus, the radius of the "red" circumference will be $[DC]=R$. In virtue of the Pythagorean Theorem in the triangles $triangle DGE$ and $triangle ECF$ respectively
beginalign* GE^2=CF^2&=DE^2-DG^2=(R+r)^2-(R-r)^2=4Rr\ &=CE^2-EF^2=(2R-r)^2-r^2=4R^2-4Rrendalign*
Hence
$$4Rr=4R^2-4Rriff colorblue2r=R$$
The radius of the yellow circumference is, consequently, one-fourth of the radius of the black circle.
answered Mar 28 at 21:08
Dr. MathvaDr. Mathva
3,190529
answered Mar 28 at 21:08
Dr. MathvaDr. Mathva
3,190529
answered Mar 28 at 21:08
Dr. MathvaDr. Mathva
3,190529
answered Mar 28 at 21:08
Dr. MathvaDr. Mathva
3,190529
3,190529
$begingroup$
:) already in my comment and in my hint below
$endgroup$
– Matteo
Mar 28 at 21:09
$begingroup$
Oh, see... We had the same idea...
$endgroup$
– Dr. Mathva
Mar 28 at 21:10
add a comment |
$begingroup$
:) already in my comment and in my hint below
$endgroup$
– Matteo
Mar 28 at 21:09
$begingroup$
Oh, see... We had the same idea...
$endgroup$
– Dr. Mathva
Mar 28 at 21:10
$begingroup$
:) already in my comment and in my hint below
$endgroup$
– Matteo
Mar 28 at 21:09
$begingroup$
:) already in my comment and in my hint below
$endgroup$
– Matteo
Mar 28 at 21:09
$begingroup$
Oh, see... We had the same idea...
$endgroup$
– Dr. Mathva
Mar 28 at 21:10
$begingroup$
Oh, see... We had the same idea...
$endgroup$
– Dr. Mathva
Mar 28 at 21:10
add a comment |
$begingroup$
HINT
Consider the Figure below and note that
$overlineO_2H = fracR2-r$;
$overlineO_2O_3 = fracR2 + r$;
$overlineHO_1 = r$;
$overlineO_1O_3 = R-r$;
Use Pythagorean theoerm on $triangle O_2HO_3$ and on $triangle O_1HO_3$ to determine $overlineHO_3$ in two ways and thus get an equation, which, once solved, will give you the desider result of $r$ as a function of $R$.
answered Mar 28 at 21:03
MatteoMatteo
1,302313
$endgroup$
add a comment |
$begingroup$
HINT
Consider the Figure below and note that
$overlineO_2H = fracR2-r$;
$overlineO_2O_3 = fracR2 + r$;
$overlineHO_1 = r$;
$overlineO_1O_3 = R-r$;
Use Pythagorean theoerm on $triangle O_2HO_3$ and on $triangle O_1HO_3$ to determine $overlineHO_3$ in two ways and thus get an equation, which, once solved, will give you the desider result of $r$ as a function of $R$.
answered Mar 28 at 21:03
MatteoMatteo
1,302313
$endgroup$
add a comment |
$begingroup$
HINT
Consider the Figure below and note that
$overlineO_2H = fracR2-r$;
$overlineO_2O_3 = fracR2 + r$;
$overlineHO_1 = r$;
$overlineO_1O_3 = R-r$;
Use Pythagorean theoerm on $triangle O_2HO_3$ and on $triangle O_1HO_3$ to determine $overlineHO_3$ in two ways and thus get an equation, which, once solved, will give you the desider result of $r$ as a function of $R$.
answered Mar 28 at 21:03
MatteoMatteo
1,302313
$endgroup$
HINT
Consider the Figure below and note that
$overlineO_2H = fracR2-r$;
$overlineO_2O_3 = fracR2 + r$;
$overlineHO_1 = r$;
$overlineO_1O_3 = R-r$;
Use Pythagorean theoerm on $triangle O_2HO_3$ and on $triangle O_1HO_3$ to determine $overlineHO_3$ in two ways and thus get an equation, which, once solved, will give you the desider result of $r$ as a function of $R$.
answered Mar 28 at 21:03
MatteoMatteo
1,302313
answered Mar 28 at 21:03
MatteoMatteo
1,302313
answered Mar 28 at 21:03
MatteoMatteo
1,302313
answered Mar 28 at 21:03
MatteoMatteo
1,302313
1,302313
add a comment |
add a comment |
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$begingroup$
Draw a line passing through the centers of the black and of the yellow circle and a line connecting the centers of the yellow and red circles and then use Pythagorean Theorem. Does that work?
$endgroup$
– Matteo
Mar 28 at 20:41