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Finding lengths when circles and squares tangents.

How can we prove the locus is a circle?Do the tangents of two circles define concentric circles?Circles and tangentsCircles and tangents and circumcirclesPlane-geometry problem with circles and tangentsFinding the euclidean centers of the geodesics AB, AC, and BCConceptual question about coordinate systems?Find all tangents between circlesTwo Touching Ellipses - Tangents, Centres and CollinearitySine of angle between direct common tangents of two circles?

6

$begingroup$

Should one approach by coordinates or by euclidean geometry?

By pure geometry, I am not able to solve.

geometry circle tangent-line

share|cite|improve this question

edited 5 hours ago

Anirban Niloy

573118

asked 5 hours ago

mavericmaveric

79412

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can find the ratio between the radius of the large semicircle and the radius of the small circle by applying the Pythagorean theorem to triangle $AEF$. After that, similar triangles will finish the job.
  $endgroup$
  – FredH
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  at corner $B$ construct a square with diagonal $BG=3$. Use similar triangles to show the side of this square $= a/2$ where $AB=2a$. This shows $a=3sqrt{2}$. Similarly at corner $D$ to get the value of $x=2aleft.sqrt{2}right/3$
  $endgroup$
  – Lozenges
  3 hours ago
  
  
  

  
  
  
  

add a comment | 

6

$begingroup$

Should one approach by coordinates or by euclidean geometry?

By pure geometry, I am not able to solve.

geometry circle tangent-line

share|cite|improve this question

edited 5 hours ago

Anirban Niloy

573118

asked 5 hours ago

mavericmaveric

79412

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can find the ratio between the radius of the large semicircle and the radius of the small circle by applying the Pythagorean theorem to triangle $AEF$. After that, similar triangles will finish the job.
  $endgroup$
  – FredH
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  at corner $B$ construct a square with diagonal $BG=3$. Use similar triangles to show the side of this square $= a/2$ where $AB=2a$. This shows $a=3sqrt{2}$. Similarly at corner $D$ to get the value of $x=2aleft.sqrt{2}right/3$
  $endgroup$
  – Lozenges
  3 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Should one approach by coordinates or by euclidean geometry?

By pure geometry, I am not able to solve.

geometry circle tangent-line

share|cite|improve this question

edited 5 hours ago

Anirban Niloy

573118

asked 5 hours ago

mavericmaveric

79412

$endgroup$

share|cite|improve this question

edited 5 hours ago

Anirban Niloy

573118

asked 5 hours ago

mavericmaveric

79412

share|cite|improve this question

edited 5 hours ago

Anirban Niloy

573118

asked 5 hours ago

mavericmaveric

79412

edited 5 hours ago

Anirban Niloy

573118

edited 5 hours ago

Anirban Niloy

573118

asked 5 hours ago

mavericmaveric

79412

asked 5 hours ago

mavericmaveric

79412

2 Answers
2

active

oldest

votes

3

$begingroup$

Assume the larger radius is 1 first (the square's side is 2). Then the smaller radius $r=FB$ satisfies
$$sqrt{(1+r)^2-1}+r=2$$
from which we solve and obtain $r=frac23$.

Setting up coordinates such that $A$ is the origin, we find $G=(3/2,1/2),GB=sqrt2/2$ and $x=frac{2sqrt2}3$. Since $GB=3$ in the picture, scaling yields the desired $x$ of
$$frac{3cdot2sqrt2/3}{sqrt2/2}=4$$

share|cite|improve this answer

answered 5 hours ago

Parcly TaxelParcly Taxel

43k1372101

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I solved it by another way, but got the same answer.
  $endgroup$
  – Michael Rozenberg
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Anirban Niloy Yes, of course. Post it. I don't want to post my solution because it still is very ugly. I used a trigonometry. I see some nice fact, but I still don't see how to use it.
  $endgroup$
  – Michael Rozenberg
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelRozenberg Thank you for your opinion. Same case of mine. I used a lot of trigonometry but at last reached to conclusion. But my solution is too broad and I think it won't be acceptable in some extent.
  $endgroup$
  – Anirban Niloy
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Anirban Niloy We can prove that $measuredangle ECF=45^{circ}$ and $HG^2=DH^2+BG^2,$ but I don't see how to use it for a simple solution.
  $endgroup$
  – Michael Rozenberg
  32 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelRozenberg You're looking for a simple solution but still now I 'm trying to prove $HG^2 = DH^2 + BG^2$. 😅😅😅Pursue your effort and you'll get something simple solution soon than I.
  $endgroup$
  – Anirban Niloy
  1 min ago
  

  
  
  
  

add a comment | 

2

$begingroup$

As we see in the above diagram, $ABCD$ is a square and two semi circles with its center $E$ and $F$ respectively. Let place the point $Q$ in such that $triangle QEF$ is a right angled triangle and draw two altitude lines $MH$ and $GL$ from two vertices $H$ and $G$ respectively.

Again, denote the side of the sqaure = $x$ and the radius of small semi circle = $r'$. So, the radius of larger semi circle = $r = frac{x}{2}$.

From $triangle QEF$, we get

$EQ^2 + QF^2 = EF^2$

$(x-r')^2 + (frac{x}{2})^2 = (frac{x}{2} + r')^2$

$x^2 -2xr' +r'^2 + frac{x^2}{4} = frac{x^2}{4} + xr' + r'^2 implies x^2 = 3xr' implies x = 3r'$

Hence, $r' = frac{x}{3} implies r' = frac{2r}{3}$

$BD$ is the diagonal of the square $ABCD$ and $angle CBD = angle LBG = 45^circ$. So, here we get that $triangle GLB$ is an isosceles triangle.

Now, by the pythagorian theorem from $triangle GLB$,

$GL^2 + LB^2 = GB^2 implies GL^2 + GL^2 = 3^2 implies 2GL^2 = 9 implies GL = frac{3}{sqrt2}$

Next, $triangle CFB sim triangle CGL$ and from both tne triangle it can be written that

$frac{BC}{BF} = frac{x}{y} = frac{3y}{y} = 3$

and similarly,

$frac{CL}{GL} = 3 implies CL = 3 × frac{3}{sqrt2} implies CL = frac{9}{sqrt2}$

From that, $CB = CL + LB = CL + GL = frac{9}{sqrt2} + frac{3}{sqrt2} = frac{12}{sqrt2} = 6sqrt2$. So, the side of the square $ABCD$ = $x$ = $6sqrt2$

After that,$triangle CDE sim triangle HME$ and triangle BAD sim triangle HMD$. From the similarity of first two triangles, we get

$frac{CD}{HM} = frac{DE}{ME}$

Likewise from the next similarity,

$frac{AB}{HM} = frac{AD}{MD} implies frac{CD}{HM} = frac{CD}{MD}$.

So, we can write that

$frac{DE}{ME} = frac{CD}{MD}$

$frac{r}{a} = frac{2r}{r-a}$......(By denoting $ME = a$)

$2ra = r^2-ra implies 6sqrt2x = (3sqrt2)^2 -3sqrt2x implies 6sqrt2x = 18 - 3sqrt2x = 9sqrt2x = 18 implies x = frac{18}{9sqrt2} implies x = sqrt2$

Then, $DM= 3sqrt2 - a = 3sqrt2 - sqrt2 = 2sqrt2$

Notice that, $triangle DMH$ is an isosceles triangle and so,

$DH^2 = 2DM^2 = 2(2sqrt2)^2 = 2×8 = 16$

And finally, we get $DH = sqrt16 = 4$.

It could have been solved by any easier effort. But I made it very difficult. I hope that you will understand or if you have any problem, please let me know.

share|cite|improve this answer

answered 2 hours ago

Anirban NiloyAnirban Niloy

573118

$endgroup$

add a comment | 

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3

$begingroup$

Assume the larger radius is 1 first (the square's side is 2). Then the smaller radius $r=FB$ satisfies
$$sqrt{(1+r)^2-1}+r=2$$
from which we solve and obtain $r=frac23$.

Setting up coordinates such that $A$ is the origin, we find $G=(3/2,1/2),GB=sqrt2/2$ and $x=frac{2sqrt2}3$. Since $GB=3$ in the picture, scaling yields the desired $x$ of
$$frac{3cdot2sqrt2/3}{sqrt2/2}=4$$

share|cite|improve this answer

answered 5 hours ago

Parcly TaxelParcly Taxel

43k1372101

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I solved it by another way, but got the same answer.
  $endgroup$
  – Michael Rozenberg
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Anirban Niloy Yes, of course. Post it. I don't want to post my solution because it still is very ugly. I used a trigonometry. I see some nice fact, but I still don't see how to use it.
  $endgroup$
  – Michael Rozenberg
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelRozenberg Thank you for your opinion. Same case of mine. I used a lot of trigonometry but at last reached to conclusion. But my solution is too broad and I think it won't be acceptable in some extent.
  $endgroup$
  – Anirban Niloy
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Anirban Niloy We can prove that $measuredangle ECF=45^{circ}$ and $HG^2=DH^2+BG^2,$ but I don't see how to use it for a simple solution.
  $endgroup$
  – Michael Rozenberg
  32 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelRozenberg You're looking for a simple solution but still now I 'm trying to prove $HG^2 = DH^2 + BG^2$. 😅😅😅Pursue your effort and you'll get something simple solution soon than I.
  $endgroup$
  – Anirban Niloy
  1 min ago
  

  
  
  
  

add a comment | 

3

$begingroup$

Assume the larger radius is 1 first (the square's side is 2). Then the smaller radius $r=FB$ satisfies
$$sqrt{(1+r)^2-1}+r=2$$
from which we solve and obtain $r=frac23$.

Setting up coordinates such that $A$ is the origin, we find $G=(3/2,1/2),GB=sqrt2/2$ and $x=frac{2sqrt2}3$. Since $GB=3$ in the picture, scaling yields the desired $x$ of
$$frac{3cdot2sqrt2/3}{sqrt2/2}=4$$

share|cite|improve this answer

answered 5 hours ago

Parcly TaxelParcly Taxel

43k1372101

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I solved it by another way, but got the same answer.
  $endgroup$
  – Michael Rozenberg
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Anirban Niloy Yes, of course. Post it. I don't want to post my solution because it still is very ugly. I used a trigonometry. I see some nice fact, but I still don't see how to use it.
  $endgroup$
  – Michael Rozenberg
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelRozenberg Thank you for your opinion. Same case of mine. I used a lot of trigonometry but at last reached to conclusion. But my solution is too broad and I think it won't be acceptable in some extent.
  $endgroup$
  – Anirban Niloy
  3 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Anirban Niloy We can prove that $measuredangle ECF=45^{circ}$ and $HG^2=DH^2+BG^2,$ but I don't see how to use it for a simple solution.
  $endgroup$
  – Michael Rozenberg
  32 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelRozenberg You're looking for a simple solution but still now I 'm trying to prove $HG^2 = DH^2 + BG^2$. 😅😅😅Pursue your effort and you'll get something simple solution soon than I.
  $endgroup$
  – Anirban Niloy
  1 min ago
  

  
  
  
  

add a comment | 

$begingroup$

Assume the larger radius is 1 first (the square's side is 2). Then the smaller radius $r=FB$ satisfies
$$sqrt{(1+r)^2-1}+r=2$$
from which we solve and obtain $r=frac23$.

Setting up coordinates such that $A$ is the origin, we find $G=(3/2,1/2),GB=sqrt2/2$ and $x=frac{2sqrt2}3$. Since $GB=3$ in the picture, scaling yields the desired $x$ of
$$frac{3cdot2sqrt2/3}{sqrt2/2}=4$$

share|cite|improve this answer

answered 5 hours ago

Parcly TaxelParcly Taxel

43k1372101

$endgroup$

share|cite|improve this answer

answered 5 hours ago

Parcly TaxelParcly Taxel

43k1372101

answered 5 hours ago

Parcly TaxelParcly Taxel

43k1372101

answered 5 hours ago

Parcly TaxelParcly Taxel

43k1372101

2

$begingroup$

As we see in the above diagram, $ABCD$ is a square and two semi circles with its center $E$ and $F$ respectively. Let place the point $Q$ in such that $triangle QEF$ is a right angled triangle and draw two altitude lines $MH$ and $GL$ from two vertices $H$ and $G$ respectively.

Again, denote the side of the sqaure = $x$ and the radius of small semi circle = $r'$. So, the radius of larger semi circle = $r = frac{x}{2}$.

From $triangle QEF$, we get

$EQ^2 + QF^2 = EF^2$

$(x-r')^2 + (frac{x}{2})^2 = (frac{x}{2} + r')^2$

$x^2 -2xr' +r'^2 + frac{x^2}{4} = frac{x^2}{4} + xr' + r'^2 implies x^2 = 3xr' implies x = 3r'$

Hence, $r' = frac{x}{3} implies r' = frac{2r}{3}$

$BD$ is the diagonal of the square $ABCD$ and $angle CBD = angle LBG = 45^circ$. So, here we get that $triangle GLB$ is an isosceles triangle.

Now, by the pythagorian theorem from $triangle GLB$,

$GL^2 + LB^2 = GB^2 implies GL^2 + GL^2 = 3^2 implies 2GL^2 = 9 implies GL = frac{3}{sqrt2}$

Next, $triangle CFB sim triangle CGL$ and from both tne triangle it can be written that

$frac{BC}{BF} = frac{x}{y} = frac{3y}{y} = 3$

and similarly,

$frac{CL}{GL} = 3 implies CL = 3 × frac{3}{sqrt2} implies CL = frac{9}{sqrt2}$

From that, $CB = CL + LB = CL + GL = frac{9}{sqrt2} + frac{3}{sqrt2} = frac{12}{sqrt2} = 6sqrt2$. So, the side of the square $ABCD$ = $x$ = $6sqrt2$

After that,$triangle CDE sim triangle HME$ and triangle BAD sim triangle HMD$. From the similarity of first two triangles, we get

$frac{CD}{HM} = frac{DE}{ME}$

Likewise from the next similarity,

$frac{AB}{HM} = frac{AD}{MD} implies frac{CD}{HM} = frac{CD}{MD}$.

So, we can write that

$frac{DE}{ME} = frac{CD}{MD}$

$frac{r}{a} = frac{2r}{r-a}$......(By denoting $ME = a$)

$2ra = r^2-ra implies 6sqrt2x = (3sqrt2)^2 -3sqrt2x implies 6sqrt2x = 18 - 3sqrt2x = 9sqrt2x = 18 implies x = frac{18}{9sqrt2} implies x = sqrt2$

Then, $DM= 3sqrt2 - a = 3sqrt2 - sqrt2 = 2sqrt2$

Notice that, $triangle DMH$ is an isosceles triangle and so,

$DH^2 = 2DM^2 = 2(2sqrt2)^2 = 2×8 = 16$

And finally, we get $DH = sqrt16 = 4$.

It could have been solved by any easier effort. But I made it very difficult. I hope that you will understand or if you have any problem, please let me know.

share|cite|improve this answer

answered 2 hours ago

Anirban NiloyAnirban Niloy

573118

$endgroup$

add a comment | 

2

$begingroup$

As we see in the above diagram, $ABCD$ is a square and two semi circles with its center $E$ and $F$ respectively. Let place the point $Q$ in such that $triangle QEF$ is a right angled triangle and draw two altitude lines $MH$ and $GL$ from two vertices $H$ and $G$ respectively.

Again, denote the side of the sqaure = $x$ and the radius of small semi circle = $r'$. So, the radius of larger semi circle = $r = frac{x}{2}$.

From $triangle QEF$, we get

$EQ^2 + QF^2 = EF^2$

$(x-r')^2 + (frac{x}{2})^2 = (frac{x}{2} + r')^2$

$x^2 -2xr' +r'^2 + frac{x^2}{4} = frac{x^2}{4} + xr' + r'^2 implies x^2 = 3xr' implies x = 3r'$

Hence, $r' = frac{x}{3} implies r' = frac{2r}{3}$

$BD$ is the diagonal of the square $ABCD$ and $angle CBD = angle LBG = 45^circ$. So, here we get that $triangle GLB$ is an isosceles triangle.

Now, by the pythagorian theorem from $triangle GLB$,

$GL^2 + LB^2 = GB^2 implies GL^2 + GL^2 = 3^2 implies 2GL^2 = 9 implies GL = frac{3}{sqrt2}$

Next, $triangle CFB sim triangle CGL$ and from both tne triangle it can be written that

$frac{BC}{BF} = frac{x}{y} = frac{3y}{y} = 3$

and similarly,

$frac{CL}{GL} = 3 implies CL = 3 × frac{3}{sqrt2} implies CL = frac{9}{sqrt2}$

From that, $CB = CL + LB = CL + GL = frac{9}{sqrt2} + frac{3}{sqrt2} = frac{12}{sqrt2} = 6sqrt2$. So, the side of the square $ABCD$ = $x$ = $6sqrt2$

After that,$triangle CDE sim triangle HME$ and triangle BAD sim triangle HMD$. From the similarity of first two triangles, we get

$frac{CD}{HM} = frac{DE}{ME}$

Likewise from the next similarity,

$frac{AB}{HM} = frac{AD}{MD} implies frac{CD}{HM} = frac{CD}{MD}$.

So, we can write that

$frac{DE}{ME} = frac{CD}{MD}$

$frac{r}{a} = frac{2r}{r-a}$......(By denoting $ME = a$)

$2ra = r^2-ra implies 6sqrt2x = (3sqrt2)^2 -3sqrt2x implies 6sqrt2x = 18 - 3sqrt2x = 9sqrt2x = 18 implies x = frac{18}{9sqrt2} implies x = sqrt2$

Then, $DM= 3sqrt2 - a = 3sqrt2 - sqrt2 = 2sqrt2$

Notice that, $triangle DMH$ is an isosceles triangle and so,

$DH^2 = 2DM^2 = 2(2sqrt2)^2 = 2×8 = 16$

And finally, we get $DH = sqrt16 = 4$.

It could have been solved by any easier effort. But I made it very difficult. I hope that you will understand or if you have any problem, please let me know.

share|cite|improve this answer

answered 2 hours ago

Anirban NiloyAnirban Niloy

573118

$endgroup$

add a comment | 

$begingroup$

As we see in the above diagram, $ABCD$ is a square and two semi circles with its center $E$ and $F$ respectively. Let place the point $Q$ in such that $triangle QEF$ is a right angled triangle and draw two altitude lines $MH$ and $GL$ from two vertices $H$ and $G$ respectively.

Again, denote the side of the sqaure = $x$ and the radius of small semi circle = $r'$. So, the radius of larger semi circle = $r = frac{x}{2}$.

From $triangle QEF$, we get

$EQ^2 + QF^2 = EF^2$

$(x-r')^2 + (frac{x}{2})^2 = (frac{x}{2} + r')^2$

$x^2 -2xr' +r'^2 + frac{x^2}{4} = frac{x^2}{4} + xr' + r'^2 implies x^2 = 3xr' implies x = 3r'$

Hence, $r' = frac{x}{3} implies r' = frac{2r}{3}$

$BD$ is the diagonal of the square $ABCD$ and $angle CBD = angle LBG = 45^circ$. So, here we get that $triangle GLB$ is an isosceles triangle.

Now, by the pythagorian theorem from $triangle GLB$,

$GL^2 + LB^2 = GB^2 implies GL^2 + GL^2 = 3^2 implies 2GL^2 = 9 implies GL = frac{3}{sqrt2}$

Next, $triangle CFB sim triangle CGL$ and from both tne triangle it can be written that

$frac{BC}{BF} = frac{x}{y} = frac{3y}{y} = 3$

and similarly,

$frac{CL}{GL} = 3 implies CL = 3 × frac{3}{sqrt2} implies CL = frac{9}{sqrt2}$

From that, $CB = CL + LB = CL + GL = frac{9}{sqrt2} + frac{3}{sqrt2} = frac{12}{sqrt2} = 6sqrt2$. So, the side of the square $ABCD$ = $x$ = $6sqrt2$

After that,$triangle CDE sim triangle HME$ and triangle BAD sim triangle HMD$. From the similarity of first two triangles, we get

$frac{CD}{HM} = frac{DE}{ME}$

Likewise from the next similarity,

$frac{AB}{HM} = frac{AD}{MD} implies frac{CD}{HM} = frac{CD}{MD}$.

So, we can write that

$frac{DE}{ME} = frac{CD}{MD}$

$frac{r}{a} = frac{2r}{r-a}$......(By denoting $ME = a$)

$2ra = r^2-ra implies 6sqrt2x = (3sqrt2)^2 -3sqrt2x implies 6sqrt2x = 18 - 3sqrt2x = 9sqrt2x = 18 implies x = frac{18}{9sqrt2} implies x = sqrt2$

Then, $DM= 3sqrt2 - a = 3sqrt2 - sqrt2 = 2sqrt2$

Notice that, $triangle DMH$ is an isosceles triangle and so,

$DH^2 = 2DM^2 = 2(2sqrt2)^2 = 2×8 = 16$

And finally, we get $DH = sqrt16 = 4$.

It could have been solved by any easier effort. But I made it very difficult. I hope that you will understand or if you have any problem, please let me know.

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answered 2 hours ago

Anirban NiloyAnirban Niloy

573118

$endgroup$

share|cite|improve this answer

answered 2 hours ago

Anirban NiloyAnirban Niloy

573118

answered 2 hours ago

Anirban NiloyAnirban Niloy

573118

answered 2 hours ago

Anirban NiloyAnirban Niloy

573118