In the figure, a quarter circle, a semicircle and a circle are mutually tangent inside a square of side...
$begingroup$
In the figure, a quarter circle, a semicircle and a circle are mutually tangent inside a square of side length $2$. Find the radius of the circle.
I first assumed that when a vertical line is drawn from the radius of the semicircle, that line would be tangent to the smallest circle and it would mean that the radius is $frac{1}{4}$, but the correct answer was $frac{2}{9}$. I also tried using coordinate geometry, but I got stuck because I did not know how to get the equation of the smallest circle.
geometry
edited May 7 at 12:43
Asaf Karagila♦
311k33445777
asked May 6 at 11:32
suklaysuklay
1236
$endgroup$
add a comment |
$begingroup$
In the figure, a quarter circle, a semicircle and a circle are mutually tangent inside a square of side length $2$. Find the radius of the circle.
I first assumed that when a vertical line is drawn from the radius of the semicircle, that line would be tangent to the smallest circle and it would mean that the radius is $frac{1}{4}$, but the correct answer was $frac{2}{9}$. I also tried using coordinate geometry, but I got stuck because I did not know how to get the equation of the smallest circle.
geometry
edited May 7 at 12:43
Asaf Karagila♦
311k33445777
asked May 6 at 11:32
suklaysuklay
1236
$endgroup$
$begingroup$
$2/9$ seems to be very much too large simply from eyeballing the figure. Should it be the diameter of the circle rather than its radius?
$endgroup$
– Henning Makholm
May 6 at 11:41
$begingroup$
@HenningMakholm: I've confirmed that $2/9$ is correct. I submit that it also looks plausible. The diameter of the circle looks to be slightly less than the radius of the semicircle, and the diameter of the semicircle looks to be about half the side of the square (in fact, it's exactly half). So, the radius of the circle should be just shy of $1/4=0.25$; and $2/9=0.222ldots$ is in that ballpark.
$endgroup$
– Blue
May 6 at 12:00
2
$begingroup$
@Blue: Ah, sorry, I missed that the outer square has side length $2$ rather than $1$.
$endgroup$
– Henning Makholm
May 6 at 13:05
1
$begingroup$
Sounds like the start of a joke...
$endgroup$
– Asaf Karagila♦
May 7 at 12:43
add a comment |
$begingroup$
In the figure, a quarter circle, a semicircle and a circle are mutually tangent inside a square of side length $2$. Find the radius of the circle.
I first assumed that when a vertical line is drawn from the radius of the semicircle, that line would be tangent to the smallest circle and it would mean that the radius is $frac{1}{4}$, but the correct answer was $frac{2}{9}$. I also tried using coordinate geometry, but I got stuck because I did not know how to get the equation of the smallest circle.
geometry
edited May 7 at 12:43
Asaf Karagila♦
311k33445777
asked May 6 at 11:32
suklaysuklay
1236
$endgroup$
In the figure, a quarter circle, a semicircle and a circle are mutually tangent inside a square of side length $2$. Find the radius of the circle.
I first assumed that when a vertical line is drawn from the radius of the semicircle, that line would be tangent to the smallest circle and it would mean that the radius is $frac{1}{4}$, but the correct answer was $frac{2}{9}$. I also tried using coordinate geometry, but I got stuck because I did not know how to get the equation of the smallest circle.
geometry
geometry
edited May 7 at 12:43
Asaf Karagila♦
311k33445777
asked May 6 at 11:32
suklaysuklay
1236
edited May 7 at 12:43
Asaf Karagila♦
311k33445777
asked May 6 at 11:32
suklaysuklay
1236
edited May 7 at 12:43
Asaf Karagila♦
311k33445777
edited May 7 at 12:43
Asaf Karagila♦
311k33445777
edited May 7 at 12:43
Asaf Karagila♦
311k33445777
311k33445777
asked May 6 at 11:32
suklaysuklay
1236
asked May 6 at 11:32
suklaysuklay
1236
asked May 6 at 11:32
suklaysuklay
1236
1236
$begingroup$
$2/9$ seems to be very much too large simply from eyeballing the figure. Should it be the diameter of the circle rather than its radius?
$endgroup$
– Henning Makholm
May 6 at 11:41
$begingroup$
@HenningMakholm: I've confirmed that $2/9$ is correct. I submit that it also looks plausible. The diameter of the circle looks to be slightly less than the radius of the semicircle, and the diameter of the semicircle looks to be about half the side of the square (in fact, it's exactly half). So, the radius of the circle should be just shy of $1/4=0.25$; and $2/9=0.222ldots$ is in that ballpark.
$endgroup$
– Blue
May 6 at 12:00
2
$begingroup$
@Blue: Ah, sorry, I missed that the outer square has side length $2$ rather than $1$.
$endgroup$
– Henning Makholm
May 6 at 13:05
1
$begingroup$
Sounds like the start of a joke...
$endgroup$
– Asaf Karagila♦
May 7 at 12:43
add a comment |
$begingroup$
$2/9$ seems to be very much too large simply from eyeballing the figure. Should it be the diameter of the circle rather than its radius?
$endgroup$
– Henning Makholm
May 6 at 11:41
$begingroup$
@HenningMakholm: I've confirmed that $2/9$ is correct. I submit that it also looks plausible. The diameter of the circle looks to be slightly less than the radius of the semicircle, and the diameter of the semicircle looks to be about half the side of the square (in fact, it's exactly half). So, the radius of the circle should be just shy of $1/4=0.25$; and $2/9=0.222ldots$ is in that ballpark.
$endgroup$
– Blue
May 6 at 12:00
2
$begingroup$
@Blue: Ah, sorry, I missed that the outer square has side length $2$ rather than $1$.
$endgroup$
– Henning Makholm
May 6 at 13:05
1
$begingroup$
Sounds like the start of a joke...
$endgroup$
– Asaf Karagila♦
May 7 at 12:43
$begingroup$
$2/9$ seems to be very much too large simply from eyeballing the figure. Should it be the diameter of the circle rather than its radius?
$endgroup$
– Henning Makholm
May 6 at 11:41
$begingroup$
$2/9$ seems to be very much too large simply from eyeballing the figure. Should it be the diameter of the circle rather than its radius?
$endgroup$
– Henning Makholm
May 6 at 11:41
$begingroup$
@HenningMakholm: I've confirmed that $2/9$ is correct. I submit that it also looks plausible. The diameter of the circle looks to be slightly less than the radius of the semicircle, and the diameter of the semicircle looks to be about half the side of the square (in fact, it's exactly half). So, the radius of the circle should be just shy of $1/4=0.25$; and $2/9=0.222ldots$ is in that ballpark.
$endgroup$
– Blue
May 6 at 12:00
$begingroup$
@HenningMakholm: I've confirmed that $2/9$ is correct. I submit that it also looks plausible. The diameter of the circle looks to be slightly less than the radius of the semicircle, and the diameter of the semicircle looks to be about half the side of the square (in fact, it's exactly half). So, the radius of the circle should be just shy of $1/4=0.25$; and $2/9=0.222ldots$ is in that ballpark.
$endgroup$
– Blue
May 6 at 12:00
2
2
$begingroup$
@Blue: Ah, sorry, I missed that the outer square has side length $2$ rather than $1$.
$endgroup$
– Henning Makholm
May 6 at 13:05
$begingroup$
@Blue: Ah, sorry, I missed that the outer square has side length $2$ rather than $1$.
$endgroup$
– Henning Makholm
May 6 at 13:05
1
1
$begingroup$
Sounds like the start of a joke...
$endgroup$
– Asaf Karagila♦
May 7 at 12:43
$begingroup$
Sounds like the start of a joke...
$endgroup$
– Asaf Karagila♦
May 7 at 12:43
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Look at the picture:
From $triangle ABE$ we have $(2+r)^2= 2^2+(2-r)^2$ so $r=1/2$. From $square ECGF$ we have $CG^2=(1/2+s)^2-(1/2-s)^2= 2s$. From $square ADGF$ we have $GD^2= (2+s)^2-(2-s)^2= 8s$. So $2=CG+GD=3sqrt 2sqrt s$, hence $s=2/9$.
answered May 6 at 12:15
SMMSMM
3,678513
$endgroup$
1
$begingroup$
It's surprising to me that the length of the square's sides are an integer multiple of the circle's radius.
$endgroup$
– BlueRaja - Danny Pflughoeft
May 6 at 16:10
$begingroup$
Could you explain the calculation of $CG^2$ and $GD^2$? What principle are you invoking here? You appear to have applied some formula (perhaps some standard formula that applies to trapezia), but it is unknown to me.
$endgroup$
– Hammerite
May 6 at 16:26
1
$begingroup$
@Hammerite If $H$ is the foot of the perpendicular from $F$ to $EC$, then $CG=FH$, and then apply Pythagorean theorem on $triangle FHE$ to find $FH$. And the same trick for $GD$.
$endgroup$
– SMM
May 6 at 16:37
$begingroup$
@BlueRaja-DannyPflughoeft Yes, in general $r=a/4$ and $s=a/9$, where $a$ is the side of the square.
$endgroup$
– SMM
May 6 at 16:46
$begingroup$
@SMM: Yes, those values are obvious from this answer, but that gives no intuition as to why the multiple should be an integer. I think Blue's answer below gives that, though.
$endgroup$
– BlueRaja - Danny Pflughoeft
May 6 at 21:17
add a comment |
$begingroup$
@SMM's proof is nicely self-contained. Here's one that invokes the Descartes "Kissing Circles" theorem, simply because everyone should be aware of that result.
Let the quarter-, semi-, and full-circles have radius $a$, $b$, $c$, respectively.
From the right triangle, we have
$$a^2+(a-b)^2=(a+b)^2 quadtoquad a=4b tag{1}$$
Considering the side of the square a circle of curvature $0$, that special case of the Kissing Circles theorem implies
$$frac1{c} = frac{1}{a}+frac{1}{b}pm 2sqrt{frac{1}{a}cdotfrac{1}{b}} = frac{5}{4b}pm 2sqrt{frac{1}{4b^2}} = frac{5pm 4}{4b}quadtoquad c = frac49 b ;text{or}; 4b;text{(extraneous}) tag{2}$$
Then, with $a=2$, we have $b=1/2$, so that $c=2/9$. $square$
answered May 6 at 12:33
BlueBlue
50.7k1070162
$endgroup$
add a comment |
$begingroup$
let the side of the square be $a$.
Let's find the radius x of the semicircle
We have $$(a+x)^2 = a^2 + (a-x)^2$$
$$ x=frac{a}{4} $$
Now, a lemma.
If circles of radiuses R and r are touching externally, then the length of their common tangent is $2sqrt{Rr}$
Proof of the lemma: draw the common tangent and radiuses as in the figure. There is a right trapezium, so we get $(R+r)^2 = h^2 + (R-r)^2$, from where $h = 2 sqrt{Rr}$.
Now, let's use the lemma. Let $y$ be the radius of the small circle.
We have $$a = 2sqrt{ay} + 2 sqrt{frac{a}{4}y}$$
$$sqrt{a} = 3 sqrt{y} $$
$$y = frac{a}{9}$$
answered May 6 at 12:19
liaombroliaombro
1,086718
$endgroup$
$begingroup$
I like this one because it taught me something useful.
$endgroup$
– richard1941
May 7 at 19:03
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Look at the picture:
From $triangle ABE$ we have $(2+r)^2= 2^2+(2-r)^2$ so $r=1/2$. From $square ECGF$ we have $CG^2=(1/2+s)^2-(1/2-s)^2= 2s$. From $square ADGF$ we have $GD^2= (2+s)^2-(2-s)^2= 8s$. So $2=CG+GD=3sqrt 2sqrt s$, hence $s=2/9$.
answered May 6 at 12:15
SMMSMM
3,678513
$endgroup$
1
$begingroup$
It's surprising to me that the length of the square's sides are an integer multiple of the circle's radius.
$endgroup$
– BlueRaja - Danny Pflughoeft
May 6 at 16:10
$begingroup$
Could you explain the calculation of $CG^2$ and $GD^2$? What principle are you invoking here? You appear to have applied some formula (perhaps some standard formula that applies to trapezia), but it is unknown to me.
$endgroup$
– Hammerite
May 6 at 16:26
1
$begingroup$
@Hammerite If $H$ is the foot of the perpendicular from $F$ to $EC$, then $CG=FH$, and then apply Pythagorean theorem on $triangle FHE$ to find $FH$. And the same trick for $GD$.
$endgroup$
– SMM
May 6 at 16:37
$begingroup$
@BlueRaja-DannyPflughoeft Yes, in general $r=a/4$ and $s=a/9$, where $a$ is the side of the square.
$endgroup$
– SMM
May 6 at 16:46
$begingroup$
@SMM: Yes, those values are obvious from this answer, but that gives no intuition as to why the multiple should be an integer. I think Blue's answer below gives that, though.
$endgroup$
– BlueRaja - Danny Pflughoeft
May 6 at 21:17
add a comment |
$begingroup$
Look at the picture:
From $triangle ABE$ we have $(2+r)^2= 2^2+(2-r)^2$ so $r=1/2$. From $square ECGF$ we have $CG^2=(1/2+s)^2-(1/2-s)^2= 2s$. From $square ADGF$ we have $GD^2= (2+s)^2-(2-s)^2= 8s$. So $2=CG+GD=3sqrt 2sqrt s$, hence $s=2/9$.
answered May 6 at 12:15
SMMSMM
3,678513
$endgroup$
1
$begingroup$
It's surprising to me that the length of the square's sides are an integer multiple of the circle's radius.
$endgroup$
– BlueRaja - Danny Pflughoeft
May 6 at 16:10
$begingroup$
Could you explain the calculation of $CG^2$ and $GD^2$? What principle are you invoking here? You appear to have applied some formula (perhaps some standard formula that applies to trapezia), but it is unknown to me.
$endgroup$
– Hammerite
May 6 at 16:26
1
$begingroup$
@Hammerite If $H$ is the foot of the perpendicular from $F$ to $EC$, then $CG=FH$, and then apply Pythagorean theorem on $triangle FHE$ to find $FH$. And the same trick for $GD$.
$endgroup$
– SMM
May 6 at 16:37
$begingroup$
@BlueRaja-DannyPflughoeft Yes, in general $r=a/4$ and $s=a/9$, where $a$ is the side of the square.
$endgroup$
– SMM
May 6 at 16:46
$begingroup$
@SMM: Yes, those values are obvious from this answer, but that gives no intuition as to why the multiple should be an integer. I think Blue's answer below gives that, though.
$endgroup$
– BlueRaja - Danny Pflughoeft
May 6 at 21:17
add a comment |
$begingroup$
Look at the picture:
From $triangle ABE$ we have $(2+r)^2= 2^2+(2-r)^2$ so $r=1/2$. From $square ECGF$ we have $CG^2=(1/2+s)^2-(1/2-s)^2= 2s$. From $square ADGF$ we have $GD^2= (2+s)^2-(2-s)^2= 8s$. So $2=CG+GD=3sqrt 2sqrt s$, hence $s=2/9$.
answered May 6 at 12:15
SMMSMM
3,678513
$endgroup$
Look at the picture:
From $triangle ABE$ we have $(2+r)^2= 2^2+(2-r)^2$ so $r=1/2$. From $square ECGF$ we have $CG^2=(1/2+s)^2-(1/2-s)^2= 2s$. From $square ADGF$ we have $GD^2= (2+s)^2-(2-s)^2= 8s$. So $2=CG+GD=3sqrt 2sqrt s$, hence $s=2/9$.
answered May 6 at 12:15
SMMSMM
3,678513
answered May 6 at 12:15
SMMSMM
3,678513
answered May 6 at 12:15
SMMSMM
3,678513
answered May 6 at 12:15
SMMSMM
3,678513
3,678513
1
$begingroup$
It's surprising to me that the length of the square's sides are an integer multiple of the circle's radius.
$endgroup$
– BlueRaja - Danny Pflughoeft
May 6 at 16:10
$begingroup$
Could you explain the calculation of $CG^2$ and $GD^2$? What principle are you invoking here? You appear to have applied some formula (perhaps some standard formula that applies to trapezia), but it is unknown to me.
$endgroup$
– Hammerite
May 6 at 16:26
1
$begingroup$
@Hammerite If $H$ is the foot of the perpendicular from $F$ to $EC$, then $CG=FH$, and then apply Pythagorean theorem on $triangle FHE$ to find $FH$. And the same trick for $GD$.
$endgroup$
– SMM
May 6 at 16:37
$begingroup$
@BlueRaja-DannyPflughoeft Yes, in general $r=a/4$ and $s=a/9$, where $a$ is the side of the square.
$endgroup$
– SMM
May 6 at 16:46
$begingroup$
@SMM: Yes, those values are obvious from this answer, but that gives no intuition as to why the multiple should be an integer. I think Blue's answer below gives that, though.
$endgroup$
– BlueRaja - Danny Pflughoeft
May 6 at 21:17
add a comment |
1
$begingroup$
It's surprising to me that the length of the square's sides are an integer multiple of the circle's radius.
$endgroup$
– BlueRaja - Danny Pflughoeft
May 6 at 16:10
$begingroup$
Could you explain the calculation of $CG^2$ and $GD^2$? What principle are you invoking here? You appear to have applied some formula (perhaps some standard formula that applies to trapezia), but it is unknown to me.
$endgroup$
– Hammerite
May 6 at 16:26
1
$begingroup$
@Hammerite If $H$ is the foot of the perpendicular from $F$ to $EC$, then $CG=FH$, and then apply Pythagorean theorem on $triangle FHE$ to find $FH$. And the same trick for $GD$.
$endgroup$
– SMM
May 6 at 16:37
$begingroup$
@BlueRaja-DannyPflughoeft Yes, in general $r=a/4$ and $s=a/9$, where $a$ is the side of the square.
$endgroup$
– SMM
May 6 at 16:46
$begingroup$
@SMM: Yes, those values are obvious from this answer, but that gives no intuition as to why the multiple should be an integer. I think Blue's answer below gives that, though.
$endgroup$
– BlueRaja - Danny Pflughoeft
May 6 at 21:17
1
1
$begingroup$
It's surprising to me that the length of the square's sides are an integer multiple of the circle's radius.
$endgroup$
– BlueRaja - Danny Pflughoeft
May 6 at 16:10
$begingroup$
It's surprising to me that the length of the square's sides are an integer multiple of the circle's radius.
$endgroup$
– BlueRaja - Danny Pflughoeft
May 6 at 16:10
$begingroup$
Could you explain the calculation of $CG^2$ and $GD^2$? What principle are you invoking here? You appear to have applied some formula (perhaps some standard formula that applies to trapezia), but it is unknown to me.
$endgroup$
– Hammerite
May 6 at 16:26
$begingroup$
Could you explain the calculation of $CG^2$ and $GD^2$? What principle are you invoking here? You appear to have applied some formula (perhaps some standard formula that applies to trapezia), but it is unknown to me.
$endgroup$
– Hammerite
May 6 at 16:26
1
1
$begingroup$
@Hammerite If $H$ is the foot of the perpendicular from $F$ to $EC$, then $CG=FH$, and then apply Pythagorean theorem on $triangle FHE$ to find $FH$. And the same trick for $GD$.
$endgroup$
– SMM
May 6 at 16:37
$begingroup$
@Hammerite If $H$ is the foot of the perpendicular from $F$ to $EC$, then $CG=FH$, and then apply Pythagorean theorem on $triangle FHE$ to find $FH$. And the same trick for $GD$.
$endgroup$
– SMM
May 6 at 16:37
$begingroup$
@BlueRaja-DannyPflughoeft Yes, in general $r=a/4$ and $s=a/9$, where $a$ is the side of the square.
$endgroup$
– SMM
May 6 at 16:46
$begingroup$
@BlueRaja-DannyPflughoeft Yes, in general $r=a/4$ and $s=a/9$, where $a$ is the side of the square.
$endgroup$
– SMM
May 6 at 16:46
$begingroup$
@SMM: Yes, those values are obvious from this answer, but that gives no intuition as to why the multiple should be an integer. I think Blue's answer below gives that, though.
$endgroup$
– BlueRaja - Danny Pflughoeft
May 6 at 21:17
$begingroup$
@SMM: Yes, those values are obvious from this answer, but that gives no intuition as to why the multiple should be an integer. I think Blue's answer below gives that, though.
$endgroup$
– BlueRaja - Danny Pflughoeft
May 6 at 21:17
add a comment |
$begingroup$
@SMM's proof is nicely self-contained. Here's one that invokes the Descartes "Kissing Circles" theorem, simply because everyone should be aware of that result.
Let the quarter-, semi-, and full-circles have radius $a$, $b$, $c$, respectively.
From the right triangle, we have
$$a^2+(a-b)^2=(a+b)^2 quadtoquad a=4b tag{1}$$
Considering the side of the square a circle of curvature $0$, that special case of the Kissing Circles theorem implies
$$frac1{c} = frac{1}{a}+frac{1}{b}pm 2sqrt{frac{1}{a}cdotfrac{1}{b}} = frac{5}{4b}pm 2sqrt{frac{1}{4b^2}} = frac{5pm 4}{4b}quadtoquad c = frac49 b ;text{or}; 4b;text{(extraneous}) tag{2}$$
Then, with $a=2$, we have $b=1/2$, so that $c=2/9$. $square$
answered May 6 at 12:33
BlueBlue
50.7k1070162
$endgroup$
add a comment |
$begingroup$
@SMM's proof is nicely self-contained. Here's one that invokes the Descartes "Kissing Circles" theorem, simply because everyone should be aware of that result.
Let the quarter-, semi-, and full-circles have radius $a$, $b$, $c$, respectively.
From the right triangle, we have
$$a^2+(a-b)^2=(a+b)^2 quadtoquad a=4b tag{1}$$
Considering the side of the square a circle of curvature $0$, that special case of the Kissing Circles theorem implies
$$frac1{c} = frac{1}{a}+frac{1}{b}pm 2sqrt{frac{1}{a}cdotfrac{1}{b}} = frac{5}{4b}pm 2sqrt{frac{1}{4b^2}} = frac{5pm 4}{4b}quadtoquad c = frac49 b ;text{or}; 4b;text{(extraneous}) tag{2}$$
Then, with $a=2$, we have $b=1/2$, so that $c=2/9$. $square$
answered May 6 at 12:33
BlueBlue
50.7k1070162
$endgroup$
add a comment |
$begingroup$
@SMM's proof is nicely self-contained. Here's one that invokes the Descartes "Kissing Circles" theorem, simply because everyone should be aware of that result.
Let the quarter-, semi-, and full-circles have radius $a$, $b$, $c$, respectively.
From the right triangle, we have
$$a^2+(a-b)^2=(a+b)^2 quadtoquad a=4b tag{1}$$
Considering the side of the square a circle of curvature $0$, that special case of the Kissing Circles theorem implies
$$frac1{c} = frac{1}{a}+frac{1}{b}pm 2sqrt{frac{1}{a}cdotfrac{1}{b}} = frac{5}{4b}pm 2sqrt{frac{1}{4b^2}} = frac{5pm 4}{4b}quadtoquad c = frac49 b ;text{or}; 4b;text{(extraneous}) tag{2}$$
Then, with $a=2$, we have $b=1/2$, so that $c=2/9$. $square$
answered May 6 at 12:33
BlueBlue
50.7k1070162
$endgroup$
@SMM's proof is nicely self-contained. Here's one that invokes the Descartes "Kissing Circles" theorem, simply because everyone should be aware of that result.
Let the quarter-, semi-, and full-circles have radius $a$, $b$, $c$, respectively.
From the right triangle, we have
$$a^2+(a-b)^2=(a+b)^2 quadtoquad a=4b tag{1}$$
Considering the side of the square a circle of curvature $0$, that special case of the Kissing Circles theorem implies
$$frac1{c} = frac{1}{a}+frac{1}{b}pm 2sqrt{frac{1}{a}cdotfrac{1}{b}} = frac{5}{4b}pm 2sqrt{frac{1}{4b^2}} = frac{5pm 4}{4b}quadtoquad c = frac49 b ;text{or}; 4b;text{(extraneous}) tag{2}$$
Then, with $a=2$, we have $b=1/2$, so that $c=2/9$. $square$
answered May 6 at 12:33
BlueBlue
50.7k1070162
answered May 6 at 12:33
BlueBlue
50.7k1070162
answered May 6 at 12:33
BlueBlue
50.7k1070162
answered May 6 at 12:33
BlueBlue
50.7k1070162
50.7k1070162
add a comment |
add a comment |
$begingroup$
let the side of the square be $a$.
Let's find the radius x of the semicircle
We have $$(a+x)^2 = a^2 + (a-x)^2$$
$$ x=frac{a}{4} $$
Now, a lemma.
If circles of radiuses R and r are touching externally, then the length of their common tangent is $2sqrt{Rr}$
Proof of the lemma: draw the common tangent and radiuses as in the figure. There is a right trapezium, so we get $(R+r)^2 = h^2 + (R-r)^2$, from where $h = 2 sqrt{Rr}$.
Now, let's use the lemma. Let $y$ be the radius of the small circle.
We have $$a = 2sqrt{ay} + 2 sqrt{frac{a}{4}y}$$
$$sqrt{a} = 3 sqrt{y} $$
$$y = frac{a}{9}$$
answered May 6 at 12:19
liaombroliaombro
1,086718
$endgroup$
$begingroup$
I like this one because it taught me something useful.
$endgroup$
– richard1941
May 7 at 19:03
add a comment |
$begingroup$
let the side of the square be $a$.
Let's find the radius x of the semicircle
We have $$(a+x)^2 = a^2 + (a-x)^2$$
$$ x=frac{a}{4} $$
Now, a lemma.
If circles of radiuses R and r are touching externally, then the length of their common tangent is $2sqrt{Rr}$
Proof of the lemma: draw the common tangent and radiuses as in the figure. There is a right trapezium, so we get $(R+r)^2 = h^2 + (R-r)^2$, from where $h = 2 sqrt{Rr}$.
Now, let's use the lemma. Let $y$ be the radius of the small circle.
We have $$a = 2sqrt{ay} + 2 sqrt{frac{a}{4}y}$$
$$sqrt{a} = 3 sqrt{y} $$
$$y = frac{a}{9}$$
answered May 6 at 12:19
liaombroliaombro
1,086718
$endgroup$
$begingroup$
I like this one because it taught me something useful.
$endgroup$
– richard1941
May 7 at 19:03
add a comment |
$begingroup$
let the side of the square be $a$.
Let's find the radius x of the semicircle
We have $$(a+x)^2 = a^2 + (a-x)^2$$
$$ x=frac{a}{4} $$
Now, a lemma.
If circles of radiuses R and r are touching externally, then the length of their common tangent is $2sqrt{Rr}$
Proof of the lemma: draw the common tangent and radiuses as in the figure. There is a right trapezium, so we get $(R+r)^2 = h^2 + (R-r)^2$, from where $h = 2 sqrt{Rr}$.
Now, let's use the lemma. Let $y$ be the radius of the small circle.
We have $$a = 2sqrt{ay} + 2 sqrt{frac{a}{4}y}$$
$$sqrt{a} = 3 sqrt{y} $$
$$y = frac{a}{9}$$
answered May 6 at 12:19
liaombroliaombro
1,086718
$endgroup$
let the side of the square be $a$.
Let's find the radius x of the semicircle
We have $$(a+x)^2 = a^2 + (a-x)^2$$
$$ x=frac{a}{4} $$
Now, a lemma.
If circles of radiuses R and r are touching externally, then the length of their common tangent is $2sqrt{Rr}$
Proof of the lemma: draw the common tangent and radiuses as in the figure. There is a right trapezium, so we get $(R+r)^2 = h^2 + (R-r)^2$, from where $h = 2 sqrt{Rr}$.
Now, let's use the lemma. Let $y$ be the radius of the small circle.
We have $$a = 2sqrt{ay} + 2 sqrt{frac{a}{4}y}$$
$$sqrt{a} = 3 sqrt{y} $$
$$y = frac{a}{9}$$
answered May 6 at 12:19
liaombroliaombro
1,086718
answered May 6 at 12:19
liaombroliaombro
1,086718
answered May 6 at 12:19
liaombroliaombro
1,086718
answered May 6 at 12:19
liaombroliaombro
1,086718
1,086718
$begingroup$
I like this one because it taught me something useful.
$endgroup$
– richard1941
May 7 at 19:03
add a comment |
$begingroup$
I like this one because it taught me something useful.
$endgroup$
– richard1941
May 7 at 19:03
$begingroup$
I like this one because it taught me something useful.
$endgroup$
– richard1941
May 7 at 19:03
$begingroup$
I like this one because it taught me something useful.
$endgroup$
– richard1941
May 7 at 19:03
add a comment |
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$begingroup$
$2/9$ seems to be very much too large simply from eyeballing the figure. Should it be the diameter of the circle rather than its radius?
$endgroup$
– Henning Makholm
May 6 at 11:41
$begingroup$
@HenningMakholm: I've confirmed that $2/9$ is correct. I submit that it also looks plausible. The diameter of the circle looks to be slightly less than the radius of the semicircle, and the diameter of the semicircle looks to be about half the side of the square (in fact, it's exactly half). So, the radius of the circle should be just shy of $1/4=0.25$; and $2/9=0.222ldots$ is in that ballpark.
$endgroup$
– Blue
May 6 at 12:00
2
$begingroup$
@Blue: Ah, sorry, I missed that the outer square has side length $2$ rather than $1$.
$endgroup$
– Henning Makholm
May 6 at 13:05
1
$begingroup$
Sounds like the start of a joke...
$endgroup$
– Asaf Karagila♦
May 7 at 12:43