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minor formatting fix – making parens around a fraction big enough and a middle fraction line a bit longer
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The fraction is

$\frac{2}{3}$

Without loss of generality let the radius of the smaller circle equal 1.enter image description here




Edit:

Bubbler has observed in a comment that from here it is faster to continue like this enter image description here

We have $JA=JD=r$, and since JH bisects the chord AB, $\angle{JHA}=90^\circ$.

We also have

$$JA^2=HA^2+(HI^2+IJ^2)$$ $$r^2=1^2+(1^2+(1+\sqrt{3}-r)^2)$$ Solving for r, $$r^2=1^2+1^2+r^2-2 \sqrt{3} r-2 r+2 \sqrt{3}+4$$ $$r=\frac{6+2\sqrt{3}}{2+2\sqrt{3}}=\sqrt{3}$$




Since $\angle HIG=90^\circ$ and $\arcsin(\frac{1}{2})=30^\circ$, we haveenter image description here

By the pythagorean theorem, ${HD}^2=HI^2+ID^2=1^2+(1+\sqrt{3})^2=5+2 \sqrt{3}$enter image description here

Now label the centre of the large circle J. We have $JA=JD=r$, and since JH bisects the chord AB, $\angle JHA=90^\circ $.enter image description here By the pythagorean theorem $$JH^2+1^2=r^2$$

And by the cosine rule $$JH^2=JD^2+HD^2-2\cdot HD\cdot JD\cdot\cos(IDH)$$ $$JH^2=(r^2)+(5+2\sqrt{3})-2\cdot r\cdot \sqrt{5+2\sqrt{3}} \cos(IDH)$$ Combining these 2 equations, $$JH^2=r^2-1^2=(r^2)+(5+2\sqrt{3})-2\cdot r\cdot \sqrt{5+2\sqrt{3}} \cos(IDH)$$ Solving for r produces $$r=\frac{6+2\sqrt{3}}{2\sqrt{5+2\sqrt{3}}\cdot \cos(IDH)}$$ We can use the pythagorean theorem to show that $\cos(IDH)=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{5+2\sqrt{3}}}$ which now gives $$r=\frac{6+2\sqrt{3}}{2\sqrt{5+2\sqrt{3}}\cdot \frac{\sqrt{4+2\sqrt{3}}}{\sqrt{5+2\sqrt{3}}}}=\frac{6+2\sqrt{3}}{2\sqrt{4+2\sqrt{3}}}$$ Squaring both sides,

$$r^2=(\frac{6+2\sqrt{3}}{2\sqrt{4+2\sqrt{3}}})^2=\frac{48+24\sqrt{3}}{4(4+2\sqrt{3})}=3$$$$r^2=\left(\frac{6+2\sqrt{3}}{2\sqrt{4+2\sqrt{3}}}\right)^2=\frac{48+24\sqrt{3}}{4(4+2\sqrt{3})}=3$$ $$r^2=3$$ $$r=\sqrt{3}$$

And the ratio of the areas is

$$\frac{\frac{2\cdot \pi \cdot (1^2)}{2}}{\frac{\pi \cdot (\sqrt{3})^2}{2}}=\frac{2}{3}$$$$\frac{\ \frac{2\cdot \pi \cdot (1^2)}{2}\ }{\ \frac{\pi \cdot (\sqrt{3})^2}{2}\ }=\frac{2}{3}$$

The fraction is

$\frac{2}{3}$

Without loss of generality let the radius of the smaller circle equal 1.enter image description here




Edit:

Bubbler has observed in a comment that from here it is faster to continue like this enter image description here

We have $JA=JD=r$, and since JH bisects the chord AB, $\angle{JHA}=90^\circ$.

We also have

$$JA^2=HA^2+(HI^2+IJ^2)$$ $$r^2=1^2+(1^2+(1+\sqrt{3}-r)^2)$$ Solving for r, $$r^2=1^2+1^2+r^2-2 \sqrt{3} r-2 r+2 \sqrt{3}+4$$ $$r=\frac{6+2\sqrt{3}}{2+2\sqrt{3}}=\sqrt{3}$$




Since $\angle HIG=90^\circ$ and $\arcsin(\frac{1}{2})=30^\circ$, we haveenter image description here

By the pythagorean theorem, ${HD}^2=HI^2+ID^2=1^2+(1+\sqrt{3})^2=5+2 \sqrt{3}$enter image description here

Now label the centre of the large circle J. We have $JA=JD=r$, and since JH bisects the chord AB, $\angle JHA=90^\circ $.enter image description here By the pythagorean theorem $$JH^2+1^2=r^2$$

And by the cosine rule $$JH^2=JD^2+HD^2-2\cdot HD\cdot JD\cdot\cos(IDH)$$ $$JH^2=(r^2)+(5+2\sqrt{3})-2\cdot r\cdot \sqrt{5+2\sqrt{3}} \cos(IDH)$$ Combining these 2 equations, $$JH^2=r^2-1^2=(r^2)+(5+2\sqrt{3})-2\cdot r\cdot \sqrt{5+2\sqrt{3}} \cos(IDH)$$ Solving for r produces $$r=\frac{6+2\sqrt{3}}{2\sqrt{5+2\sqrt{3}}\cdot \cos(IDH)}$$ We can use the pythagorean theorem to show that $\cos(IDH)=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{5+2\sqrt{3}}}$ which now gives $$r=\frac{6+2\sqrt{3}}{2\sqrt{5+2\sqrt{3}}\cdot \frac{\sqrt{4+2\sqrt{3}}}{\sqrt{5+2\sqrt{3}}}}=\frac{6+2\sqrt{3}}{2\sqrt{4+2\sqrt{3}}}$$ Squaring both sides,

$$r^2=(\frac{6+2\sqrt{3}}{2\sqrt{4+2\sqrt{3}}})^2=\frac{48+24\sqrt{3}}{4(4+2\sqrt{3})}=3$$ $$r^2=3$$ $$r=\sqrt{3}$$

And the ratio of the areas is

$$\frac{\frac{2\cdot \pi \cdot (1^2)}{2}}{\frac{\pi \cdot (\sqrt{3})^2}{2}}=\frac{2}{3}$$

The fraction is

$\frac{2}{3}$

Without loss of generality let the radius of the smaller circle equal 1.enter image description here




Edit:

Bubbler has observed in a comment that from here it is faster to continue like this enter image description here

We have $JA=JD=r$, and since JH bisects the chord AB, $\angle{JHA}=90^\circ$.

We also have

$$JA^2=HA^2+(HI^2+IJ^2)$$ $$r^2=1^2+(1^2+(1+\sqrt{3}-r)^2)$$ Solving for r, $$r^2=1^2+1^2+r^2-2 \sqrt{3} r-2 r+2 \sqrt{3}+4$$ $$r=\frac{6+2\sqrt{3}}{2+2\sqrt{3}}=\sqrt{3}$$




Since $\angle HIG=90^\circ$ and $\arcsin(\frac{1}{2})=30^\circ$, we haveenter image description here

By the pythagorean theorem, ${HD}^2=HI^2+ID^2=1^2+(1+\sqrt{3})^2=5+2 \sqrt{3}$enter image description here

Now label the centre of the large circle J. We have $JA=JD=r$, and since JH bisects the chord AB, $\angle JHA=90^\circ $.enter image description here By the pythagorean theorem $$JH^2+1^2=r^2$$

And by the cosine rule $$JH^2=JD^2+HD^2-2\cdot HD\cdot JD\cdot\cos(IDH)$$ $$JH^2=(r^2)+(5+2\sqrt{3})-2\cdot r\cdot \sqrt{5+2\sqrt{3}} \cos(IDH)$$ Combining these 2 equations, $$JH^2=r^2-1^2=(r^2)+(5+2\sqrt{3})-2\cdot r\cdot \sqrt{5+2\sqrt{3}} \cos(IDH)$$ Solving for r produces $$r=\frac{6+2\sqrt{3}}{2\sqrt{5+2\sqrt{3}}\cdot \cos(IDH)}$$ We can use the pythagorean theorem to show that $\cos(IDH)=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{5+2\sqrt{3}}}$ which now gives $$r=\frac{6+2\sqrt{3}}{2\sqrt{5+2\sqrt{3}}\cdot \frac{\sqrt{4+2\sqrt{3}}}{\sqrt{5+2\sqrt{3}}}}=\frac{6+2\sqrt{3}}{2\sqrt{4+2\sqrt{3}}}$$ Squaring both sides,

$$r^2=\left(\frac{6+2\sqrt{3}}{2\sqrt{4+2\sqrt{3}}}\right)^2=\frac{48+24\sqrt{3}}{4(4+2\sqrt{3})}=3$$ $$r^2=3$$ $$r=\sqrt{3}$$

And the ratio of the areas is

$$\frac{\ \frac{2\cdot \pi \cdot (1^2)}{2}\ }{\ \frac{\pi \cdot (\sqrt{3})^2}{2}\ }=\frac{2}{3}$$

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user37842
user37842

The fraction is

$\frac{2}{3}$

Without loss of generality let the radius of the smaller circle equal 1.enter image description here




Edit:

Bubbler has observed in a comment that from here it is faster to continue like this enter image description here

We have $JA=JD=r$, and since JH bisects the chord AB, $\angle{JHA}=90^\circ$.

We also have

$$JA^2=HA^2+(HI^2+IJ^2)$$ $$r^2=1^2+(1^2+(1+\sqrt{3}-r)^2)$$ Solving for r, $$r^2=1^2+1^2+r^2-2 \sqrt{3} r-2 r+2 \sqrt{3}+4$$ $$r=\frac{6+2\sqrt{3}}{2+2\sqrt{3}}=\sqrt{3}$$




Since $\angle HIG=90^\circ$ and $\arcsin(\frac{1}{2})=30^\circ$, we haveenter image description here

By the pythagorean theorem, ${HD}^2=HI^2+ID^2=1^2+(1+\sqrt{3})^2=5+2 \sqrt{3}$enter image description here

Now label the centre of the large circle J. We have $JA=JD=r$, and since JH bisects the chord AB, $\angle JHA=90^\circ $.enter image description here By the pythagorean theorem $$JH^2+1^2=r^2$$

And by the cosine rule $$JH^2=JD^2+HD^2-2\cdot HD\cdot JD\cdot\cos(IDH)$$ $$JH^2=(r^2)+(5+2\sqrt{3})-2\cdot r\cdot \sqrt{5+2\sqrt{3}} \cos(IDH)$$ Combining these 2 equations, $$JH^2=r^2-1^2=(r^2)+(5+2\sqrt{3})-2\cdot r\cdot \sqrt{5+2\sqrt{3}} \cos(IDH)$$ Solving for r produces $$r=\frac{6+2\sqrt{3}}{2\sqrt{5+2\sqrt{3}}\cdot \cos(IDH)}$$ We can use the pythagorean theorem to show that $\cos(IDH)=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{5+2\sqrt{3}}}$ which now gives $$r=\frac{6+2\sqrt{3}}{2\sqrt{5+2\sqrt{3}}\cdot \frac{\sqrt{4+2\sqrt{3}}}{\sqrt{5+2\sqrt{3}}}}=\frac{6+2\sqrt{3}}{2\sqrt{4+2\sqrt{3}}}$$ Squaring both sides,

$$r^2=(\frac{6+2\sqrt{3}}{2\sqrt{4+2\sqrt{3}}})^2=\frac{48+24\sqrt{3}}{4(4+2\sqrt{3})}=3$$ $$r^2=3$$ $$r=\sqrt{3}$$

And the ratio of the areas is

$$\frac{\frac{2\cdot \pi \cdot (1^2)}{2}}{\frac{\pi \cdot (\sqrt{3})^2}{2}}=\frac{2}{3}$$

The fraction is

$\frac{2}{3}$

Without loss of generality let the radius of the smaller circle equal 1.enter image description here

Since $\angle HIG=90^\circ$ and $\arcsin(\frac{1}{2})=30^\circ$, we haveenter image description here

By the pythagorean theorem, ${HD}^2=HI^2+ID^2=1^2+(1+\sqrt{3})^2=5+2 \sqrt{3}$enter image description here

Now label the centre of the large circle J. We have $JA=JD=r$, and since JH bisects the chord AB, $\angle JHA=90^\circ $.enter image description here By the pythagorean theorem $$JH^2+1^2=r^2$$

And by the cosine rule $$JH^2=JD^2+HD^2-2\cdot HD\cdot JD\cdot\cos(IDH)$$ $$JH^2=(r^2)+(5+2\sqrt{3})-2\cdot r\cdot \sqrt{5+2\sqrt{3}} \cos(IDH)$$ Combining these 2 equations, $$JH^2=r^2-1^2=(r^2)+(5+2\sqrt{3})-2\cdot r\cdot \sqrt{5+2\sqrt{3}} \cos(IDH)$$ Solving for r produces $$r=\frac{6+2\sqrt{3}}{2\sqrt{5+2\sqrt{3}}\cdot \cos(IDH)}$$ We can use the pythagorean theorem to show that $\cos(IDH)=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{5+2\sqrt{3}}}$ which now gives $$r=\frac{6+2\sqrt{3}}{2\sqrt{5+2\sqrt{3}}\cdot \frac{\sqrt{4+2\sqrt{3}}}{\sqrt{5+2\sqrt{3}}}}=\frac{6+2\sqrt{3}}{2\sqrt{4+2\sqrt{3}}}$$ Squaring both sides,

$$r^2=(\frac{6+2\sqrt{3}}{2\sqrt{4+2\sqrt{3}}})^2=\frac{48+24\sqrt{3}}{4(4+2\sqrt{3})}=3$$ $$r^2=3$$ $$r=\sqrt{3}$$

And the ratio of the areas is

$$\frac{\frac{2\cdot \pi \cdot (1^2)}{2}}{\frac{\pi \cdot (\sqrt{3})^2}{2}}=\frac{2}{3}$$

The fraction is

$\frac{2}{3}$

Without loss of generality let the radius of the smaller circle equal 1.enter image description here




Edit:

Bubbler has observed in a comment that from here it is faster to continue like this enter image description here

We have $JA=JD=r$, and since JH bisects the chord AB, $\angle{JHA}=90^\circ$.

We also have

$$JA^2=HA^2+(HI^2+IJ^2)$$ $$r^2=1^2+(1^2+(1+\sqrt{3}-r)^2)$$ Solving for r, $$r^2=1^2+1^2+r^2-2 \sqrt{3} r-2 r+2 \sqrt{3}+4$$ $$r=\frac{6+2\sqrt{3}}{2+2\sqrt{3}}=\sqrt{3}$$




Since $\angle HIG=90^\circ$ and $\arcsin(\frac{1}{2})=30^\circ$, we haveenter image description here

By the pythagorean theorem, ${HD}^2=HI^2+ID^2=1^2+(1+\sqrt{3})^2=5+2 \sqrt{3}$enter image description here

Now label the centre of the large circle J. We have $JA=JD=r$, and since JH bisects the chord AB, $\angle JHA=90^\circ $.enter image description here By the pythagorean theorem $$JH^2+1^2=r^2$$

And by the cosine rule $$JH^2=JD^2+HD^2-2\cdot HD\cdot JD\cdot\cos(IDH)$$ $$JH^2=(r^2)+(5+2\sqrt{3})-2\cdot r\cdot \sqrt{5+2\sqrt{3}} \cos(IDH)$$ Combining these 2 equations, $$JH^2=r^2-1^2=(r^2)+(5+2\sqrt{3})-2\cdot r\cdot \sqrt{5+2\sqrt{3}} \cos(IDH)$$ Solving for r produces $$r=\frac{6+2\sqrt{3}}{2\sqrt{5+2\sqrt{3}}\cdot \cos(IDH)}$$ We can use the pythagorean theorem to show that $\cos(IDH)=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{5+2\sqrt{3}}}$ which now gives $$r=\frac{6+2\sqrt{3}}{2\sqrt{5+2\sqrt{3}}\cdot \frac{\sqrt{4+2\sqrt{3}}}{\sqrt{5+2\sqrt{3}}}}=\frac{6+2\sqrt{3}}{2\sqrt{4+2\sqrt{3}}}$$ Squaring both sides,

$$r^2=(\frac{6+2\sqrt{3}}{2\sqrt{4+2\sqrt{3}}})^2=\frac{48+24\sqrt{3}}{4(4+2\sqrt{3})}=3$$ $$r^2=3$$ $$r=\sqrt{3}$$

And the ratio of the areas is

$$\frac{\frac{2\cdot \pi \cdot (1^2)}{2}}{\frac{\pi \cdot (\sqrt{3})^2}{2}}=\frac{2}{3}$$

added 29 characters in body
Source Link
user37842
user37842

The fraction is

$\frac{2}{3}$

Without loss of generality let the radius of the smaller circle equal 1.enter image description here

Since $\angle HIG=90^\circ$ and $\arcsin(\frac{1}{2})=30$$\arcsin(\frac{1}{2})=30^\circ$, we haveenter image description here

By the cosine rulepythagorean theorem, ${HD}^2={1}^2+{2}^2-2\cdot 1\cdot 2\cdot\cos(150)=5+2\cdot \sqrt{3}$${HD}^2=HI^2+ID^2=1^2+(1+\sqrt{3})^2=5+2 \sqrt{3}$enter image description here

Now label the centre of the large circle J, and $\angle JHA=90 $. We also have $JA=JD=r$, and since JH bisects the chord AB, $\angle JHA=90^\circ $.enter image description here By the pythagorean theorem $$JH^2+1^2=r^2$$

And by the cosine rule $$JH^2=JD^2+HD^2-2\cdot HD\cdot JD\cdot\cos(IDH)$$ $$JH^2=(r^2)+(5+2\sqrt{3})-2\cdot r\cdot \sqrt{5+2\sqrt{3}} \cos(IDH)$$ Combining these 2 equations, $$JH^2=r^2-1^2=(r^2)+(5+2\sqrt{3})-2\cdot r\cdot \sqrt{5+2\sqrt{3}} \cos(IDH)$$ Solving for r produces $$r=\frac{6+2\sqrt{3}}{2\sqrt{5+2\sqrt{3}}\cdot \cos(IDH)}$$ We can use the pythagorean theorem to show that $\cos(IDH)=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{5+2\sqrt{3}}}$ which now gives $$r=\frac{6+2\sqrt{3}}{2\sqrt{5+2\sqrt{3}}\cdot \frac{\sqrt{4+2\sqrt{3}}}{\sqrt{5+2\sqrt{3}}}}=\frac{6+2\sqrt{3}}{2\sqrt{4+2\sqrt{3}}}$$ Squaring both sides,

$$r^2=(\frac{6+2\sqrt{3}}{2\sqrt{4+2\sqrt{3}}})^2=\frac{48+24\sqrt{3}}{4(4+2\sqrt{3})}=3$$ $$r^2=3$$ $$r=\sqrt{3}$$

And the ratio of the areas is

$$\frac{\frac{2\cdot \pi \cdot (1^2)}{2}}{\frac{\pi \cdot (\sqrt{3})^2}{2}}=\frac{2}{3}$$

The fraction is

$\frac{2}{3}$

Without loss of generality let the radius of the smaller circle equal 1.enter image description here

Since $\angle HIG=90^\circ$ and $\arcsin(\frac{1}{2})=30$, we haveenter image description here

By the cosine rule, ${HD}^2={1}^2+{2}^2-2\cdot 1\cdot 2\cdot\cos(150)=5+2\cdot \sqrt{3}$enter image description here

Now label the centre of the large circle J, and $\angle JHA=90 $. We also have $JA=JD=r$enter image description here By the pythagorean theorem $$JH^2+1^2=r^2$$

And by the cosine rule $$JH^2=JD^2+HD^2-2\cdot HD\cdot JD\cdot\cos(IDH)$$ $$JH^2=(r^2)+(5+2\sqrt{3})-2\cdot r\cdot \sqrt{5+2\sqrt{3}} \cos(IDH)$$ Combining these 2 equations, $$JH^2=r^2-1^2=(r^2)+(5+2\sqrt{3})-2\cdot r\cdot \sqrt{5+2\sqrt{3}} \cos(IDH)$$ Solving for r produces $$r=\frac{6+2\sqrt{3}}{2\sqrt{5+2\sqrt{3}}\cdot \cos(IDH)}$$ We can use the pythagorean theorem to show that $\cos(IDH)=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{5+2\sqrt{3}}}$ which now gives $$r=\frac{6+2\sqrt{3}}{2\sqrt{5+2\sqrt{3}}\cdot \frac{\sqrt{4+2\sqrt{3}}}{\sqrt{5+2\sqrt{3}}}}=\frac{6+2\sqrt{3}}{2\sqrt{4+2\sqrt{3}}}$$ Squaring both sides,

$$r^2=(\frac{6+2\sqrt{3}}{2\sqrt{4+2\sqrt{3}}})^2=\frac{48+24\sqrt{3}}{4(4+2\sqrt{3})}=3$$ $$r^2=3$$ $$r=\sqrt{3}$$

And the ratio of the areas is

$$\frac{\frac{2\cdot \pi \cdot (1^2)}{2}}{\frac{\pi \cdot (\sqrt{3})^2}{2}}=\frac{2}{3}$$

The fraction is

$\frac{2}{3}$

Without loss of generality let the radius of the smaller circle equal 1.enter image description here

Since $\angle HIG=90^\circ$ and $\arcsin(\frac{1}{2})=30^\circ$, we haveenter image description here

By the pythagorean theorem, ${HD}^2=HI^2+ID^2=1^2+(1+\sqrt{3})^2=5+2 \sqrt{3}$enter image description here

Now label the centre of the large circle J. We have $JA=JD=r$, and since JH bisects the chord AB, $\angle JHA=90^\circ $.enter image description here By the pythagorean theorem $$JH^2+1^2=r^2$$

And by the cosine rule $$JH^2=JD^2+HD^2-2\cdot HD\cdot JD\cdot\cos(IDH)$$ $$JH^2=(r^2)+(5+2\sqrt{3})-2\cdot r\cdot \sqrt{5+2\sqrt{3}} \cos(IDH)$$ Combining these 2 equations, $$JH^2=r^2-1^2=(r^2)+(5+2\sqrt{3})-2\cdot r\cdot \sqrt{5+2\sqrt{3}} \cos(IDH)$$ Solving for r produces $$r=\frac{6+2\sqrt{3}}{2\sqrt{5+2\sqrt{3}}\cdot \cos(IDH)}$$ We can use the pythagorean theorem to show that $\cos(IDH)=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{5+2\sqrt{3}}}$ which now gives $$r=\frac{6+2\sqrt{3}}{2\sqrt{5+2\sqrt{3}}\cdot \frac{\sqrt{4+2\sqrt{3}}}{\sqrt{5+2\sqrt{3}}}}=\frac{6+2\sqrt{3}}{2\sqrt{4+2\sqrt{3}}}$$ Squaring both sides,

$$r^2=(\frac{6+2\sqrt{3}}{2\sqrt{4+2\sqrt{3}}})^2=\frac{48+24\sqrt{3}}{4(4+2\sqrt{3})}=3$$ $$r^2=3$$ $$r=\sqrt{3}$$

And the ratio of the areas is

$$\frac{\frac{2\cdot \pi \cdot (1^2)}{2}}{\frac{\pi \cdot (\sqrt{3})^2}{2}}=\frac{2}{3}$$

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user37842
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