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Consider 3 circles of radii $r_1=r$, $r_2=2r$ and $r_3=3r$, where $r=1/(2\sqrt{7\pi})$, so that their total area $s$ is $$ s=\pi(r_1^2+r_2^2+r_3^2)=\frac12. $$

Descartes' theorem asserts that if three circles of radii $r_1$, $r_2$ and $r_3$ are pairwise externally tangent to each other and circumscribed by the circle of radius $R$, then $$ \frac1R=2\sqrt{\frac1{r_1r_2}+\frac1{r_2r_3}+\frac1{r_3r_1}} -\biggl(\frac1{r_1}+\frac1{r_2}+\frac1{r_3}\biggr). $$ In our case we find from this formula that $$ R=\frac3{\sqrt{7\pi}}, $$ so that the area $S$ of the circumscribing circle is $$ S=\pi R^2=\frac97>1. $$ This implies that the three given circles of total area $1/2$ cannot be put inside a circle of total area 1 without intersections.

Edit. I have to agree that my geometric intuition is too weak to notify an obvious non-applicability of Decartes' theorem. As TonyK mentions in his comment, one can fit these three circles in a circle of radius $r_2+r_3$, and the circle of radius $r_1$ fits in one of the gaps, without touching the enclosing circle.

Without trying to correct the above solution I indicate another choice: $r_1=0.99q$, $r_2=r_3=2q$, where $q$ is chosen in such a way that the total area is again $1/2$. Decartes' theorem produces the circumscribing circle of area $>1.008$. There is still an option to put two large circles along a diameter of circle of total area 1 and see whether there is a room for the smaller one. This definitely means that more geometry is involved...

Final edit (hopefully). As roland-bacher mentions, the three circles again fit a circle of radius $r_2+r_3$. After the two large equal circles are inscribed, there is a room for another circle of radius $2r_2/3$ (well, one can still have the space for the other of radius $2r_3/3$, but I do not care of more than 3 circles inscribed). The corresponding geometric picture involves the right triangles with sides $1$, $4/3$, and $5/3$, a nice appearance of the Pythagorean triple $3^2+4^2=5^2$.

Is it true that this ($r_1=2/3$, $r_2=r_3=1$, total area $s=22\pi/9$) is the worth case of inscribing 3 circles into the circle of radius $R=2$ (area $S=4\pi$)? The quality of this inscription is $S/s=18/11$. In other words, can we replace areas $1/2$ and $1$ in the original problem by $11/18$ and $1$ respectively, if at least three circles are inscribed?

I have to apologize for my unsuccessful attempt. You still have a chance to enjoy the beauty and difficulty of the original problem. I thank TonyK and roland-bacher for pointing out my mistakes in geometry.

After passions calmed down, I can put back my old unsuccessful attempt. At least I was quite enthusiastic at that time about the problem, until I have got downvotes and seen some vague ideas of others (they are still here) as answers.

Post as it was on May 12, 2010. Consider 3 circles of radii $r_1=r$, $r_2=2r$ and $r_3=3r$, where $r=1/(2\sqrt{7\pi})$, so that their total area $s$ is $$ s=\pi(r_1^2+r_2^2+r_3^2)=\frac12. $$

Descartes' theorem asserts that if three circles of radii $r_1$, $r_2$ and $r_3$ are pairwise externally tangent to each other and circumscribed by the circle of radius $R$, then $$ \frac1R=2\sqrt{\frac1{r_1r_2}+\frac1{r_2r_3}+\frac1{r_3r_1}} -\biggl(\frac1{r_1}+\frac1{r_2}+\frac1{r_3}\biggr). $$ In our case we find from this formula that $$ R=\frac3{\sqrt{7\pi}}, $$ so that the area $S$ of the circumscribing circle is $$ S=\pi R^2=\frac97>1. $$ This implies that the three given circles of total area $1/2$ cannot be put inside a circle of total area 1 without intersections.

Edit. I have to agree that my geometric intuition is too weak to notify an obvious non-applicability of Decartes' theorem. As Tony mentions in his comment, one can fit these three circles in a circle of radius $r_2+r_3$, and the circle of radius $r_1$ fits in one of the gaps, without touching the enclosing circle.

Without trying to correct the above solution I indicate another choice: $r_1=0.99q$, $r_2=r_3=2q$, where $q$ is chosen in such a way that the total area is again $1/2$. Decartes' theorem produces the circumscribing circle of area $>1.008$. There is still an option to put two large circles along a diameter of circle of total area 1 and see whether there is a room for the smaller one. This definitely means that more geometry is involved...

Final edit (hopefully). As Roland mentions, the three circles again fit a circle of radius $r_2+r_3$. After the two large equal circles are inscribed, there is a room for another circle of radius $2r_2/3$ (well, one can still have the space for the other of radius $2r_3/3$, but I do not care of more than 3 circles inscribed). The corresponding geometric picture involves the right triangles with sides $1$, $4/3$, and $5/3$, a nice appearance of the Pythagorean triple $3^2+4^2=5^2$.

Is it true that this ($r_1=2/3$, $r_2=r_3=1$, total area $s=22\pi/9$) is the worth case of inscribing 3 circles into the circle of radius $R=2$ (area $S=4\pi$)? The quality of this inscription is $S/s=18/11$. In other words, can we replace areas $1/2$ and $1$ in the original problem by $11/18$ and $1$ respectively, if at least three circles are inscribed?

I have to apologize for my unsuccessful attempt. You still have a chance to enjoy the beauty and difficulty of the original problem. I thank Tony and Roland for pointing out my mistakes in geometry.

Consider 3 circles of radii $r_1=r$, $r_2=2r$ and $r_3=3r$, where $r=1/(2\sqrt{7\pi})$, so that their total area $s$ is $$ s=\pi(r_1^2+r_2^2+r_3^2)=\frac12. $$

Descartes' theorem asserts that if three circles of radii $r_1$, $r_2$ and $r_3$ are pairwise externally tangent to each other and circumscribed by the circle of radius $R$, then $$ \frac1R=2\sqrt{\frac1{r_1r_2}+\frac1{r_2r_3}+\frac1{r_3r_1}} -\biggl(\frac1{r_1}+\frac1{r_2}+\frac1{r_3}\biggr). $$ In our case we find from this formula that $$ R=\frac3{\sqrt{7\pi}}, $$ so that the area $S$ of the circumscribing circle is $$ S=\pi R^2=\frac97>1. $$ This implies that the three given circles of total area $1/2$ cannot be put inside a circle of total area 1 without intersections.

Edit. I have to agree that my geometric intuition is too weak to notify an obvious non-applicability of Decartes' theorem. As TonyK mentions in his comment, one can fit these three circles in a circle of radius $r_2+r_3$, and the circle of radius $r_1$ fits in one of the gaps, without touching the enclosing circle.

Without trying to correct the above solution I indicate another choice: $r_1=0.99q$, $r_2=r_3=2q$, where $q$ is chosen in such a way that the total area is again $1/2$. Decartes' theorem produces the circumscribing circle of area $>1.008$. There is still an option to put two large circles along a diameter of circle of total area 1 and see whether there is a room for the smaller one. This definitely means that more geometry is involved...

Final edit (hopefully). As roland-bacher mentions, the three circles again fit a circle of radius $r_2+r_3$. After the two large equal circles are inscribed, there is a room for another circle of radius $2r_2/3$ (well, one can still have the space for the other of radius $2r_3/3$, but I do not care of more than 3 circles inscribed). The corresponding geometric picture involves the right triangles with sides $1$, $4/3$, and $5/3$, a nice appearance of the Pythagorean triple $3^2+4^2=5^2$.

Is it true that this ($r_1=2/3$, $r_2=r_3=1$, total area $s=22\pi/9$) is the worth case of inscribing 3 circles into the circle of radius $R=2$ (area $S=4\pi$)? The quality of this inscription is $S/s=18/11$.

I have to apologize for my unsuccessful attempt. You still have a chance to enjoy the beauty and difficulty of the original problem. I thank TonyK and roland-bacher for pointing out my mistakes in geometry.

Consider 3 circles of radii $r_1=r$, $r_2=2r$ and $r_3=3r$, where $r=1/(2\sqrt{7\pi})$, so that their total area $s$ is $$ s=\pi(r_1^2+r_2^2+r_3^2)=\frac12. $$

Descartes' theorem asserts that if three circles of radii $r_1$, $r_2$ and $r_3$ are pairwise externally tangent to each other and circumscribed by the circle of radius $R$, then $$ \frac1R=2\sqrt{\frac1{r_1r_2}+\frac1{r_2r_3}+\frac1{r_3r_1}} -\biggl(\frac1{r_1}+\frac1{r_2}+\frac1{r_3}\biggr). $$ In our case we find from this formula that $$ R=\frac3{\sqrt{7\pi}}, $$ so that the area $S$ of the circumscribing circle is $$ S=\pi R^2=\frac97>1. $$ This implies that the three given circles of total area $1/2$ cannot be put inside a circle of total area 1 without intersections.

Edit. I have to agree that my geometric intuition is too weak to notify an obvious non-applicability of Decartes' theorem. As TonyK mentions in his comment, one can fit these three circles in a circle of radius $r_2+r_3$, and the circle of radius $r_1$ fits in one of the gaps, without touching the enclosing circle.

Without trying to correct the above solution I indicate another choice: $r_1=0.99q$, $r_2=r_3=2q$, where $q$ is chosen in such a way that the total area is again $1/2$. Decartes' theorem produces the circumscribing circle of area $>1.008$. There is still an option to put two large circles along a diameter of circle of total area 1 and see whether there is a room for the smaller one. This definitely means that more geometry is involved...

Final edit (hopefully). As roland-bacher mentions, the three circles again fit a circle of radius $r_2+r_3$. After the two large equal circles are inscribed, there is a room for another circle of radius $2r_2/3$ (well, one can still have the space for the other of radius $2r_3/3$, but I do not care of more than 3 circles inscribed). The corresponding geometric picture involves the right triangles with sides $1$, $4/3$, and $5/3$, a nice appearance of the Pythagorean triple $3^2+4^2=5^2$.

Is it true that this ($r_1=2/3$, $r_2=r_3=1$, total area $s=22\pi/9$) is the worth case of inscribing 3 circles into the circle of radius $R=2$ (area $S=4\pi$)? The quality of this inscription is $S/s=18/11$. In other words, can we replace areas $1/2$ and $1$ in the original problem by $11/18$ and $1$ respectively, if at least three circles are inscribed?

I have to apologize for my unsuccessful attempt. You still have a chance to enjoy the beauty and difficulty of the original problem. I thank TonyK and roland-bacher for pointing out my mistakes in geometry.

Consider 3 circles of radii $r_1=r$, $r_2=2r$ and $r_3=3r$, where $r=1/(2\sqrt{7\pi})$, so that their total area $s$ is $$ s=\pi(r_1^2+r_2^2+r_3^2)=\frac12. $$

Descartes' theorem asserts that if three circles of radii $r_1$, $r_2$ and $r_3$ are pairwise externally tangent to each other and circumscribed by the circle of radius $R$, then $$ \frac1R=2\sqrt{\frac1{r_1r_2}+\frac1{r_2r_3}+\frac1{r_3r_1}} -\biggl(\frac1{r_1}+\frac1{r_2}+\frac1{r_3}\biggr). $$ In our case we find from this formula that $$ R=\frac3{\sqrt{7\pi}}, $$ so that the area $S$ of the circumscribing circle is $$ S=\pi R^2=\frac97>1. $$ This implies that the three given circles of total area $1/2$ cannot be put inside a circle of total area 1 without intersections.

Edit. I have to agree that my geometric intuition is too weak to notify an obvious non-applicability of Decartes' theorem. As TonyK mentions in his comment, one can fit these three circles in a circle of radius $r_2+r_3$, and the circle of radius $r_1$ fits in one of the gaps, without touching the enclosing circle.

Without trying to correct the above solution I indicate another choice: $r_1=0.99q$, $r_2=r_3=2q$, where $q$ is chosen in such a way that the total area is again $1/2$. Decartes' theorem produces the circumscribing circle of area $>1.008$. There is still an option to put two large circles along a diameter of circle of total area 1 and see whether there is a room for the smaller one. This definitely means that more geometry is involved...

Consider 3 circles of radii $r_1=r$, $r_2=2r$ and $r_3=3r$, where $r=1/(2\sqrt{7\pi})$, so that their total area $s$ is $$ s=\pi(r_1^2+r_2^2+r_3^2)=\frac12. $$

Descartes' theorem asserts that if three circles of radii $r_1$, $r_2$ and $r_3$ are pairwise externally tangent to each other and circumscribed by the circle of radius $R$, then $$ \frac1R=2\sqrt{\frac1{r_1r_2}+\frac1{r_2r_3}+\frac1{r_3r_1}} -\biggl(\frac1{r_1}+\frac1{r_2}+\frac1{r_3}\biggr). $$ In our case we find from this formula that $$ R=\frac3{\sqrt{7\pi}}, $$ so that the area $S$ of the circumscribing circle is $$ S=\pi R^2=\frac97>1. $$ This implies that the three given circles of total area $1/2$ cannot be put inside a circle of total area 1 without intersections.

Edit. I have to agree that my geometric intuition is too weak to notify an obvious non-applicability of Decartes' theorem. As TonyK mentions in his comment, one can fit these three circles in a circle of radius $r_2+r_3$, and the circle of radius $r_1$ fits in one of the gaps, without touching the enclosing circle.

Without trying to correct the above solution I indicate another choice: $r_1=0.99q$, $r_2=r_3=2q$, where $q$ is chosen in such a way that the total area is again $1/2$. Decartes' theorem produces the circumscribing circle of area $>1.008$. There is still an option to put two large circles along a diameter of circle of total area 1 and see whether there is a room for the smaller one. This definitely means that more geometry is involved...

Final edit (hopefully). As roland-bacher mentions, the three circles again fit a circle of radius $r_2+r_3$. After the two large equal circles are inscribed, there is a room for another circle of radius $2r_2/3$ (well, one can still have the space for the other of radius $2r_3/3$, but I do not care of more than 3 circles inscribed). The corresponding geometric picture involves the right triangles with sides $1$, $4/3$, and $5/3$, a nice appearance of the Pythagorean triple $3^2+4^2=5^2$.

Is it true that this ($r_1=2/3$, $r_2=r_3=1$, total area $s=22\pi/9$) is the worth case of inscribing 3 circles into the circle of radius $R=2$ (area $S=4\pi$)? The quality of this inscription is $S/s=18/11$.

I have to apologize for my unsuccessful attempt. You still have a chance to enjoy the beauty and difficulty of the original problem. I thank TonyK and roland-bacher for pointing out my mistakes in geometry.