di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Geometry


Question: A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.

This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.

Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.

<img_4001>

Circle $A$ has curvature 2; circle $B$ has curvature 1 .

<img_3649>

Circle $A$ has curvature 2; circle $B$ has curvature -1 .

Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then

$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$

where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Given $A, B, C, D$ as above with $s=a+b+c+d$, if there is a second circle $A^{\prime}$ with curvature $a^{\prime}$ also tangent to $B, C$, and $D$. We can describe $A$ and $A^{\prime}$ as conjugate circles. Use Descartes' Circle Formula to show that $a^{\prime}=2 s-3 a$ and therefore $s^{\prime}=a^{\prime}+b+c+d=$ $3 s-4 a$.

Solution:
The equation $(x+b+c+d)^{2}=2\left(x^{2}+b^{2}+c^{2}+d^{2}\right)$ is quadratic with two solutions. Call them $a$ and $a^{\prime}$. These are the curvatures of the two circles which are tangent to circles with curvatures $b, c$, and $d$. Rewrite the equation in standard form to obtain $x^{2}-2(b+c+d) x+$ $\ldots=0$. Using the sum of the roots formula, $a+a^{\prime}=2(b+c+d)=2(s-a)$. So $a^{\prime}=2 s-3 a$, and therefore

$$
\begin{aligned}
s^{\prime} & =a^{\prime}+b+c+d \\
& =2 s-3 a+s-a \\
& =3 s-4 a .
\end{aligned}
$$
Subfield: Geometry


Question: A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.

This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.

Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.

![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=304&width=455&top_left_y=1886&top_left_x=347)

Circle $A$ has curvature 2; circle $B$ has curvature 1 .

![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=301&width=307&top_left_y=1888&top_left_x=1386)

Circle $A$ has curvature 2; circle $B$ has curvature -1 .

Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then

$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$

where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Show that by area, $12 \%$ of the kingdom is sold on day 2 .

Solution:
On day 2, six plots are sold: two with curvature 15 from the configuration $(2,2,3,15)$, and four with curvature 6 from the configuration $(-1,2,3,6)$. The total area sold on day 2 is therefore

$$
2 \cdot \frac{\pi}{15^{2}}+4 \cdot \frac{\pi}{6^{2}}=\frac{3}{25} \pi
$$

which is exactly $12 \%$ of the unit circle.

Final answer: 证明题，略
Subfield: Geometry


Question: A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.

This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.

Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.

<img_4001>

Circle $A$ has curvature 2; circle $B$ has curvature 1 .

<img_3649>

Circle $A$ has curvature 2; circle $B$ has curvature -1 .

Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then

$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$

where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Show that by area, $12 \%$ of the kingdom is sold on day 2 .

Solution:
On day 2, six plots are sold: two with curvature 15 from the configuration $(2,2,3,15)$, and four with curvature 6 from the configuration $(-1,2,3,6)$. The total area sold on day 2 is therefore

$$
2 \cdot \frac{\pi}{15^{2}}+4 \cdot \frac{\pi}{6^{2}}=\frac{3}{25} \pi
$$

which is exactly $12 \%$ of the unit circle.
Subfield: Geometry


Question: A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.

This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.

Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.

<img_4001>

Circle $A$ has curvature 2; circle $B$ has curvature 1 .

<img_3649>

Circle $A$ has curvature 2; circle $B$ has curvature -1 .

Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then

$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$

where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Show that the plots sold on day 3 have mean curvature of 23.

Solution:
Because 18 plots were sold on day 3 , the mean curvature is

$$
\frac{2(38+38+35)+4(23+14+11)}{18}=23 .
$$
Subfield: Geometry


Question: A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.

This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.

Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.

![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=304&width=455&top_left_y=1886&top_left_x=347)

Circle $A$ has curvature 2; circle $B$ has curvature 1 .

![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=301&width=307&top_left_y=1888&top_left_x=1386)

Circle $A$ has curvature 2; circle $B$ has curvature -1 .

Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then

$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$

where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Show that the plots sold on day 3 have mean curvature of 23.

Solution:
Because 18 plots were sold on day 3 , the mean curvature is

$$
\frac{2(38+38+35)+4(23+14+11)}{18}=23 .
$$

Final answer: 证明题，略
Subfield: Geometry


Question: A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.

This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.

Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.

<img_4001>

Circle $A$ has curvature 2; circle $B$ has curvature 1 .

<img_3649>

Circle $A$ has curvature 2; circle $B$ has curvature -1 .

Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then

$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$

where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Prove that the curvature of each circular plot is an integer.

Solution:
Proceed by induction. The base case, that all curvatures prior to day 2 are integers. Using the formula $a^{\prime}=2 s-3 a$, if $a, b, c, d$, and $s$ are integers on day $n$, then $a^{\prime}, b^{\prime}, c^{\prime}$, and $d^{\prime}$ are integer curvatures on day $n+1$, proving inductively that all curvatures are integers.
Subfield: Geometry


Question: A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.

This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.

Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.

![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=304&width=455&top_left_y=1886&top_left_x=347)

Circle $A$ has curvature 2; circle $B$ has curvature 1 .

![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=301&width=307&top_left_y=1888&top_left_x=1386)

Circle $A$ has curvature 2; circle $B$ has curvature -1 .

Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then

$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$

where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Prove that the curvature of each circular plot is an integer.

Solution:
Proceed by induction. The base case, that all curvatures prior to day 2 are integers. Using the formula $a^{\prime}=2 s-3 a$, if $a, b, c, d$, and $s$ are integers on day $n$, then $a^{\prime}, b^{\prime}, c^{\prime}$, and $d^{\prime}$ are integer curvatures on day $n+1$, proving inductively that all curvatures are integers.

Final answer: 证明题，略
Subfield: Geometry


Question: Let $A B C D$ be a parallelogram with $\angle A B C$ obtuse. Let $\overline{B E}$ be the altitude to side $\overline{A D}$ of $\triangle A B D$. Let $X$ be the point of intersection of $\overline{A C}$ and $\overline{B E}$, and let $F$ be the point of intersection of $\overline{A B}$ and $\overleftrightarrow{D X}$. If $B C=30, C D=13$, and $B E=12$, compute the ratio $\frac{A C}{A F}$.

Solution:
Extend $\overline{A D}$ to a point $M$ such that $\overline{C M} \| \overline{B E}$ as shown below.

<img_3958>

Because $C D=A B=13$ and $B E=12=C M, A E=D M=5$. Then $A C=\sqrt{35^{2}+12^{2}}=$ $\sqrt{1369}=37$. Because $\overline{E X} \| \overline{C M}, X E / C M=A E / A M=\frac{1}{7}$. Thus $E X=\frac{12}{7}$ and $X B=\frac{72}{7}$, from which $E X / X B=\frac{1}{6}$. Apply Menelaus's Theorem to $\triangle A E B$ and Menelaus line $\overline{F D}$ :

$$
\begin{aligned}
\frac{A D}{E D} \cdot \frac{E X}{X B} \cdot \frac{B F}{F A} & =1 \\
\frac{30}{25} \cdot \frac{1}{6} \cdot \frac{13-F A}{F A} & =1 \\
\frac{13-F A}{F A} & =5 .
\end{aligned}
$$

Thus $F A=\frac{13}{6}$. The desired ratio is:

$$
\frac{37}{13 / 6}=\frac{\mathbf{2 2 2}}{\mathbf{1 3}}
$$
After calculating $A C$ as above, draw $\overline{B D}$, intersecting $\overline{A C}$ at $Y$. Because the diagonals of a parallelogram bisect each other, $D Y=Y B$. Then apply Ceva's Theorem to $\triangle A B D$ and concurrent cevians $\overline{A Y}, \overline{B E}, \overline{D F}$ :

$$
\begin{aligned}
& \frac{A E}{E D} \cdot \frac{D Y}{Y B} \cdot \frac{B F}{F A}=1 \\
& \frac{5}{25} \cdot 1 \cdot \frac{13-F A}{F A}=1
\end{aligned}
$$

Thus $F A=\frac{13}{6}$, and the desired ratio is $\frac{\mathbf{2 2 2}}{\mathbf{1 3}}$.
By AA similarity, note that $\triangle A F X \sim \triangle C D X$ and $\triangle A E X \sim \triangle C B X$. Thus $\frac{A F}{C D}=\frac{A X}{X C}=\frac{A E}{C B}$. Thus $\frac{A F}{13}=\frac{A E}{C B}=\frac{5}{30}$, so $A F=\frac{13}{6}$, and the answer follows after calculating $A C$, as in the first solution.

Final answer: $\frac{222}{13}$
Subfield: Number Theory


Question: Compute the sum of all positive two-digit factors of $2^{32}-1$.

Solution:
Using the difference of squares, $2^{32}-1=\left(2^{16}-1\right)\left(2^{16}+1\right)$. The second factor, $2^{16}+1$, is the Fermat prime 65537 , so continue with the first factor:

$$
\begin{aligned}
2^{16}-1 & =\left(2^{8}+1\right)\left(2^{8}-1\right) \\
2^{8}-1 & =\left(2^{4}+1\right)\left(2^{4}-1\right) \\
2^{4}-1 & =15=3 \cdot 5
\end{aligned}
$$



Because the problem does not specify that the two-digit factors must be prime, the possible two-digit factors are $17,3 \cdot 17=51,5 \cdot 17=85$ and $3 \cdot 5=15$, for a sum of $17+51+85+15=\mathbf{1 6 8}$.

Final answer: 168
Subfield: Algebra


Question: Compute all ordered pairs of real numbers $(x, y)$ that satisfy both of the equations:

$$
x^{2}+y^{2}=6 y-4 x+12 \quad \text { and } \quad 4 y=x^{2}+4 x+12
$$

Solution:
Rearrange the terms in the first equation to yield $x^{2}+4 x+12=6 y-y^{2}+24$, so that the two equations together yield $4 y=6 y-y^{2}+24$, or $y^{2}-2 y-24=0$, from which $y=6$ or $y=-4$. If $y=6$, then $x^{2}+4 x+12=24$, from which $x=-6$ or $x=2$. If $y=-4$, then $x^{2}+4 x+12=-16$, which has no real solutions because $x^{2}+4 x+12=(x+2)^{2}+8 \geq 8$ for all real $x$. So there are two ordered pairs satisfying the system, namely $(-6,6)$ and $(2,6)$.

Final answer: $(-6,6)$, $(2,6)$
Subfield: Algebra


Question: Define $\log ^{*}(n)$ to be the smallest number of times the log function must be iteratively applied to $n$ to get a result less than or equal to 1 . For example, $\log ^{*}(1000)=2$ since $\log 1000=3$ and $\log (\log 1000)=\log 3=0.477 \ldots \leq 1$. Let $a$ be the smallest integer such that $\log ^{*}(a)=3$. Compute the number of zeros in the base 10 representation of $a$.

Solution:
If $\log ^{*}(a)=3$, then $\log (\log (\log (a))) \leq 1$ and $\log (\log (a))>1$. If $\log (\log (a))>1$, then $\log (a)>10$ and $a>10^{10}$. Because the problem asks for the smallest such $a$ that is an integer, choose $a=10^{10}+1=10,000,000,001$, which has 9 zeros.

Final answer: 9