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


Question: In the isosceles trapezoid $A B C D$, $A B=C D=x$. The area of the trapezoid is 80 and the circle with centre $O$ and radius 4 is tangent to the four sides of the trapezoid. Determine the value of $x$.

<img_3972>

Solution:
Using the tangent properties of a circle, the lengths of line segments are as shown on the diagram.

Area of trapezoid $A B C D=\frac{1}{2}(8)(B C+A D)$

$$
\begin{aligned}
& =4(2 b+2 x-2 b) \\
& =8 x .
\end{aligned}
$$

<img_3854>

Thus, $8 x=80$.

Therefore, $x=10$.

Final answer: 10
Subfield: Geometry


Question: In the isosceles trapezoid $A B C D$, $A B=C D=x$. The area of the trapezoid is 80 and the circle with centre $O$ and radius 4 is tangent to the four sides of the trapezoid. Determine the value of $x$.

![](https://cdn.mathpix.com/cropped/2023_12_21_b1dc13dda57202a677cdg-1.jpg?height=266&width=333&top_left_y=2182&top_left_x=1362)

Solution:
Using the tangent properties of a circle, the lengths of line segments are as shown on the diagram.

Area of trapezoid $A B C D=\frac{1}{2}(8)(B C+A D)$

$$
\begin{aligned}
& =4(2 b+2 x-2 b) \\
& =8 x .
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2023_12_21_4eb15a30463877039083g-1.jpg?height=287&width=482&top_left_y=1250&top_left_x=1250)

Thus, $8 x=80$.

Therefore, $x=10$.

Final answer: 10
Subfield: Geometry


Question: In parallelogram $A B C D, A B=a$ and $B C=b$, where $a>b$. The points of intersection of the angle bisectors are the vertices of quadrilateral $P Q R S$.

![](https://cdn.mathpix.com/cropped/2023_12_21_df527143fd6a00cd2aabg-1.jpg?height=268&width=599&top_left_y=224&top_left_x=1205)
Prove that $P Q R S$ is a rectangle.

Solution:
In a parallelogram opposite angles are equal.

Since $D F$ and $B E$ bisect the two angles, let $\angle A D F=\angle C D F=\angle A B E=\angle C B E$

$$
=x \text { (in degrees) }
$$

Also $\angle C D F=\angle A F D=x$ (alternate angles)

Let $\angle D A M=\angle B A M=\angle D C N=\angle B C N=y$ (in degrees)



For any parallelogram, any two consecutive angles add to $180^{\circ}, \therefore 2 x+2 y=180$ or, $x+y=90$.

![](https://cdn.mathpix.com/cropped/2023_12_21_2940ec3d1fef53148a02g-1.jpg?height=345&width=707&top_left_y=234&top_left_x=1121)

Therefore in $\triangle P A F, \angle A P F=90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2023_12_21_2940ec3d1fef53148a02g-1.jpg?height=222&width=407&top_left_y=637&top_left_x=1119)

Using similar reasoning and properties of parallel lines we get right angles at $Q, R$ and $S$.

Thus $P Q R S$ is a rectangle.

Final answer: 证明题，略
Subfield: Geometry


Question: In parallelogram $A B C D, A B=a$ and $B C=b$, where $a>b$. The points of intersection of the angle bisectors are the vertices of quadrilateral $P Q R S$.

<img_3974>
Prove that $P Q R S$ is a rectangle.

Solution:
In a parallelogram opposite angles are equal.

Since $D F$ and $B E$ bisect the two angles, let $\angle A D F=\angle C D F=\angle A B E=\angle C B E$

$$
=x \text { (in degrees) }
$$

Also $\angle C D F=\angle A F D=x$ (alternate angles)

Let $\angle D A M=\angle B A M=\angle D C N=\angle B C N=y$ (in degrees)



For any parallelogram, any two consecutive angles add to $180^{\circ}, \therefore 2 x+2 y=180$ or, $x+y=90$.

<img_3560>

Therefore in $\triangle P A F, \angle A P F=90^{\circ}$.

<img_3544>

Using similar reasoning and properties of parallel lines we get right angles at $Q, R$ and $S$.

Thus $P Q R S$ is a rectangle.
Subfield: Geometry


Question: In parallelogram $A B C D, A B=a$ and $B C=b$, where $a>b$. The points of intersection of the angle bisectors are the vertices of quadrilateral $P Q R S$.

![](https://cdn.mathpix.com/cropped/2023_12_21_df527143fd6a00cd2aabg-1.jpg?height=268&width=599&top_left_y=224&top_left_x=1205)
Prove that $P R=a-b$.

Solution:
Since $A M$ is a bisector of $\angle D A B$, let $\angle D A M=\angle B A M=y$.

Also, $\angle D M A=y$ (alternate angles)

This implies that $\triangle A D M$ is isosceles.

Using the same reasoning in $\triangle C B N$, we see that it is also isosceles and so the diagram may now be labelled as:

![](https://cdn.mathpix.com/cropped/2023_12_21_2940ec3d1fef53148a02g-1.jpg?height=431&width=718&top_left_y=1481&top_left_x=622)

$A N=a-b$

Thus $\triangle A D M$ and $\triangle C B N$ are identical isosceles triangles.

Also, $A M \| N C$ (corresponding angles)

or, $A P \| N R$.

By using properties of isosceles triangles (or congruency), $A P=N R$ implying that $A P R N$ is a parallelogram.

Thus $A N=P R$ and since $A N=a-b, P R=a-b$ (as required)

Final answer: 证明题，略
Subfield: Geometry


Question: In parallelogram $A B C D, A B=a$ and $B C=b$, where $a>b$. The points of intersection of the angle bisectors are the vertices of quadrilateral $P Q R S$.

<img_3974>
Prove that $P R=a-b$.

Solution:
Since $A M$ is a bisector of $\angle D A B$, let $\angle D A M=\angle B A M=y$.

Also, $\angle D M A=y$ (alternate angles)

This implies that $\triangle A D M$ is isosceles.

Using the same reasoning in $\triangle C B N$, we see that it is also isosceles and so the diagram may now be labelled as:

<img_3709>

$A N=a-b$

Thus $\triangle A D M$ and $\triangle C B N$ are identical isosceles triangles.

Also, $A M \| N C$ (corresponding angles)

or, $A P \| N R$.

By using properties of isosceles triangles (or congruency), $A P=N R$ implying that $A P R N$ is a parallelogram.

Thus $A N=P R$ and since $A N=a-b, P R=a-b$ (as required)
Subfield: Number Theory


Question: A permutation of the integers $1,2, \ldots, n$ is a listing of these integers in some order. For example, $(3,1,2)$ and $(2,1,3)$ are two different permutations of the integers 1, 2, 3. A permutation $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ of the integers $1,2, \ldots, n$ is said to be "fantastic" if $a_{1}+a_{2}+\ldots+a_{k}$ is divisible by $k$, for each $k$ from 1 to $n$. For example, $(3,1,2)$ is a fantastic permutation of $1,2,3$ because 3 is divisible by $1,3+1$ is divisible by 2 , and $3+1+2$ is divisible by 3 . However, $(2,1,3)$ is not fantastic because $2+1$ is not divisible by 2 .
Show that no fantastic permutation exists for $n=2000$.

Solution:
In our consideration of whether there is a fantastic permutation for $n=2000$, we start by looking at the 2000th position.

Using our definition of fantastic permutation, it is necessary that $2000 \mid(1+2+3+\cdots+2000)$.

Since $1+2+3+\cdots+2000=\frac{(2000)(2001)}{2}=(1000)(2001)$, it is required that $2000 \mid 1000(2001)$.

This is not possible and so no fantastic permutation exists for $n=2000$.
Subfield: Number Theory


Question: A permutation of the integers $1,2, \ldots, n$ is a listing of these integers in some order. For example, $(3,1,2)$ and $(2,1,3)$ are two different permutations of the integers 1, 2, 3. A permutation $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ of the integers $1,2, \ldots, n$ is said to be "fantastic" if $a_{1}+a_{2}+\ldots+a_{k}$ is divisible by $k$, for each $k$ from 1 to $n$. For example, $(3,1,2)$ is a fantastic permutation of $1,2,3$ because 3 is divisible by $1,3+1$ is divisible by 2 , and $3+1+2$ is divisible by 3 . However, $(2,1,3)$ is not fantastic because $2+1$ is not divisible by 2 .
Does a fantastic permutation exist for $n=2001$ ? Explain.

Solution:
The sum of the integers from 1 to 2001 is $\frac{(2001)(2002)}{2}=(2001)(1001)$ which is divisible by 2001. If $t_{1}, t_{2}, \ldots, t_{2001}$ is a fantastic permutation, when we remove $t_{2001}$ from the above sum, and what remains must be divisible by 2000 .

We now consider $t_{1}+t_{2}+\cdots+t_{2000}$ and determine what integer is not included in the permutation.

$$
\begin{aligned}
t_{1}+t_{2}+\cdots+t_{2000} & =\frac{(2001)(2002)}{2}-t_{2001} \\
& =(1001)(2001)-t_{2001}
\end{aligned}
$$

Since $(1001)(2001)=2003001, t_{2001}$ must be a number of the form $k 001$ where $k$ is odd. The only integer less than or equal to 2001 with this property is 1001 . Therefore $t_{2001}=1001$.

So the sum up to $t_{2000}$ is $2003001-1001=2002000$.

When we remove $t_{2000}$ we must get a multiple of 1999 .

The largest multiple of 1999 less than 2002000 is (1999)(1001) $=2000$ 999. This would make $t_{2000}=2002000-2000999=1001$ which is impossible since $t_{2000} \neq t_{2001}$. If we choose lesser multiples of 1999 to subtract from 2002000 we will get values of $t_{2000}$ which are greater than 2001, which is also not possible.

Thus, a fantastic permutation is not possible for $n=2001$.
Subfield: Geometry


Question: An equilateral triangle $A B C$ has side length 2 . A square, $P Q R S$, is such that $P$ lies on $A B, Q$ lies on $B C$, and $R$ and $S$ lie on $A C$ as shown. The points $P, Q, R$, and $S$ move so that $P, Q$ and $R$ always remain on the sides of the triangle and $S$ moves from $A C$ to $A B$ through the interior of the triangle. If the points $P, Q, R$ and $S$ always form the vertices of a square, show that the path traced out by $S$ is a straight line parallel to $B C$.

![](https://cdn.mathpix.com/cropped/2023_12_21_df527143fd6a00cd2aabg-1.jpg?height=372&width=387&top_left_y=1126&top_left_x=1346)

Solution:
Let $\angle R Q C=\theta$ and from $S$ draw a line perpendicular to the base at $P$.

Then $\angle T Q B=180-(90+\theta)=90-\theta$.

Let $s$ be the length of the side of the square.

From $R$ draw a line perpendicular to $B C$ at $D$ and then through $S$ draw a line parallel to $B C$. From $R$ draw a line perpendicular to this line at E.

![](https://cdn.mathpix.com/cropped/2023_12_21_216628493d6885b0bc57g-1.jpg?height=482&width=490&top_left_y=1190&top_left_x=1273)

From $\triangle R Q D, R D=s \sin \theta$.

Since $\angle Q R D=90-\theta$ then $\angle S R E=\theta$.

From $\triangle S E R, E R=s \cos \theta$.

The perpendicular distance from $S$ to $B C$ is $R D+E R=s \sin \theta+s \cos \theta$ which we must now show is a constant.

We can now take each of the lengths $D C, D Q, P F, F B$ and express them in terms of $s$.

From $\triangle R D C$ which is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, $\frac{D C}{R D}=\frac{1}{\sqrt{3}}$.

Since $R D=s \sin \theta$ (from above)

$$
D C=\frac{1}{\sqrt{3}}(s \sin \theta)=\frac{\sqrt{3}}{3} s \sin \theta
$$



From $\Delta R D Q, \frac{Q D}{R Q}=\cos \theta, Q D=s \cos \theta$.

From $\triangle T F Q, \sin \theta=\frac{F Q}{s}$ and $\cos \theta=\frac{T F}{s}$. or, $F Q=s \sin \theta$ and $T F=s \sin \theta$.

From $\triangle T F B, \frac{B F}{T F}=\frac{1}{\sqrt{3}}, B F=\frac{1}{\sqrt{3}} T F=\frac{1}{\sqrt{3}} s \cos \theta=\frac{\sqrt{3}}{3} s \cos \theta$.

Since $D C+Q D+F Q+B F=2, \frac{\sqrt{3}}{3} s \sin \theta+s \cos \theta+s \sin \theta+\frac{\sqrt{3}}{3} s \cos \theta=2$.

$$
\begin{aligned}
\frac{\sqrt{3}}{3}(s \cos \theta+s \sin \theta)+(s \cos \theta+s \sin \theta) & =2 \\
s \cos \theta+s \sin \theta & =\frac{2}{\left(\frac{\sqrt{3}}{3}+1\right)}
\end{aligned}
$$

Thus $s \cos \theta+s \sin \theta$ is a constant and the path traced out by $S$ is a straight line parallel to $B C$.

Final answer: 证明题，略
Subfield: Geometry


Question: An equilateral triangle $A B C$ has side length 2 . A square, $P Q R S$, is such that $P$ lies on $A B, Q$ lies on $B C$, and $R$ and $S$ lie on $A C$ as shown. The points $P, Q, R$, and $S$ move so that $P, Q$ and $R$ always remain on the sides of the triangle and $S$ moves from $A C$ to $A B$ through the interior of the triangle. If the points $P, Q, R$ and $S$ always form the vertices of a square, show that the path traced out by $S$ is a straight line parallel to $B C$.

<img_3173>

Solution:
Let $\angle R Q C=\theta$ and from $S$ draw a line perpendicular to the base at $P$.

Then $\angle T Q B=180-(90+\theta)=90-\theta$.

Let $s$ be the length of the side of the square.

From $R$ draw a line perpendicular to $B C$ at $D$ and then through $S$ draw a line parallel to $B C$. From $R$ draw a line perpendicular to this line at E.

<img_3178>

From $\triangle R Q D, R D=s \sin \theta$.

Since $\angle Q R D=90-\theta$ then $\angle S R E=\theta$.

From $\triangle S E R, E R=s \cos \theta$.

The perpendicular distance from $S$ to $B C$ is $R D+E R=s \sin \theta+s \cos \theta$ which we must now show is a constant.

We can now take each of the lengths $D C, D Q, P F, F B$ and express them in terms of $s$.

From $\triangle R D C$ which is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, $\frac{D C}{R D}=\frac{1}{\sqrt{3}}$.

Since $R D=s \sin \theta$ (from above)

$$
D C=\frac{1}{\sqrt{3}}(s \sin \theta)=\frac{\sqrt{3}}{3} s \sin \theta
$$



From $\Delta R D Q, \frac{Q D}{R Q}=\cos \theta, Q D=s \cos \theta$.

From $\triangle T F Q, \sin \theta=\frac{F Q}{s}$ and $\cos \theta=\frac{T F}{s}$. or, $F Q=s \sin \theta$ and $T F=s \sin \theta$.

From $\triangle T F B, \frac{B F}{T F}=\frac{1}{\sqrt{3}}, B F=\frac{1}{\sqrt{3}} T F=\frac{1}{\sqrt{3}} s \cos \theta=\frac{\sqrt{3}}{3} s \cos \theta$.

Since $D C+Q D+F Q+B F=2, \frac{\sqrt{3}}{3} s \sin \theta+s \cos \theta+s \sin \theta+\frac{\sqrt{3}}{3} s \cos \theta=2$.

$$
\begin{aligned}
\frac{\sqrt{3}}{3}(s \cos \theta+s \sin \theta)+(s \cos \theta+s \sin \theta) & =2 \\
s \cos \theta+s \sin \theta & =\frac{2}{\left(\frac{\sqrt{3}}{3}+1\right)}
\end{aligned}
$$

Thus $s \cos \theta+s \sin \theta$ is a constant and the path traced out by $S$ is a straight line parallel to $B C$.
Subfield: Geometry


Question: In the diagram, points $P(p, 4), B(10,0)$, and $O(0,0)$ are shown. If $\triangle O P B$ is right-angled at $P$, determine all possible values of $p$.

<img_3857>

Solution:
Since $\angle O P B=90^{\circ}$, then $O P$ and $P B$ are perpendicular, so the product of their slopes is -1 .

The slope of $O P$ is $\frac{4-0}{p-0}=\frac{4}{p}$ and the slope of $P B$ is $\frac{4-0}{p-10}=\frac{4}{p-10}$.

Therefore, we need

$$
\begin{aligned}
\frac{4}{p} \cdot \frac{4}{p-10} & =-1 \\
16 & =-p(p-10) \\
p^{2}-10 p+16 & =0 \\
(p-2)(p-8) & =0
\end{aligned}
$$

and so $p=2$ or $p=8$. Since each these steps is reversible, then $\triangle O P B$ is right-angled precisely when $p=2$ and $p=8$.
Since $\triangle O P B$ is right-angled at $P$, then $O P^{2}+P B^{2}=O B^{2}$ by the Pythagorean Theorem. Note that $O B=10$ since $O$ has coordinates $(0,0)$ and $B$ has coordinates $(10,0)$.

Also, $O P^{2}=(p-0)^{2}+(4-0)^{2}=p^{2}+16$ and $P B^{2}=(10-p)^{2}+(4-0)^{2}=p^{2}-20 p+116$. Therefore,

$$
\begin{aligned}
\left(p^{2}+16\right)+\left(p^{2}-20 p+116\right) & =10^{2} \\
2 p^{2}-20 p+32 & =0 \\
p^{2}-10 p+16 & =0
\end{aligned}
$$



and so $(p-2)(p-8)=0$, or $p=2$ or $p=8$. Since each these steps is reversible, then $\triangle O P B$ is right-angled precisely when $p=2$ and $p=8$.

Final answer: 2,8
Subfield: Geometry


Question: In the diagram, points $P(p, 4), B(10,0)$, and $O(0,0)$ are shown. If $\triangle O P B$ is right-angled at $P$, determine all possible values of $p$.

![](https://cdn.mathpix.com/cropped/2023_12_21_6612ae75d8ef169f4be5g-1.jpg?height=377&width=477&top_left_y=722&top_left_x=1255)

Solution:
Since $\angle O P B=90^{\circ}$, then $O P$ and $P B$ are perpendicular, so the product of their slopes is -1 .

The slope of $O P$ is $\frac{4-0}{p-0}=\frac{4}{p}$ and the slope of $P B$ is $\frac{4-0}{p-10}=\frac{4}{p-10}$.

Therefore, we need

$$
\begin{aligned}
\frac{4}{p} \cdot \frac{4}{p-10} & =-1 \\
16 & =-p(p-10) \\
p^{2}-10 p+16 & =0 \\
(p-2)(p-8) & =0
\end{aligned}
$$

and so $p=2$ or $p=8$. Since each these steps is reversible, then $\triangle O P B$ is right-angled precisely when $p=2$ and $p=8$.
Since $\triangle O P B$ is right-angled at $P$, then $O P^{2}+P B^{2}=O B^{2}$ by the Pythagorean Theorem. Note that $O B=10$ since $O$ has coordinates $(0,0)$ and $B$ has coordinates $(10,0)$.

Also, $O P^{2}=(p-0)^{2}+(4-0)^{2}=p^{2}+16$ and $P B^{2}=(10-p)^{2}+(4-0)^{2}=p^{2}-20 p+116$. Therefore,

$$
\begin{aligned}
\left(p^{2}+16\right)+\left(p^{2}-20 p+116\right) & =10^{2} \\
2 p^{2}-20 p+32 & =0 \\
p^{2}-10 p+16 & =0
\end{aligned}
$$



and so $(p-2)(p-8)=0$, or $p=2$ or $p=8$. Since each these steps is reversible, then $\triangle O P B$ is right-angled precisely when $p=2$ and $p=8$.

Final answer: 2,8
Subfield: Combinatorics


Question: Thurka bought some stuffed goats and some toy helicopters. She paid a total of $\$ 201$. She did not buy partial goats or partial helicopters. Each stuffed goat cost $\$ 19$ and each toy helicopter cost $\$ 17$. How many of each did she buy?

Solution:
Suppose that Thurka bought $x$ goats and $y$ helicopters.

Then $19 x+17 y=201$.

Since $x$ and $y$ are non-negative integers, then $19 x \leq 201$ so $x \leq 10$.

If $x=10$, then $17 y=201-19 x=11$, which does not have an integer solution because 11 is not divisible by 17 .

If $x=9$, then $17 y=201-19 x=30$, which does not have an integer solution.

If $x=8$, then $17 y=201-19 x=49$, which does not have an integer solution.

If $x=7$, then $17 y=201-19 x=68$, so $y=4$.

Therefore, $19(7)+17(4)=201$, and so Thurka buys 7 goats and 4 helicopters.

(We can check that $x=0,1,2,3,4,5,6$ do not give values of $y$ that work.)

Final answer: 7,4
Subfield: Algebra


Question: Determine all real values of $x$ for which $(x+8)^{4}=(2 x+16)^{2}$.

Solution:
Manipulating algebraically,

$$
\begin{aligned}
(x+8)^{4} & =(2 x+16)^{2} \\
(x+8)^{4}-2^{2}(x+8)^{2} & =0 \\
(x+8)^{2}\left((x+8)^{2}-2^{2}\right) & =0 \\
(x+8)^{2}((x+8)+2)((x+8)-2) & =0 \\
(x+8)^{2}(x+10)(x+6) & =0
\end{aligned}
$$

Therefore, $x=-8$ or $x=-10$ or $x=-6$.
Manipulating algebraically,

$$
\begin{aligned}
(x+8)^{4} & =(2 x+16)^{2} \\
(x+8)^{4}-2^{2}(x+8)^{2} & =0 \\
(x+8)^{2}\left((x+8)^{2}-2^{2}\right) & =0 \\
(x+8)^{2}\left(x^{2}+16 x+64-4\right) & =0 \\
(x+8)^{2}\left(x^{2}+16 x+60\right) & =0 \\
(x+8)^{2}(x+10)(x+6) & =0
\end{aligned}
$$

Therefore, $x=-8$ or $x=-10$ or $x=-6$.
Since $(x+8)^{4}=(2 x+16)^{2}$, then $(x+8)^{2}=2 x+16$ or $(x+8)^{2}=-(2 x+16)$.

From the first equation, $x^{2}+16 x+64=2 x+16$ or $x^{2}+14 x+48=0$ or $(x+6)(x+8)=0$. From the second equation, $x^{2}+16 x+64=-2 x-16$ or $x^{2}+18 x+80=0$ or $(x+10)(x+8)=0$.

Therefore, $x=-8$ or $x=-10$ or $x=-6$.

Final answer: -6,-8,-10
Subfield: Algebra


Question: If $f(x)=2 x+1$ and $g(f(x))=4 x^{2}+1$, determine an expression for $g(x)$.

Solution:
We use the fact that $g(x)=g\left(f\left(f^{-1}(x)\right)\right)$.

Since $f(x)=2 x+1$, then to determine $f^{-1}(x)$ we solve $x=2 y+1$ for $y$ to get $2 y=x-1$ or $y=\frac{1}{2}(x-1)$. Thus, $f^{-1}(x)=\frac{1}{2}(x-1)$.

Since $g(f(x))=4 x^{2}+1$, then

$$
\begin{aligned}
g(x) & =g\left(f\left(f^{-1}(x)\right)\right) \\
& =g\left(f\left(\frac{1}{2}(x-1)\right)\right) \\
& =4\left(\frac{1}{2}(x-1)\right)^{2}+1 \\
& =4 \cdot \frac{1}{4}(x-1)^{2}+1 \\
& =(x-1)^{2}+1 \\
& =x^{2}-2 x+2
\end{aligned}
$$
We use the expressions for $f(x)$ and $g(f(x))$ to construct $g(x)$.

Since $f(x)$ is linear and $g(f(x))$ is quadratic, then it is likely that $g(x)$ is also quadratic.

Since $f(x)=2 x+1$, then $(f(x))^{2}=4 x^{2}+4 x+1$.

Since $g(f(x))$ has no term involving $x$, then we subtract $2 f(x)$ (to remove the $4 x$ term) to get

$$
(f(x))^{2}-2 f(x)=\left(4 x^{2}+4 x+1\right)-2(2 x+1)=4 x^{2}-1
$$

To get $g(f(x))$ from this, we add 2 to get $4 x^{2}+1$.

Therefore, $g(f(x))=(f(x))^{2}-2 f(x)+2$, and so an expression for $g(x)$ is $x^{2}-2 x+2$.
We use the expressions for $f(x)$ and $g(f(x))$ to construct $g(x)$.

Since $f(x)$ is linear and $g(f(x))$ is quadratic, then it is likely that $g(x)$ is also quadratic.

Suppose that $g(x)=a x^{2}+b x+c$ for some real numbers $a, b, c$.

Then

$$
\begin{aligned}
g(f(x)) & =g(2 x+1) \\
& =a(2 x+1)^{2}+b(2 x+1)+c \\
& =a\left(4 x^{2}+4 x+1\right)+b(2 x+1)+c \\
& =4 a x^{2}+(4 a+2 b) x+(a+b+c)
\end{aligned}
$$

Since we are told that $g(f(x))=4 x^{2}+1$, then we can compare coefficients to deduce that $4 a=4$ and $4 a+2 b=0$ and $a+b+c=1$.

From the first equation, $a=1$.

From the second equation, $b=-2 a=-2$.

From the third equation, $c=1-a-b=2$.

Therefore, an expression for $g(x)$ is $x^{2}-2 x+2$.

Final answer: $g(x)=x^2-2x+2$
Subfield: Number Theory


Question: A geometric sequence has 20 terms.

The sum of its first two terms is 40 .

The sum of its first three terms is 76 .

The sum of its first four terms is 130 .

Determine how many of the terms in the sequence are integers.

(A geometric sequence is a sequence in which each term after the first is obtained from the previous term by multiplying it by a constant. For example, $3,6,12$ is a geometric sequence with three terms.)

Solution:
Since the sum of the first two terms is 40 and the sum of the first three terms is 76, then the third term is $76-40=36$.

Since the sum of the first three terms is 76 and the sum of the first four terms is 130, then the fourth term is $130-76=54$.

Since the third term is 36 and the fourth term is 54 , then the common ratio in the geometric sequence is $\frac{54}{36}=\frac{3}{2}$.

Therefore, the fifth term is $54 \cdot \frac{3}{2}=81$ and the sixth term is $81 \cdot \frac{3}{2}=\frac{243}{2}$.



Also, the second term is $36 \div \frac{3}{2}=36 \cdot \frac{2}{3}=24$ and the first term is $24 \div \frac{3}{2}=24 \cdot \frac{2}{3}=16$.

Thus, the first six terms of the sequence are $16,24,36,54,81, \frac{243}{2}$.

Since the first term equals $2^{4}$ and the common ratio is $\frac{3}{2}$, then the $n$th term in the sequence is $2^{4}\left(\frac{3}{2}\right)^{n-1}=\frac{3^{n-1}}{2^{n-5}}$.

When $n \geq 6$, this is a fraction whose numerator is odd and whose denominator is even, and so, when $n \geq 6$, the $n$th term is not an integer. (An odd integer is never divisible by an even integer.)

Therefore, there will be 5 integers in the sequence.
Suppose that $a$ is the first term and $r$ is the common ratio between consecutive terms (so that $a r$ is the second term, $a r^{2}$ is the third term, and so on).

From the given information, $a+a r=40$ and $a+a r+a r^{2}=76$ and $a+a r+a r^{2}+a r^{3}=130$.

Subtracting the first equation from the second, we obtain $a r^{2}=36$.

Subtracting the second equation from the third, we obtain $a r^{3}=54$.

Since $a r^{3}=54$ and $a r^{2}=36$, then $r=\frac{a r^{3}}{a r^{2}}=\frac{54}{36}=\frac{3}{2}$.

Since $a r^{2}=36$ and $r=\frac{3}{2}$, then $a\left(\frac{3}{2}\right)^{2}=36$ or $\frac{9}{4} a=36$ or $a=\frac{4}{9} \cdot 36=16$.

Since $a=16$ and $r=\frac{3}{2}$, then the first six terms of the sequence are 16, 24, 36, 54, 81, $\frac{243}{2}$. Since the first term equals $2^{4}$ and the common ratio is $\frac{3}{2}$, then the $n$th term in the sequence is $2^{4}\left(\frac{3}{2}\right)^{n-1}=\frac{3^{n-1}}{2^{n-5}}$.

When $n \geq 6$, this is a fraction whose numerator is odd and whose denominator is even, and so, when $n \geq 6$, the $n$th term is not an integer. (An odd integer is never divisible by an even integer.)

Therefore, there will be 5 integers in the sequence.

Final answer: 5
Subfield: Geometry


Question: A snail's shell is formed from six triangular sections, as shown. Each triangle has interior angles of $30^{\circ}, 60^{\circ}$ and $90^{\circ}$. If $A B$ has a length of $1 \mathrm{~cm}$, what is the length of $A H$, in $\mathrm{cm}$ ?

<img_3884>

Solution:
In a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, the ratio of the side opposite the $90^{\circ}$ to the side opposite the $60^{\circ}$ angle is $2: \sqrt{3}$.

Note that each of $\triangle A B C, \triangle A C D, \triangle A D E, \triangle A E F, \triangle A F G$, and $\triangle A G H$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.

Therefore, $\frac{A H}{A G}=\frac{A G}{A F}=\frac{A F}{A E}=\frac{A E}{A D}=\frac{A D}{A C}=\frac{A C}{A B}=\frac{2}{\sqrt{3}}$.

Thus, $A H=\frac{2}{\sqrt{3}} A G=\left(\frac{2}{\sqrt{3}}\right)^{2} A F=\left(\frac{2}{\sqrt{3}}\right)^{3} A E=\left(\frac{2}{\sqrt{3}}\right)^{4} A D=\left(\frac{2}{\sqrt{3}}\right)^{5} A C=\left(\frac{2}{\sqrt{3}}\right)^{6} A B$.

(In other words, to get from $A B=1$ to the length of $A H$, we multiply by the "scaling factor" $\frac{2}{\sqrt{3}}$ six times.)

Therefore, $A H=\left(\frac{2}{\sqrt{3}}\right)^{6}=\frac{64}{27}$.

Final answer: $\frac{64}{27}$