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


Question: A circle with its centre on the $y$-axis intersects the graph of $y=|x|$ at the origin, $O$, and exactly two other distinct points, $A$ and $B$, as shown. Prove that the ratio of the area of triangle $A B O$ to the area of the circle is always $1: \pi$.

<img_3332>

Solution:
Since both the circle with its centre on the $y$-axis and the graph of $y=|x|$ are symmetric about the $y$-axis, then for each point of intersection between these two graphs, there should be a corresponding point of intersection symmetrically located across the $y$-axis. Thus, since there are exactly three points of intersection, then one of these points must be on the $y$-axis, ie. has $x$-coordinate 0 . Since this point is on the graph of $y=|x|$, then this point must be $(0,0)$.

Since the circle has centre on the $y$-axis (say, has coordinates $(0, b)$ ), then its radius is equal to $b$ (and $b$ must be positive for there to be three points of intersection). So the circle has equation $x^{2}+(y-b)^{2}=b^{2}$. Where are the other two points of intersection? We consider the points with $x$ positive and use symmetry to get the other point of intersection.



When $x \geq 0$, then $y=|x|$ has equation $y=x$. Substituting into the equation of the circle,

$$
\begin{aligned}
x^{2}+(x-b)^{2} & =b^{2} \\
2 x^{2}-2 b x & =0 \\
2 x(x-b) & =0
\end{aligned}
$$

Therefore, the points of intersection are $(0,0)$ and $(b, b)$ on the positive side of the $y$-axis, and so at the point $(-b, b)$ on the negative side of the $y$ axis.

Thus the points $O, A$ and $B$ are the points $(0,0)$, $(b, b)$ and $(-b, b)$.

<img_3620>

Since the radius of the circle is $b$, then the area of the circle is $\pi b^{2}$.

Triangle $O A B$ has a base from $(-b, b)$ to $(b, b)$ of length $2 b$, and a height from the line $y=b$ to the point $(0,0)$ of length $b$, and so an area of $\frac{1}{2} b(2 b)=b^{2}$.

Therefore, the ratio of the area of the triangle to the area of the circle is $b^{2}: \pi b^{2}=1: \pi$.
Subfield: Geometry


Question: A circle with its centre on the $y$-axis intersects the graph of $y=|x|$ at the origin, $O$, and exactly two other distinct points, $A$ and $B$, as shown. Prove that the ratio of the area of triangle $A B O$ to the area of the circle is always $1: \pi$.

![](https://cdn.mathpix.com/cropped/2023_12_21_8c2daaa43e7e62803ac0g-1.jpg?height=412&width=483&top_left_y=1062&top_left_x=1336)

Solution:
Since both the circle with its centre on the $y$-axis and the graph of $y=|x|$ are symmetric about the $y$-axis, then for each point of intersection between these two graphs, there should be a corresponding point of intersection symmetrically located across the $y$-axis. Thus, since there are exactly three points of intersection, then one of these points must be on the $y$-axis, ie. has $x$-coordinate 0 . Since this point is on the graph of $y=|x|$, then this point must be $(0,0)$.

Since the circle has centre on the $y$-axis (say, has coordinates $(0, b)$ ), then its radius is equal to $b$ (and $b$ must be positive for there to be three points of intersection). So the circle has equation $x^{2}+(y-b)^{2}=b^{2}$. Where are the other two points of intersection? We consider the points with $x$ positive and use symmetry to get the other point of intersection.



When $x \geq 0$, then $y=|x|$ has equation $y=x$. Substituting into the equation of the circle,

$$
\begin{aligned}
x^{2}+(x-b)^{2} & =b^{2} \\
2 x^{2}-2 b x & =0 \\
2 x(x-b) & =0
\end{aligned}
$$

Therefore, the points of intersection are $(0,0)$ and $(b, b)$ on the positive side of the $y$-axis, and so at the point $(-b, b)$ on the negative side of the $y$ axis.

Thus the points $O, A$ and $B$ are the points $(0,0)$, $(b, b)$ and $(-b, b)$.

![](https://cdn.mathpix.com/cropped/2023_12_21_d64fac5709fb52027973g-1.jpg?height=401&width=613&top_left_y=510&top_left_x=1233)

Since the radius of the circle is $b$, then the area of the circle is $\pi b^{2}$.

Triangle $O A B$ has a base from $(-b, b)$ to $(b, b)$ of length $2 b$, and a height from the line $y=b$ to the point $(0,0)$ of length $b$, and so an area of $\frac{1}{2} b(2 b)=b^{2}$.

Therefore, the ratio of the area of the triangle to the area of the circle is $b^{2}: \pi b^{2}=1: \pi$.

Final answer: 证明题，略
Subfield: Geometry


Question: In the diagram, triangle $A B C$ has a right angle at $B$ and $M$ is the midpoint of $B C$. A circle is drawn using $B C$ as its diameter. $P$ is the point of intersection of the circle with $A C$. The tangent to the circle at $P$ cuts $A B$ at $Q$. Prove that $Q M$ is parallel to $A C$.

<img_3405>

Solution:
Since $M$ is the midpoint of a diameter of the circle, $M$ is the centre of the circle.

Join $P$ to $M$. Since $Q P$ is tangent to the circle, $P M$ is perpendicular to $Q P$.

Since $P M$ and $B M$ are both radii of the circle, then $P M=M B$.

<img_3432>

Therefore, $\triangle Q P M$ and $\triangle Q B M$ are congruent (Hypotenuse - Side).

Thus, let $\angle M Q B=\angle M Q P=\theta$. So $\angle Q M B=\angle Q M P=90^{\circ}-\theta$

Then $\angle P M C=180^{\circ}-\angle P M Q-\angle B M Q=180^{\circ}-\left(90^{\circ}-\theta\right)-\left(90^{\circ}-\theta\right)=2 \theta$.

But $\triangle P M C$ is isosceles with $P M=M C$ since $P M$ and $M C$ are both radii.

Therefore, $\angle C P M=\frac{1}{2}\left(180^{\circ}-\angle P M C\right)=90^{\circ}-\theta$.

But then $\angle C P M=\angle P M Q$, and since $P M$ is a transversal between $A C$ and $Q M$, then $Q M$ is parallel to $A C$ because of equal alternating angles.
Join $M$ to $P$ and $B$ to $P$.

Since $Q P$ and $Q B$ are tangents to the circle coming from the same point, they have the same length. Since $Q M$ joins the point of intersection of the tangents to the centre of the circle, then by symmetry, $\angle P Q M=\angle B Q M$ and $\angle P M Q=\angle B M Q$. So let $\angle P Q M=\angle B Q M=x$ and $\angle P M Q=\angle B M Q=y$.

<img_3687>

Looking at $\triangle Q M B$, we see that $x+y=90^{\circ}$, since $\triangle Q M B$ is right-angled.

Now if we consider the chord $P B$, we see that its central angle is $2 y$, so any angle that it subtends on the circle (eg. $\angle P C B$ ) is equal to $y$.

Thus, $\angle A C B=\angle Q M B$, so $Q M$ is parallel to $A C$.
Join $P B$.

Since $Q P$ is tangent to the circle, then by the Tangent-Chord Theorem, $\angle Q P B=\angle P C B=x$ (ie. the inscribed angle of a chord is equal to the angle between the tangent and chord.

<img_3658>

Since $B C$ is a diameter of the circle, then $\angle C P B=90^{\circ}$ and so $\angle A P B=90^{\circ}$, whence

$\angle A P Q=90^{\circ}-\angle Q P B=90^{\circ}-x$.

Looking at $\triangle A B C$, we see that $\angle P A Q=90^{\circ}-x$, so $\angle P A Q=\angle A P Q$, and so $A Q=Q P$.

But $Q P$ and $Q B$ are both tangents to the circle $(Q B$ is tangent since it is perpendicular to a radius), so $Q P=Q B$.

But then $A Q=Q B$ and $B M=M C$, so $Q$ is the midpoint of $A B$ and $M$ is the midpoint of $B C$. Thus we can conclude that $Q M$ is parallel to $A C$.

(To justify this last statement, we can show very easily that $\triangle Q B M$ is similar to $\triangle A B C$, and so show that $\angle C A B=\angle M Q B$.)
Subfield: Algebra


Question: A positive integer $n$ is called "savage" if the integers $\{1,2,\dots,n\}$ can be partitioned into three sets $A, B$ and $C$ such that

i) the sum of the elements in each of $A, B$, and $C$ is the same,

ii) $A$ contains only odd numbers,

iii) $B$ contains only even numbers, and

iv) C contains every multiple of 3 (and possibly other numbers).
Show that 8 is a savage integer.

Solution:
To show that 8 is a savage integer, we must partition the set $\{1,2,3,4,5,6,7,8\}$ according to the given criteria.

Since the sum of the integers from 1 to 8 is 36 , then the sum of the elements in each of the sets $A, B$, and $C$ must be 12 .

$C$ must contain both 3 and 6 .

$A$ can contain only the numbers $1,5,7$, and may not contain all of these.

$B$ can contain only the numbers $2,4,8$, and may not contain all of these.

So if we let $C=\{1,2,3,6\}, A=\{5,7\}$ and $B=\{4,8\}$, then these sets have the desired properties.

Therefore, 8 is a savage integer.
Subfield: Combinatorics


Question: Tanner has two identical dice. Each die has six faces which are numbered 2, 3, 5, $7,11,13$. When Tanner rolls the two dice, what is the probability that the sum of the numbers on the top faces is a prime number?

Solution:
We make a table of the 36 possible combinations of rolls and the resulting sums:

|  | 2 | 3 | 5 | 7 | 11 | 13 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 2 | 4 | 5 | 7 | 9 | 13 | 15 |
| 3 | 5 | 6 | 8 | 10 | 14 | 16 |
| 5 | 7 | 8 | 10 | 12 | 16 | 18 |
| 7 | 9 | 10 | 12 | 14 | 18 | 20 |
| 11 | 13 | 14 | 16 | 18 | 22 | 24 |
| 13 | 15 | 16 | 18 | 20 | 24 | 26 |

Of the 36 entries in the table, 6 are prime numbers (two entries each of 5, 7 and 13).

Therefore, the probability that the sum is a prime number is $\frac{6}{36}$ or $\frac{1}{6}$.

(Note that each sum is at least 4 and so must be odd to be prime. Since odd plus odd equals even, then the only possibilities that really need to be checked are even plus odd and odd plus even (that is, the first row and first column of the table).)

Final answer: $\frac{1}{6}$
Subfield: Geometry


Question: In the diagram, $V$ is the vertex of the parabola with equation $y=-x^{2}+4 x+1$. Also, $A$ and $B$ are the points of intersection of the parabola and the line with equation $y=-x+1$. Determine the value of $A V^{2}+B V^{2}-A B^{2}$.

![](https://cdn.mathpix.com/cropped/2023_12_21_859e4fcfbb4e54a90d70g-1.jpg?height=431&width=521&top_left_y=847&top_left_x=1298)

Solution:
First, we find the coordinates of $V$.

To do this, we use the given equation for the parabola and complete the square:

$y=-x^{2}+4 x+1=-\left(x^{2}-4 x-1\right)=-\left(x^{2}-4 x+2^{2}-2^{2}-1\right)=-\left((x-2)^{2}-5\right)=-(x-2)^{2}+5$

Therefore, the coordinates of the vertex $V$ are $(2,5)$.

Next, we find the coordinates of $A$ and $B$.

Note that $A$ and $B$ are the points of intersection of the line with equation $y=-x+1$ and the parabola with equation $y=-x^{2}+4 x+1$.

We equate $y$-values to obtain $-x+1=-x^{2}+4 x+1$ or $x^{2}-5 x=0$ or $x(x-5)=0$.

Therefore, $x=0$ or $x=5$.

If $x=0$, then $y=-x+1=1$, and so $A$ (which is on the $y$-axis) has coordinates $(0,1)$.

If $x=5$, then $y=-x+1=-4$, and so $B$ has coordinates $(5,-4)$.



We now have the points $V(2,5), A(0,1), B(5,-4)$.

This gives

$$
\begin{aligned}
A V^{2} & =(0-2)^{2}+(1-5)^{2}=20 \\
B V^{2} & =(5-2)^{2}+(-4-5)^{2}=90 \\
A B^{2} & =(0-5)^{2}+(1-(-4))^{2}=50
\end{aligned}
$$

and so $A V^{2}+B V^{2}-A B^{2}=20+90-50=60$.

Final answer: 60
Subfield: Geometry


Question: In the diagram, $V$ is the vertex of the parabola with equation $y=-x^{2}+4 x+1$. Also, $A$ and $B$ are the points of intersection of the parabola and the line with equation $y=-x+1$. Determine the value of $A V^{2}+B V^{2}-A B^{2}$.

<img_3223>

Solution:
First, we find the coordinates of $V$.

To do this, we use the given equation for the parabola and complete the square:

$y=-x^{2}+4 x+1=-\left(x^{2}-4 x-1\right)=-\left(x^{2}-4 x+2^{2}-2^{2}-1\right)=-\left((x-2)^{2}-5\right)=-(x-2)^{2}+5$

Therefore, the coordinates of the vertex $V$ are $(2,5)$.

Next, we find the coordinates of $A$ and $B$.

Note that $A$ and $B$ are the points of intersection of the line with equation $y=-x+1$ and the parabola with equation $y=-x^{2}+4 x+1$.

We equate $y$-values to obtain $-x+1=-x^{2}+4 x+1$ or $x^{2}-5 x=0$ or $x(x-5)=0$.

Therefore, $x=0$ or $x=5$.

If $x=0$, then $y=-x+1=1$, and so $A$ (which is on the $y$-axis) has coordinates $(0,1)$.

If $x=5$, then $y=-x+1=-4$, and so $B$ has coordinates $(5,-4)$.



We now have the points $V(2,5), A(0,1), B(5,-4)$.

This gives

$$
\begin{aligned}
A V^{2} & =(0-2)^{2}+(1-5)^{2}=20 \\
B V^{2} & =(5-2)^{2}+(-4-5)^{2}=90 \\
A B^{2} & =(0-5)^{2}+(1-(-4))^{2}=50
\end{aligned}
$$

and so $A V^{2}+B V^{2}-A B^{2}=20+90-50=60$.

Final answer: 60
Subfield: Geometry


Question: In the diagram, $A B C$ is a quarter of a circular pizza with centre $A$ and radius $20 \mathrm{~cm}$. The piece of pizza is placed on a circular pan with $A, B$ and $C$ touching the circumference of the pan, as shown. What fraction of the pan is covered by the piece of pizza?

<img_3564>

Solution:
Since $A B C$ is a quarter of a circular pizza with centre $A$ and radius $20 \mathrm{~cm}$, then $A C=A B=20 \mathrm{~cm}$.

We are also told that $\angle C A B=90^{\circ}$ (one-quarter of $360^{\circ}$ ).

Since $\angle C A B=90^{\circ}$ and $A, B$ and $C$ are all on the circumference of the circle, then $C B$ is a diameter of the pan. (This is a property of circles: if $X, Y$ and $Z$ are three points on a circle with $\angle Z X Y=90^{\circ}$, then $Y Z$ must be a diameter of the circle.)

Since $\triangle C A B$ is right-angled and isosceles, then $C B=\sqrt{2} A C=20 \sqrt{2} \mathrm{~cm}$.

Therefore, the radius of the circular plate is $\frac{1}{2} C B$ or $10 \sqrt{2} \mathrm{~cm}$.

Thus, the area of the circular pan is $\pi(10 \sqrt{2} \mathrm{~cm})^{2}=200 \pi \mathrm{cm}^{2}$.

The area of the slice of pizza is one-quarter of the area of a circle with radius $20 \mathrm{~cm}$, or $\frac{1}{4} \pi(20 \mathrm{~cm})^{2}=100 \pi \mathrm{cm}^{2}$.

Finally, the fraction of the pan that is covered is the area of the slice of pizza divided by the area of the pan, or $\frac{100 \pi \mathrm{cm}^{2}}{200 \pi \mathrm{cm}^{2}}=\frac{1}{2}$.

Final answer: $\frac{1}{2}$
Subfield: Geometry


Question: In the diagram, $A B C$ is a quarter of a circular pizza with centre $A$ and radius $20 \mathrm{~cm}$. The piece of pizza is placed on a circular pan with $A, B$ and $C$ touching the circumference of the pan, as shown. What fraction of the pan is covered by the piece of pizza?

![](https://cdn.mathpix.com/cropped/2023_12_21_859e4fcfbb4e54a90d70g-1.jpg?height=401&width=415&top_left_y=1336&top_left_x=1256)

Solution:
Since $A B C$ is a quarter of a circular pizza with centre $A$ and radius $20 \mathrm{~cm}$, then $A C=A B=20 \mathrm{~cm}$.

We are also told that $\angle C A B=90^{\circ}$ (one-quarter of $360^{\circ}$ ).

Since $\angle C A B=90^{\circ}$ and $A, B$ and $C$ are all on the circumference of the circle, then $C B$ is a diameter of the pan. (This is a property of circles: if $X, Y$ and $Z$ are three points on a circle with $\angle Z X Y=90^{\circ}$, then $Y Z$ must be a diameter of the circle.)

Since $\triangle C A B$ is right-angled and isosceles, then $C B=\sqrt{2} A C=20 \sqrt{2} \mathrm{~cm}$.

Therefore, the radius of the circular plate is $\frac{1}{2} C B$ or $10 \sqrt{2} \mathrm{~cm}$.

Thus, the area of the circular pan is $\pi(10 \sqrt{2} \mathrm{~cm})^{2}=200 \pi \mathrm{cm}^{2}$.

The area of the slice of pizza is one-quarter of the area of a circle with radius $20 \mathrm{~cm}$, or $\frac{1}{4} \pi(20 \mathrm{~cm})^{2}=100 \pi \mathrm{cm}^{2}$.

Finally, the fraction of the pan that is covered is the area of the slice of pizza divided by the area of the pan, or $\frac{100 \pi \mathrm{cm}^{2}}{200 \pi \mathrm{cm}^{2}}=\frac{1}{2}$.

Final answer: $\frac{1}{2}$
Subfield: Geometry


Question: The deck $A B$ of a sailboat is $8 \mathrm{~m}$ long. Rope extends at an angle of $60^{\circ}$ from $A$ to the top $(M)$ of the mast of the boat. More rope extends at an angle of $\theta$ from $B$ to a point $P$ that is $2 \mathrm{~m}$ below $M$, as shown. Determine the height $M F$ of the mast, in terms of $\theta$.

<img_3738>

Solution:
Suppose that the length of $A F$ is $x \mathrm{~m}$.

Since the length of $A B$ is $8 \mathrm{~m}$, then the length of $F B$ is $(8-x) \mathrm{m}$.

Since $\triangle M A F$ is right-angled and has an angle of $60^{\circ}$, then it is $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.

Therefore, $M F=\sqrt{3} A F$, since $M F$ is opposite the $60^{\circ}$ angle and $A F$ is opposite the $30^{\circ}$ angle.

Thus, $M F=\sqrt{3} x \mathrm{~m}$.

Since $M P=2 \mathrm{~m}$, then $P F=M F-M P=(\sqrt{3} x-2) \mathrm{m}$.

We can now look at $\triangle B F P$ which is right-angled at $F$.

We have

$$
\tan \theta=\frac{P F}{F B}=\frac{(\sqrt{3} x-2) \mathrm{m}}{(8-x) \mathrm{m}}=\frac{\sqrt{3} x-2}{8-x}
$$

Therefore, $(8-x) \tan \theta=\sqrt{3} x-2$ or $8 \tan \theta+2=\sqrt{3} x+(\tan \theta) x$.

This gives $8 \tan \theta+2=x(\sqrt{3}+\tan \theta)$ or $x=\frac{8 \tan \theta+2}{\tan \theta+\sqrt{3}}$.

Finally, $M F=\sqrt{3} x=\frac{8 \sqrt{3} \tan \theta+2 \sqrt{3}}{\tan \theta+\sqrt{3}} \mathrm{~m}$.



Final answer: $\frac{8 \sqrt{3} \tan \theta+2 \sqrt{3}}{\tan \theta+\sqrt{3}}$ \mathrm{~m}
Subfield: Geometry


Question: The deck $A B$ of a sailboat is $8 \mathrm{~m}$ long. Rope extends at an angle of $60^{\circ}$ from $A$ to the top $(M)$ of the mast of the boat. More rope extends at an angle of $\theta$ from $B$ to a point $P$ that is $2 \mathrm{~m}$ below $M$, as shown. Determine the height $M F$ of the mast, in terms of $\theta$.

![](https://cdn.mathpix.com/cropped/2023_12_21_859e4fcfbb4e54a90d70g-1.jpg?height=377&width=371&top_left_y=1774&top_left_x=1254)

Solution:
Suppose that the length of $A F$ is $x \mathrm{~m}$.

Since the length of $A B$ is $8 \mathrm{~m}$, then the length of $F B$ is $(8-x) \mathrm{m}$.

Since $\triangle M A F$ is right-angled and has an angle of $60^{\circ}$, then it is $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.

Therefore, $M F=\sqrt{3} A F$, since $M F$ is opposite the $60^{\circ}$ angle and $A F$ is opposite the $30^{\circ}$ angle.

Thus, $M F=\sqrt{3} x \mathrm{~m}$.

Since $M P=2 \mathrm{~m}$, then $P F=M F-M P=(\sqrt{3} x-2) \mathrm{m}$.

We can now look at $\triangle B F P$ which is right-angled at $F$.

We have

$$
\tan \theta=\frac{P F}{F B}=\frac{(\sqrt{3} x-2) \mathrm{m}}{(8-x) \mathrm{m}}=\frac{\sqrt{3} x-2}{8-x}
$$

Therefore, $(8-x) \tan \theta=\sqrt{3} x-2$ or $8 \tan \theta+2=\sqrt{3} x+(\tan \theta) x$.

This gives $8 \tan \theta+2=x(\sqrt{3}+\tan \theta)$ or $x=\frac{8 \tan \theta+2}{\tan \theta+\sqrt{3}}$.

Finally, $M F=\sqrt{3} x=\frac{8 \sqrt{3} \tan \theta+2 \sqrt{3}}{\tan \theta+\sqrt{3}} \mathrm{~m}$.

Final answer: $\frac{8 \sqrt{3} \tan \theta+2 \sqrt{3}}{\tan \theta+\sqrt{3}} \mathrm{~m}$
Subfield: Algebra


Question: If $\frac{1}{\cos x}-\tan x=3$, what is the numerical value of $\sin x$ ?

Solution:
Beginning with the given equation, we have

$$
\begin{aligned}
\frac{1}{\cos x}-\tan x & =3 \\
\frac{1}{\cos x}-\frac{\sin x}{\cos x} & =3 \\
1-\sin x & =3 \cos x \quad(\text { since } \cos x \neq 0) \\
(1-\sin x)^{2} & =9 \cos ^{2} x \quad \text { (squaring both sides) } \\
1-2 \sin x+\sin ^{2} x & =9\left(1-\sin ^{2} x\right) \\
10 \sin ^{2} x-2 \sin x-8 & =0 \\
5 \sin ^{2} x-\sin x-4 & =0 \\
(5 \sin x+4)(\sin x-1) & =0
\end{aligned}
$$

Therefore, $\sin x=-\frac{4}{5}$ or $\sin x=1$.

If $\sin x=1$, then $\cos x=0$ and $\tan x$ is undefined, which is inadmissible in the original equation.

Therefore, $\sin x=-\frac{4}{5}$.

(We can check that if $\sin x=-\frac{4}{5}$, then $\cos x= \pm \frac{3}{5}$ and the possibility that $\cos x=\frac{3}{5}$ satisfies the original equation, since in this case $\frac{1}{\cos x}=\frac{5}{3}$ and $\tan x=-\frac{4}{3}$ and the difference between these fractions is 3 .)

Final answer: $-\frac{4}{5}$
Subfield: Algebra


Question: Determine all linear functions $f(x)=a x+b$ such that if $g(x)=f^{-1}(x)$ for all values of $x$, then $f(x)-g(x)=44$ for all values of $x$. (Note: $f^{-1}$ is the inverse function of $f$.)

Solution:
Since $f(x)=a x+b$, we can determine an expression for $g(x)=f^{-1}(x)$ by letting $y=f(x)$ to obtain $y=a x+b$. We then interchange $x$ and $y$ to obtain $x=a y+b$ which we solve for $y$ to obtain $a y=x-b$ or $y=\frac{x}{a}-\frac{b}{a}$.

Therefore, $f^{-1}(x)=\frac{x}{a}-\frac{b}{a}$.

Note that $a \neq 0$. (This makes sense since the function $f(x)=b$ has a graph which is a horizontal line, and so cannot be invertible.)

Therefore, the equation $f(x)-g(x)=44$ becomes $(a x+b)-\left(\frac{x}{a}-\frac{b}{a}\right)=44$ or $\left(a-\frac{1}{a}\right) x+\left(b+\frac{b}{a}\right)=44=0 x+44$, and this equation is true for all $x$.

We can proceed in two ways.

Method \#1: Comparing coefficients

Since the equation

$$
\left(a-\frac{1}{a}\right) x+\left(b+\frac{b}{a}\right)=0 x+44
$$

is true for all $x$, then the coefficients of the linear expression on the left side must match the coefficients of the linear expression on the right side.

Therefore, $a-\frac{1}{a}=0$ and $b+\frac{b}{a}=44$.

From the first of these equations, we obtain $a=\frac{1}{a}$ or $a^{2}=1$, which gives $a=1$ or $a=-1$. If $a=1$, the equation $b+\frac{b}{a}=44$ becomes $b+b=44$, which gives $b=22$.



If $a=-1$, the equation $b+\frac{b}{a}=44$ becomes $b-b=44$, which is not possible.

Therefore, we must have $a=1$ and $b=22$, and so $f(x)=x+22$.

Method \#2: Trying specific values for $x$

Since the equation

$$
\left(a-\frac{1}{a}\right) x+\left(b+\frac{b}{a}\right)=0 x+44
$$

is true for all values of $x$, then it must be true for any specific values of $x$ that we choose.

Choosing $x=0$, we obtain $0+\left(b+\frac{b}{a}\right)=44$ or $b+\frac{b}{a}=44$.

Choosing $x=b$, we obtain $\left(a-\frac{1}{a}\right) b+\left(b+\frac{b}{a}\right)=44$ or $a b+b=44$.

We can rearrange the first of these equations to get $\frac{a b+b}{a}=44$.

Using the second equation, we obtain $\frac{44}{a}=44$ or $a=1$.

Since $a=1$, then $a b+b=44$ gives $2 b=44$ or $b=22$.

Thus, $f(x)=x+22$.

In summary, the only linear function $f$ for which the given equation is true for all $x$ is $f(x)=x+22$.

Final answer: $f(x)=x+22$
Subfield: Number Theory


Question: Determine all pairs $(a, b)$ of positive integers for which $a^{3}+2 a b=2013$.

Solution:
First, we factor the left side of the given equation to obtain $a\left(a^{2}+2 b\right)=2013$.

Next, we factor the integer 2013 as $2013=3 \times 671=3 \times 11 \times 61$. Note that each of 3,11 and 61 is prime, so we can factor 2013 no further. (We can find the factors of 3 and 11 using tests for divisibility by 3 and 11, or by systematic trial and error.)

Since $2013=3 \times 11 \times 61$, then the positive divisors of 2013 are

$$
1,3,11,33,61,183,671,2013
$$

Since $a$ and $b$ are positive integers, then $a$ and $a^{2}+2 b$ are both positive integers.

Since $a$ and $b$ are positive integers, then $a^{2} \geq a$ and $2 b>0$, so $a^{2}+2 b>a$.

Since $a\left(a^{2}+2 b\right)=2013$, then $a$ and $a^{2}+2 b$ must be a divisor pair of 2013 (that is, a pair of positive integers whose product is 2013) with $a<a^{2}+2 b$.

We make a table of the possibilities:

| $a$ | $a^{2}+2 b$ | $2 b$ | $b$ |
| :---: | :---: | :---: | :---: |
| 1 | 2013 | 2012 | 1006 |
| 3 | 671 | 662 | 331 |
| 11 | 183 | 62 | 31 |
| 33 | 61 | -1028 | N/A |

Note that the last case is not possible, since $b$ must be positive.

Therefore, the three pairs of positive integers that satisfy the equation are $(1,1006)$, $(3,331),(11,31)$.

(We can verify by substitution that each is a solution of the original equation.)

Final answer: $(1,1006),(3,331),(11,31)$