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


Question: In the diagram, the ferris wheel has a diameter of $18 \mathrm{~m}$ and rotates at a constant rate. When Kolapo rides the ferris wheel and is at its lowest point, he is $1 \mathrm{~m}$ above the ground. When Kolapo is at point $P$ that is $16 \mathrm{~m}$ above the ground and is rising, it takes him 4 seconds to reach the highest point, $T$. He continues to travel for another 8 seconds reaching point $Q$. Determine Kolapo's height above the ground when he reaches point $Q$.

![](https://cdn.mathpix.com/cropped/2023_12_21_040d04737e70e698092cg-1.jpg?height=439&width=591&top_left_y=2125&top_left_x=1258)

Solution:
Let $O$ be the centre of the ferris wheel and $B$ the lowest point on the wheel.

Since the radius of the ferris wheel is $9 \mathrm{~m}$ (half of the diameter of $18 \mathrm{~m}$ ) and $B$ is $1 \mathrm{~m}$ above the ground, then $O$ is $9+1=10 \mathrm{~m}$ above the ground.

Let $\angle T O P=\theta$.

![](https://cdn.mathpix.com/cropped/2023_12_21_09e23e682ca3ac23cf59g-1.jpg?height=441&width=555&top_left_y=384&top_left_x=880)

Since the ferris wheel rotates at a constant speed, then in 8 seconds, the angle through which the wheel rotates is twice the angle through which it rotates in 4 seconds. In other words, $\angle T O Q=2 \theta$.

Draw a perpendicular from $P$ to $R$ on $T B$ and from $Q$ to $G$ on $T B$.

Since $P$ is $16 \mathrm{~m}$ above the ground and $O$ is $10 \mathrm{~m}$ above the ground, then $O R=6 \mathrm{~m}$.

Since $O P$ is a radius of the circle, then $O P=9 \mathrm{~m}$.

Looking at right-angled $\triangle O R P$, we see that $\cos \theta=\frac{O R}{O P}=\frac{6}{9}=\frac{2}{3}$.

Since $\cos \theta=\frac{2}{3}<\frac{1}{\sqrt{2}}=\cos \left(45^{\circ}\right)$, then $\theta>45^{\circ}$.

This means that $2 \theta>90^{\circ}$, which means that $Q$ is below the horizontal diameter through $O$ and so $G$ is below $O$.

Since $\angle T O Q=2 \theta$, then $\angle Q O G=180^{\circ}-2 \theta$.

Kolapo's height above the ground at $Q$ equals $1 \mathrm{~m}$ plus the length of $B G$.

Now $B G=O B-O G$. We know that $O B=9 \mathrm{~m}$.

Also, considering right-angled $\triangle Q O G$, we have

$$
O G=O Q \cos (\angle Q O G)=9 \cos \left(180^{\circ}-2 \theta\right)=-9 \cos (2 \theta)=-9\left(2 \cos ^{2} \theta-1\right)
$$

Since $\cos \theta=\frac{2}{3}$, then $O G=-9\left(2\left(\frac{2}{3}\right)^{2}-1\right)=-9\left(\frac{8}{9}-1\right)=1 \mathrm{~m}$.

Therefore, $B G=9-1=8 \mathrm{~m}$ and so $Q$ is $1+8=9 \mathrm{~m}$ above the ground.

Final answer: 9 m
Subfield: Algebra


Question: Determine all real values of $x$ such that

$$
\log _{5 x+9}\left(x^{2}+6 x+9\right)+\log _{x+3}\left(5 x^{2}+24 x+27\right)=4
$$

Solution:
We manipulate the given equation into a sequence of equivalent equations:

$$
\begin{array}{rll}
\log _{5 x+9}\left(x^{2}+6 x+9\right)+\log _{x+3}\left(5 x^{2}+24 x+27\right) & =4 & \\
\frac{\log \left(x^{2}+6 x+9\right)}{\log (5 x+9)}+\frac{\log \left(5 x^{2}+24 x+27\right)}{\log (x+3)} & =4 & \text { (using the "change of base" formula) } \\
\frac{\log \left((x+3)^{2}\right)}{\log (5 x+9)}+\frac{\log ((5 x+9)(x+3))}{\log (x+3)} & =4 & \text { (factoring) } \\
\frac{2 \log (x+3)}{\log (5 x+9)}+\frac{\log (5 x+9)+\log (x+3)}{\log (x+3)} & =4 & \text { (using logarithm rules) } \\
2\left(\frac{\log (x+3)}{\log (5 x+9)}\right)+\frac{\log (5 x+9)}{\log (x+3)}+\frac{\log (x+3)}{\log (x+3)} & =4 & \text { (rearranging fractions) }
\end{array}
$$



Making the substitution $t=\frac{\log (x+3)}{\log (5 x+9)}$, we obtain successively

$$
\begin{aligned}
2 t+\frac{1}{t}+1 & =4 \\
2 t^{2}+1+t & =4 t \\
2 t^{2}-3 t+1 & =0 \\
(2 t-1)(t-1) & =0
\end{aligned}
$$

Therefore, $t=1$ or $t=\frac{1}{2}$.

If $\frac{\log (x+3)}{\log (5 x+9)}=1$, then $\log (x+3)=\log (5 x+9)$ or $x+3=5 x+9$, which gives $4 x=-6$ or $x=-\frac{3}{2}$.

If $\frac{\log (x+3)}{\log (5 x+9)}=\frac{1}{2}$, then $2 \log (x+3)=\log (5 x+9)$ or $\log \left((x+3)^{2}\right)=\log (5 x+9)$ or $(x+3)^{2}=5 x+9$.

Here, $x^{2}+6 x+9=5 x+9$ or $x^{2}+x=0$ or $x(x+1)=0$, and so $x=0$ or $x=-1$.

Therefore, there are three possible values for $x: x=0, x=-1$ and $x=-\frac{3}{2}$.

We should check each of these in the original equation.

If $x=0$, the left side of the original equation is $\log _{9} 9+\log _{3} 27=1+3=4$.

If $x=-1$, the left side of the original equation is $\log _{4} 4+\log _{2} 8=1+3=4$.

If $x=-\frac{3}{2}$, the left side of the original equation is $\log _{3 / 2}(9 / 4)+\log _{3 / 2}(9 / 4)=2+2=4$.

Therefore, the solutions are $x=0,-1,-\frac{3}{2}$.

Final answer: $0,-1,-\frac{3}{2}$
Subfield: Combinatorics


Question: An auditorium has a rectangular array of chairs. There are exactly 14 boys seated in each row and exactly 10 girls seated in each column. If exactly 3 chairs are empty, prove that there are at least 567 chairs in the auditorium.

Solution:
Suppose that the auditorium with these properties has $r$ rows and $c$ columns of chairs.

Then there are $r c$ chairs in total.

Each chair is empty, is occupied by a boy, or is occupied by a girl.

Since there are 14 boys in each row, then there are $14 r$ chairs occupied by boys.

Since there are 10 girls in each column, then there are 10c chairs occupied by girls.

Since there are exactly 3 empty chairs, then the total number of chairs can also be written as $14 r+10 c+3$.

Therefore, $r c=14 r+10 c+3$.

We proceed to find all pairs of positive integers $r$ and $c$ that satisfy this equation. We note that since there are 14 boys in each row, then there must be at least 14 columns (that is, $c \geq 14$ ) and since there are 10 girls in each column, then there must be at least 10 rows (that is, $r \geq 10$ ).

Manipulating the equation,

$$
\begin{aligned}
r c & =14 r+10 c+3 \\
r c-14 r & =10 c+3 \\
r(c-14) & =10 c+3 \\
r & =\frac{10 c+3}{c-14} \\
r & =\frac{10 c-140+143}{c-14} \\
r & =\frac{10 c-140}{c-14}+\frac{143}{c-14} \\
r & =10+\frac{143}{c-14}
\end{aligned}
$$



Since $r$ is an integer, then $10+\frac{143}{c-14}$ is an integer, so $\frac{143}{c-14}$ must be an integer.

Therefore, $c-14$ is a divisor of 143 . Since $c \geq 14$, then $c-14 \geq 0$, so $c-14$ is a positive divisor of 143 .

Since $143=11 \times 13$, then its positive divisors are 1,11,13, 143 .

We make a table of the possible values of $c-14$ along with the resulting values of $c, r$ (calculated using $r=10+\frac{143}{c-14}$ ) and $r c$ :

| $c-14$ | $c$ | $r$ | $r c$ |
| :---: | :---: | :---: | :---: |
| 1 | 15 | 153 | 2295 |
| 11 | 25 | 23 | 575 |
| 13 | 27 | 21 | 567 |
| 143 | 157 | 11 | 1727 |

Therefore, the four possible values for $r c$ are 567,575,1727,2295. That is, the smallest possible number of chairs in the auditorium is 567 .
Subfield: Combinatorics


Question: For each positive integer $N$, an Eden sequence from $\{1,2,3, \ldots, N\}$ is defined to be a sequence that satisfies the following conditions:

(i) each of its terms is an element of the set of consecutive integers $\{1,2,3, \ldots, N\}$,

(ii) the sequence is increasing, and

(iii) the terms in odd numbered positions are odd and the terms in even numbered positions are even.

For example, the four Eden sequences from $\{1,2,3\}$ are

$$
\begin{array}{llll}
1 & 3 & 1,2 & 1,2,3
\end{array}
$$
Determine the number of Eden sequences from $\{1,2,3,4,5\}$.

Solution:
The Eden sequences from $\{1,2,3,4,5\}$ are

$$
135 \quad 5 \quad 1,2 \quad 1,4 \quad 3,4 \quad 1,2,3 \quad 1,2,5 \quad 1,4,5 \quad 3,4,5 \quad 1,2,3,4 \quad 1,2,3,4,5
$$

There are 12 such sequences.

We present a brief justification of why these are all of the sequences.

* An Eden sequence of length 1 consists of a single odd integer. The possible choices are 1 and 3 and 5 .
* An Eden sequence of length 2 consists of an odd integer followed by a larger even integer. Since the only possible even integers here are 2 and 4 , then the possible sequences are 1, 2 and 1, 4 and 3,4 .
* An Eden sequence of length 3 starts with an Eden sequence of length 2 and appends (that is, adds to the end) a larger odd integer. Starting with 1,2, we form 1,2,3 and $1,2,5$. Starting with 1,4 , we form $1,4,5$. Starting with 3,4 , we form $3,4,5$.
* An Eden sequence of length 4 starts with an Eden sequence of length 3 and appends a larger even integer. Since 2 and 4 are the only possible even integers, then the only possible sequence here is $1,2,3,4$.
* An Eden sequence of length 5 from $\{1,2,3,4,5\}$ must include all 5 elements, so is $1,2,3,4,5$.

Final answer: 12
Subfield: Geometry


Question: In the diagram, the circle with centre $C(1,1)$ passes through the point $O(0,0)$, intersects the $y$-axis at $A$, and intersects the $x$-axis at $B(2,0)$. Determine, with justification, the coordinates of $A$ and the area of the part of the circle that lies in the first quadrant.

<img_3733>

Solution:
Since $\angle A O B=90^{\circ}, A B$ is a diameter of the circle.

Join $A B$.

<img_3988>

Since $C$ is the centre of the circle and $A B$ is a diameter, then $C$ is the midpoint of $A B$, so $A$ has coordinates $(0,2)$.

Therefore, the area of the part of the circle inside the first quadrant is equal to the area of $\triangle A O B$ plus the area of the semi-circle above $A B$.

The radius of the circle is equal to the distance from $C$ to $B$, or $\sqrt{(1-2)^{2}+(1-0)^{2}}=\sqrt{2}$, so the area of the semi-circle is $\frac{1}{2} \pi(\sqrt{2})^{2}=\pi$.

The area of $\triangle A O B$ is $\frac{1}{2}(O B)(A O)=\frac{1}{2}(2)(2)=2$.

Thus, the area of the part of the circle inside the first quadrant is $\pi+2$.
Since $\angle A O B=90^{\circ}, A B$ is a diameter of the circle.

Join $A B$.

<img_3957>

Since $C$ is the centre of the circle and $A B$ is a diameter, then $C$ is the midpoint of $A B$, so $A$ has coordinates $(0,2)$.

Thus, $A O=B O$.

We "complete the square" by adding point $D(2,2)$, which is on the circle, by symmetry.



<img_3605>

The area of the square is 4 .

The radius of the circle is equal to the distance from $C$ to $B$, or $\sqrt{(1-2)^{2}+(1-0)^{2}}=\sqrt{2}$, so the area of the circle is $\pi(\sqrt{2})^{2}=2 \pi$.

The area of the portion of the circle outside the square is thus $2 \pi-4$. This area is divided into four equal sections (each of area $\frac{1}{4}(2 \pi-4)=\frac{1}{2} \pi-1$ ), two of which are the only portions of the circle outside the first quadrant.

Therefore, the area of the part of the circle inside the the first quadrant is $2 \pi-2\left(\frac{1}{2} \pi-1\right)=$ $\pi+2$.

Two additional ways to find the coordinates of $A$ :

$*$ The length of $O C$ is $\sqrt{1^{2}+1^{2}}=\sqrt{2}$.

Since $C$ is the centre of the circle and $O$ lies on the circle, then the circle has radius $\sqrt{2}$.

Since the circle has centre $(1,1)$ and radius $\sqrt{2}$, its equation is $(x-1)^{2}+(y-1)^{2}=2$. To find the coordinates of $A$, we substitute $x=0$ to obtain $(0-1)^{2}+(y-1)^{2}=2$ or $(y-1)^{2}=1$, and so $y=0$ or $y=2$.

Since $y=0$ gives us the point $O$, then $y=2$ gives us $A$, ie. $A$ has coordinates $(0,2)$.

* Since $O$ and $A$ are both on the circle and each has a horizontal distance of 1 from $C$, then their vertical distances from $C$ must be same, ie. must each be 1.

Thus, $A$ has coordinates $(0,2)$.

Final answer: $(0,2),\pi+2$
Subfield: Combinatorics


Question: If $a$ is chosen randomly from the set $\{1,2,3,4,5\}$ and $b$ is chosen randomly from the set $\{6,7,8\}$, what is the probability that $a^{b}$ is an even number?

Solution:
Since there are 5 choices for $a$ and 3 choices for $b$, there are fifteen possible ways of choosing $a$ and $b$.

If $a$ is even, $a^{b}$ is even; if $a$ is odd, $a^{b}$ is odd.

So the choices of $a$ and $b$ which give an even value for $a^{b}$ are those where $a$ is even, or 6 of the choices (since there are two even choices for $a$ and three ways of choosing $b$ for each of these). (Notice that in fact the value of $b$ does not affect whether $a^{b}$ is even or odd, so the probability depends only on the choice of $a$.)

Thus, the probability is $\frac{6}{15}=\frac{2}{5}$.

Final answer: $\frac{2}{5}$
Subfield: Combinatorics


Question: A bag contains some blue and some green hats. On each turn, Julia removes one hat without looking, with each hat in the bag being equally likely to be chosen. If it is green, she adds a blue hat into the bag from her supply of extra hats, and if it is blue, she adds a green hat to the bag. The bag initially contains 4 blue hats and 2 green hats. What is the probability that the bag again contains 4 blue hats and 2 green hats after two turns?

Solution:
Starting with 4 blue hats and 2 green hats, the probability that Julia removes a blue hat is $\frac{4}{6}=\frac{2}{3}$. The result would be 3 blue hats and 3 green hats, since a blue hat is replaced with a green hat.

In order to return to 4 blue hats and 2 green hats from 3 blue and 3 green, Julia would need remove a green hat (which would be replaced by a blue hat). The probability of her



removing a green hat from 3 blue and 3 green is $\frac{3}{6}=\frac{1}{2}$.

Summarizing, the probability of choosing a blue hat and then a green hat is $\frac{2}{3} \times \frac{1}{2}=\frac{1}{3}$.

Starting with 4 blue hats and 2 green hats, the probability that Julia removes a green hat is $\frac{2}{6}=\frac{1}{3}$. The result would be 5 blue hats and 1 green hat, since a green hat is replaced with a blue hat.

In order to return to 4 blue hats and 2 green hats from 5 blue and 1 green, Julia would need remove a blue hat (which would be replaced by a green hat). The probability of her removing a green hat from 5 blue and 1 green is $\frac{5}{6}$.

Summarizing, the probability of choosing a green hat and then a blue hat is $\frac{1}{3} \times \frac{5}{6}=\frac{5}{18}$.

These are the only two ways to return to 4 blue hats and 2 green hats after two turns removing a blue hat then a green, or removing a green then a blue.

Therefore, the total probability of returning to 4 blue hats and 2 green hats after two turns is $\frac{1}{3}+\frac{5}{18}=\frac{11}{18}$.

Final answer: $\frac{11}{18}$
Subfield: Combinatorics


Question: Suppose that, for some angles $x$ and $y$,

$$
\begin{aligned}
& \sin ^{2} x+\cos ^{2} y=\frac{3}{2} a \\
& \cos ^{2} x+\sin ^{2} y=\frac{1}{2} a^{2}
\end{aligned}
$$

Determine the possible value(s) of $a$.

Solution:
Adding the two equations, we obtain

$$
\begin{aligned}
\sin ^{2} x+\cos ^{2} x+\sin ^{2} y+\cos ^{2} y & =\frac{3}{2} a+\frac{1}{2} a^{2} \\
2 & =\frac{3}{2} a+\frac{1}{2} a^{2} \\
4 & =3 a+a^{2} \\
0 & =a^{2}+3 a-4 \\
0 & =(a+4)(a-1)
\end{aligned}
$$

and so $a=-4$ or $a=1$.

However, $a=-4$ is impossible, since this would give $\sin ^{2} x+\cos ^{2} y=-6$, whose left side is non-negative and whose right side is negative.

Therefore, the only possible value for $a$ is $a=1$.

(We can check that angles $x=90^{\circ}$ and $y=45^{\circ}$ give $\sin ^{2} x+\cos ^{2} y=\frac{3}{2}$ and $\cos ^{2} x+\sin ^{2} y=$ $\frac{1}{2}$, so $a=1$ is indeed possible.)

Final answer: 1
Subfield: Geometry


Question: Survivors on a desert island find a piece of plywood $(A B C)$ in the shape of an equilateral triangle with sides of length $2 \mathrm{~m}$. To shelter their goat from the sun, they place edge $B C$ on the ground, lift corner $A$, and put in a vertical post $P A$ which is $h \mathrm{~m}$ long above ground. When the sun is directly overhead, the shaded region $(\triangle P B C)$ on the ground directly underneath the plywood is an isosceles triangle with largest angle $(\angle B P C)$ equal to $120^{\circ}$. Determine the value of $h$, to the nearest centimetre.

![](https://cdn.mathpix.com/cropped/2023_12_21_09a69cbcf9a187d86706g-1.jpg?height=317&width=376&top_left_y=1603&top_left_x=1449)

Solution:
From the given information, $P C=P B$.

If we can calculate the length of $P C$, we can calculate the value of $h$, since we already know the length of $A C$.

Now $\triangle C P B$ is isosceles with $P C=P B, B C=2$ and $\angle B P C=120^{\circ}$.

Since $\triangle C P B$ is isosceles, $\angle P C B=\angle P B C=30^{\circ}$.

![](https://cdn.mathpix.com/cropped/2023_12_21_6944921028a13ba96297g-1.jpg?height=206&width=463&top_left_y=2274&top_left_x=934)

Join $P$ to the midpoint, $M$, of $B C$.

Then $P M$ is perpendicular to $B C$, since $\triangle P C B$ is isosceles.



Therefore, $\triangle P M C$ is right-angled, has $\angle P C M=30^{\circ}$ and has $C M=1$.

Thus, $P C=\frac{2}{\sqrt{3}}$.

(There are many other techniques that we can use to calculate the length of $P C$.)

Returning to $\triangle A P C$, we see $A P^{2}=A C^{2}-P C^{2}$ or $h^{2}=2^{2}-\left(\frac{2}{\sqrt{3}}\right)^{2}=4-\frac{4}{3}=\frac{8}{3}$, and so $h=\sqrt{\frac{8}{3}}=2 \sqrt{\frac{2}{3}}=\frac{2 \sqrt{6}}{3} \approx 1.630$.

Therefore, the height is approximately $1.63 \mathrm{~m}$ or $163 \mathrm{~cm}$.

Final answer: 163 cm
Subfield: Geometry


Question: Survivors on a desert island find a piece of plywood $(A B C)$ in the shape of an equilateral triangle with sides of length $2 \mathrm{~m}$. To shelter their goat from the sun, they place edge $B C$ on the ground, lift corner $A$, and put in a vertical post $P A$ which is $h \mathrm{~m}$ long above ground. When the sun is directly overhead, the shaded region $(\triangle P B C)$ on the ground directly underneath the plywood is an isosceles triangle with largest angle $(\angle B P C)$ equal to $120^{\circ}$. Determine the value of $h$, to the nearest centimetre.

<img_3293>

Solution:
From the given information, $P C=P B$.

If we can calculate the length of $P C$, we can calculate the value of $h$, since we already know the length of $A C$.

Now $\triangle C P B$ is isosceles with $P C=P B, B C=2$ and $\angle B P C=120^{\circ}$.

Since $\triangle C P B$ is isosceles, $\angle P C B=\angle P B C=30^{\circ}$.

<img_3969>

Join $P$ to the midpoint, $M$, of $B C$.

Then $P M$ is perpendicular to $B C$, since $\triangle P C B$ is isosceles.



Therefore, $\triangle P M C$ is right-angled, has $\angle P C M=30^{\circ}$ and has $C M=1$.

Thus, $P C=\frac{2}{\sqrt{3}}$.

(There are many other techniques that we can use to calculate the length of $P C$.)

Returning to $\triangle A P C$, we see $A P^{2}=A C^{2}-P C^{2}$ or $h^{2}=2^{2}-\left(\frac{2}{\sqrt{3}}\right)^{2}=4-\frac{4}{3}=\frac{8}{3}$, and so $h=\sqrt{\frac{8}{3}}=2 \sqrt{\frac{2}{3}}=\frac{2 \sqrt{6}}{3} \approx 1.630$.

Therefore, the height is approximately $1.63 \mathrm{~m}$ or $163 \mathrm{~cm}$.

Final answer: 163 cm
Subfield: Algebra


Question: The sequence $2,5,10,50,500, \ldots$ is formed so that each term after the second is the product of the two previous terms. The 15 th term ends with exactly $k$ zeroes. What is the value of $k$ ?

Solution:
We calculate the first 15 terms, writing each as an integer times a power of 10:

$$
\begin{gathered}
2,5,10,5 \times 10,5 \times 10^{2}, 5^{2} \times 10^{3}, 5^{3} \times 10^{5}, 5^{5} \times 10^{8}, 5^{8} \times 10^{13}, 5^{13} \times 10^{21}, 5^{21} \times 10^{34} \\
5^{34} \times 10^{55}, 5^{55} \times 10^{89}, 5^{89} \times 10^{144}, 5^{144} \times 10^{233}
\end{gathered}
$$

Since the 15 th term equals an odd integer times $10^{233}$, then the 15 th term ends with 233 zeroes.
To obtain the 6 th term, we calculate $50 \times 500=25 \times 1000$.

Each of the 4th and 5th terms equals an odd integer followed by a number of zeroes, so the 6th term also equals an odd integer followed by a number of zeroes, where the number of zeroes is the sum of the numbers of zeroes at the ends of the 4th and 5th terms.

This pattern will continue. Thus, starting with the 6th term, the number of zeroes at the end of the term will be the sum of the number of zeroes at the ends of the two previous terms.

This tells us that, starting with the 4th term, the number of zeroes at the ends of the terms is

$$
1,2,3,5,8,13,21,34,55,89,144,233
$$

Therefore, the 15 th term ends with 233 zeroes.

Final answer: 233
Subfield: Algebra


Question: Suppose that $a, b, c$ are three consecutive terms in an arithmetic sequence. Prove that $a^{2}-b c, b^{2}-a c$, and $c^{2}-a b$ are also three consecutive terms in an arithmetic sequence.

(An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant. For example, $3,5,7$ is an arithmetic sequence with three terms.)

Solution:
Since $a, b$ and $c$ are consecutive terms in an arithmetic sequence, then $b=a+d$ and $c=a+2 d$ for some number $d$.

Therefore,

$$
\begin{aligned}
& a^{2}-b c=a^{2}-(a+d)(a+2 d)=a^{2}-a^{2}-3 a d-2 d^{2}=-3 a d-2 d^{2} \\
& b^{2}-a c=(a+d)^{2}-a(a+2 d)=a^{2}+2 a d+d^{2}-a^{2}-2 a d=d^{2} \\
& c^{2}-a b=(a+2 d)^{2}-a(a+d)=a^{2}+4 a d+4 d^{2}-a^{2}-a d=3 a d+4 d^{2}
\end{aligned}
$$

Thus,

$$
\left(b^{2}-a c\right)-\left(a^{2}-b c\right)=d^{2}-\left(-3 a d-2 d^{2}\right)=3 d^{2}+3 a d
$$

and

$$
\left(c^{2}-a b\right)-\left(b^{2}-a c\right)=\left(3 a d+4 d^{2}\right)-d^{2}=3 d^{2}+3 a d
$$

Therefore, $\left(b^{2}-a c\right)-\left(a^{2}-b c\right)=\left(c^{2}-a b\right)-\left(b^{2}-a c\right)$, so the sequence $a^{2}-b c, b^{2}-a c$ and $c^{2}-a b$ is arithmetic.
Since $a, b$ and $c$ are consecutive terms in an arithmetic sequence, then $a=b-d$ and $c=b+d$ for some number $d$.

Therefore,

$$
\begin{aligned}
& a^{2}-b c=(b-d)^{2}-b(b+d)=b^{2}-2 b d+d^{2}-b^{2}-b d=-3 b d+d^{2} \\
& b^{2}-a c=b^{2}-(b-d)(b+d)=b^{2}-b^{2}+d^{2}=d^{2} \\
& c^{2}-a b=(b+d)^{2}-(b-d) b=b^{2}+2 b d+d^{2}-b^{2}+b d=3 b d+d^{2}
\end{aligned}
$$

Thus,

$$
\left(b^{2}-a c\right)-\left(a^{2}-b c\right)=d^{2}-\left(-3 b d+d^{2}\right)=3 b d
$$

and

$$
\left(c^{2}-a b\right)-\left(b^{2}-a c\right)=\left(3 b d+d^{2}\right)-d^{2}=3 b d
$$

Therefore, $\left(b^{2}-a c\right)-\left(a^{2}-b c\right)=\left(c^{2}-a b\right)-\left(b^{2}-a c\right)$, so the sequence $a^{2}-b c, b^{2}-a c$ and $c^{2}-a b$ is arithmetic.
To show that $a^{2}-b c, b^{2}-a c$ and $c^{2}-a b$ form an arithmetic sequence, we can show that $\left(c^{2}-a b\right)+\left(a^{2}-b c\right)=2\left(b^{2}-a c\right)$.

Since $a, b$ and $c$ form an arithmetic sequence, then $a+c=2 b$.

Now

$$
\begin{aligned}
\left(c^{2}-a b\right)+\left(a^{2}-b c\right) & =c^{2}+a^{2}-b(a+c) \\
& =c^{2}+a^{2}+2 a c-b(a+c)-2 a c \\
& =(c+a)^{2}-b(a+c)-2 a c \\
& =(c+a)(a+c-b)-2 a c \\
& =2 b(2 b-b)-2 a c \\
& =2 b^{2}-2 a c \\
& =2\left(b^{2}-a c\right)
\end{aligned}
$$

as required.
Subfield: Algebra


Question: If $\log _{2} x-2 \log _{2} y=2$, determine $y$, as a function of $x$

Solution:
We use logarithm rules to rearrange the equation to solve for $y$ :

$$
\begin{aligned}
\log _{2} x-2 \log _{2} y & =2 \\
\log _{2} x-\log _{2}\left(y^{2}\right) & =2 \\
\log _{2}\left(\frac{x}{y^{2}}\right) & =2 \\
\frac{x}{y^{2}} & =2^{2} \\
\frac{1}{4} x & =y^{2} \\
y & = \pm \frac{1}{2} \sqrt{x}
\end{aligned}
$$

But since the domain of the $\log _{2}$ function is all positive real numbers, we must have $x>0$ and $y>0$, so we can reject the negative square root to obtain

$$
y=\frac{1}{2} \sqrt{x}, \quad x>0
$$

Final answer: $\frac{1}{2},\sqrt{x}$