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


Question: Let $k$ be a positive integer with $k \geq 2$. Two bags each contain $k$ balls, labelled with the positive integers from 1 to $k$. André removes one ball from each bag. (In each bag, each ball is equally likely to be chosen.) Define $P(k)$ to be the probability that the product of the numbers on the two balls that he chooses is divisible by $k$.
Determine, with justification, a polynomial $f(n)$ for which

- $P(n) \geq \frac{f(n)}{n^{2}}$ for all positive integers $n$ with $n \geq 2$, and
- $P(n)=\frac{f(n)}{n^{2}}$ for infinitely many positive integers $n$ with $n \geq 2$.

(A polynomial $f(x)$ is an algebraic expression of the form $f(x)=a_{m} x^{m}+a_{m-1} x^{m-1}+\cdots+a_{1} x+a_{0}$ for some integer $m \geq 0$ and for some real numbers $a_{m}, a_{m-1}, \ldots, a_{1}, a_{0}$.)

Solution:
Let $n$ be a positive integer with $n \geq 2$.

Consider $f(n)=2 n-1$. This is a polynomial in $n$.

We demonstrate that $P(n) \geq \frac{2 n-1}{n^{2}}$ for all positive integers $n$ with $n \geq 2$ and that $P(n)=\frac{2 n-1}{n^{2}}$ for infinitely many positive integers $n$ with $n \geq 2$.

Suppose that there are two bags, each containing $n$ balls labelled from 1 to $n$.

Since there are $n$ balls in each bag, there are $n^{2}$ pairs of balls that can be chosen.

Let $a$ be the number on the first ball chosen and $b$ be the number on the second ball chosen.

The pairs

$$
(a, b)=(1, n),(2, n), \ldots,(n-1, n),(n, n),(n, n-1), \ldots,(n, 2),(n, 1)
$$

each have the property that $a b$ is divisible by $n$.

There are $(n-1)+1+(n-1)=2 n-1$ of these pairs.

Therefore, at least $2 n-1$ of the pairs of balls that can be chosen have labels whose product is divisible by $n$.

Since there are $n^{2}$ pairs that can be chosen and the number of pairs of balls with the desired property is at least $2 n-1$, then $P(n) \geq \frac{2 n-1}{n^{2}}$.

This proves the first part of what we needed to prove.



Next, suppose that $n=p$, a prime number.

For $a b$ to be divisible by $p$, then either $a$ is divisible by $p$ or $b$ is divisible by $p$ (or both). (This property is not true when $p$ is not a prime number; for example, $2 \cdot 2$ is divisible by 4 even though neither factor is divisible by 4.)

Since $1 \leq a \leq p$ and $1 \leq b \leq p$, then if either $a$ is divisible by $p$ or $b$ is divisible by $p$ (or both), we must have either $a=p$ or $b=p$, or both.

In other words, $a b$ is divisible by $p$ exactly when $(a, b)$ is in the list

$$
(1, p),(2, p), \ldots,(p-1, p),(p, p),(p, p-1), \ldots,(p, 2),(p, 1)
$$

There are $2 p-1$ pairs in this list and these are the only pairs for which $a b$ is divisible by $p$.

Therefore, $P(n)=\frac{2 n-1}{n^{2}}$ when $n$ is a prime number.

Since there are infinitely many prime numbers, then $P(n)=\frac{2 n-1}{n^{2}}$ for infinitely many positive integers $n$ with $n \geq 2$.

Thus, $f(n)=2 n-1$ is a polynomial with the desired properties.
Subfield: Combinatorics


Question: Let $k$ be a positive integer with $k \geq 2$. Two bags each contain $k$ balls, labelled with the positive integers from 1 to $k$. André removes one ball from each bag. (In each bag, each ball is equally likely to be chosen.) Define $P(k)$ to be the probability that the product of the numbers on the two balls that he chooses is divisible by $k$.
Prove there exists a positive integer $m$ for which $P(m)>\frac{2016}{m}$.

Solution:
Let $N=2^{k}$, where $k$ is a positive integer with $k \geq 2$.

Suppose that there are two bags, each containing $N$ balls labelled from 1 to $N$.

Since there are $N$ balls in each bag, there are $N^{2}$ pairs of balls that can be chosen.

Let $a$ be the number on the first ball chosen and $b$ be the number on the second ball chosen.

Let $j$ be a positive integer with $1 \leq j \leq k-1$.

Consider pairs of the form $(a, b)=\left(2^{j} x, 2^{k-j} y\right)$ where $x$ and $y$ are odd positive integers that keep $a$ and $b$ in the desired range.

Note that, in each case, $a b=\left(2^{j} x\right)\left(2^{k-j} y\right)=2^{k} x y$ which is divisible by $N=2^{k}$.

Since $1 \leq a \leq 2^{k}$, then $1 \leq 2^{j} x \leq 2^{k}$ and so $x \leq 2^{k-j}$.

Since half of the positive integers from 1 to $2^{k-j}$ are odd, then there are $\frac{1}{2} 2^{k-j}=2^{k-j-1}$ choices for $x$.

Similarly, there are $2^{k-(k-j)-1}=2^{j-1}$ choices for $y$.

Note that each choice of $x$ and $y$ gives a unique pair $(a, b)$.

For any fixed $j$, there are $2^{k-j-1}$ choices for $x$ and $2^{j-1}$ choices for $y$.

Thus, there are $2^{k-j-1} \cdot 2^{j-1}=2^{k-2}$ choices of this form for the pair $(a, b)$.

So for a fixed $j$ with $1 \leq j \leq k-1$, this method gives $2^{k-2}$ pairs $(a, b)$ for which $a b$ is divisible by $N$.

Since there are $k-1$ different values for $j$, then there are at least $(k-1) 2^{k-2}$ pairs $(a, b)$ for which $a b$ is divisible by $N$. (Note that two pairs that come from different values of $j$ will in fact be different since the number of factors of 2 in their values of $a$ will be different.) Since there are $N^{2}$ choices for $(a, b)$, then

$$
P(N) \geq \frac{(k-1) 2^{k-2}}{N^{2}}=\frac{(k-1) 2^{k} 2^{-2}}{N^{2}}=\frac{k-1}{4} \cdot \frac{1}{N}
$$

When $\frac{k-1}{4}>2016$, we will have $P(N)>2016 \cdot \frac{1}{N}$.

The inequality $\frac{k-1}{4}>2016$ is equivalent to $k-1>8064$ or $k>8065$.

We want to show that there exists a positive integer $m$ with $P(m)>\frac{2016}{m}$.

Set $m=2^{8066}$.

By the work above, $P(m) \geq \frac{8065}{4} \cdot \frac{1}{m}>\frac{2016}{m}$, as required.
Subfield: Geometry


Question: In rectangle $A B C D, F$ is on diagonal $B D$ so that $A F$ is perpendicular to $B D$. Also, $B C=30, C D=40$ and $A F=x$. Determine the value of $x$.

<img_3965>

Solution:
Since $A B C D$ is a rectangle, then $A B=C D=40$ and $A D=B C=30$.

By the Pythagorean Theorem, $B D^{2}=A D^{2}+A B^{2}$ and since $B D>0$, then

$$
B D=\sqrt{30^{2}+40^{2}}=\sqrt{900+1600}=\sqrt{2500}=50
$$

We calculate the area of $\triangle A D B$ is two different ways.

First, using $A B$ as base and $A D$ as height, we obtain an area of $\frac{1}{2}(40)(30)=600$.

Next, using $D B$ as base and $A F$ as height, we obtain an area of $\frac{1}{2}(50) x=25 x$.

We must have $25 x=600$ and so $x=\frac{600}{25}=24$.
Since $A B C D$ is a rectangle, then $A B=C D=40$ and $A D=B C=30$.

By the Pythagorean Theorem, $B D^{2}=A D^{2}+A B^{2}$ and since $B D>0$, then

$$
B D=\sqrt{30^{2}+40^{2}}=\sqrt{900+1600}=\sqrt{2500}=50
$$

Since $\triangle D A B$ is right-angled at $A$, then $\sin (\angle A D B)=\frac{A B}{B D}=\frac{40}{50}=\frac{4}{5}$.

But $\triangle A D F$ is right-angled at $F$ and $\angle A D F=\angle A D B$.

Therefore, $\sin (\angle A D F)=\frac{A F}{A D}=\frac{x}{30}$.

Thus, $\frac{x}{30}=\frac{4}{5}$ and so $x=\frac{4}{5}(30)=24$.

Solution 3

Since $A B C D$ is a rectangle, then $A B=C D=40$ and $A D=B C=30$.

By the Pythagorean Theorem, $B D^{2}=A D^{2}+A B^{2}$ and since $B D>0$, then

$$
B D=\sqrt{30^{2}+40^{2}}=\sqrt{900+1600}=\sqrt{2500}=50
$$

Note that $\triangle B F A$ is similar to $\triangle B A D$, since each is right-angled and they share a common angle at $B$.

Thus, $\frac{A F}{A B}=\frac{A D}{B D}$ and so $\frac{x}{30}=\frac{40}{50}$ which gives $x=\frac{30(40)}{50}=24$.

Final answer: 24
Subfield: Geometry


Question: In rectangle $A B C D, F$ is on diagonal $B D$ so that $A F$ is perpendicular to $B D$. Also, $B C=30, C D=40$ and $A F=x$. Determine the value of $x$.

![](https://cdn.mathpix.com/cropped/2023_12_21_a9a67760d94609d98542g-1.jpg?height=271&width=380&top_left_y=604&top_left_x=1401)

Solution:
Since $A B C D$ is a rectangle, then $A B=C D=40$ and $A D=B C=30$.

By the Pythagorean Theorem, $B D^{2}=A D^{2}+A B^{2}$ and since $B D>0$, then

$$
B D=\sqrt{30^{2}+40^{2}}=\sqrt{900+1600}=\sqrt{2500}=50
$$

We calculate the area of $\triangle A D B$ is two different ways.

First, using $A B$ as base and $A D$ as height, we obtain an area of $\frac{1}{2}(40)(30)=600$.

Next, using $D B$ as base and $A F$ as height, we obtain an area of $\frac{1}{2}(50) x=25 x$.

We must have $25 x=600$ and so $x=\frac{600}{25}=24$.
Since $A B C D$ is a rectangle, then $A B=C D=40$ and $A D=B C=30$.

By the Pythagorean Theorem, $B D^{2}=A D^{2}+A B^{2}$ and since $B D>0$, then

$$
B D=\sqrt{30^{2}+40^{2}}=\sqrt{900+1600}=\sqrt{2500}=50
$$

Since $\triangle D A B$ is right-angled at $A$, then $\sin (\angle A D B)=\frac{A B}{B D}=\frac{40}{50}=\frac{4}{5}$.

But $\triangle A D F$ is right-angled at $F$ and $\angle A D F=\angle A D B$.

Therefore, $\sin (\angle A D F)=\frac{A F}{A D}=\frac{x}{30}$.

Thus, $\frac{x}{30}=\frac{4}{5}$ and so $x=\frac{4}{5}(30)=24$.

Solution 3

Since $A B C D$ is a rectangle, then $A B=C D=40$ and $A D=B C=30$.

By the Pythagorean Theorem, $B D^{2}=A D^{2}+A B^{2}$ and since $B D>0$, then

$$
B D=\sqrt{30^{2}+40^{2}}=\sqrt{900+1600}=\sqrt{2500}=50
$$

Note that $\triangle B F A$ is similar to $\triangle B A D$, since each is right-angled and they share a common angle at $B$.

Thus, $\frac{A F}{A B}=\frac{A D}{B D}$ and so $\frac{x}{30}=\frac{40}{50}$ which gives $x=\frac{30(40)}{50}=24$.

Final answer: 24
Subfield: Combinatorics


Question: In an arithmetic sequence, the first term is 1 and the last term is 19 . The sum of all the terms in the sequence is 70 . How many terms does the sequence have? (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, 9 is an arithmetic sequence with four terms.)

Solution:
The sum of the terms in an arithmetic sequence is equal to the average of the first and last terms times the number of terms.

If $n$ is the number of terms in the sequence, then $\frac{1}{2}(1+19) n=70$ or $10 n=70$ and so $n=7$.
Let $n$ be the number of terms in the sequence and $d$ the common difference.

Since the first term is 1 and the $n$th term equals 19 , then $1+(n-1) d=19$ and so $(n-1) d=18$.

Since the sum of the terms in the sequence is 70 , then $\frac{1}{2} n(1+1+(n-1) d)=70$.

Thus, $\frac{1}{2} n(2+18)=70$ or $10 n=70$ and so $n=7$.

Final answer: 7
Subfield: Algebra


Question: Suppose that $a(x+b(x+3))=2(x+6)$ for all values of $x$. Determine $a$ and $b$.

Solution:
Since the given equation is true for all values of $x$, then it is true for any particular value of $x$ that we try.

If $x=-3$, the equation becomes $a(-3+b(0))=2(3)$ or $-3 a=6$ and so $a=-2$.

If $x=0$, the equation becomes $-2(0+b(3))=2(6)$ or $-6 b=12$ and so $b=-2$.

Therefore, $a=-2$ and $b=-2$.
We expand both sides of the equation:

$$
\begin{aligned}
a(x+b(x+3)) & =2(x+6) \\
a(x+b x+3 b) & =2 x+12 \\
a x+a b x+3 a b & =2 x+12 \\
(a+a b) x+3 a b & =2 x+12
\end{aligned}
$$

Since this equation is true for all values of $x$, then the coefficients on the left side and right side must be equal, so $a+a b=2$ and $3 a b=12$.

From the second equation, $a b=4$ so the first equation becomes $a+4=2$ or $a=-2$.

Since $a b=4$, then $-2 b=4$ and so $b=-2$.

Thus, $a=b=-2$.

Final answer: -2,-2
Subfield: Geometry


Question: In the diagram, $\triangle A B C$ is right-angled at $C$. Also, $2 \sin B=3 \tan A$. Determine the measure of angle $A$.

<img_3840>

Solution:
Since $\triangle A B C$ is right-angled at $C$, then $\sin B=\cos A$.

Therefore, $2 \cos A=3 \tan A=\frac{3 \sin A}{\cos A}$ or $2 \cos ^{2} A=3 \sin A$.

Using the fact that $\cos ^{2} A=1-\sin ^{2} A$, this becomes $2-2 \sin ^{2} A=3 \sin A$

or $2 \sin ^{2} A+3 \sin A-2=0$ or $(2 \sin A-1)(\sin A+2)=0$.

Since $\sin A$ is between -1 and 1 , then $\sin A=\frac{1}{2}$.

Since $A$ is an acute angle, then $A=30^{\circ}$.
Since $\triangle A B C$ is right-angled at $C$, then $\sin B=\frac{b}{c}$ and $\tan A=\frac{a}{b}$.

Thus, the given equation is $\frac{2 b}{c}=\frac{3 a}{b}$ or $2 b^{2}=3 a c$.

Using the Pythagorean Theorem, $b^{2}=c^{2}-a^{2}$ and so we obtain $2 c^{2}-2 a^{2}=3 a c$ or $2 c^{2}-3 a c-2 a^{2}=0$.

Factoring, we obtain $(c-2 a)(2 c+a)=0$.

Since $a$ and $c$ must both be positive, then $c=2 a$.

Since $\triangle A B C$ is right-angled, the relation $c=2 a$ means that $\triangle A B C$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, with $A=30^{\circ}$.

Final answer: $30^{\circ}$
Subfield: Geometry


Question: In the diagram, $\triangle A B C$ is right-angled at $C$. Also, $2 \sin B=3 \tan A$. Determine the measure of angle $A$.

![](https://cdn.mathpix.com/cropped/2023_12_21_a9a67760d94609d98542g-1.jpg?height=285&width=374&top_left_y=1535&top_left_x=1320)

Solution:
Since $\triangle A B C$ is right-angled at $C$, then $\sin B=\cos A$.

Therefore, $2 \cos A=3 \tan A=\frac{3 \sin A}{\cos A}$ or $2 \cos ^{2} A=3 \sin A$.

Using the fact that $\cos ^{2} A=1-\sin ^{2} A$, this becomes $2-2 \sin ^{2} A=3 \sin A$

or $2 \sin ^{2} A+3 \sin A-2=0$ or $(2 \sin A-1)(\sin A+2)=0$.

Since $\sin A$ is between -1 and 1 , then $\sin A=\frac{1}{2}$.

Since $A$ is an acute angle, then $A=30^{\circ}$.
Since $\triangle A B C$ is right-angled at $C$, then $\sin B=\frac{b}{c}$ and $\tan A=\frac{a}{b}$.

Thus, the given equation is $\frac{2 b}{c}=\frac{3 a}{b}$ or $2 b^{2}=3 a c$.

Using the Pythagorean Theorem, $b^{2}=c^{2}-a^{2}$ and so we obtain $2 c^{2}-2 a^{2}=3 a c$ or $2 c^{2}-3 a c-2 a^{2}=0$.

Factoring, we obtain $(c-2 a)(2 c+a)=0$.

Since $a$ and $c$ must both be positive, then $c=2 a$.

Since $\triangle A B C$ is right-angled, the relation $c=2 a$ means that $\triangle A B C$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, with $A=30^{\circ}$.

Final answer: $30^{\circ}$
Subfield: Combinatorics


Question: An integer $n$, with $100 \leq n \leq 999$, is chosen at random. What is the probability that the sum of the digits of $n$ is 24 ?

Solution:
The number of integers between 100 and 999 inclusive is $999-100+1=900$.

An integer $n$ in this range has three digits, say $a, b$ and $c$, with the hundreds digit equal to $a$.

Note that $0 \leq b \leq 9$ and $0 \leq c \leq 9$ and $1 \leq a \leq 9$.

To have $a+b+c=24$, then the possible triples for $a, b, c$ in some order are $9,9,6 ; 9,8,7$; $8,8,8$. (There cannot be three 9's. If there are two 9's, the the other digit equals 6 . If there is one 9 , the second and third digits add to 15 but are both less than 9 , so must equal 8 and 7 . If there are zero 9's, the maximum for each digit is 8 , and so each digt must be 8 in order for the sum of all three to equal 24.)

If the digits are 9, 9 and 6, there are 3 arrangements: 996, 969, 699.



If the digits are 9, 8 and 7, there are 6 arrangements: 987, 978, 897, 879, 798, 789.

If the digits are 8,8 and 8 , there is only 1 arrangement: 888 .

Therefore, there are $3+6+1=10$ integers $n$ in the range 100 to 999 with the sum of the digits of $n$ equal to 24 .

The required probability equals the number of possible values of $n$ with the sum of digits equal to 24 divided by the total number of integers in the range, or $\frac{10}{900}=\frac{1}{90}$.

Final answer: $\frac{1}{90}$
Subfield: Geometry


Question: Alice drove from town $E$ to town $F$ at a constant speed of $60 \mathrm{~km} / \mathrm{h}$. Bob drove from $F$ to $E$ along the same road also at a constant speed. They started their journeys at the same time and passed each other at point $G$.

![](https://cdn.mathpix.com/cropped/2023_12_21_a9a67760d94609d98542g-1.jpg?height=157&width=482&top_left_y=2141&top_left_x=884)

Alice drove from $G$ to $F$ in 45 minutes. Bob drove from $G$ to $E$ in 20 minutes. Determine Bob's constant speed.

Solution:
Since Alice drives at $60 \mathrm{~km} / \mathrm{h}$, then she drives $1 \mathrm{~km}$ every minute.

Since Alice drove from $G$ to $F$ in 45 minutes, then the distance from $G$ to $F$ is $45 \mathrm{~km}$.

Let the distance from $E$ to $G$ be $d \mathrm{~km}$ and let Bob's speed be $B \mathrm{~km} / \mathrm{h}$.

Since Bob drove from $G$ to $E$ in 20 minutes (or $\frac{1}{3}$ of an hour), then $\frac{d}{B}=\frac{1}{3}$. Thus, $d=\frac{1}{3} B$.

The time that it took Bob to drive from $F$ to $G$ was $\frac{45}{B}$ hours.

The time that it took Alice to drive from $E$ to $G$ was $\frac{d}{60}$ hours.

Since the time that it took each of Alice and Bob to reach $G$ was the same, then $\frac{d}{60}=\frac{45}{B}$

and so $B d=45(60)=2700$.

Thus, $B\left(\frac{1}{3} B\right)=2700$ so $B^{2}=8100$ or $B=90$ since $B>0$.

Therefore, Bob's speed was $90 \mathrm{~km} / \mathrm{h}$.

Final answer: 90 km/h
Subfield: Geometry


Question: Alice drove from town $E$ to town $F$ at a constant speed of $60 \mathrm{~km} / \mathrm{h}$. Bob drove from $F$ to $E$ along the same road also at a constant speed. They started their journeys at the same time and passed each other at point $G$.

<img_4044>

Alice drove from $G$ to $F$ in 45 minutes. Bob drove from $G$ to $E$ in 20 minutes. Determine Bob's constant speed.

Solution:
Since Alice drives at $60 \mathrm{~km} / \mathrm{h}$, then she drives $1 \mathrm{~km}$ every minute.

Since Alice drove from $G$ to $F$ in 45 minutes, then the distance from $G$ to $F$ is $45 \mathrm{~km}$.

Let the distance from $E$ to $G$ be $d \mathrm{~km}$ and let Bob's speed be $B \mathrm{~km} / \mathrm{h}$.

Since Bob drove from $G$ to $E$ in 20 minutes (or $\frac{1}{3}$ of an hour), then $\frac{d}{B}=\frac{1}{3}$. Thus, $d=\frac{1}{3} B$.

The time that it took Bob to drive from $F$ to $G$ was $\frac{45}{B}$ hours.

The time that it took Alice to drive from $E$ to $G$ was $\frac{d}{60}$ hours.

Since the time that it took each of Alice and Bob to reach $G$ was the same, then $\frac{d}{60}=\frac{45}{B}$

and so $B d=45(60)=2700$.

Thus, $B\left(\frac{1}{3} B\right)=2700$ so $B^{2}=8100$ or $B=90$ since $B>0$.

Therefore, Bob's speed was $90 \mathrm{~km} / \mathrm{h}$.

Final answer: 90 km/h
Subfield: Geometry


Question: The parabola $y=x^{2}-2 x+4$ is translated $p$ units to the right and $q$ units down. The $x$-intercepts of the resulting parabola are 3 and 5 . What are the values of $p$ and $q$ ?

Solution:
Completing the square on the original parabola, we obtain

$$
y=x^{2}-2 x+4=x^{2}-2 x+1-1+4=(x-1)^{2}+3
$$

Therefore, the vertex of the original parabola is $(1,3)$.

Since the new parabola is a translation of the original parabola and has $x$-intercepts 3 and 5 , then its equation is $y=1(x-3)(x-5)=x^{2}-8 x+15$.

Completing the square here, we obtain

$$
y=x^{2}-8 x+15=x^{2}-8 x+16-16+15=(x-4)^{2}-1
$$

Therefore, the vertex of the new parabola is $(4,-1)$.

Thus, the point $(1,3)$ is translated $p$ units to the right and $q$ units down to reach $(4,-1)$, so $p=3$ and $q=4$.

Final answer: 3,4
Subfield: Geometry


Question: In the diagram, $A B$ is tangent to the circle with centre $O$ and radius $r$. The length of $A B$ is $p$. Point $C$ is on the circle and $D$ is inside the circle so that $B C D$ is a straight line, as shown. If $B C=C D=D O=q$, prove that $q^{2}+r^{2}=p^{2}$.

<img_3250>

Solution:
Join $O$ to $A, B$ and $C$.

<img_3419>

Since $A B$ is tangent to the circle at $A$, then $\angle O A B=90^{\circ}$.

By the Pythagorean Theorem in $\triangle O A B$, we get $O A^{2}+A B^{2}=O B^{2}$ or $r^{2}+p^{2}=O B^{2}$.

In $\triangle O D C$, we have $O D=D C=q$ and $O C=r$.

By the cosine law,

$$
\begin{aligned}
O C^{2} & =O D^{2}+D C^{2}-2(O D)(D C) \cos (\angle O D C) \\
r^{2} & =q^{2}+q^{2}-2 q^{2} \cos (\angle O D C) \\
\cos (\angle O D C) & =\frac{2 q^{2}-r^{2}}{2 q^{2}}
\end{aligned}
$$

In $\triangle O D B$, we have $\angle O D B=\angle O D C$.

Thus, using the cosine law again,

$$
\begin{aligned}
O B^{2} & =O D^{2}+D B^{2}-2(O D)(D B) \cos (\angle O D B) \\
& =q^{2}+(2 q)^{2}-2(q)(2 q)\left(\frac{2 q^{2}-r^{2}}{2 q^{2}}\right) \\
& =q^{2}+4 q^{2}-2\left(2 q^{2}-r^{2}\right) \\
& =q^{2}+2 r^{2}
\end{aligned}
$$

So $O B^{2}=r^{2}+p^{2}=q^{2}+2 r^{2}$, which gives $p^{2}=q^{2}+r^{2}$, as required.
Subfield: Geometry


Question: In the diagram, $A B$ is tangent to the circle with centre $O$ and radius $r$. The length of $A B$ is $p$. Point $C$ is on the circle and $D$ is inside the circle so that $B C D$ is a straight line, as shown. If $B C=C D=D O=q$, prove that $q^{2}+r^{2}=p^{2}$.

![](https://cdn.mathpix.com/cropped/2023_12_21_b301ea0eb47ddd66a1f3g-1.jpg?height=314&width=353&top_left_y=1274&top_left_x=1344)

Solution:
Join $O$ to $A, B$ and $C$.

![](https://cdn.mathpix.com/cropped/2023_12_21_dc1d6915b9aeb263a0c9g-1.jpg?height=306&width=338&top_left_y=1332&top_left_x=991)

Since $A B$ is tangent to the circle at $A$, then $\angle O A B=90^{\circ}$.

By the Pythagorean Theorem in $\triangle O A B$, we get $O A^{2}+A B^{2}=O B^{2}$ or $r^{2}+p^{2}=O B^{2}$.

In $\triangle O D C$, we have $O D=D C=q$ and $O C=r$.

By the cosine law,

$$
\begin{aligned}
O C^{2} & =O D^{2}+D C^{2}-2(O D)(D C) \cos (\angle O D C) \\
r^{2} & =q^{2}+q^{2}-2 q^{2} \cos (\angle O D C) \\
\cos (\angle O D C) & =\frac{2 q^{2}-r^{2}}{2 q^{2}}
\end{aligned}
$$

In $\triangle O D B$, we have $\angle O D B=\angle O D C$.

Thus, using the cosine law again,

$$
\begin{aligned}
O B^{2} & =O D^{2}+D B^{2}-2(O D)(D B) \cos (\angle O D B) \\
& =q^{2}+(2 q)^{2}-2(q)(2 q)\left(\frac{2 q^{2}-r^{2}}{2 q^{2}}\right) \\
& =q^{2}+4 q^{2}-2\left(2 q^{2}-r^{2}\right) \\
& =q^{2}+2 r^{2}
\end{aligned}
$$

So $O B^{2}=r^{2}+p^{2}=q^{2}+2 r^{2}$, which gives $p^{2}=q^{2}+r^{2}$, as required.

Final answer: 证明题，略
Subfield: Algebra


Question: If $\log _{2} x,\left(1+\log _{4} x\right)$, and $\log _{8} 4 x$ are consecutive terms of a geometric sequence, determine the possible values of $x$.

(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:
First, we convert each of the logarithms to a logarithm with base 2:

$$
\begin{aligned}
1+\log _{4} x & =1+\frac{\log _{2} x}{\log _{2} 4}=1+\frac{\log _{2} x}{2}=1+\frac{1}{2} \log _{2} x \\
\log _{8} 4 x & =\frac{\log _{2} 4 x}{\log _{2} 8}=\frac{\log _{2} 4+\log _{2} x}{3}=\frac{2}{3}+\frac{1}{3} \log _{2} x
\end{aligned}
$$

Let $y=\log _{2} x$. Then the three terms are $y, 1+\frac{1}{2} y$, and $\frac{2}{3}+\frac{1}{3} y$. Since these three are in geometric sequence, then

$$
\begin{aligned}
\frac{y}{1+\frac{1}{2} y} & =\frac{1+\frac{1}{2} y}{\frac{2}{3}+\frac{1}{3} y} \\
y\left(\frac{2}{3}+\frac{1}{3} y\right) & =\left(1+\frac{1}{2} y\right)^{2} \\
\frac{2}{3} y+\frac{1}{3} y^{2} & =1+y+\frac{1}{4} y^{2} \\
8 y+4 y^{2} & =12+12 y+3 y^{2} \\
y^{2}-4 y-12 & =0 \\
(y-6)(y+2) & =0
\end{aligned}
$$

Therefore, $y=\log _{2} x=6$ or $y=\log _{2} x=-2$, which gives $x=2^{6}=64$ or $x=2^{-2}=\frac{1}{4}$.

Final answer: $64,\frac{1}{4}$