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


Question: Suppose that $n>5$ and that the numbers $t_{1}, t_{2}, t_{3}, \ldots, t_{n-2}, t_{n-1}, t_{n}$ form an arithmetic sequence with $n$ terms. If $t_{3}=5, t_{n-2}=95$, and the sum of all $n$ terms is 1000 , what is the value of $n$ ?

(An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant, called the common difference. For example, $3,5,7,9$ are the first four terms of an arithmetic sequence.)

Solution:
Since the sequence $t_{1}, t_{2}, t_{3}, \ldots, t_{n-2}, t_{n-1}, t_{n}$ is arithmetic, then

$$
t_{1}+t_{n}=t_{2}+t_{n-1}=t_{3}+t_{n-2}
$$

This is because, if $d$ is the common difference, we have $t_{2}=t_{1}+d$ and $t_{n-1}=t_{n}-d$, as well as having $t_{3}=t_{1}+2 d$ and $t_{n-2}=t_{n}-2 d$.

Since the sum of all $n$ terms is 1000, using one formula for the sum of an arithmetic sequence gives

$$
\begin{aligned}
\frac{n}{2}\left(t_{1}+t_{n}\right) & =1000 \\
n\left(t_{1}+t_{n}\right) & =2000 \\
n\left(t_{3}+t_{n-2}\right) & =2000 \\
n(5+95) & =2000
\end{aligned}
$$

and so $n=20$.
Suppose that the arithmetic sequence with $n$ terms has first term $a$ and common difference $d$.

Then $t_{3}=a+2 d=5$ and $t_{n-2}=a+(n-3) d=95$.

Since the sum of the $n$ terms equals 1000, then

$$
\frac{n}{2}(2 a+(n-1) d)=1000
$$

Adding the equations $a+2 d=5$ and $a+(n-3) d=95$, we obtain $2 a+(n-1) d=100$.

Substituting, we get $\frac{n}{2}(100)=1000$ from which we obtain $n=20$.

Final answer: 20
Subfield: Algebra


Question: Suppose that $a$ and $r$ are real numbers. A geometric sequence with first term $a$ and common ratio $r$ has 4 terms. The sum of this geometric sequence is $6+6 \sqrt{2}$. A second geometric sequence has the same first term $a$ and the same common ratio $r$, but has 8 terms. The sum of this second geometric sequence is $30+30 \sqrt{2}$. Determine all possible values for $a$.

(A geometric sequence is a sequence in which each term after the first is obtained from the previous term by multiplying it by a non-zero constant, called the common ratio. For example, $3,-6,12,-24$ are the first four terms of a geometric sequence.)

Solution:
Since the sum of a geometric sequence with first term $a$, common ratio $r$ and 4 terms is $6+6 \sqrt{2}$, then

$$
a+a r+a r^{2}+a r^{3}=6+6 \sqrt{2}
$$

Since the sum of a geometric sequence with first term $a$, common ratio $r$ and 8 terms is $30+30 \sqrt{2}$, then

$$
a+a r+a r^{2}+a r^{3}+a r^{4}+a r^{5}+a r^{6}+a r^{7}=30+30 \sqrt{2}
$$

But

$$
\begin{aligned}
a+a r & +a r^{2}+a r^{3}+a r^{4}+a r^{5}+a r^{6}+a r^{7} \\
& =\left(a+a r+a r^{2}+a r^{3}\right)+r^{4}\left(a+a r+a r^{2}+a r^{3}\right) \\
& =\left(1+r^{4}\right)\left(a+a r+a r^{2}+a r^{3}\right)
\end{aligned}
$$



Therefore,

$$
\begin{aligned}
30+30 \sqrt{2} & =\left(1+r^{4}\right)(6+6 \sqrt{2}) \\
\frac{30+30 \sqrt{2}}{6+6 \sqrt{2}} & =1+r^{4} \\
5 & =1+r^{4} \\
r^{4} & =4 \\
r^{2} & =2 \quad\left(\text { since } r^{2}>0\right) \\
r & = \pm \sqrt{2}
\end{aligned}
$$

If $r=\sqrt{2}$,

$a+a r+a r^{2}+a r^{3}=a+\sqrt{2} a+a(\sqrt{2})^{2}+a(\sqrt{2})^{3}=a+\sqrt{2} a+2 a+2 \sqrt{2} a=a(3+3 \sqrt{2})$

Since $a+a r+a r^{2}+a r^{3}=6+6 \sqrt{2}$, then $a(3+3 \sqrt{2})=6+6 \sqrt{2}$ and so $a=\frac{6+6 \sqrt{2}}{3+3 \sqrt{2}}=2$.

If $r=-\sqrt{2}$,

$a+a r+a r^{2}+a r^{3}=a-\sqrt{2} a+a(-\sqrt{2})^{2}+a(-\sqrt{2})^{3}=a-\sqrt{2} a+2 a-2 \sqrt{2} a=a(3-3 \sqrt{2})$

Since $a+a r+a r^{2}+a r^{3}=6+6 \sqrt{2}$, then $a(3-3 \sqrt{2})=6+6 \sqrt{2}$ and so

$$
a=\frac{6+6 \sqrt{2}}{3-3 \sqrt{2}}=\frac{2+2 \sqrt{2}}{1-\sqrt{2}}=\frac{(2+2 \sqrt{2})(1+\sqrt{2})}{(1-\sqrt{2})(1+\sqrt{2})}=\frac{2+2 \sqrt{2}+2 \sqrt{2}+4}{1-2}=-6-4 \sqrt{2}
$$

Therefore, the possible values of $a$ are $a=2$ and $a=-6-4 \sqrt{2}$.

An alternate way of arriving at the equation $1+r^{4}=5$ is to use the formula for the sum of a geometric sequence twice to obtain

$$
\frac{a\left(1-r^{4}\right)}{1-r}=6+6 \sqrt{2} \quad \frac{a\left(1-r^{8}\right)}{1-r}=30+30 \sqrt{2}
$$

assuming that $r \neq 1$. (Can you explain why $r \neq 1$ and $r^{4} \neq 1$ without knowing already that $r= \pm \sqrt{2}$ ?)

Dividing the second equation by the first, we obtain

$$
\frac{a\left(1-r^{8}\right)}{1-r} \cdot \frac{1-r}{a\left(1-r^{4}\right)}=\frac{30+30 \sqrt{2}}{6+6 \sqrt{2}}
$$

which gives

$$
\frac{1-r^{8}}{1-r^{4}}=5
$$

Since $1-r^{8}=\left(1+r^{4}\right)\left(1-r^{4}\right)$, we obtain $1+r^{4}=5$. We then can proceed as above.

Final answer: $a=2$, $a=-6-4 \sqrt{2}$
Subfield: Combinatorics


Question: A bag contains 3 green balls, 4 red balls, and no other balls. Victor removes balls randomly from the bag, one at a time, and places them on a table. Each ball in the bag is equally likely to be chosen each time that he removes a ball. He stops removing balls when there are two balls of the same colour on the table. What is the probability that, when he stops, there is at least 1 red ball and at least 1 green ball on the table?

Solution:
Victor stops when there are either 2 green balls on the table or 2 red balls on the table. If the first 2 balls that Victor removes are the same colour, Victor will stop.

If the first 2 balls that Victor removes are different colours, Victor does not yet stop, but when he removes a third ball, its colour must match the colour of one of the first 2 balls and so Victor does stop.

Therefore, the probability that he stops with at least 1 red ball and 1 green ball on the table is equal to the probability that the first 2 balls that he removes are different colours. Also, the probability that the first 2 balls that he removes are different colours is equal to 1 minus the probability that the first 2 balls that he removes are the same colour.

The probability that the first two balls that Victor draws are both green is $\frac{3}{7} \cdot \frac{2}{6}$ because for the first ball there are 7 balls in the bag, 3 of which are green and for the second ball there are 6 balls in the bag, 2 of which are green.

The probability that the first two balls that Victor draws are both red is $\frac{4}{7} \cdot \frac{3}{6}$ because for the first ball there are 7 balls in the bag, 4 of which are red and for the second ball there are 6 balls in the bag, 3 of which are red.

Thus, the probability that the first two balls that Victor removes are the same colour is

$$
\frac{3}{7} \cdot \frac{2}{6}+\frac{4}{7} \cdot \frac{3}{6}=\frac{1}{7}+\frac{2}{7}=\frac{3}{7}
$$

This means that the desired probability is $1-\frac{3}{7}=\frac{4}{7}$.

Final answer: $\frac{4}{7}$
Subfield: Algebra


Question: Suppose that $f(a)=2 a^{2}-3 a+1$ for all real numbers $a$ and $g(b)=\log _{\frac{1}{2}} b$ for all $b>0$. Determine all $\theta$ with $0 \leq \theta \leq 2 \pi$ for which $f(g(\sin \theta))=0$.

Solution:
Using the definition of $f$, the following equations are equivalent:

$$
\begin{aligned}
f(a) & =0 \\
2 a^{2}-3 a+1 & =0 \\
(a-1)(2 a-1) & =0
\end{aligned}
$$

Therefore, $f(a)=0$ exactly when $a=1$ or $a=\frac{1}{2}$.

Thus, $f(g(\sin \theta))=0$ exactly when $g(\sin \theta)=1$ or $g(\sin \theta)=\frac{1}{2}$.

Using the definition of $g$,

- $g(b)=1$ exactly when $\log _{\frac{1}{2}} b=1$, which gives $b=\left(\frac{1}{2}\right)^{1}=\frac{1}{2}$, and
- $g(b)=1 / 2$ exactly when $\log _{\frac{1}{2}} b=1 / 2$, which gives $b=\left(\frac{1}{2}\right)^{1 / 2}=\frac{1}{\sqrt{2}}$.

Therefore, $f(g(\sin \theta))=0$ exactly when $\sin \theta=\frac{1}{2}$ or $\sin \theta=\frac{1}{\sqrt{2}}$.

Since $0 \leq \theta \leq 2 \pi$, the solutions are $\theta=\frac{1}{6} \pi, \frac{5}{6} \pi, \frac{1}{4} \pi, \frac{3}{4} \pi$.

Final answer: $\frac{1}{6} \pi, \frac{5}{6} \pi, \frac{1}{4} \pi, \frac{3}{4} \pi$
Subfield: Number Theory


Question: Prove that the integer $\frac{(1 !)(2 !)(3 !) \cdots(398 !)(399 !)(400 !)}{200 !}$ is a perfect square. (In this fraction, the numerator is the product of the factorials of the integers from 1 to 400 , inclusive.)

Solution:
Let $N=\frac{(1 !)(2 !)(3 !) \cdots(398 !)(399 !)(400 !)}{200 !}$.

For each integer $k$ from 1 to 200 , inclusive, we rewrite $(2 k)$ ! as $2 k \cdot(2 k-1)$ !.

Therefore, $(2 k-1) !(2 k) !=(2 k-1) ! \cdot 2 k \cdot(2 k-1) !=2 k((2 k-1) !)^{2}$.

(In particular, $(1 !)(2 !)=2(1 !)^{2},(3 !)(4 !)=4(3 !)^{2}$, and so on.)

Thus,

$$
N=\frac{2(1 !)^{2} \cdot 4(3 !)^{2} \cdots \cdot 398(397 !)^{2} \cdot 400(399 !)^{2}}{200 !}
$$

Re-arranging the numerator of the expression, we obtain

$$
N=\frac{(1 !)^{2}(3 !)^{2} \cdots(397 !)^{2}(399 !)^{2} \cdot(2 \cdot 4 \cdot \cdots 398 \cdot 400)}{200 !}
$$

We can now re-write $2 \cdot 4 \cdot \cdots 398 \cdot 400$ as $(2 \cdot 1) \cdot(2 \cdot 2) \cdots(2 \cdot 199) \cdot(2 \cdot 200)$.

Since there are 200 sets of parentheses, we obtain

$$
N=\frac{(1 !)^{2}(3 !)^{2} \cdots(397 !)^{2}(399 !)^{2} \cdot 2^{200} \cdot(1 \cdot 2 \cdots \cdots 199 \cdot 200)}{200 !}
$$

Since $1 \cdot 2 \cdot \cdots \cdot 199 \cdot 200=200$ !, we can conclude that

$$
N=2^{200}(1 !)^{2}(3 !)^{2} \cdots(397 !)^{2}(399 !)^{2}
$$

Therefore,

$$
\sqrt{N}=2^{100}(1 !)(3 !) \cdots(397 !)(399 !)
$$

which is a product of integers and thus an integer itself.

Since $\sqrt{N}$ is an integer, $N$ is a perfect square, as required.
Subfield: Number Theory


Question: Suppose that $a=5$ and $b=4$. Determine all pairs of integers $(K, L)$ for which $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$.

Solution:
When $a=5$ and $b=4$, we obtain $a^{2}+b^{2}-a b=5^{2}+4^{2}-5 \cdot 4=21$.

Therefore, we want to find all pairs of integers $(K, L)$ with $K^{2}+3 L^{2}=21$.

If $L=0$, then $L^{2}=0$, which gives $K^{2}=21$ which has no integer solutions.

If $L= \pm 1$, then $L^{2}=1$, which gives $K^{2}=18$ which has no integer solutions.

If $L= \pm 2$, then $L^{2}=4$, which gives $K^{2}=9$ which gives $K= \pm 3$.

If $L= \pm 3$, then $L^{2}=9$. Since $3 L^{2}=27>21$, then there are no real solutions for $K$.

Similarly, if $L^{2}>9$, there are no real solutions for $K$.

Therefore, the solutions are $(K, L)=(3,2),(-3,2),(3,-2),(-3,-2)$.

Final answer: $(3,2),(-3,2),(3,-2),(-3,-2)$
Subfield: Number Theory


Question: Prove that, for all integers $K$ and $L$, there is at least one pair of integers $(a, b)$ for which $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$.

Solution:
Suppose that $K$ and $L$ are integers.

Then

$$
\begin{aligned}
& (K+L)^{2}+(K-L)^{2}-(K+L)(K-L) \\
& \quad=\left(K^{2}+2 K L+L^{2}\right)+\left(K^{2}-2 K L+L^{2}\right)-\left(K^{2}-L^{2}\right) \\
& \quad=K^{2}+3 L^{2}
\end{aligned}
$$

Therefore, the integers $a=K+L$ and $b=K-L$ satisfy the equation $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$, and so for all integers $K$ and $L$, there is at least one pair of integers $(a, b)$ that satisfy the equation.

How could we come up with this? One way to do this would be trying some small values



of $K$ and $L$, calculating $K^{2}+3 L^{2}$ and using this to make a guess, which can then be proven algebraically as above. In particular, here are some values:

| $K$ | $L$ | $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$ | $a$ | $b$ |
| :---: | :---: | :---: | :---: | :---: |
| 1 | 1 | 4 | 2 | 0 |
| 2 | 1 | 7 | 3 | 1 |
| 3 | 1 | 12 | 4 | 2 |
| 1 | 2 | 13 | 3 | -1 |
| 2 | 2 | 16 | 4 | 0 |
| 3 | 2 | 21 | 5 | 1 |

The columns for $a$ and $b$ might lead us to guess that $a=K+L$ and $b=K-L$, which we proved above does in fact work.
Subfield: Number Theory


Question: Prove that, for all integers $a$ and $b$, there is at least one pair of integers $(K, L)$ for which $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$.

Solution:
Suppose that $a$ and $b$ are integers.

If $a$ is even, then $\frac{a}{2}$ is an integer and so

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

Thus, if $K=\frac{a}{2}-b$ and $L=\frac{a}{2}$, we have $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$.

If $b$ is even, then $\frac{b}{2}$ is an integer and so a similar algebraic argument shows that

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

and so if $K=\frac{b}{2}-a$ and $L=\frac{b}{2}$, we have $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$.

If $a$ and $b$ are both odd, then $a+b$ and $a-b$ are both even, which means that $\frac{a+b}{2}$ and $\frac{a-b}{2}$ are both integers, and so

$\left(\frac{a+b}{2}\right)^{2}+3\left(\frac{a-b}{2}\right)^{2}=\frac{a^{2}+2 a b+b^{2}}{4}+\frac{3 a^{2}-6 a b+3 b^{2}}{4}=\frac{4 a^{2}+4 b^{2}-4 a b}{4}=a^{2}+b^{2}-a b$

Thus, if $K=\frac{a+b}{2}$ and $L=\frac{a-b}{2}$, we have $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$.

Therefore, in all cases, for all integers $a$ and $b$, there is at least one pair of integers $(K, L)$ with $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$.

Now, trying some small cases might help us make a guess of possible expressions for $K$ and $L$ in terms of $a$ and $b$ :

| $a$ | $b$ | $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$ | $K$ | $L$ |
| :---: | :---: | :---: | :---: | :---: |
| 1 | 1 | 1 | 1 | 0 |
| 2 | 1 | 3 | 0 | 1 |
| 3 | 1 | 7 | 2 | 1 |
| 4 | 1 | 13 | 1 | 2 |
| 1 | 2 | 3 | 0 | 1 |
| 2 | 2 | 4 | 1 | 1 |
| 3 | 2 | 7 | 2 | 1 |
| 4 | 2 | 12 | 3 | 1 |
| 5 | 3 | 19 | 4 | 1 |



While there might not initially seem to be useful patterns here, re-arranging the rows and adding some duplicates might help show a pattern:

| $a$ | $b$ | $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$ | $K$ | $L$ |
| :---: | :---: | :---: | :---: | :---: |
| 2 | 1 | 3 | 0 | 1 |
| 4 | 1 | 13 | 1 | 2 |
| 2 | 2 | 4 | 1 | 1 |
| 4 | 2 | 12 | 3 | 1 |
| 1 | 2 | 3 | 0 | 1 |
| 3 | 2 | 7 | 2 | 1 |
| 4 | 2 | 12 | 3 | 1 |
| 1 | 1 | 1 | 1 | 0 |
| 3 | 1 | 7 | 2 | 1 |
| 5 | 3 | 19 | 4 | 1 |
Subfield: Algebra


Question: A circular disc is divided into 36 sectors. A number is written in each sector. When three consecutive sectors contain $a, b$ and $c$ in that order, then $b=a c$. If the number 2 is placed in one of the sectors and the number 3 is placed in one of the adjacent sectors, as shown, what is the sum of the 36 numbers on the disc?

![](https://cdn.mathpix.com/cropped/2023_12_21_ff40fe20a2d5d1ed9823g-1.jpg?height=572&width=577&top_left_y=272&top_left_x=1164)

Solution:
We are told that when $a, b$ and $c$ are the numbers in consecutive sectors, then $b=a c$. This means that if $a$ and $b$ are the numbers in consecutive sectors, then the number in the next sector is $c=\frac{b}{a}$. (That is, each number is equal to the previous number divided by the one before that.)

Starting with the given 2 and 3 and proceeding clockwise, we obtain

$$
2,3, \quad \frac{3}{2}, \frac{3 / 2}{3}=\frac{1}{2}, \frac{1 / 2}{3 / 2}=\frac{1}{3}, \frac{1 / 3}{1 / 2}=\frac{2}{3}, \frac{2 / 3}{1 / 3}=2, \frac{2}{2 / 3}=3, \quad \frac{3}{2}, \ldots
$$

After the first 6 terms, the first 2 terms ( 2 and 3) reappear, and so the first 6 terms will repeat again. (This is because each term comes from the previous two terms, so when two consecutive terms reappear, then the following terms are the same as when these two consecutive terms appeared earlier.)

Since there are 36 terms in total, then the 6 terms repeat exactly $\frac{36}{6}=6$ times.

Therefore, the sum of the 36 numbers is $6\left(2+3+\frac{3}{2}+\frac{1}{2}+\frac{1}{3}+\frac{2}{3}\right)=6(2+3+2+1)=48$.

Final answer: 48
Subfield: Algebra


Question: A circular disc is divided into 36 sectors. A number is written in each sector. When three consecutive sectors contain $a, b$ and $c$ in that order, then $b=a c$. If the number 2 is placed in one of the sectors and the number 3 is placed in one of the adjacent sectors, as shown, what is the sum of the 36 numbers on the disc?

<img_3397>

Solution:
We are told that when $a, b$ and $c$ are the numbers in consecutive sectors, then $b=a c$. This means that if $a$ and $b$ are the numbers in consecutive sectors, then the number in the next sector is $c=\frac{b}{a}$. (That is, each number is equal to the previous number divided by the one before that.)

Starting with the given 2 and 3 and proceeding clockwise, we obtain

$$
2,3, \quad \frac{3}{2}, \frac{3 / 2}{3}=\frac{1}{2}, \frac{1 / 2}{3 / 2}=\frac{1}{3}, \frac{1 / 3}{1 / 2}=\frac{2}{3}, \frac{2 / 3}{1 / 3}=2, \frac{2}{2 / 3}=3, \quad \frac{3}{2}, \ldots
$$

After the first 6 terms, the first 2 terms ( 2 and 3) reappear, and so the first 6 terms will repeat again. (This is because each term comes from the previous two terms, so when two consecutive terms reappear, then the following terms are the same as when these two consecutive terms appeared earlier.)

Since there are 36 terms in total, then the 6 terms repeat exactly $\frac{36}{6}=6$ times.

Therefore, the sum of the 36 numbers is $6\left(2+3+\frac{3}{2}+\frac{1}{2}+\frac{1}{3}+\frac{2}{3}\right)=6(2+3+2+1)=48$.

Final answer: 48
Subfield: Algebra


Question: Determine all values of $x$ for which $0<\frac{x^{2}-11}{x+1}<7$.

Solution:
We consider two cases: $x>-1$ (that is, $x+1>0$ ) and $x<-1$ (that is, $x+1<0$ ). Note that $x \neq-1$.

Case 1: $x>-1$

We take the given inequality $0<\frac{x^{2}-11}{x+1}<7$ and multiply through by $x+1$, which is positive, to obtain $0<x^{2}-11<7 x+7$.

Thus, $x^{2}-11>0$ and $x^{2}-11<7 x+7$.

From the first, we obtain $x^{2}>11$ and so $x>\sqrt{11}$ or $x<-\sqrt{11}$.

Since $x>-1$, then $x>\sqrt{11}$. (Note that $-\sqrt{11}<-1$.)

From the second, we obtain $x^{2}-7 x-18<0$ or $(x-9)(x+2)<0$. Thus, $-2<x<9$. (Since $y=x^{2}-7 x-18$ represents a parabola opening upwards, its $y$-values are negative between its $x$-intercepts.)

Since $x>-1$ and $-2<x<9$, then $-1<x<9$.

Since $x>\sqrt{11}$ and $-1<x<9$, then the solution in this case is $\sqrt{11}<x<9$.

Case 2: $x<-1$

We take the given inequality $0<\frac{x^{2}-11}{x+1}<7$ and multiply through by $x+1$, which is negative, to obtain $0>x^{2}-11>7 x+7$.

Thus, $x^{2}-11<0$ and $x^{2}-11>7 x+7$.

From the first, we obtain $x^{2}<11$ and so $-\sqrt{11}<x<\sqrt{11}$.

Since $x<-1$ and $-\sqrt{11}<x<\sqrt{11}$, then $-\sqrt{11}<x<-1$.

From the second, we obtain $x^{2}-7 x-18>0$ or $(x-9)(x+2)>0$. Thus, $x<-2$ or $x>9$. (Since $y=x^{2}-7 x-18$ represents a parabola opening upwards, its $y$-values are positive outside its $x$-intercepts.)

Since $x<-1$, we obtain $x<-2$.

Since $-\sqrt{11}<x<-1$ and $x<-2$, then the solution in this case is $-\sqrt{11}<x<-2$.

In summary, the values of $x$ for which $0<\frac{x^{2}-11}{x+1}<7$ those $x$ with $-\sqrt{11}<x<-2$ and those $x$ with $\sqrt{11}<x<9$.

Final answer: $(-\sqrt{11},-2)\cup (\sqrt{11},9)$
Subfield: Geometry


Question: In the diagram, $A C D F$ is a rectangle with $A C=200$ and $C D=50$. Also, $\triangle F B D$ and $\triangle A E C$ are congruent triangles which are right-angled at $B$ and $E$, respectively. What is the area of the shaded region?

![](https://cdn.mathpix.com/cropped/2023_12_21_ff40fe20a2d5d1ed9823g-1.jpg?height=282&width=593&top_left_y=995&top_left_x=1167)

Solution:
Join $B E$.

![](https://cdn.mathpix.com/cropped/2023_12_21_7fa7e5676b9e8a3529e8g-1.jpg?height=279&width=607&top_left_y=245&top_left_x=865)

Since $\triangle F B D$ is congruent to $\triangle A E C$, then $F B=A E$.

Since $\triangle F A B$ and $\triangle A F E$ are each right-angled, share a common side $A F$ and have equal hypotenuses $(F B=A E)$, then these triangles are congruent, and so $A B=F E$.

Now $B A F E$ has two right angles at $A$ and $F$ (so $A B$ and $F E$ are parallel) and has equal sides $A B=F E$ so must be a rectangle.

This means that $B C D E$ is also a rectangle.

Now the diagonals of a rectangle partition it into four triangles of equal area. (Diagonal $A E$ of the rectangle splits the rectangle into two congruent triangles, which have equal area. The diagonals bisect each other, so the four smaller triangles all have equal area.) Since $\frac{1}{4}$ of rectangle $A B E F$ is shaded and $\frac{1}{4}$ of rectangle $B C D E$ is shaded, then $\frac{1}{4}$ of the total area is shaded. (If the area of $A B E F$ is $x$ and the area of $B C D E$ is $y$, then the total shaded area is $\frac{1}{4} x+\frac{1}{4} y$, which is $\frac{1}{4}$ of the total area $x+y$.)

Since $A C=200$ and $C D=50$, then the area of rectangle $A C D F$ is $200(50)=10000$, so the total shaded area is $\frac{1}{4}(10000)=2500$.

Final answer: 2500
Subfield: Geometry


Question: In the diagram, $A C D F$ is a rectangle with $A C=200$ and $C D=50$. Also, $\triangle F B D$ and $\triangle A E C$ are congruent triangles which are right-angled at $B$ and $E$, respectively. What is the area of the shaded region?

<img_3742>

Solution:
Join $B E$.

<img_3698>

Since $\triangle F B D$ is congruent to $\triangle A E C$, then $F B=A E$.

Since $\triangle F A B$ and $\triangle A F E$ are each right-angled, share a common side $A F$ and have equal hypotenuses $(F B=A E)$, then these triangles are congruent, and so $A B=F E$.

Now $B A F E$ has two right angles at $A$ and $F$ (so $A B$ and $F E$ are parallel) and has equal sides $A B=F E$ so must be a rectangle.

This means that $B C D E$ is also a rectangle.

Now the diagonals of a rectangle partition it into four triangles of equal area. (Diagonal $A E$ of the rectangle splits the rectangle into two congruent triangles, which have equal area. The diagonals bisect each other, so the four smaller triangles all have equal area.) Since $\frac{1}{4}$ of rectangle $A B E F$ is shaded and $\frac{1}{4}$ of rectangle $B C D E$ is shaded, then $\frac{1}{4}$ of the total area is shaded. (If the area of $A B E F$ is $x$ and the area of $B C D E$ is $y$, then the total shaded area is $\frac{1}{4} x+\frac{1}{4} y$, which is $\frac{1}{4}$ of the total area $x+y$.)

Since $A C=200$ and $C D=50$, then the area of rectangle $A C D F$ is $200(50)=10000$, so the total shaded area is $\frac{1}{4}(10000)=2500$.

Final answer: 2500
Subfield: Algebra


Question: The numbers $a_{1}, a_{2}, a_{3}, \ldots$ form an arithmetic sequence with $a_{1} \neq a_{2}$. The three numbers $a_{1}, a_{2}, a_{6}$ form a geometric sequence in that order. Determine all possible positive integers $k$ for which the three numbers $a_{1}, a_{4}, a_{k}$ also form a geometric sequence in that order.

(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 are the first four terms of an arithmetic sequence.

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

Solution:
Suppose that the arithmetic sequence $a_{1}, a_{2}, a_{3}, \ldots$ has first term $a$ and common difference $d$.

Then, for each positive integer $n, a_{n}=a+(n-1) d$.

Since $a_{1}=a$ and $a_{2}=a+d$ and $a_{1} \neq a_{2}$, then $d \neq 0$.

Since $a_{1}, a_{2}, a_{6}$ form a geometric sequence in that order, then $\frac{a_{2}}{a_{1}}=\frac{a_{6}}{a_{2}}$ or $\left(a_{2}\right)^{2}=a_{1} a_{6}$.

Substituting, we obtain

$$
\begin{aligned}
(a+d)^{2} & =a(a+5 d) \\
a^{2}+2 a d+d^{2} & =a^{2}+5 a d \\
d^{2} & =3 a d \\
d & =3 a \quad(\text { since } d \neq 0)
\end{aligned}
$$

Therefore, $a_{n}=a+(n-1) d=a+(n-1)(3 a)=(3 n-2) a$ for each $n \geq 1$.

Thus, $a_{4}=(3(4)-2) a=10 a$, and $a_{k}=(3 k-2) a$. (Note that $a_{1}=(3(1)-2) a=a$.)

For $a_{1}, a_{4}, a_{k}$ to also form a geometric sequence then, as above, $\left(a_{4}\right)^{2}=a_{1} a_{k}$, and so

$$
\begin{aligned}
(10 a)^{2} & =(a)((3 k-2) a) \\
100 a^{2} & =(3 k-2) a^{2}
\end{aligned}
$$

Since $d \neq 0$ and $d=3 a$, then $a \neq 0$.

Since $100 a^{2}=(3 k-2) a^{2}$ and $a \neq 0$, then $100=3 k-2$ and so $3 k=102$ or $k=34$.

Checking, we note that $a_{1}=a, a_{4}=10 a$ and $a_{34}=100 a$ which form a geometric sequence with common ratio 10 .

Therefore, the only possible value of $k$ is $k=34$.

Final answer: 34