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


Question: Determine all values of $k$ for which the points $A(1,2), B(11,2)$ and $C(k, 6)$ form a right-angled triangle.

Solution:
$\triangle A B C$ is right-angled exactly when one of the following statements is true:

- $A B$ is perpendicular to $B C$, or
- $A B$ is perpendicular to $A C$, or
- $A C$ is perpendicular to $B C$.

Since $A(1,2)$ and $B(11,2)$ share a $y$-coordinate, then $A B$ is horizontal.

For $A B$ and $B C$ to be perpendicular, $B C$ must be vertical.

Thus, $B(11,2)$ and $C(k, 6)$ must have the same $x$-coordinate, and so $k=11$.

For $A B$ and $A C$ to be perpendicular, $A C$ must be vertical.

Thus, $A(1,2)$ and $C(k, 6)$ must have the same $x$-coordinate, and so $k=1$.



For $A C$ to be perpendicular to $B C$, their slopes must have a product of -1 .

The slope of $A C$ is $\frac{6-2}{k-1}$, which equals $\frac{4}{k-1}$.

The slope of $B C$ is $\frac{6-2}{k-11}$, which equals $\frac{4}{k-11}$.

Thus, $A C$ and $B C$ are perpendicular when $\frac{4}{k-1} \cdot \frac{4}{k-11}=-1$.

Assuming that $k \neq 1$ and $k \neq 11$, we manipulate to obtain $16=-(k-1)(k-11)$ or $16=-k^{2}+12 k-11$ or $k^{2}-12 k+27=0$.

Factoring, we obtain $(k-3)(k-9)=0$ and so $A C$ and $B C$ are perpendicular when $k=3$ or $k=9$.

In summary, $\triangle A B C$ is right-angled when $k$ equals one of $1,3,9,11$.
$\triangle A B C$ is right-angled exactly when its three side lengths satisfy the Pythagorean Theorem in some orientation. That is, $\triangle A B C$ is right-angled exactly when $A B^{2}+B C^{2}=A C^{2}$ or $A B^{2}+A C^{2}=B C^{2}$ or $A C^{2}+B C^{2}=A B^{2}$.

Using $A(1,2)$ and $B(11,2)$, we obtain $A B^{2}=(11-1)^{2}+(2-2)^{2}=100$.

Using $A(1,2)$ and $C(k, 6)$, we obtain $A C^{2}=(k-1)^{2}+(6-2)^{2}=(k-1)^{2}+16$.

Using $B(11,2)$ and $C(k, 6)$, we obtain $B C^{2}=(k-11)^{2}+(6-2)^{2}=(k-11)^{2}+16$.

Using the Pythagorean relationships above, $\triangle A B C$ is right-angled when one of the following is true:

(i)

$$
\begin{aligned}
100+\left((k-11)^{2}+16\right) & =(k-1)^{2}+16 \\
100+k^{2}-22 k+121+16 & =k^{2}-2 k+1+16 \\
220 & =20 k \\
k & =11
\end{aligned}
$$

(ii)

$$
\begin{aligned}
100+\left((k-1)^{2}+16\right) & =(k-11)^{2}+16 \\
100+k^{2}-2 k+1+16 & =k^{2}-22 k+121+16 \\
20 k & =20 \\
k & =1
\end{aligned}
$$

(iii)

$$
\begin{aligned}
\left((k-1)^{2}+16\right)+\left((k-11)^{2}+16\right) & =100 \\
k^{2}-2 k+1+16+k^{2}-22 k+121+16 & =100 \\
2 k^{2}-24 k+54 & =0 \\
k^{2}-12 k+27 & =0 \\
(k-3)(k-9) & =0
\end{aligned}
$$

and so $k=3$ or $k=9$.

In summary, $\triangle A B C$ is right-angled when $k$ equals one of $1,3,9,11$.

Final answer: 1,3,9,11
Subfield: Geometry


Question: The diagram shows two hills that meet at $O$. One hill makes a $30^{\circ}$ angle with the horizontal and the other hill makes a $45^{\circ}$ angle with the horizontal. Points $A$ and $B$ are on the hills so that $O A=O B=20 \mathrm{~m}$. Vertical poles $B D$ and $A C$ are connected by a straight cable $C D$. If $A C=6 \mathrm{~m}$, what is the length of $B D$ for which $C D$ is as short as possible?

![](https://cdn.mathpix.com/cropped/2023_12_21_c71fc327224b5fdf6510g-1.jpg?height=287&width=437&top_left_y=2095&top_left_x=1256)

Solution:
Extend $C A$ and $D B$ downwards until they meet the horizontal through $O$ at $P$ and $Q$, respectively.

![](https://cdn.mathpix.com/cropped/2023_12_21_3361192d5aa7e57927a4g-1.jpg?height=287&width=423&top_left_y=239&top_left_x=951)

Since $C A$ and $D B$ are vertical, then $\angle C P O=\angle D Q O=90^{\circ}$.

Since $O A=20 \mathrm{~m}$, then $A P=O A \sin 30^{\circ}=(20 \mathrm{~m}) \cdot \frac{1}{2}=10 \mathrm{~m}$.

Since $O B=20 \mathrm{~m}$, then $B Q=O B \sin 45^{\circ}=(20 \mathrm{~m}) \cdot \frac{1}{\sqrt{2}}=10 \sqrt{2} \mathrm{~m}$.

Since $A C=6 \mathrm{~m}$, then $C P=A C+A P=16 \mathrm{~m}$.

For $C D$ to be as short as possible and given that $C$ is fixed, then it must be the case that $C D$ is horizontal:

If $C D$ were not horizontal, then suppose that $X$ is on $D Q$, possibly extended, so that $C X$ is horizontal.

![](https://cdn.mathpix.com/cropped/2023_12_21_3361192d5aa7e57927a4g-1.jpg?height=282&width=421&top_left_y=981&top_left_x=955)

Then $\angle C X D=90^{\circ}$ and so $\triangle C X D$ is right-angled with hypotenuse $C D$.

In this case, $C D$ is longer than $C X$ or $X D$.

In particular, $C D>C X$, which means that if $D$ were at $X$, then $C D$ would be shorter.

In other words, a horizontal $C D$ makes $C D$ as short as possible.

When $C D$ is horizontal, $C D Q P$ is a rectangle, since it has two vertical and two horizontal sides. Thus, $D Q=C P=16 \mathrm{~m}$.

Finally, this means that $B D=D Q-B Q=(16-10 \sqrt{2}) \mathrm{m}$.

Final answer: $(16-10 \sqrt{2})$ m
Subfield: Geometry


Question: The diagram shows two hills that meet at $O$. One hill makes a $30^{\circ}$ angle with the horizontal and the other hill makes a $45^{\circ}$ angle with the horizontal. Points $A$ and $B$ are on the hills so that $O A=O B=20 \mathrm{~m}$. Vertical poles $B D$ and $A C$ are connected by a straight cable $C D$. If $A C=6 \mathrm{~m}$, what is the length of $B D$ for which $C D$ is as short as possible?

<img_3434>

Solution:
Extend $C A$ and $D B$ downwards until they meet the horizontal through $O$ at $P$ and $Q$, respectively.

<img_3828>

Since $C A$ and $D B$ are vertical, then $\angle C P O=\angle D Q O=90^{\circ}$.

Since $O A=20 \mathrm{~m}$, then $A P=O A \sin 30^{\circ}=(20 \mathrm{~m}) \cdot \frac{1}{2}=10 \mathrm{~m}$.

Since $O B=20 \mathrm{~m}$, then $B Q=O B \sin 45^{\circ}=(20 \mathrm{~m}) \cdot \frac{1}{\sqrt{2}}=10 \sqrt{2} \mathrm{~m}$.

Since $A C=6 \mathrm{~m}$, then $C P=A C+A P=16 \mathrm{~m}$.

For $C D$ to be as short as possible and given that $C$ is fixed, then it must be the case that $C D$ is horizontal:

If $C D$ were not horizontal, then suppose that $X$ is on $D Q$, possibly extended, so that $C X$ is horizontal.

<img_3255>

Then $\angle C X D=90^{\circ}$ and so $\triangle C X D$ is right-angled with hypotenuse $C D$.

In this case, $C D$ is longer than $C X$ or $X D$.

In particular, $C D>C X$, which means that if $D$ were at $X$, then $C D$ would be shorter.

In other words, a horizontal $C D$ makes $C D$ as short as possible.

When $C D$ is horizontal, $C D Q P$ is a rectangle, since it has two vertical and two horizontal sides. Thus, $D Q=C P=16 \mathrm{~m}$.

Finally, this means that $B D=D Q-B Q=(16-10 \sqrt{2}) \mathrm{m}$.

Final answer: $(16-10 \sqrt{2})$ m
Subfield: Algebra


Question: If $\cos \theta=\tan \theta$, determine all possible values of $\sin \theta$, giving your answer(s) as simplified exact numbers.

Solution:
Since $\tan \theta=\frac{\sin \theta}{\cos \theta}$, then we assume that $\cos \theta \neq 0$.

Therefore, we obtain the following equivalent equations:

$$
\begin{aligned}
\cos \theta & =\tan \theta \\
\cos \theta & =\frac{\sin \theta}{\cos \theta} \\
\cos ^{2} \theta & =\sin \theta \\
1-\sin ^{2} \theta & =\sin \theta \\
0 & =\sin ^{2} \theta+\sin \theta-1
\end{aligned}
$$

Let $u=\sin \theta$. This quadratic equation becomes $u^{2}+u-1=0$

By the quadratic formula, $u=\frac{-1 \pm \sqrt{1^{2}-4(1)(-1)}}{2(1)}=\frac{-1 \pm \sqrt{5}}{2}$.

Therefore, $\sin \theta=\frac{-1+\sqrt{5}}{2} \approx 0.62$ or $\sin \theta=\frac{-1-\sqrt{5}}{2} \approx-1.62$.

Since $-1 \leq \sin \theta \leq 1$, then the second solution is inadmissible. Thus, $\sin \theta=\frac{-1+\sqrt{5}}{2}$.

Final answer: $\frac{-1+\sqrt{5}}{2}$
Subfield: Algebra


Question: Linh is driving at $60 \mathrm{~km} / \mathrm{h}$ on a long straight highway parallel to a train track. Every 10 minutes, she is passed by a train travelling in the same direction as she is. These trains depart from the station behind her every 3 minutes and all travel at the same constant speed. What is the constant speed of the trains, in $\mathrm{km} / \mathrm{h}$ ?

Solution:
Suppose that the trains are travelling at $v \mathrm{~km} / \mathrm{h}$.

Consider two consecutive points in time at which the car is passed by a train.

Since these points are 10 minutes apart, and 10 minutes equals $\frac{1}{6}$ hour, and the car travels at $60 \mathrm{~km} / \mathrm{h}$, then the car travels $(60 \mathrm{~km} / \mathrm{h}) \cdot\left(\frac{1}{6} \mathrm{~h}\right)=10 \mathrm{~km}$.

During these 10 minutes, each train travels $\frac{1}{6} v \mathrm{~km}$, since its speed is $v \mathrm{~km} / \mathrm{h}$.

At the first instance, Train A and the car are next to each other.

At this time, Train B is " 3 minutes" behind Train A.

<img_4020>

Since 3 minutes is $\frac{1}{20}$ hour, then Train B is $\frac{1}{20} v \mathrm{~km}$ behind Train A and the car.

Therefore, the distance from the location of Train B at the first instance to the location where it passes the car is $\left(\frac{1}{20} v+10\right) \mathrm{km}$.

But this distance also equals $\frac{1}{6} v \mathrm{~km}$, since Train B travels for 10 minutes.

Thus, $\frac{1}{6} v=\frac{1}{20} v+10$ or $\frac{10}{60} v-\frac{3}{60} v=10$ and so $\frac{7}{60} v=10$ or $v=\frac{600}{7}$.

Therefore, the trains are travelling at $\frac{600}{7} \mathrm{~km} / \mathrm{h}$.
Suppose that the trains are travelling at $v \mathrm{~km} / \mathrm{h}$.

Consider the following three points in time: the instant when the car and Train A are next to each other, the instant when Train B is at the same location that the car and Train A were at in the previous instant, and the instant when the car and Train B are next to each other.

<img_3611>

From the first instant to the second, Train B "catches up" to where Train A was, so this must take a total of 3 minutes, because the trains leave the station 3 minutes apart.

Since 3 minutes equals $\frac{3}{60}$ hour and the car travels at $60 \mathrm{~km} / \mathrm{h}$, then the car travels $(60 \mathrm{~km} / \mathrm{h}) \cdot\left(\frac{3}{60} \mathrm{~h}\right)=3 \mathrm{~km}$ between these two instants.

From the first instant to the third, 10 minutes passes, since these are consecutive points at which the car is passed by trains. In 10 minutes, the car travels $10 \mathrm{~km}$.

Therefore, between the second and third instants, $10-3=7$ minutes pass. During these 7 minutes, Train B travels $10 \mathrm{~km}$.

Since 7 minutes equals $\frac{7}{60}$ hour, then $v \mathrm{~km} / \mathrm{h}=\frac{10 \mathrm{~km}}{7 / 60 \mathrm{~h}}=\frac{600}{7} \mathrm{~km} / \mathrm{h}$, and so the trains are travelling at $\frac{600}{7} \mathrm{~km} / \mathrm{h}$.

Final answer: $\frac{600}{7}$ km/h
Subfield: Algebra


Question: Determine all pairs $(a, b)$ of real numbers that satisfy the following system of equations:

$$
\begin{aligned}
\sqrt{a}+\sqrt{b} & =8 \\
\log _{10} a+\log _{10} b & =2
\end{aligned}
$$

Give your answer(s) as pairs of simplified exact numbers.

Solution:
From the first equation, we note that $a \geq 0$ and $b \geq 0$, since the argument of a square root must be non-negative.

From the second equation, we note that $a>0$ and $b>0$, since the argument of a logarithm must be positive.

Combining these restrictions, we see that $a>0$ and $b>0$.

From the equation $\log _{10} a+\log _{10} b=2$, we obtain $\log _{10}(a b)=2$ and so $a b=10^{2}=100$. From the first equation, obtain

$$
\begin{aligned}
(\sqrt{a}+\sqrt{b})^{2} & =8^{2} \\
a+2 \sqrt{a b}+b & =64 \\
a+2 \sqrt{100}+b & =64 \\
a+b & =64-2 \sqrt{100}=44
\end{aligned}
$$

Since $a+b=44$, then $b=44-a$.

Since $a b=100$, then $a(44-a)=100$ or $44 a-a^{2}=100$ and so $0=a^{2}-44 a+100$.

By the quadratic formula,

$$
a=\frac{44 \pm \sqrt{44^{2}-4(1)(100)}}{2 \cdot 1}=\frac{44 \pm \sqrt{1536}}{2}=\frac{44 \pm 16 \sqrt{6}}{2}=22 \pm 8 \sqrt{6}
$$

Since $b=44-a$, then $b=44-(22 \pm 8 \sqrt{6})=22 \mp 8 \sqrt{6}$.

Therefore, $(a, b)=(22+8 \sqrt{6}, 22-8 \sqrt{6})$ or $(a, b)=(22-8 \sqrt{6}, 22+8 \sqrt{6})$.

(We note that $22+8 \sqrt{6}>0$ and $22-8 \sqrt{6}>0$, so the initial restrictions on $a$ and $b$ are satisfied.)

Final answer: $(22+8 \sqrt{6}, 22-8 \sqrt{6})$,$(22-8 \sqrt{6}, 22+8 \sqrt{6})$
Subfield: Geometry


Question: In the diagram, line segments $A C$ and $D F$ are tangent to the circle at $B$ and $E$, respectively. Also, $A F$ intersects the circle at $P$ and $R$, and intersects $B E$ at $Q$, as shown. If $\angle C A F=35^{\circ}, \angle D F A=30^{\circ}$, and $\angle F P E=25^{\circ}$, determine the measure of $\angle P E Q$.

<img_3970>

Solution:
Let $\angle P E Q=\theta$.

Join $P$ to $B$.

We use the fact that the angle between a tangent to a circle and a chord in that circle that passes through the point of tangency equals the angle inscribed by that chord. We prove this fact below.

More concretely, $\angle D E P=\angle P B E$ (using the chord $E P$ and the tangent through $E$ ) and $\angle A B P=\angle P E Q=\theta$ (using the chord $B P$ and the tangent through $B$ ).

Now $\angle D E P$ is exterior to $\triangle F E P$ and so $\angle D E P=\angle F P E+\angle E F P=25^{\circ}+30^{\circ}$, and so $\angle P B E=\angle D E P=55^{\circ}$.

Furthermore, $\angle A Q B$ is an exterior angle of $\triangle P Q E$.

Thus, $\angle A Q B=\angle Q P E+\angle P E Q=25^{\circ}+\theta$.

<img_3991>

In $\triangle A B Q$, we have $\angle B A Q=35^{\circ}, \angle A B Q=\theta+55^{\circ}$, and $\angle A Q B=25^{\circ}+\theta$.

Thus, $35^{\circ}+\left(\theta+55^{\circ}\right)+\left(25^{\circ}+\theta\right)=180^{\circ}$ or $115^{\circ}+2 \theta=180^{\circ}$, and so $2 \theta=65^{\circ}$.

Therefore $\angle P E Q=\theta=\frac{1}{2}\left(65^{\circ}\right)=32.5^{\circ}$.

As an addendum, we prove that the angle between a tangent to a circle and a chord in that circle that passes through the point of tangency equals the angle inscribed by that chord.

Consider a circle with centre $O$ and a chord $X Y$, with tangent $Z X$ meeting the circle at $X$. We prove that if $Z X$ is tangent to the circle, then $\angle Z X Y$ equals $\angle X W Y$ whenever $W$ is a point on the circle on the opposite side of $X Y$ as $X Z$ (that is, the angle subtended by $X Y$ on the opposite side of the circle).

We prove this in the case that $\angle Z X Y$ is acute. The cases where $\angle Z X Y$ is a right angle or an obtuse angle are similar.

Draw diameter $X O V$ and join $V Y$.

<img_3622>

Since $\angle Z X Y$ is acute, points $V$ and $W$ are on the same arc of chord $X Y$.

This means that $\angle X V Y=\angle X W Y$, since they are angles subtended by the same chord.

Since $O X$ is a radius and $X Z$ is a tangent, then $\angle O X Z=90^{\circ}$.

Thus, $\angle O X Y+\angle Z X Y=90^{\circ}$.

Since $X V$ is a diameter, then $\angle X Y V=90^{\circ}$.

From $\triangle X Y V$, we see that $\angle X V Y+\angle V X Y=90^{\circ}$.

But $\angle O X Y+\angle Z X Y=90^{\circ}$ and $\angle X V Y+\angle V X Y=90^{\circ}$ and $\angle O X Y=\angle V X Y$ tells us that $\angle Z X Y=\angle X V Y$.

This gives us that $\angle Z X Y=\angle X W Y$, as required.

Final answer: $32.5^{\circ}$
Subfield: Combinatorics


Question: A permutation of a list of numbers is an ordered arrangement of the numbers in that list. For example, $3,2,4,1,6,5$ is a permutation of $1,2,3,4,5,6$. We can write this permutation as $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$, where $a_{1}=3, a_{2}=2, a_{3}=4, a_{4}=1, a_{5}=6$, and $a_{6}=5$.
Determine the average value of

$$
\left|a_{1}-a_{2}\right|+\left|a_{3}-a_{4}\right|
$$

over all permutations $a_{1}, a_{2}, a_{3}, a_{4}$ of $1,2,3,4$.

Solution:
There are 4 ! $=4 \cdot 3 \cdot 2 \cdot 1=24$ permutations of $1,2,3,4$.

This is because there are 4 possible choices for $a_{1}$, and for each of these there are 3 possible choices for $a_{2}$, and for each of these there are 2 possible choices for $a_{3}$, and then 1 possible choice for $a_{4}$.

Consider the permutation $a_{1}=1, a_{2}=2, a_{3}=3, a_{4}=4$. (We write this as $1,2,3,4$.)

Here, $\left|a_{1}-a_{2}\right|+\left|a_{3}-a_{4}\right|=|1-2|+|3-4|=1+1=2$.

This value is the same as the value for each of $2,1,3,4$ and $1,2,4,3$ and $2,1,4,3$ and $3,4,1,2$ and 4,3,1,2 and 3,4,2,1 and 4,3,2,1.

Consider the permutation $1,3,2,4$.

Here, $\left|a_{1}-a_{2}\right|+\left|a_{3}-a_{4}\right|=|1-3|+|2-4|=2+2=4$.

This value is the same as the value for each of $3,1,2,4$ and $1,3,4,2$ and $3,1,4,2$ and $2,4,1,3$ and 4,2,1,3 and 2,4,3,1 and 4,2,3,1.

Consider the permutation $1,4,2,3$.

Here, $\left|a_{1}-a_{2}\right|+\left|a_{3}-a_{4}\right|=|1-4|+|2-3|=3+1=4$.

This value is the same as the value for each of 4,1,2,3 and 1,4,3,2 and 4,1,3,2 and 2,3,1,4 and $3,2,1,4$ and $2,3,4,1$ and $3,2,4,1$.

This accounts for all 24 permutations.

Therefore, the average value is $\frac{2 \cdot 8+4 \cdot 8+4 \cdot 8}{24}=\frac{80}{24}=\frac{10}{3}$.

Final answer: $\frac{10}{3}$
Subfield: Combinatorics


Question: A permutation of a list of numbers is an ordered arrangement of the numbers in that list. For example, $3,2,4,1,6,5$ is a permutation of $1,2,3,4,5,6$. We can write this permutation as $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$, where $a_{1}=3, a_{2}=2, a_{3}=4, a_{4}=1, a_{5}=6$, and $a_{6}=5$.
Determine the average value of

$$
a_{1}-a_{2}+a_{3}-a_{4}+a_{5}-a_{6}+a_{7}
$$

over all permutations $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}$ of $1,2,3,4,5,6,7$.

Solution:
There are $7 !=7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$ permutations of $1,2,3,4,5,6,7$, because there are 7 choices for $a_{1}$, then 6 choices for $a_{2}$, and so on.

We determine the average value of $a_{1}-a_{2}+a_{3}-a_{4}+a_{5}-a_{6}+a_{7}$ over all of these permutations by determining the sum of all 7 ! values of this expression and dividing by $7 !$.

To determine the sum of all 7 ! values, we determine the sum of the values of $a_{1}$ in each of these expressions and call this total $s_{1}$, the sum of the values of $a_{2}$ in each of these expressions and call this total $s_{2}$, and so on.

The sum of the 7 ! values of the original expression must equal $s_{1}-s_{2}+s_{3}-s_{4}+s_{5}-s_{6}+s_{7}$. This uses the fact that, when adding, the order in which we add the same set of numbers does not matter.

By symmetry, the sums of the values of $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}$ will all be equal. That is, $s_{1}=s_{2}=s_{3}=s_{4}=s_{5}=s_{6}=s_{7}$.

This means that the desired average value equals

$$
\frac{s_{1}-s_{2}+s_{3}-s_{4}+s_{5}-s_{6}+s_{7}}{7 !}=\frac{\left(s_{1}+s_{3}+s_{5}+s_{7}\right)-\left(s_{2}+s_{4}+s_{6}\right)}{7 !}=\frac{4 s_{1}-3 s_{1}}{7 !}=\frac{s_{1}}{7 !}
$$

So we need to determine the value of $s_{1}$.

Now $a_{1}$ can equal each of $1,2,3,4,5,6,7$.

If $a_{1}=1$, there are 6 ! combinations of values for $a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}$, since there are still 6 choices for $a_{2}, 5$ for $a_{3}$, and so on.

Similarly, there are 6 ! combinations with $a_{1}$ equal to each of $2,3,4,5,6,7$.

Thus, $s_{1}=1 \cdot 6 !+2 \cdot 6 !+3 \cdot 6 !+4 \cdot 6 !+5 \cdot 6 !+6 \cdot 6 !+7 \cdot 6 !=6 !(1+2+3+4+5+6+7)=28(6 !)$.

Therefore, the average value of the expression is $\frac{28(6 !)}{7 !}=\frac{28(6 !)}{7(6 !)}=\frac{28}{7}=4$.

Final answer: 4
Subfield: Number Theory


Question: Consider a set $S$ that contains $m \geq 4$ elements, each of which is a positive integer and no two of which are equal. We call $S$ boring if it contains four distinct integers $a, b, c, d$ such that $a+b=c+d$. We call $S$ exciting if it is not boring. For example, $\{2,4,6,8,10\}$ is boring since $4+8=2+10$. Also, $\{1,5,10,25,50\}$ is exciting.
Prove that, if $S$ is an exciting set of $m \geq 4$ positive integers, then $S$ contains an integer greater than or equal to $\frac{m^{2}-m}{4}$.

Solution:
Suppose that $S$ is an exciting set that contains exactly $m$ elements.

There are $\left(\begin{array}{c}m \\ 2\end{array}\right)=\frac{m(m-1)}{2}$ pairs of elements of $S$.

Since $S$ is exciting, the sums of these pairs of elements are all distinct positive integers.

This means that the largest of these sums is greater than or equal to $\frac{m(m-1)}{2}$.

When two numbers add to $\frac{m(m-1)}{2}$ or greater, then at least one of them must be at least as large as $\frac{1}{2} \cdot \frac{m(m-1)}{2}=\frac{m^{2}-m}{4}$.

Therefore, there is an element of $S$ that is greater than or equal to $\frac{m^{2}-m}{4}$.
Subfield: Geometry


Question: If $0^{\circ}<x<90^{\circ}$ and $3 \sin (x)-\cos \left(15^{\circ}\right)=0$, what is the value of $x$ to the nearest tenth of a degree?

Solution:
Rearranging the equation,

$$
\begin{aligned}
3 \sin (x) & =\cos \left(15^{\circ}\right) \\
\sin (x) & =\frac{1}{3} \cos \left(15^{\circ}\right) \\
\sin (x) & \approx 0.3220
\end{aligned}
$$

Using a calculator, $x \approx 18.78^{\circ}$. To the nearest tenth of a degree, $x=18.8^{\circ}$.

Final answer: $18.8^{\circ}$
Subfield: Geometry


Question: In the diagram, $\triangle A B C$ is right-angled at $B$ and $A C=20$. If $\sin C=\frac{3}{5}$, what is the length of side $B C$ ?

![](https://cdn.mathpix.com/cropped/2023_12_21_1f6e0e123b672977527ag-1.jpg?height=228&width=296&top_left_y=1629&top_left_x=1424)

Solution:
Since $\sin C=\frac{A B}{A C}$, then $A B=A C \sin C=20\left(\frac{3}{5}\right)=12$.

By Pythagoras, $B C^{2}=A C^{2}-A B^{2}=20^{2}-12^{2}=256$ or $B C=16$.
Using the standard trigonometric ratios, $B C=A C \cos C$.

Since $\sin C=\frac{3}{5}$, then $\cos ^{2} C=1-\sin ^{2} C=1-\frac{9}{25}=\frac{16}{25}$ or $\cos C=\frac{4}{5}$. (Notice that $\cos C$ is positive since angle $C$ is acute in triangle $A B C$.)

Therefore, $B C=20\left(\frac{4}{5}\right)=16$.

Final answer: 16
Subfield: Geometry


Question: In the diagram, $\triangle A B C$ is right-angled at $B$ and $A C=20$. If $\sin C=\frac{3}{5}$, what is the length of side $B C$ ?

<img_4028>

Solution:
Since $\sin C=\frac{A B}{A C}$, then $A B=A C \sin C=20\left(\frac{3}{5}\right)=12$.

By Pythagoras, $B C^{2}=A C^{2}-A B^{2}=20^{2}-12^{2}=256$ or $B C=16$.
Using the standard trigonometric ratios, $B C=A C \cos C$.

Since $\sin C=\frac{3}{5}$, then $\cos ^{2} C=1-\sin ^{2} C=1-\frac{9}{25}=\frac{16}{25}$ or $\cos C=\frac{4}{5}$. (Notice that $\cos C$ is positive since angle $C$ is acute in triangle $A B C$.)

Therefore, $B C=20\left(\frac{4}{5}\right)=16$.

Final answer: 16
Subfield: Geometry


Question: A helicopter is flying due west over level ground at a constant altitude of $222 \mathrm{~m}$ and at a constant speed. A lazy, stationary goat, which is due west of the helicopter, takes two measurements of the angle between the ground and the helicopter. The first measurement the goat makes is $6^{\circ}$ and the second measurement, which he makes 1 minute later, is $75^{\circ}$. If the helicopter has not yet passed over the goat, as shown, how fast is the helicopter travelling to the nearest kilometre per hour?

<img_3140>

Solution:
Let $G$ be the point where the goat is standing, $H$ the position of the helicopter when the goat first measures the angle, $P$ the point directly below the helicopter at this time, $J$ the position of the helicopter one minute later, and $Q$ the point directly below the helicopter at this time.

<img_3329>

Using the initial position of the helicopter, $\tan \left(6^{\circ}\right)=\frac{H P}{P G}$ or $P G=\frac{222}{\tan \left(6^{\circ}\right)} \approx 2112.19 \mathrm{~m}$.

Using the second position of the helicopter, $\tan \left(75^{\circ}\right)=\frac{J Q}{Q G}$ or $Q G=\frac{222}{\tan \left(75^{\circ}\right)} \approx 59.48 \mathrm{~m}$.

So in the one minute that has elapsed, the helicopter has travelled

$2112.19 \mathrm{~m}-59.48 \mathrm{~m}=2052.71 \mathrm{~m}$ or $2.0527 \mathrm{~km}$.

Therefore, in one hour, the helicopter will travel $60(2.0527)=123.162 \mathrm{~km}$.

Thus, the helicopter is travelling $123 \mathrm{~km} / \mathrm{h}$.

Final answer: 123 km/h