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


Question: Determine all pairs of angles $(x, y)$ with $0^{\circ} \leq x<180^{\circ}$ and $0^{\circ} \leq y<180^{\circ}$ that satisfy the following system of equations:

$$
\begin{aligned}
\log _{2}(\sin x \cos y) & =-\frac{3}{2} \\
\log _{2}\left(\frac{\sin x}{\cos y}\right) & =\frac{1}{2}
\end{aligned}
$$

Solution:
Using logarithm laws, the given equations are equivalent to

$$
\begin{aligned}
& \log _{2}(\sin x)+\log _{2}(\cos y)=-\frac{3}{2} \\
& \log _{2}(\sin x)-\log _{2}(\cos y)=\frac{1}{2}
\end{aligned}
$$

Adding these two equations, we obtain $2 \log _{2}(\sin x)=-1$ which gives $\log _{2}(\sin x)=-\frac{1}{2}$ and so $\sin x=2^{-1 / 2}=\frac{1}{2^{1 / 2}}=\frac{1}{\sqrt{2}}$.

Since $0^{\circ} \leq x<180^{\circ}$, then $x=45^{\circ}$ or $x=135^{\circ}$.

Since $\log _{2}(\sin x)+\log _{2}(\cos y)=-\frac{3}{2}$ and $\log _{2}(\sin x)=-\frac{1}{2}$, then $\log _{2}(\cos y)=-1$, which gives $\cos y=2^{-1}=\frac{1}{2}$.

Since $0^{\circ} \leq y<180^{\circ}$, then $y=60^{\circ}$.

Therefore, $(x, y)=\left(45^{\circ}, 60^{\circ}\right)$ or $(x, y)=\left(135^{\circ}, 60^{\circ}\right)$.
First, we note that $2^{1 / 2}=\sqrt{2}$ and $2^{-3 / 2}=\frac{1}{2^{3 / 2}}=\frac{1}{2^{1} 2^{1 / 2}}=\frac{1}{2 \sqrt{2}}$.

From the given equations, we obtain

$$
\begin{aligned}
\sin x \cos y & =2^{-3 / 2}=\frac{1}{2 \sqrt{2}} \\
\frac{\sin x}{\cos y} & =2^{1 / 2}=\sqrt{2}
\end{aligned}
$$

Multiplying these two equations together, we obtain $(\sin x)^{2}=\frac{1}{2}$ which gives $\sin x= \pm \frac{1}{\sqrt{2}}$.

Since $0^{\circ} \leq x<180^{\circ}$, it must be the case that $\sin x \geq 0$ and so $\sin x=\frac{1}{\sqrt{2}}$.

Since $0^{\circ} \leq x<180^{\circ}$, we obtain $x=45^{\circ}$ or $x=135^{\circ}$.

Since $\sin x \cos y=\frac{1}{2 \sqrt{2}}$ and $\sin x=\frac{1}{\sqrt{2}}$, we obtain $\cos y=\frac{1}{2}$.

Since $0^{\circ} \leq y<180^{\circ}$, then $y=60^{\circ}$.

Therefore, $(x, y)=\left(45^{\circ}, 60^{\circ}\right)$ or $(x, y)=\left(135^{\circ}, 60^{\circ}\right)$.

Final answer: $(45^{\circ}, 60^{\circ}),(135^{\circ}, 60^{\circ})$
Subfield: Geometry


Question: Three microphones $A, B$ and $C$ are placed on a line such that $A$ is $1 \mathrm{~km}$ west of $B$ and $C$ is $2 \mathrm{~km}$ east of $B$. A large explosion occurs at a point $P$ not on this line. Each of the three microphones receives the sound. The sound travels at $\frac{1}{3} \mathrm{~km} / \mathrm{s}$. Microphone $B$ receives the sound first, microphone $A$ receives the sound $\frac{1}{2}$ s later, and microphone $C$ receives it $1 \mathrm{~s}$ after microphone $A$. Determine the distance from microphone $B$ to the explosion at $P$.

Solution:
Throughout this solution, we will mostly not include units, but will assume that all lengths are in kilometres, all times are in seconds, and all speeds are in kilometres per second.

We place the points in the coordinate plane with $B$ at $(0,0), A$ on the negative $x$-axis, and $C$ on the positive $x$-axis.

We put $A$ at $(-1,0)$ and $C$ at $(2,0)$.

Suppose that $P$ has coordinates $(x, y)$ and that the distance from $P$ to $B$ is $d \mathrm{~km}$.

<img_4023>

Since the sound arrives at $A \frac{1}{2} \mathrm{~s}$ after arriving at $B$ and sound travels at $\frac{1}{3} \mathrm{~km} / \mathrm{s}$, then $A$ is $\left(\frac{1}{2} \mathrm{~s}\right) \cdot\left(\frac{1}{3} \mathrm{~km} / \mathrm{s}\right)=\frac{1}{6} \mathrm{~km}$ farther from $P$ than $B$ is.

Thus, the distance from $P$ to $A$ is $\left(d+\frac{1}{6}\right) \mathrm{km}$.

Since the sound arrives at $C$ an additional 1 second later, then $C$ is an additional $\frac{1}{3} \mathrm{~km}$ farther, and so is $\left(d+\frac{1}{6}\right) \mathrm{km}+\left(\frac{1}{3} \mathrm{~km}\right)=\left(d+\frac{1}{2}\right) \mathrm{km}$ from $P$.

Since the distance from $P$ to $B$ is $d \mathrm{~km}$, then $(x-0)^{2}+(y-0)^{2}=d^{2}$.

Since the distance from $P$ to $A$ is $\left(d+\frac{1}{6}\right) \mathrm{km}$, then $(x+1)^{2}+(y-0)^{2}=\left(d+\frac{1}{6}\right)^{2}$.

Since the distance from $P$ to $C$ is $\left(d+\frac{1}{2}\right) \mathrm{km}$, then $(x-2)^{2}+(y-0)^{2}=\left(d+\frac{1}{2}\right)^{2}$.

When these equations are expanded and simplified, we obtain

$$
\begin{aligned}
x^{2}+y^{2} & =d^{2} \\
x^{2}+2 x+1+y^{2} & =d^{2}+\frac{1}{3} d+\frac{1}{36} \\
x^{2}-4 x+4+y^{2} & =d^{2}+d+\frac{1}{4}
\end{aligned}
$$

Subtracting the first equation from the second, we obtain

$$
2 x+1=\frac{1}{3} d+\frac{1}{36}
$$

Subtracting the first equation from the third, we obtain

$$
-4 x+4=d+\frac{1}{4}
$$

Therefore,

$$
\begin{aligned}
2(2 x+1)+(-4 x+4) & =2\left(\frac{1}{3} d+\frac{1}{36}\right)+\left(d+\frac{1}{4}\right) \\
6 & =\frac{2}{3} d+\frac{1}{18}+d+\frac{1}{4} \\
216 & =24 d+2+36 d+9 \quad \text { (multiplying by } 36) \\
205 & =60 d \\
d & =\frac{41}{12}
\end{aligned}
$$

Therefore, the distance from $B$ to $P$ is $\frac{41}{12} \mathrm{~km}$.

Final answer: $\frac{41}{12}$ km
Subfield: Number Theory


Question: An L shape is made by adjoining three congruent squares. The L is subdivided into four smaller L shapes, as shown. Each of the resulting L's is subdivided in this same way. After the third round of subdivisions, how many L's of the smallest size are there?
<img_3486>

Solution:
After each round, each L shape is divided into 4 smaller $\mathrm{L}$ shapes.

This means that the number of $\mathrm{L}$ shapes increases by a factor of 4 after each round.

After 1 round, there are $4 \mathrm{~L}$ shapes.

After 2 rounds, there are $4^{2}=16$ L's of the smallest size.

After 3 rounds, there are $4^{3}=64$ L's of the smallest size.

Final answer: 64
Subfield: Number Theory


Question: An L shape is made by adjoining three congruent squares. The L is subdivided into four smaller L shapes, as shown. Each of the resulting L's is subdivided in this same way. After the third round of subdivisions, how many L's of the smallest size are there?
![](https://cdn.mathpix.com/cropped/2023_12_21_68ed10fc1401a8e73334g-1.jpg?height=226&width=472&top_left_y=1372&top_left_x=890)

Solution:
After each round, each L shape is divided into 4 smaller $\mathrm{L}$ shapes.

This means that the number of $\mathrm{L}$ shapes increases by a factor of 4 after each round.

After 1 round, there are $4 \mathrm{~L}$ shapes.

After 2 rounds, there are $4^{2}=16$ L's of the smallest size.

After 3 rounds, there are $4^{3}=64$ L's of the smallest size.

Final answer: 64
Subfield: Number Theory


Question: Kerry has a list of $n$ integers $a_{1}, a_{2}, \ldots, a_{n}$ satisfying $a_{1} \leq a_{2} \leq \ldots \leq a_{n}$. Kerry calculates the pairwise sums of all $m=\frac{1}{2} n(n-1)$ possible pairs of integers in her list and orders these pairwise sums as $s_{1} \leq s_{2} \leq \ldots \leq s_{m}$. For example, if Kerry's list consists of the three integers $1,2,4$, the three pairwise sums are $3,5,6$.


Suppose that $n=4$ and that the 6 pairwise sums are $s_{1}=8, s_{2}=104, s_{3}=106$, $s_{4}=110, s_{5}=112$, and $s_{6}=208$. Determine two possible lists $a_{1}, a_{2}, a_{3}, a_{4}$ that Kerry could have.

Solution:
Here, the pairwise sums of the numbers $a_{1} \leq a_{2} \leq a_{3} \leq a_{4}$ are $s_{1} \leq s_{2} \leq s_{3} \leq s_{4} \leq s_{5} \leq s_{6}$. The six pairwise sums of the numbers in the list can be expressed as

$$
a_{1}+a_{2}, a_{1}+a_{3}, a_{1}+a_{4}, a_{2}+a_{3}, a_{2}+a_{4}, a_{3}+a_{4}
$$

Since $a_{1} \leq a_{2} \leq a_{3} \leq a_{4}$, then the smallest sum must be the sum of the two smallest numbers. Thus, $s_{1}=a_{1}+a_{2}$.

Similarly, the largest sum must be the sum of the two largest numbers, and so $s_{6}=a_{3}+a_{4}$. Since $a_{1} \leq a_{2} \leq a_{3} \leq a_{4}$, then the second smallest sum is $a_{1}+a_{3}$. This is because $a_{1}+a_{3}$ is no greater than each of the four sums $a_{1}+a_{4}, a_{2}+a_{3}, a_{2}+a_{4}$, and $a_{3}+a_{4}$ :

Since $a_{3} \leq a_{4}$, then $a_{1}+a_{3} \leq a_{1}+a_{4}$.

Since $a_{1} \leq a_{2}$, then $a_{1}+a_{3} \leq a_{2}+a_{3}$.

Since $a_{1} \leq a_{2}$ and $a_{3} \leq a_{4}$, then $a_{1}+a_{3} \leq a_{2}+a_{4}$.

Since $a_{1} \leq a_{4}$, then $a_{1}+a_{3} \leq a_{3}+a_{4}$.

Thus, $s_{2}=a_{1}+a_{3}$.

Using a similar argument, $s_{5}=a_{2}+a_{4}$.

So far, we have $s_{1}=a_{1}+a_{2}$ and $s_{2}=a_{1}+a_{3}$ and $s_{5}=a_{2}+a_{4}$ and $s_{6}=a_{3}+a_{4}$.

This means that $s_{3}$ and $s_{4}$ equal $a_{1}+a_{4}$ and $a_{2}+a_{3}$ in some order.

It turns out that either order is possible.

Case 1: $s_{3}=a_{1}+a_{4}$ and $s_{4}=a_{2}+a_{3}$

Here, $a_{1}+a_{2}=8$ and $a_{1}+a_{3}=104$ and $a_{2}+a_{3}=110$.

Adding these three equations gives

$$
\left(a_{1}+a_{2}\right)+\left(a_{1}+a_{3}\right)+\left(a_{2}+a_{3}\right)=8+104+110
$$

and so $2 a_{1}+2 a_{2}+2 a_{3}=222$ or $a_{1}+a_{2}+a_{3}=111$.

Since $a_{2}+a_{3}=110$, then $a_{1}=\left(a_{1}+a_{2}+a_{3}\right)-\left(a_{2}+a_{3}\right)=111-110=1$.

Since $a_{1}=1$ and $a_{1}+a_{2}=8$, then $a_{2}=7$.

Since $a_{1}=1$ and $a_{1}+a_{3}=104$, then $a_{3}=103$.

Since $a_{3}=103$ and $a_{3}+a_{4}=208$, then $a_{4}=105$.

Thus, $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=(1,7,103,105)$.

Case 2: $s_{3}=a_{2}+a_{3}$ and $s_{4}=a_{1}+a_{4}$

Here, $a_{1}+a_{2}=8$ and $a_{1}+a_{3}=104$ and $a_{2}+a_{3}=106$.

Using the same process, $a_{1}+a_{2}+a_{3}=109$.

From this, we obtain $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=(3,5,101,107)$.

Therefore, Kerry's two possible lists are 1,7,103, 105 and 3, 5, 101, 107.

Final answer: 1,7,103, 105 and 3, 5, 101, 107
Subfield: Number Theory


Question: Kerry has a list of $n$ integers $a_{1}, a_{2}, \ldots, a_{n}$ satisfying $a_{1} \leq a_{2} \leq \ldots \leq a_{n}$. Kerry calculates the pairwise sums of all $m=\frac{1}{2} n(n-1)$ possible pairs of integers in her list and orders these pairwise sums as $s_{1} \leq s_{2} \leq \ldots \leq s_{m}$. For example, if Kerry's list consists of the three integers $1,2,4$, the three pairwise sums are $3,5,6$.


Suppose that $n=4$ and that the 6 pairwise sums are $s_{1}=8, s_{2}=104, s_{3}=106$, $s_{4}=110, s_{5}=112$, and $s_{6}=208$. Determine two possible lists $(a_{1}, a_{2}, a_{3}, a_{4})$ that Kerry could have.

Solution:
Here, the pairwise sums of the numbers $a_{1} \leq a_{2} \leq a_{3} \leq a_{4}$ are $s_{1} \leq s_{2} \leq s_{3} \leq s_{4} \leq s_{5} \leq s_{6}$. The six pairwise sums of the numbers in the list can be expressed as

$$
a_{1}+a_{2}, a_{1}+a_{3}, a_{1}+a_{4}, a_{2}+a_{3}, a_{2}+a_{4}, a_{3}+a_{4}
$$

Since $a_{1} \leq a_{2} \leq a_{3} \leq a_{4}$, then the smallest sum must be the sum of the two smallest numbers. Thus, $s_{1}=a_{1}+a_{2}$.

Similarly, the largest sum must be the sum of the two largest numbers, and so $s_{6}=a_{3}+a_{4}$. Since $a_{1} \leq a_{2} \leq a_{3} \leq a_{4}$, then the second smallest sum is $a_{1}+a_{3}$. This is because $a_{1}+a_{3}$ is no greater than each of the four sums $a_{1}+a_{4}, a_{2}+a_{3}, a_{2}+a_{4}$, and $a_{3}+a_{4}$ :

Since $a_{3} \leq a_{4}$, then $a_{1}+a_{3} \leq a_{1}+a_{4}$.

Since $a_{1} \leq a_{2}$, then $a_{1}+a_{3} \leq a_{2}+a_{3}$.

Since $a_{1} \leq a_{2}$ and $a_{3} \leq a_{4}$, then $a_{1}+a_{3} \leq a_{2}+a_{4}$.

Since $a_{1} \leq a_{4}$, then $a_{1}+a_{3} \leq a_{3}+a_{4}$.

Thus, $s_{2}=a_{1}+a_{3}$.

Using a similar argument, $s_{5}=a_{2}+a_{4}$.

So far, we have $s_{1}=a_{1}+a_{2}$ and $s_{2}=a_{1}+a_{3}$ and $s_{5}=a_{2}+a_{4}$ and $s_{6}=a_{3}+a_{4}$.

This means that $s_{3}$ and $s_{4}$ equal $a_{1}+a_{4}$ and $a_{2}+a_{3}$ in some order.

It turns out that either order is possible.

Case 1: $s_{3}=a_{1}+a_{4}$ and $s_{4}=a_{2}+a_{3}$

Here, $a_{1}+a_{2}=8$ and $a_{1}+a_{3}=104$ and $a_{2}+a_{3}=110$.

Adding these three equations gives

$$
\left(a_{1}+a_{2}\right)+\left(a_{1}+a_{3}\right)+\left(a_{2}+a_{3}\right)=8+104+110
$$

and so $2 a_{1}+2 a_{2}+2 a_{3}=222$ or $a_{1}+a_{2}+a_{3}=111$.

Since $a_{2}+a_{3}=110$, then $a_{1}=\left(a_{1}+a_{2}+a_{3}\right)-\left(a_{2}+a_{3}\right)=111-110=1$.

Since $a_{1}=1$ and $a_{1}+a_{2}=8$, then $a_{2}=7$.

Since $a_{1}=1$ and $a_{1}+a_{3}=104$, then $a_{3}=103$.

Since $a_{3}=103$ and $a_{3}+a_{4}=208$, then $a_{4}=105$.

Thus, $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=(1,7,103,105)$.

Case 2: $s_{3}=a_{2}+a_{3}$ and $s_{4}=a_{1}+a_{4}$

Here, $a_{1}+a_{2}=8$ and $a_{1}+a_{3}=104$ and $a_{2}+a_{3}=106$.

Using the same process, $a_{1}+a_{2}+a_{3}=109$.

From this, we obtain $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=(3,5,101,107)$.

Therefore, Kerry's two possible lists are 1,7,103, 105 and 3, 5, 101, 107.



Final answer: (1,7,103, 105), (3, 5, 101, 107)
Subfield: Geometry


Question: Jimmy is baking two large identical triangular cookies, $\triangle A B C$ and $\triangle D E F$. Each cookie is in the shape of an isosceles right-angled triangle. The length of the shorter sides of each of these triangles is $20 \mathrm{~cm}$. He puts the cookies on a rectangular baking tray so that $A, B, D$, and $E$ are at the vertices of the rectangle, as shown. If the distance between parallel sides $A C$ and $D F$ is $4 \mathrm{~cm}$, what is the width $B D$ of the tray?

<img_3349>

Solution:
We note that $B D=B C+C D$ and that $B C=20 \mathrm{~cm}$, so we need to determine $C D$.

We draw a line from $C$ to $P$ on $F D$ so that $C P$ is perpendicular to $D F$.

Since $A C$ and $D F$ are parallel, then $C P$ is also perpendicular to $A C$.

The distance between $A C$ and $D F$ is $4 \mathrm{~cm}$, so $C P=4 \mathrm{~cm}$.

Since $\triangle A B C$ is isosceles and right-angled, then $\angle A C B=45^{\circ}$.

<img_3723>

Thus, $\angle P C D=180^{\circ}-\angle A C B-\angle P C A=180^{\circ}-45^{\circ}-90^{\circ}=45^{\circ}$.

Since $\triangle C P D$ is right-angled at $P$ and $\angle P C D=45^{\circ}$, then $\triangle C P D$ is also an isosceles right-angled triangle.

Therefore, $C D=\sqrt{2} C P=4 \sqrt{2} \mathrm{~cm}$.

Finally, $B D=B C+C D=(20+4 \sqrt{2}) \mathrm{cm}$.

Final answer: $(20+4 \sqrt{2})$ cm
Subfield: Geometry


Question: Jimmy is baking two large identical triangular cookies, $\triangle A B C$ and $\triangle D E F$. Each cookie is in the shape of an isosceles right-angled triangle. The length of the shorter sides of each of these triangles is $20 \mathrm{~cm}$. He puts the cookies on a rectangular baking tray so that $A, B, D$, and $E$ are at the vertices of the rectangle, as shown. If the distance between parallel sides $A C$ and $D F$ is $4 \mathrm{~cm}$, what is the width $B D$ of the tray?

![](https://cdn.mathpix.com/cropped/2023_12_21_ac18afada1a016e79e7bg-1.jpg?height=387&width=480&top_left_y=278&top_left_x=1343)

Solution:
We note that $B D=B C+C D$ and that $B C=20 \mathrm{~cm}$, so we need to determine $C D$.

We draw a line from $C$ to $P$ on $F D$ so that $C P$ is perpendicular to $D F$.

Since $A C$ and $D F$ are parallel, then $C P$ is also perpendicular to $A C$.

The distance between $A C$ and $D F$ is $4 \mathrm{~cm}$, so $C P=4 \mathrm{~cm}$.

Since $\triangle A B C$ is isosceles and right-angled, then $\angle A C B=45^{\circ}$.

![](https://cdn.mathpix.com/cropped/2023_12_21_3c1abeb8d8eb989cf095g-1.jpg?height=388&width=469&top_left_y=1733&top_left_x=1530)

Thus, $\angle P C D=180^{\circ}-\angle A C B-\angle P C A=180^{\circ}-45^{\circ}-90^{\circ}=45^{\circ}$.

Since $\triangle C P D$ is right-angled at $P$ and $\angle P C D=45^{\circ}$, then $\triangle C P D$ is also an isosceles right-angled triangle.

Therefore, $C D=\sqrt{2} C P=4 \sqrt{2} \mathrm{~cm}$.

Finally, $B D=B C+C D=(20+4 \sqrt{2}) \mathrm{cm}$.

Final answer: $(20+4 \sqrt{2})$ cm
Subfield: Algebra


Question: Determine all values of $x$ for which $\frac{x^{2}+x+4}{2 x+1}=\frac{4}{x}$.

Solution:
Manipulating the given equation and noting that $x \neq 0$ and $x \neq-\frac{1}{2}$ since neither denominator can equal 0 , we obtain

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

Therefore, $x=-1$ or $x=2$ or $x=-2$. We can check by substitution that each satisfies the original equation.

Final answer: $-1$,$2$,$-2$
Subfield: Number Theory


Question: Determine the number of positive divisors of 900, including 1 and 900, that are perfect squares. (A positive divisor of 900 is a positive integer that divides exactly into 900.)

Solution:
Since $900=30^{2}$ and $30=2 \times 3 \times 5$, then $900=2^{2} 3^{2} 5^{2}$.

The positive divisors of 900 are those integers of the form $d=2^{a} 3^{b} 5^{c}$, where each of $a, b, c$ is 0,1 or 2 .

For $d$ to be a perfect square, the exponent on each prime factor in the prime factorization of $d$ must be even.

Thus, for $d$ to be a perfect square, each of $a, b, c$ must be 0 or 2 .

There are two possibilities for each of $a, b, c$ so $2 \times 2 \times 2=8$ possibilities for $d$.

These are $2^{0} 3^{0} 5^{0}=1,2^{2} 3^{0} 5^{0}=4,2^{0} 3^{2} 5^{0}=9,2^{0} 3^{0} 5^{2}=25,2^{2} 3^{2} 5^{0}=36,2^{2} 3^{0} 5^{2}=100$, $2^{0} 3^{2} 5^{2}=225$, and $2^{2} 3^{2} 5^{2}=900$.

Thus, 8 of the positive divisors of 900 are perfect squares.
The positive divisors of 900 are

$1,2,3,4,5,6,9,10,12,15,18,20,25,30,36,45,50,60,75,90,100,150,180,225,300,450,900$

Of these, $1,4,9,25,36,100,225$, and 900 are perfect squares $\left(1^{2}, 2^{2}, 3^{2}, 5^{2}, 6^{2}, 10^{2}, 15^{2}, 30^{2}\right.$, respectively).

Thus, 8 of the positive divisors of 900 are perfect squares.

Final answer: 8
Subfield: Geometry


Question: Points $A(k, 3), B(3,1)$ and $C(6, k)$ form an isosceles triangle. If $\angle A B C=\angle A C B$, determine all possible values of $k$.

Solution:
In isosceles triangle $A B C, \angle A B C=\angle A C B$, so the sides opposite these angles $(A C$ and $A B$, respectively) are equal in length.

Since the vertices of the triangle are $A(k, 3), B(3,1)$ and $C(6, k)$, then we obtain

$$
\begin{aligned}
A C & =A B \\
\sqrt{(k-6)^{2}+(3-k)^{2}} & =\sqrt{(k-3)^{2}+(3-1)^{2}} \\
(k-6)^{2}+(3-k)^{2} & =(k-3)^{2}+(3-1)^{2} \\
(k-6)^{2}+(k-3)^{2} & =(k-3)^{2}+2^{2} \\
(k-6)^{2} & =4
\end{aligned}
$$

Thus, $k-6=2$ or $k-6=-2$, and so $k=8$ or $k=4$.

We can check by substitution that each satisfies the original equation.

Final answer: $8$,$4$
Subfield: Combinatorics


Question: A chemist has three bottles, each containing a mixture of acid and water:

- bottle A contains $40 \mathrm{~g}$ of which $10 \%$ is acid,
- bottle B contains $50 \mathrm{~g}$ of which $20 \%$ is acid, and
- bottle C contains $50 \mathrm{~g}$ of which $30 \%$ is acid.

She uses some of the mixture from each of the bottles to create a mixture with mass $60 \mathrm{~g}$ of which $25 \%$ is acid. Then she mixes the remaining contents of the bottles to create a new mixture. What percentage of the new mixture is acid?

Solution:
Bottle A contains $40 \mathrm{~g}$ of which $10 \%$ is acid.

Thus, it contains $0.1 \times 40=4 \mathrm{~g}$ of acid and $40-4=36 \mathrm{~g}$ of water.

Bottle B contains $50 \mathrm{~g}$ of which $20 \%$ is acid.

Thus, it contains $0.2 \times 50=10 \mathrm{~g}$ of acid and $50-10=40 \mathrm{~g}$ of water.

Bottle C contains $50 \mathrm{~g}$ of which $30 \%$ is acid.

Thus, it contains $0.3 \times 50=15 \mathrm{~g}$ of acid and $50-15=35 \mathrm{~g}$ of water.

In total, the three bottles contain $40+50+50=140 \mathrm{~g}$, of which $4+10+15=29 \mathrm{~g}$ is acid and $140-29=111 \mathrm{~g}$ is water.

The new mixture has mass $60 \mathrm{~g}$ of which $25 \%$ is acid.

Thus, it contains $0.25 \times 60=15 \mathrm{~g}$ of acid and $60-15=45 \mathrm{~g}$ of water.

Since the total mass in the three bottles is initially $140 \mathrm{~g}$ and the new mixture has mass $60 \mathrm{~g}$, then the remaining contents have mass $140-60=80 \mathrm{~g}$.

Since the total mass of acid in the three bottles is initially $29 \mathrm{~g}$ and the acid in the new mixture has mass $15 \mathrm{~g}$, then the acid in the remaining contents has mass $29-15=14 \mathrm{~g}$. This remaining mixture is thus $\frac{14 \mathrm{~g}}{80 \mathrm{~g}} \times 100 \%=17.5 \%$ acid.

Final answer: 17.5%
Subfield: Algebra


Question: Suppose that $x$ and $y$ are real numbers with $3 x+4 y=10$. Determine the minimum possible value of $x^{2}+16 y^{2}$.

Solution:
Since $3 x+4 y=10$, then $4 y=10-3 x$.

Therefore, when $3 x+4 y=10$,

$$
\begin{aligned}
x^{2}+16 y^{2} & =x^{2}+(4 y)^{2} \\
& =x^{2}+(10-3 x)^{2} \\
& =x^{2}+\left(9 x^{2}-60 x+100\right) \\
& =10 x^{2}-60 x+100 \\
& =10\left(x^{2}-6 x+10\right) \\
& =10\left(x^{2}-6 x+9+1\right) \\
& =10\left((x-3)^{2}+1\right) \\
& =10(x-3)^{2}+10
\end{aligned}
$$

Since $(x-3)^{2} \geq 0$, then the minimum possible value of $10(x-3)^{2}+10$ is $10(0)+10=10$. This occurs when $(x-3)^{2}=0$ or $x=3$.

Therefore, the minimum possible value of $x^{2}+16 y^{2}$ when $3 x+4 y=10$ is 10 .

Final answer: 10
Subfield: Combinatorics


Question: A bag contains 40 balls, each of which is black or gold. Feridun reaches into the bag and randomly removes two balls. Each ball in the bag is equally likely to be removed. If the probability that two gold balls are removed is $\frac{5}{12}$, how many of the 40 balls are gold?

Solution:
Suppose that the bag contains $g$ gold balls.

We assume that Feridun reaches into the bag and removes the two balls one after the other.

There are 40 possible balls that he could remove first and then 39 balls that he could remove second. In total, there are 40(39) pairs of balls that he could choose in this way.

If he removes 2 gold balls, then there are $g$ possible balls that he could remove first and then $g-1$ balls that he could remove second. In total, there are $g(g-1)$ pairs of gold balls that he could remove.

We are told that the probability of removing 2 gold balls is $\frac{5}{12}$.

Since there are $40(39)$ total pairs of balls that can be chosen and $g(g-1)$ pairs of gold balls that can be chosen in this way, then $\frac{g(g-1)}{40(39)}=\frac{5}{12}$ which is equivalent to $g(g-1)=\frac{5}{12}(40)(39)=650$.



Therefore, $g^{2}-g-650=0$ or $(g-26)(g+25)=0$, and so $g=26$ or $g=-25$.

Since $g>0$, then $g=26$, so there are 26 gold balls in the bag.
Suppose that the bag contains $g$ gold balls.

We assume that Feridun reaches into the bag and removes the two balls together.

Since there are 40 balls in the bag, there are $\left(\begin{array}{c}40 \\ 2\end{array}\right)$ pairs of balls that he could choose in this way.

Since there are $g$ gold balls in the bag, then there are $\left(\begin{array}{l}g \\ 2\end{array}\right)$ pairs of gold balls that he could choose in this way.

We are told that the probability of removing 2 gold balls is $\frac{5}{12}$.

Since there are $\left(\begin{array}{c}40 \\ 2\end{array}\right)$ pairs in total that can be chosen and $\left(\begin{array}{l}g \\ 2\end{array}\right)$ pairs of gold balls that can be chosen in this way, then $\frac{\left(\begin{array}{l}g \\ 2\end{array}\right)}{\left(\begin{array}{c}40 \\ 2\end{array}\right)}=\frac{5}{12}$ which is equivalent to $\left(\begin{array}{l}g \\ 2\end{array}\right)=\frac{5}{12}\left(\begin{array}{c}40 \\ 2\end{array}\right)$.

Since $\left(\begin{array}{l}n \\ 2\end{array}\right)=\frac{n(n-1)}{2}$, then this equation is equivalent to $\frac{g(g-1)}{2}=\frac{5}{12} \frac{40(39)}{2}=325$.

Therefore, $g(g-1)=650$ or $g^{2}-g-650=0$ or $(g-26)(g+25)=0$, and so $g=26$ or $g=-25$.

Since $g>0$, then $g=26$, so there are 26 gold balls in the bag.

Final answer: 26