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


Question: Suppose that $\triangle A B C$ is right-angled at $B$ and has $A B=n(n+1)$ and $A C=(n+1)(n+4)$, where $n$ is a positive integer. Determine the number of positive integers $n<100000$ for which the length of side $B C$ is also an integer.

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

$$
\begin{aligned}
B C^{2} & =A C^{2}-A B^{2} \\
& =((n+1)(n+4))^{2}-(n(n+1))^{2} \\
& =(n+1)^{2}(n+4)^{2}-n^{2}(n+1)^{2} \\
& =(n+1)^{2}\left((n+4)^{2}-n^{2}\right) \\
& =(n+1)^{2}\left(n^{2}+8 n+16-n^{2}\right) \\
& =(n+1)^{2}(8 n+16) \\
& =4(n+1)^{2}(2 n+4)
\end{aligned}
$$

The length of $B C$ is an integer exactly when $4(n+1)^{2}(2 n+4)$ is a perfect square.

Since $4(n+1)^{2}$ is a perfect square, then $B C$ is an integer exactly when $2 n+4$ is a perfect square.

We note that $2 n+4 \geq 6$ (since $n \geq 1)$ and that $2 n+4$ is even.

Since $n<100000$, then $6 \leq 2 n+4<200004$, and so we need to count the number of even perfect squares between 6 and 200004 .

The smallest even perfect square in this range is $4^{2}=16$.

Since $\sqrt{200004} \approx 447.2$, the largest even perfect square in this range is $446^{2}$.

Therefore, the number of even perfect squares in this range is $\frac{446}{2}-1=222$.

Thus, there are 222 positive integers $n$ for which the length of $B C$ is an integer.

Final answer: 222
Subfield: Algebra


Question: Determine all real values of $x$ for which

$$
\sqrt{\log _{2} x \cdot \log _{2}(4 x)+1}+\sqrt{\log _{2} x \cdot \log _{2}\left(\frac{x}{64}\right)+9}=4
$$

Solution:
Let $f(x)=\sqrt{\log _{2} x \cdot \log _{2}(4 x)+1}+\sqrt{\log _{2} x \cdot \log _{2}\left(\frac{x}{64}\right)+9}$.

Using logarithm laws,

$$
\begin{aligned}
\log _{2} x \cdot \log _{2}(4 x)+1 & =\log _{2} x\left(\log _{2} 4+\log _{2} x\right)+1 \\
& =\log _{2} x\left(2+\log _{2} x\right)+1 \quad\left(\text { since } 2^{2}=4\right) \\
& =\left(\log _{2} x\right)^{2}+2 \cdot \log _{2} x+1 \\
& =\left(\log _{2} x+1\right)^{2}
\end{aligned}
$$

and

$$
\begin{aligned}
\log _{2} x \cdot \log _{2}\left(\frac{x}{64}\right)+9 & =\log _{2} x\left(\log _{2} x-\log _{2} 64\right)+9 \\
& =\log _{2} x\left(\log _{2} x-6\right)+9 \quad\left(\text { since } 2^{6}=64\right) \\
& =\left(\log _{2} x\right)^{2}-6 \log _{2} x+9 \\
& =\left(\log _{2} x-3\right)^{2}
\end{aligned}
$$

Therefore,

$f(x)=\sqrt{\log _{2} x \cdot \log _{2}(4 x)+1}+\sqrt{\log _{2} x \cdot \log _{2}\left(\frac{x}{64}\right)+9}=\sqrt{\left(\log _{2} x+1\right)^{2}}+\sqrt{\left(\log _{2} x-3\right)^{2}}$

Before proceeding, we recall that if $a \leq 0$, then $\sqrt{a^{2}}=-a$ and if $a>0$, then $\sqrt{a^{2}}=a$.

When $\log _{2} x \leq-1$, we know that $\log _{2} x+1 \leq 0$ and $\log _{2} x-3<0$, and so

$$
f(x)=-\left(\log _{2} x+1\right)-\left(\log _{2} x-3\right)=2-2 \log _{2} x
$$

When $-1<\log _{2} x \leq 3$, we know that $\log _{2} x+1>0$ and $\log _{2} x-3 \leq 0$, and so

$$
f(x)=\left(\log _{2} x+1\right)-\left(\log _{2} x-3\right)=4
$$

When $\log _{2} x>3$, we know that $\log _{2} x+1 \geq 0$ and $\log _{2} x-3>0$, and so

$$
f(x)=\left(\log _{2} x+1\right)+\left(\log _{2} x-3\right)=2 \log _{2} x-2
$$

We want to find all values of $x$ for which $f(x)=4$.

When $\log _{2} x \leq-1, f(x)=2-2 \log _{2} x=4$ exactly when $\log _{2} x=-1$.

When $-1<\log _{2} x \leq 3, f(x)$ is always equal to 4 .

When $\log _{2} x>3, f(x)=2 \log _{2} x-2=4$ exactly when $\log _{2} x=3$.

Therefore, $f(x)=4$ exactly when $-1 \leq \log _{2} x \leq 3$, which is true exactly when $\frac{1}{2} \leq x \leq 8$. (It seems surprising that the solution to this equation is actually an interval of values, rather than a finite number of specific values.)

Final answer: $[\frac{1}{2}, 8]$
Subfield: Combinatorics


Question: At the Canadian Eatery with Multiple Configurations, there are round tables, around which chairs are placed. When a table has $n$ chairs around it for some integer $n \geq 3$, the chairs are labelled $1,2,3, \ldots, n-1, n$ in order around the table. A table is considered full if no more people can be seated without having two people sit in neighbouring chairs. For example, when $n=6$, full tables occur when people are seated in chairs labelled $\{1,4\}$ or $\{2,5\}$ or $\{3,6\}$ or $\{1,3,5\}$ or $\{2,4,6\}$. Thus, there are 5 different full tables when $n=6$.

<img_3455>
A full table with $6 k+5$ chairs, for some positive integer $k$, has $t$ people seated in its chairs. Determine, in terms of $k$, the number of possible values of $t$.

Solution:
Suppose that $k$ is a positive integer.

Suppose that $t$ people are seated at a table with $6 k+5$ chairs so that the table is full.

When $t$ people are seated, there are $t$ gaps. Each gap consists of either 1 or 2 chairs. (A gap with 3 or more chairs can have an additional person seated in it, so the table is not full.)

Therefore, there are between $t$ and $2 t$ empty chairs.

This means that the total number of chairs is between $t+t$ and $t+2 t$.

In other words, $2 t \leq 6 k+5 \leq 3 t$.

Since $2 t \leq 6 k+5$, then $t \leq 3 k+\frac{5}{2}$. Since $k$ and $t$ are integers, then $t \leq 3 k+2$.

We note that it is possible to seat $3 k+2$ people around the table in seats

$$
\{2,4,6, \ldots, 6 k+2,6 k+4\}
$$

This table is full becase $3 k+1$ of the gaps consist of 1 chair and 1 gap consists of 2 chairs. Since $3 t \geq 6 k+5$, then $t \geq 2 k+\frac{5}{3}$. Since $k$ and $t$ are integers, then $t \geq 2 k+2$.

We note that it is possible to seat $2 k+2$ people around the table in seats

$$
\{3,6,9, \ldots, 6 k, 6 k+3,6 k+5\}
$$

This table is full becase $2 k+1$ of the gaps consist of 2 chairs and 1 gap consists of 1 chair.

We now know that, if there are $t$ people seated at a full table with $6 k+5$ chairs, then $2 k+2 \leq t \leq 3 k+2$.

To confirm that every such value of $t$ is possible, consider a table with $t$ people, $3 t-(6 k+5)$



gaps of 1 chair, and $(6 k+5)-2 t$ gaps of 2 chairs.

From the work above, we know that $3 t \geq 6 k+5$ and so $3 t-(6 k+5) \geq 0$, and that $2 t \leq 6 k+5$ and so $(6 k+5)-2 t \geq 0$.

The total number of gaps is $3 t-(6 k+5)+(6 k+5)-2 t=t$, since there are $t$ people seated.

Finally, the total number of chairs is

$$
t+1 \cdot(3 t-(6 k+5))+2 \cdot((6 k+5)-2 t)=t+3 t-4 t-(6 k+5)+2(6 k+5)=6 k+5
$$

as expected.

This shows that every $t$ with $2 k+2 \leq t \leq 3 k+2$ can produce a full table.

Therefore, the possible values of $t$ are those integers that satisfy $2 k+2 \leq t \leq 3 k+2$.

There are $(3 k+2)-(2 k+2)+1=k+1$ possible values of $t$.

Final answer: $k+1$
Subfield: Combinatorics


Question: At the Canadian Eatery with Multiple Configurations, there are round tables, around which chairs are placed. When a table has $n$ chairs around it for some integer $n \geq 3$, the chairs are labelled $1,2,3, \ldots, n-1, n$ in order around the table. A table is considered full if no more people can be seated without having two people sit in neighbouring chairs. For example, when $n=6$, full tables occur when people are seated in chairs labelled $\{1,4\}$ or $\{2,5\}$ or $\{3,6\}$ or $\{1,3,5\}$ or $\{2,4,6\}$. Thus, there are 5 different full tables when $n=6$.

![](https://cdn.mathpix.com/cropped/2023_12_21_ed73f7f90f51d294b52fg-1.jpg?height=401&width=331&top_left_y=1304&top_left_x=1469)
A full table with $6 k+5$ chairs, for some positive integer $k$, has $t$ people seated in its chairs. Determine, in terms of $k$, the number of possible values of $t$.

Solution:
Suppose that $k$ is a positive integer.

Suppose that $t$ people are seated at a table with $6 k+5$ chairs so that the table is full.

When $t$ people are seated, there are $t$ gaps. Each gap consists of either 1 or 2 chairs. (A gap with 3 or more chairs can have an additional person seated in it, so the table is not full.)

Therefore, there are between $t$ and $2 t$ empty chairs.

This means that the total number of chairs is between $t+t$ and $t+2 t$.

In other words, $2 t \leq 6 k+5 \leq 3 t$.

Since $2 t \leq 6 k+5$, then $t \leq 3 k+\frac{5}{2}$. Since $k$ and $t$ are integers, then $t \leq 3 k+2$.

We note that it is possible to seat $3 k+2$ people around the table in seats

$$
\{2,4,6, \ldots, 6 k+2,6 k+4\}
$$

This table is full becase $3 k+1$ of the gaps consist of 1 chair and 1 gap consists of 2 chairs. Since $3 t \geq 6 k+5$, then $t \geq 2 k+\frac{5}{3}$. Since $k$ and $t$ are integers, then $t \geq 2 k+2$.

We note that it is possible to seat $2 k+2$ people around the table in seats

$$
\{3,6,9, \ldots, 6 k, 6 k+3,6 k+5\}
$$

This table is full becase $2 k+1$ of the gaps consist of 2 chairs and 1 gap consists of 1 chair.

We now know that, if there are $t$ people seated at a full table with $6 k+5$ chairs, then $2 k+2 \leq t \leq 3 k+2$.

To confirm that every such value of $t$ is possible, consider a table with $t$ people, $3 t-(6 k+5)$



gaps of 1 chair, and $(6 k+5)-2 t$ gaps of 2 chairs.

From the work above, we know that $3 t \geq 6 k+5$ and so $3 t-(6 k+5) \geq 0$, and that $2 t \leq 6 k+5$ and so $(6 k+5)-2 t \geq 0$.

The total number of gaps is $3 t-(6 k+5)+(6 k+5)-2 t=t$, since there are $t$ people seated.

Finally, the total number of chairs is

$$
t+1 \cdot(3 t-(6 k+5))+2 \cdot((6 k+5)-2 t)=t+3 t-4 t-(6 k+5)+2(6 k+5)=6 k+5
$$

as expected.

This shows that every $t$ with $2 k+2 \leq t \leq 3 k+2$ can produce a full table.

Therefore, the possible values of $t$ are those integers that satisfy $2 k+2 \leq t \leq 3 k+2$.

There are $(3 k+2)-(2 k+2)+1=k+1$ possible values of $t$.

Final answer: $k+1$
Subfield: Number Theory


Question: For every real number $x$, define $\lfloor x\rfloor$ to be equal to the greatest integer less than or equal to $x$. (We call this the "floor" of $x$.) For example, $\lfloor 4.2\rfloor=4,\lfloor 5.7\rfloor=5$, $\lfloor-3.4\rfloor=-4,\lfloor 0.4\rfloor=0$, and $\lfloor 2\rfloor=2$.
Determine the integer equal to $\left\lfloor\frac{1}{3}\right\rfloor+\left\lfloor\frac{2}{3}\right\rfloor+\left\lfloor\frac{3}{3}\right\rfloor+\ldots+\left\lfloor\frac{59}{3}\right\rfloor+\left\lfloor\frac{60}{3}\right\rfloor$. (The sum has 60 terms.)

Solution:
Since $0<\frac{1}{3}<\frac{2}{3}<1$, then $\left\lfloor\frac{1}{3}\right\rfloor=\left\lfloor\frac{2}{3}\right\rfloor=0$.

Since $1 \leq \frac{3}{3}<\frac{4}{3}<\frac{5}{3}<2$, then $\left\lfloor\frac{3}{3}\right\rfloor=\left\lfloor\frac{4}{3}\right\rfloor=\left\lfloor\frac{5}{3}\right\rfloor=1$.

These fractions can continue to be grouped in groups of 3 with the last full group of 3 satisfying $19 \leq \frac{57}{3}<\frac{58}{3}<\frac{59}{3}<20$, which means that $\left\lfloor\frac{57}{3}\right\rfloor=\left\lfloor\frac{58}{3}\right\rfloor=\left\lfloor\frac{59}{3}\right\rfloor=19$.

The last term is $\left\lfloor\frac{60}{3}\right\rfloor=\lfloor 20\rfloor=20$.



If the given sum is $S$, we obtain

$$
\begin{aligned}
S & =2 \cdot 0+3 \cdot 1+3 \cdot 2+\cdots+3 \cdot 19+1 \cdot 20 \\
& =0+3(1+2+\cdot+19)+20 \\
& =3 \cdot \frac{1}{2} \cdot 19 \cdot 20+20 \\
& =570+20 \\
& =590
\end{aligned}
$$

Final answer: 590
Subfield: Geometry


Question: In the diagram, $\triangle A B C$ is right-angled at $B$ and $\triangle A C D$ is right-angled at $A$. Also, $A B=3, B C=4$, and $C D=13$. What is the area of quadrilateral $A B C D$ ?

<img_3281>

Solution:
The area of quadrilateral $A B C D$ is the sum of the areas of $\triangle A B C$ and $\triangle A C D$.

Since $\triangle A B C$ is right-angled at $B$, its area equals $\frac{1}{2}(A B)(B C)=\frac{1}{2}(3)(4)=6$.

Since $\triangle A B C$ is right-angled at $B$, then by the Pythagorean Theorem,

$$
A C=\sqrt{A B^{2}+B C^{2}}=\sqrt{3^{2}+4^{2}}=\sqrt{25}=5
$$

because $A C>0$. (We could have also observed that $\triangle A B C$ must be a "3-4-5" triangle.) Since $\triangle A C D$ is right-angled at $A$, then by the Pythagorean Theorem,

$$
A D=\sqrt{C D^{2}-A C^{2}}=\sqrt{13^{2}-5^{2}}=\sqrt{144}=12
$$

because $A D>0$. (We could have also observed that $\triangle A C D$ must be a " $5-12-13$ " triangle.) Thus, the area of $\triangle A C D$ equals $\frac{1}{2}(A C)(A D)=\frac{1}{2}(5)(12)=30$.

Finally, the area of quadrilateral $A B C D$ is thus $6+30=36$.

Final answer: 36
Subfield: Geometry


Question: In the diagram, $\triangle A B C$ is right-angled at $B$ and $\triangle A C D$ is right-angled at $A$. Also, $A B=3, B C=4$, and $C D=13$. What is the area of quadrilateral $A B C D$ ?

![](https://cdn.mathpix.com/cropped/2023_12_21_c71fc327224b5fdf6510g-1.jpg?height=279&width=475&top_left_y=278&top_left_x=1256)

Solution:
The area of quadrilateral $A B C D$ is the sum of the areas of $\triangle A B C$ and $\triangle A C D$.

Since $\triangle A B C$ is right-angled at $B$, its area equals $\frac{1}{2}(A B)(B C)=\frac{1}{2}(3)(4)=6$.

Since $\triangle A B C$ is right-angled at $B$, then by the Pythagorean Theorem,

$$
A C=\sqrt{A B^{2}+B C^{2}}=\sqrt{3^{2}+4^{2}}=\sqrt{25}=5
$$

because $A C>0$. (We could have also observed that $\triangle A B C$ must be a "3-4-5" triangle.) Since $\triangle A C D$ is right-angled at $A$, then by the Pythagorean Theorem,

$$
A D=\sqrt{C D^{2}-A C^{2}}=\sqrt{13^{2}-5^{2}}=\sqrt{144}=12
$$

because $A D>0$. (We could have also observed that $\triangle A C D$ must be a " $5-12-13$ " triangle.) Thus, the area of $\triangle A C D$ equals $\frac{1}{2}(A C)(A D)=\frac{1}{2}(5)(12)=30$.

Finally, the area of quadrilateral $A B C D$ is thus $6+30=36$.

Final answer: 36
Subfield: Geometry


Question: Three identical rectangles $P Q R S$, WTUV and $X W V Y$ are arranged, as shown, so that $R S$ lies along $T X$. The perimeter of each of the three rectangles is $21 \mathrm{~cm}$. What is the perimeter of the whole shape?
![](https://cdn.mathpix.com/cropped/2023_12_21_c71fc327224b5fdf6510g-1.jpg?height=400&width=1510&top_left_y=610&top_left_x=294)

Solution:
Let the width of each of the identical rectangles be $a$.

In other words, $Q P=R S=T W=W X=U V=V Y=a$.

Let the height of each of the identical rectangles be $b$.

In other words, $Q R=P S=T U=W V=X Y=b$.

The perimeter of the whole shape equals

$$
Q P+P S+S X+X Y+V Y+U V+T U+T R+Q R
$$

Substituting for known lengths, we obtain

$$
a+b+S X+b+a+a+b+T R+b
$$

or $3 a+4 b+(S X+T R)$.

But $S X+T R=(T R+R S+S X)-R S=(T W+W X)-R S=a+a-a=a$.

Therefore, the perimeter of the whole shape equals $4 a+4 b$.

The perimeter of one rectangle is $2 a+2 b$, which we are told equals $21 \mathrm{~cm}$.

Finally, the perimeter of the whole shape is thus $2(2 a+2 b)$ which equals $42 \mathrm{~cm}$.

Final answer: 42 cm
Subfield: Geometry


Question: Three identical rectangles $P Q R S$, WTUV and $X W V Y$ are arranged, as shown, so that $R S$ lies along $T X$. The perimeter of each of the three rectangles is $21 \mathrm{~cm}$. What is the perimeter of the whole shape?
<img_3203>

Solution:
Let the width of each of the identical rectangles be $a$.

In other words, $Q P=R S=T W=W X=U V=V Y=a$.

Let the height of each of the identical rectangles be $b$.

In other words, $Q R=P S=T U=W V=X Y=b$.

The perimeter of the whole shape equals

$$
Q P+P S+S X+X Y+V Y+U V+T U+T R+Q R
$$

Substituting for known lengths, we obtain

$$
a+b+S X+b+a+a+b+T R+b
$$

or $3 a+4 b+(S X+T R)$.

But $S X+T R=(T R+R S+S X)-R S=(T W+W X)-R S=a+a-a=a$.

Therefore, the perimeter of the whole shape equals $4 a+4 b$.

The perimeter of one rectangle is $2 a+2 b$, which we are told equals $21 \mathrm{~cm}$.

Finally, the perimeter of the whole shape is thus $2(2 a+2 b)$ which equals $42 \mathrm{~cm}$.

Final answer: 42 cm
Subfield: Geometry


Question: One of the faces of a rectangular prism has area $27 \mathrm{~cm}^{2}$. Another face has area $32 \mathrm{~cm}^{2}$. If the volume of the prism is $144 \mathrm{~cm}^{3}$, determine the surface area of the prism in $\mathrm{cm}^{2}$.

Solution:
Suppose that the rectangular prism has dimensions $a \mathrm{~cm}$ by $b \mathrm{~cm}$ by $c \mathrm{~cm}$.

Suppose further that one of the faces that is $a \mathrm{~cm}$ by $b \mathrm{~cm}$ is the face with area $27 \mathrm{~cm}^{2}$ and that one of the faces that is $a \mathrm{~cm}$ by $c \mathrm{~cm}$ is the face with area $32 \mathrm{~cm}^{2}$. (Since every pair of non-congruent faces shares exactly one side length, there is no loss of generality in picking these particular variables for these faces.)

Therefore, $a b=27$ and $a c=32$.

Further, we are told that the volume of the prism is $144 \mathrm{~cm}^{3}$, and so $a b c=144$.



Thus, $b c=\frac{a^{2} b^{2} c^{2}}{a^{2} b c}=\frac{(a b c)^{2}}{(a b)(a c)}=\frac{144^{2}}{(27)(32)}=24$.

(We could also note that $a b c=144$ means $a^{2} b^{2} c^{2}=144^{2}$ or $(a b)(a c)(b c)=144^{2}$ and so $b c=\frac{144^{2}}{(27)(32)}$.)

In other words, the third type of face of the prism has area $24 \mathrm{~cm}^{2}$.

Thus, since the prism has two faces of each type, the surface area of the prism is equal to $2\left(27 \mathrm{~cm}^{2}+32 \mathrm{~cm}^{2}+24 \mathrm{~cm}^{2}\right)$ or $166 \mathrm{~cm}^{2}$.
Suppose that the rectangular prism has dimensions $a \mathrm{~cm}$ by $b \mathrm{~cm}$ by $c \mathrm{~cm}$.

Suppose further that one of the faces that is $a \mathrm{~cm}$ by $b \mathrm{~cm}$ is the face with area $27 \mathrm{~cm}^{2}$ and that one of the faces that is $a \mathrm{~cm}$ by $c \mathrm{~cm}$ is the face with area $32 \mathrm{~cm}^{2}$. (Since every pair of non-congruent faces shares exactly one side length, there is no loss of generality in picking these particular variables for these faces.)

Therefore, $a b=27$ and $a c=32$.

Further, we are told that the volume of the prism is $144 \mathrm{~cm}^{3}$, and so $a b c=144$.

Since $a b c=144$ and $a b=27$, then $c=\frac{144}{27}=\frac{16}{3}$.

Since $a b c=144$ and $a c=32$, then $b=\frac{144}{32}=\frac{9}{2}$.

This means that $b c=\frac{16}{3} \cdot \frac{9}{2}=24$.

In $\mathrm{cm}^{2}$, the surface area of the prism equals $2 a b+2 a c+2 b c=2(27)+2(32)+2(24)=166$. Thus, the surface area of the prism is $166 \mathrm{~cm}^{2}$.

Final answer: $166$ $cm^2$
Subfield: Algebra


Question: The equations $y=a(x-2)(x+4)$ and $y=2(x-h)^{2}+k$ represent the same parabola. What are the values of $a, h$ and $k$ ?

Solution:
We expand the right sides of the two equations, collecting like terms in each case:

$$
\begin{aligned}
& y=a(x-2)(x+4)=a\left(x^{2}+2 x-8\right)=a x^{2}+2 a x-8 a \\
& y=2(x-h)^{2}+k=2\left(x^{2}-2 h x+h^{2}\right)+k=2 x^{2}-4 h x+\left(2 h^{2}+k\right)
\end{aligned}
$$

Since these two equations represent the same parabola, then the corresponding coefficients must be equal. That is, $a=2$ and $2 a=-4 h$ and $-8 a=2 h^{2}+k$.

Since $a=2$ and $2 a=-4 h$, then $4=-4 h$ and so $h=-1$.

Since $-8 a=2 h^{2}+k$ and $a=2$ and $h=-1$, then $-16=2+k$ and so $k=-18$.

Thus, $a=2, h=-1$, and $k=-18$.
From the equation $y=a(x-2)(x+4)$, we can find the axis of symmetry by calculating the midpoint of the $x$-intercepts.

Since the $x$-intercepts are 2 and -4 , the axis of symmetry is at $x=\frac{1}{2}(2+(-4))=-1$.

Since the vertex of the parabola lies on the axis of symmetry, then the $x$-coordinate of the vertex is -1 .

To find the $y$-coordinate of the vertex, we substitute $x=-1$ back into the equation $y=a(x-2)(x+4)$ to obtain $y=a(-1-2)(-1+4)=-9 a$.

Thus, the vertex of the parabola is $(-1,-9 a)$.

Since the second equation for the same parabola is in vertex form, $y=2(x-h)^{2}+k$, we can see that the vertex is at $(h, k)$ and $a=2$.

Since $a=2$, the vertex has coordinates $(-1,-18)$, which means that $h=-1$ and $k=-18$. Thus, $a=2, h=-1$ and $k=-18$.

Final answer: $2,-1,-18$
Subfield: Algebra


Question: In an arithmetic sequence with 5 terms, the sum of the squares of the first 3 terms equals the sum of the squares of the last 2 terms. If the first term is 5 , determine all possible values of the fifth term.

(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, 3,5,7,9,11 is an arithmetic sequence with five terms.)

Solution:
Let the common difference in this arithmetic sequence be $d$.

Since the first term in the sequence is 5 , then the 5 terms are $5,5+d, 5+2 d, 5+3 d, 5+4 d$.

From the given information, $5^{2}+(5+d)^{2}+(5+2 d)^{2}=(5+3 d)^{2}+(5+4 d)^{2}$.

Manipulating, we obtain the following equivalent equations:

$$
\begin{aligned}
5^{2}+(5+d)^{2}+(5+2 d)^{2} & =(5+3 d)^{2}+(5+4 d)^{2} \\
25+\left(25+10 d+d^{2}\right)+\left(25+20 d+4 d^{2}\right) & =\left(25+30 d+9 d^{2}\right)+\left(25+40 d+16 d^{2}\right) \\
75+30 d+5 d^{2} & =50+70 d+25 d^{2} \\
0 & =20 d^{2}+40 d-25 \\
0 & =4 d^{2}+8 d-5 \\
0 & =(2 d+5)(2 d-1)
\end{aligned}
$$

Therefore, $d=-\frac{5}{2}$ or $d=\frac{1}{2}$.

These give possible fifth terms of $5+4 d=5+4\left(-\frac{5}{2}\right)=-5$ and $5+4 d=5+4\left(\frac{1}{2}\right)=7$.

(We note that, for these two values of $d$, the sequences are $5, \frac{5}{2}, 0,-\frac{5}{2},-5$ and $5, \frac{11}{2}, 6, \frac{13}{2}, 7$.)

Final answer: -5,7
Subfield: Algebra


Question: Dan was born in a year between 1300 and 1400. Steve was born in a year between 1400 and 1500. Each was born on April 6 in a year that is a perfect square. Each lived for 110 years. In what year while they were both alive were their ages both perfect squares on April 7?

Solution:
First, we determine the perfect squares between 1300 and 1400 and between 1400 and 1500.

Since $\sqrt{1300} \approx 36.06$, then the first perfect square larger than 1300 is $37^{2}=1369$.

The next perfect squares are $38^{2}=1444$ and $39^{2}=1521$.

Since Dan was born between 1300 and 1400 in a year that was a perfect square, then Dan was born in 1369.

Since Steve was born between 1400 and 1500 in a year that was a perfect square, then Steve was born in 1444.

Suppose that on April 7 in some year, Dan was $m^{2}$ years old and Steve was $n^{2}$ years old for some positive integers $m$ and $n$. Thus, Dan was $m^{2}$ years old in the year $1369+m^{2}$ and Steve was $n^{2}$ years old in the year $1444+n^{2}$.

Since these represent the same years, then $1369+m^{2}=1444+n^{2}$, or $m^{2}-n^{2}=1444-$ $1369=75$.

In other words, we want to find two perfect squares less than 110 (since their ages are less than 110) whose difference is 75.

The perfect squares less than 110 are $1,4,9,16,25,36,49,64,81,100$.

The two that differ by 75 are 100 and 25 .

Thus, $m^{2}=100$ and $n^{2}=25$.

This means that the year in which the age of each of Dan and Steve was a perfect square was the year $1369+100=1469$.

Final answer: 1469