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


Question: Consider the sequence $t_{1}=1, t_{2}=-1$ and $t_{n}=\left(\frac{n-3}{n-1}\right) t_{n-2}$ where $n \geq 3$. What is the value of $t_{1998}$ ?

Solution:
Calculating some terms, $t_{1}=1, t_{2}=-1, t_{3}=0, t_{4}=\frac{-1}{3}, t_{5}=0, t_{6}=\frac{-1}{5}$ etc.

By pattern recognition, $t_{1998}=\frac{-1}{1997}$.
$$
\begin{aligned}
t_{1998} & =\frac{1995}{1997} t_{1996}=\frac{1995}{1997} \times \frac{1993}{1995} t_{1994} \\
& =\frac{1995}{1997} \cdot \frac{1993}{1995} \cdot \frac{1991}{1993} \cdots \frac{3}{5} \cdot \frac{1}{3} t_{2} \\
& =\frac{-1}{1997}
\end{aligned}
$$

Final answer: $\frac{-1}{1997}$
Subfield: Algebra


Question: The $n$th term of an arithmetic sequence is given by $t_{n}=555-7 n$.

If $S_{n}=t_{1}+t_{2}+\ldots+t_{n}$, determine the smallest value of $n$ for which $S_{n}<0$.

Solution:
This is an arithmetic sequence in which $a=548$ and $d=-7$.

Therefore, $S_{n}=\frac{n}{2}[2(548)+(n-1)(-7)]=\frac{n}{2}[-7 n+1103]$.

We now want $\frac{n}{2}(-7 n+1103)<0$.

Since $n>0,-7 n+1103<0$

$$
n>157 \frac{4}{7}
$$

Therefore the smallest value of $n$ is 158 .
For this series we want, $\sum_{k=1}^{n} t_{k}<0$, or $\sum_{k=1}^{n}(555-7 k)<0$.

Rewriting, $555 n-7 \frac{(n)(n+1)}{2}<0$

$$
\begin{aligned}
1110 n-7 n^{2}-7 n & <0 \\
7 n^{2}-1103 n & >0 \\
\text { or, } n & >\frac{1103}{7} .
\end{aligned}
$$

The smallest value of $n$ is 158 .
We generate the series as $548,541,534, \ldots, 2,-5, \ldots,-544,-551$.

If we pair the series from front to back the sum of each pair is -3 .

Including all the pairs $548-551,541-544$ and so on there would be 79 pairs which give a sum of -237 .

If the last term, -551 , were omitted we would have a positive sum.

Therefore we need all 79 pairs or 158 terms.

Final answer: 158
Subfield: Algebra


Question: If $x$ and $y$ are real numbers, determine all solutions $(x, y)$ of the system of equations

$$
\begin{aligned}
& x^{2}-x y+8=0 \\
& x^{2}-8 x+y=0
\end{aligned}
$$

Solution:
Subtracting,

$$
\begin{array}{r}
x^{2}-x y+8=0 \\
x^{2}-8 x+y=0 \\
\hline-x y+8 x+8-y=0 \\
8(1+x)-y(1+x)=0 \\
(8-y)(1+x)=0 \\
y=8 \text { or } x=-1
\end{array}
$$



If $y=8$, both equations become $x^{2}-8 x+8=0, x=4 \pm 2 \sqrt{2}$.

If $x=-1$ both equations become $y+9=0, y=-9$.

The solutions are $(-1,-9),(4+2 \sqrt{2}, 8)$ and $(4-2 \sqrt{2}, 8)$.
If $x^{2}-x y+8=0, y=\frac{x^{2}+8}{x}$.

And $x^{2}-8 x+y=0$ implies $y=8 x-x^{2}$.

Equating, $\frac{x^{2}+8}{x}=8 x-x^{2}$

$$
\text { or, } x^{3}-7 x^{2}+8=0 \text {. }
$$

By inspection, $x=-1$ is a root.

By division, $x^{3}-7 x^{2}+8=(x+1)\left(x^{2}-8 x+8\right)$.

As before, the solutions are $(-1,-9),(4 \pm 2 \sqrt{2}, 8)$.

Final answer: $(-1,-9),(4+2 \sqrt{2}, 8),(4-2 \sqrt{2}, 8)$
Subfield: Geometry


Question: In the graph, the parabola $y=x^{2}$ has been translated to the position shown.

Prove that $d e=f$.

![](https://cdn.mathpix.com/cropped/2023_12_21_c810d95f38a712e9ddf7g-1.jpg?height=458&width=569&top_left_y=129&top_left_x=1228)

Solution:
Since the given graph is congruent to $y=x^{2}$ and has $x$-intercepts $-d$ and $e$, its general form is $y=(x+d)(x-e)$.

To find the $y$-intercept, let $x=0$. Therefore $y$-intercept $=-d e$.

We are given that the $y$-intercept is $-f$.

Therefore $-f=-d e$ or $f=d e$.

Final answer: 证明题，略
Subfield: Geometry


Question: In the graph, the parabola $y=x^{2}$ has been translated to the position shown.

Prove that $d e=f$.

<img_3273>

Solution:
Since the given graph is congruent to $y=x^{2}$ and has $x$-intercepts $-d$ and $e$, its general form is $y=(x+d)(x-e)$.

To find the $y$-intercept, let $x=0$. Therefore $y$-intercept $=-d e$.

We are given that the $y$-intercept is $-f$.

Therefore $-f=-d e$ or $f=d e$.
Subfield: Geometry


Question: In quadrilateral $K W A D$, the midpoints of $K W$ and $A D$ are $M$ and $N$ respectively. If $M N=\frac{1}{2}(A W+D K)$, prove that $WA$ is parallel to $K D$.

<img_4042>

Solution:
Establish a coordinate system with $K(0,0), D(2 a, 0)$ on the $x$-axes. Let $W$ be $(2 b, 2 c)$ and $A$ be $(2 d, 2 e)$.

Thus $M$ is $(b, c)$ and $N$ is $(a+d, e)$.

$K D$ has slope 0 and slope $W A=\frac{e-c}{d-b}$.

Since $M N=\frac{1}{2}(A W+D K)$

$$
\begin{aligned}
& \sqrt{(a+d-b)^{2}+(e-c)^{2}} \\
= & \frac{1}{2}\left(2 a+\sqrt{(2 d-2 b)^{2}+(2 e-2 c)^{2}}\right) \\
= & \frac{1}{2}\left(2 a+2 \sqrt{(d-b)^{2}+(e-c)^{2}}\right)
\end{aligned}
$$

<img_3380>

Squaring both sides gives,

$$
\begin{aligned}
& (a+d-b)^{2}+(e-c)^{2}=a^{2}+2 a \sqrt{(d-b)^{2}+(e-c)^{2}}+(d-b)^{2}+(e-c)^{2} \\
& a^{2}+2 a(d-b)+(d-b)^{2}=a^{2}+2 a \sqrt{(d-b)^{2}+(e-c)^{2}}+(d-b)^{2}
\end{aligned}
$$

Simplifying and dividing by $2 a$ we have, $d-b=\sqrt{(d-b)^{2}+(e-c)^{2}}$.

Squaring, $(d-b)^{2}=(d-b)^{2}+(e-c)^{2}$.

Therefore $(e-c)^{2}=0$ or $e=c$.

Since $e=c$ then slope of $W A$ is 0 and $K D \| A W$.
Join $A$ to $K$ and call $P$ the mid-point of $A K$.

Join $M$ to $P, N$ to $P$ and $M$ to $N$.

In $\triangle K A W, P$ and $M$ are the mid-points of $K A$ and $K W$.

Therefore, $M P=\frac{1}{2} W A$.

Similarly in $\triangle K A D, P N=\frac{1}{2} K D$.

Therefore $M P+P N=M N$.

<img_3772>

As a result $M, P$ and $N$ cannot form the vertices of a triangle but must form a straight line.

So if $M P N$ is a straight line with $M P \| W A$ and $P N \| K D$ then $W A \| K D$ as required.
We are given that $\overrightarrow{A N}=\overrightarrow{N D}$ and $\overrightarrow{W M}=\overrightarrow{M K}$.

Using vectors,

(1) $\overrightarrow{M N}=\overrightarrow{M W}+\overrightarrow{W A}+\overrightarrow{A N}$ (from quad. $M W A N$ )

(2) $\overrightarrow{M N}=\overrightarrow{M K}+\overrightarrow{K D}+\overrightarrow{D N} \quad$ (from quad. $K M N D$ )

It is also possible to write, $\overrightarrow{M N}=-\overrightarrow{M W}+\overrightarrow{K D}-\overrightarrow{A N}$,

(3) (This comes from taking statement (2) and making appropriate substitutions.)



If we add (1) and (3) we find, $2 \overrightarrow{M N}=\overrightarrow{W A}+\overrightarrow{K D}$.

But it is given that $2|\overrightarrow{M N}|=|\overrightarrow{A W}|+|\overrightarrow{D K}|$.

From these two previous statements, $\overrightarrow{M N}$ must be parallel to $\overrightarrow{W A}$ and $\overrightarrow{K D}$ otherwise $2|\overrightarrow{M N}|<|\overrightarrow{A W}|+|\overrightarrow{D K}|$.

Therefore, $W A \| K D$.
Subfield: Geometry


Question: In quadrilateral $K W A D$, the midpoints of $K W$ and $A D$ are $M$ and $N$ respectively. If $M N=\frac{1}{2}(A W+D K)$, prove that $WA$ is parallel to $K D$.

![](https://cdn.mathpix.com/cropped/2023_12_21_c810d95f38a712e9ddf7g-1.jpg?height=404&width=488&top_left_y=644&top_left_x=1271)

Solution:
Establish a coordinate system with $K(0,0), D(2 a, 0)$ on the $x$-axes. Let $W$ be $(2 b, 2 c)$ and $A$ be $(2 d, 2 e)$.

Thus $M$ is $(b, c)$ and $N$ is $(a+d, e)$.

$K D$ has slope 0 and slope $W A=\frac{e-c}{d-b}$.

Since $M N=\frac{1}{2}(A W+D K)$

$$
\begin{aligned}
& \sqrt{(a+d-b)^{2}+(e-c)^{2}} \\
= & \frac{1}{2}\left(2 a+\sqrt{(2 d-2 b)^{2}+(2 e-2 c)^{2}}\right) \\
= & \frac{1}{2}\left(2 a+2 \sqrt{(d-b)^{2}+(e-c)^{2}}\right)
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2023_12_21_821212715b4b9a6167d5g-1.jpg?height=423&width=613&top_left_y=320&top_left_x=1336)

Squaring both sides gives,

$$
\begin{aligned}
& (a+d-b)^{2}+(e-c)^{2}=a^{2}+2 a \sqrt{(d-b)^{2}+(e-c)^{2}}+(d-b)^{2}+(e-c)^{2} \\
& a^{2}+2 a(d-b)+(d-b)^{2}=a^{2}+2 a \sqrt{(d-b)^{2}+(e-c)^{2}}+(d-b)^{2}
\end{aligned}
$$

Simplifying and dividing by $2 a$ we have, $d-b=\sqrt{(d-b)^{2}+(e-c)^{2}}$.

Squaring, $(d-b)^{2}=(d-b)^{2}+(e-c)^{2}$.

Therefore $(e-c)^{2}=0$ or $e=c$.

Since $e=c$ then slope of $W A$ is 0 and $K D \| A W$.
Join $A$ to $K$ and call $P$ the mid-point of $A K$.

Join $M$ to $P, N$ to $P$ and $M$ to $N$.

In $\triangle K A W, P$ and $M$ are the mid-points of $K A$ and $K W$.

Therefore, $M P=\frac{1}{2} W A$.

Similarly in $\triangle K A D, P N=\frac{1}{2} K D$.

Therefore $M P+P N=M N$.

![](https://cdn.mathpix.com/cropped/2023_12_21_821212715b4b9a6167d5g-1.jpg?height=423&width=613&top_left_y=1439&top_left_x=1336)

As a result $M, P$ and $N$ cannot form the vertices of a triangle but must form a straight line.

So if $M P N$ is a straight line with $M P \| W A$ and $P N \| K D$ then $W A \| K D$ as required.
We are given that $\overrightarrow{A N}=\overrightarrow{N D}$ and $\overrightarrow{W M}=\overrightarrow{M K}$.

Using vectors,

(1) $\overrightarrow{M N}=\overrightarrow{M W}+\overrightarrow{W A}+\overrightarrow{A N}$ (from quad. $M W A N$ )

(2) $\overrightarrow{M N}=\overrightarrow{M K}+\overrightarrow{K D}+\overrightarrow{D N} \quad$ (from quad. $K M N D$ )

It is also possible to write, $\overrightarrow{M N}=-\overrightarrow{M W}+\overrightarrow{K D}-\overrightarrow{A N}$,

(3) (This comes from taking statement (2) and making appropriate substitutions.)



If we add (1) and (3) we find, $2 \overrightarrow{M N}=\overrightarrow{W A}+\overrightarrow{K D}$.

But it is given that $2|\overrightarrow{M N}|=|\overrightarrow{A W}|+|\overrightarrow{D K}|$.

From these two previous statements, $\overrightarrow{M N}$ must be parallel to $\overrightarrow{W A}$ and $\overrightarrow{K D}$ otherwise $2|\overrightarrow{M N}|<|\overrightarrow{A W}|+|\overrightarrow{D K}|$.

Therefore, $W A \| K D$.

Final answer: 证明题，略
Subfield: Number Theory


Question: Consider the first $2 n$ natural numbers. Pair off the numbers, as shown, and multiply the two members of each pair. Prove that there is no value of $n$ for which two of the $n$ products are equal.

![](https://cdn.mathpix.com/cropped/2023_12_21_c810d95f38a712e9ddf7g-1.jpg?height=214&width=1366&top_left_y=1205&top_left_x=431)

Solution:
The sequence is $1(2 n), 2(2 n-1), 3(2 n-2), \ldots, k(2 n-k+1), \ldots, p(2 n-p+1), \ldots, n(n+1)$.

In essence we are asking the question, 'is it possible that $k(2 n-k+1)=p(2 n-p+1)$ where $p$ and $k$ are both less than or equal to $n$ ?'

$$
\begin{aligned}
k(2 n-k+1) & =p(2 n-p+1) \\
2 n k-k^{2}+k=2 n p-p^{2}+p & \text { (supposing them to be equal) } \\
p^{2}-k^{2}+2 n k-2 n p+k-p & =0 \\
(p-k)(p+k)+2 n(k-p)+(k-p) & =0 \\
(p-k)[(p+k)-2 n-1] & =0 \\
(p-k)(p+k-2 n-1) & =0
\end{aligned}
$$

Since $p$ and $k$ are both less than or equal to $n$, it follows $p+k-2 n-1 \neq 0$. Therefore $p=k$ and they represent the same pair. Thus the required is proven.
The products are $1(2 n+1-1), 2(2 n+1-2), 3(2 n+1-3), \ldots, n(2 n+1-n)$.

Consider the function, $y=x(2 n+1-x)=-x^{2}+(2 n+1) x=f(x)$.



The graph of this function is a parabola, opening down, with its vertex at $x=n+\frac{1}{2}$.

The products are the $y$-coordinates of the points on the parabola corresponding to $x=1,2,3, \ldots, n$. Since all the points are to the left of the vertex, no two have the same $y$ coordinate.

Thus the products are distinct.

![](https://cdn.mathpix.com/cropped/2023_12_21_f4caa869f0e75f41fb17g-1.jpg?height=523&width=515&top_left_y=226&top_left_x=1336)
The sum of these numbers is $\frac{2 n(2 n+1)}{2}$ or $n(2 n+1)$.

Their average is $\frac{n(2 n+1)}{2 n}=n+\frac{1}{2}$.

The $2 n$ numbers can be rewritten as,

$$
n+\frac{1}{2}-\left(\frac{2 n-1}{2}\right), \cdots, n+\frac{1}{2}-\frac{3}{2}, n+\frac{1}{2}-\frac{1}{2}, n+\frac{1}{2}+\frac{1}{2}, n+\frac{1}{2}+\frac{3}{2}, \cdots, n+\frac{1}{2}+\left(\frac{2 n-1}{2}\right) .
$$

The product pairs, starting from the middle and working outward are

$$
\begin{gathered}
P_{1}=\left(n+\frac{1}{2}\right)^{2}-\frac{1}{4} \\
P_{2}=\left(n+\frac{1}{2}\right)^{2}-\frac{9}{4} \\
\vdots \\
P_{n}=\left(n+\frac{1}{2}\right)^{2}-\left(\frac{2 n-1}{2}\right)^{2}
\end{gathered}
$$

Each of the numbers $\left(\frac{2 k-1}{2}\right)^{2}$ is distinct for $k=1,2,3, \ldots, n$ and hence no terms of $P_{k}$ are equal.
The sequence is $1(2 n), 2(2 n-1), 3(2 n-2), \ldots, n[2 n-(n-1)]$.

This sequence has exactly $n$ terms.

When the $k$ th term is subtracted from the $(k+1)$ th term the difference is $(k+1)[2 n-k]-k[2 n-(k-1)]=2(n-k)$. Since $n>k$, this is a positive difference.

Therefore each term is greater than the term before, so no two terms are equal.

Final answer: 证明题，略
Subfield: Number Theory


Question: Consider the first $2 n$ natural numbers. Pair off the numbers, as shown, and multiply the two members of each pair. Prove that there is no value of $n$ for which two of the $n$ products are equal.

<img_3894>

Solution:
The sequence is $1(2 n), 2(2 n-1), 3(2 n-2), \ldots, k(2 n-k+1), \ldots, p(2 n-p+1), \ldots, n(n+1)$.

In essence we are asking the question, 'is it possible that $k(2 n-k+1)=p(2 n-p+1)$ where $p$ and $k$ are both less than or equal to $n$ ?'

$$
\begin{aligned}
k(2 n-k+1) & =p(2 n-p+1) \\
2 n k-k^{2}+k=2 n p-p^{2}+p & \text { (supposing them to be equal) } \\
p^{2}-k^{2}+2 n k-2 n p+k-p & =0 \\
(p-k)(p+k)+2 n(k-p)+(k-p) & =0 \\
(p-k)[(p+k)-2 n-1] & =0 \\
(p-k)(p+k-2 n-1) & =0
\end{aligned}
$$

Since $p$ and $k$ are both less than or equal to $n$, it follows $p+k-2 n-1 \neq 0$. Therefore $p=k$ and they represent the same pair. Thus the required is proven.
The products are $1(2 n+1-1), 2(2 n+1-2), 3(2 n+1-3), \ldots, n(2 n+1-n)$.

Consider the function, $y=x(2 n+1-x)=-x^{2}+(2 n+1) x=f(x)$.



The graph of this function is a parabola, opening down, with its vertex at $x=n+\frac{1}{2}$.

The products are the $y$-coordinates of the points on the parabola corresponding to $x=1,2,3, \ldots, n$. Since all the points are to the left of the vertex, no two have the same $y$ coordinate.

Thus the products are distinct.

<img_3942>
The sum of these numbers is $\frac{2 n(2 n+1)}{2}$ or $n(2 n+1)$.

Their average is $\frac{n(2 n+1)}{2 n}=n+\frac{1}{2}$.

The $2 n$ numbers can be rewritten as,

$$
n+\frac{1}{2}-\left(\frac{2 n-1}{2}\right), \cdots, n+\frac{1}{2}-\frac{3}{2}, n+\frac{1}{2}-\frac{1}{2}, n+\frac{1}{2}+\frac{1}{2}, n+\frac{1}{2}+\frac{3}{2}, \cdots, n+\frac{1}{2}+\left(\frac{2 n-1}{2}\right) .
$$

The product pairs, starting from the middle and working outward are

$$
\begin{gathered}
P_{1}=\left(n+\frac{1}{2}\right)^{2}-\frac{1}{4} \\
P_{2}=\left(n+\frac{1}{2}\right)^{2}-\frac{9}{4} \\
\vdots \\
P_{n}=\left(n+\frac{1}{2}\right)^{2}-\left(\frac{2 n-1}{2}\right)^{2}
\end{gathered}
$$

Each of the numbers $\left(\frac{2 k-1}{2}\right)^{2}$ is distinct for $k=1,2,3, \ldots, n$ and hence no terms of $P_{k}$ are equal.
The sequence is $1(2 n), 2(2 n-1), 3(2 n-2), \ldots, n[2 n-(n-1)]$.

This sequence has exactly $n$ terms.

When the $k$ th term is subtracted from the $(k+1)$ th term the difference is $(k+1)[2 n-k]-k[2 n-(k-1)]=2(n-k)$. Since $n>k$, this is a positive difference.

Therefore each term is greater than the term before, so no two terms are equal.
Subfield: Algebra


Question: The equations $x^{2}+5 x+6=0$ and $x^{2}+5 x-6=0$ each have integer solutions whereas only one of the equations in the pair $x^{2}+4 x+5=0$ and $x^{2}+4 x-5=0$ has integer solutions.
Show that if $x^{2}+p x+q=0$ and $x^{2}+p x-q=0$ both have integer solutions, then it is possible to find integers $a$ and $b$ such that $p^{2}=a^{2}+b^{2}$. (i.e. $(a, b, p)$ is a Pythagorean triple).

Solution:
We have that $x^{2}+p x+q=0$ and $x^{2}+p x-q=0$ both have integer solutions.

For $x^{2}+p x+q=0$, its roots are $\frac{-p \pm \sqrt{p^{2}-4 q}}{2}$.

In order that these roots be integers, $p^{2}-4 q$ must be a perfect square.

Therefore, $p^{2}-4 q=m^{2}$ for some positive integer $m$.

Similarly for $x^{2}+p x-q=0$, it has roots $\frac{-p \pm \sqrt{p^{2}+4 q}}{2}$ and in order that these roots be integers $p^{2}+4 q$ must be a perfect square.

Thus $p^{2}+4 q=n^{2}$ for some positive integer $n$.

Adding gives $2 p^{2}=m^{2}+n^{2}$ (with $n \geq m$ since $n^{2}=p^{2}+4 q$

$$
\left.\geq p^{2}-4 q=m^{2}\right)
$$

And so $p^{2}=\frac{1}{2} m^{2}+\frac{1}{2} n^{2}=\left(\frac{n+m}{2}\right)^{2}+\left(\frac{n-m}{2}\right)^{2}$.

We note that $m$ and $n$ have the same parity since $m^{2}=p^{2}-4 q \equiv p^{2}(\bmod 2)$ and $n^{2} \equiv p^{2}+4 q \equiv p^{2}(\bmod 2)$.

Since $\frac{n+m}{2}$ and $\frac{n-m}{2}$ are positive integers then $p^{2}=a^{2}+b^{2}$ where $a=\frac{n+m}{2}$ and $b=\frac{n-m}{2}$.
Subfield: Algebra


Question: The equations $x^{2}+5 x+6=0$ and $x^{2}+5 x-6=0$ each have integer solutions whereas only one of the equations in the pair $x^{2}+4 x+5=0$ and $x^{2}+4 x-5=0$ has integer solutions.
Determine $q$ in terms of $a$ and $b$.

Solution:
We have that $x^{2}+p x+q=0$ and $x^{2}+p x-q=0$ both have integer solutions.

For $x^{2}+p x+q=0$, its roots are $\frac{-p \pm \sqrt{p^{2}-4 q}}{2}$.

In order that these roots be integers, $p^{2}-4 q$ must be a perfect square.

Therefore, $p^{2}-4 q=m^{2}$ for some positive integer $m$.

Similarly for $x^{2}+p x-q=0$, it has roots $\frac{-p \pm \sqrt{p^{2}+4 q}}{2}$ and in order that these roots be integers $p^{2}+4 q$ must be a perfect square.

Thus $p^{2}+4 q=n^{2}$ for some positive integer $n$.

Adding gives $2 p^{2}=m^{2}+n^{2}$ (with $n \geq m$ since $n^{2}=p^{2}+4 q$

$$
\left.\geq p^{2}-4 q=m^{2}\right)
$$

And so $p^{2}=\frac{1}{2} m^{2}+\frac{1}{2} n^{2}=\left(\frac{n+m}{2}\right)^{2}+\left(\frac{n-m}{2}\right)^{2}$.

We note that $m$ and $n$ have the same parity since $m^{2}=p^{2}-4 q \equiv p^{2}(\bmod 2)$ and $n^{2} \equiv p^{2}+4 q \equiv p^{2}(\bmod 2)$.

Since $\frac{n+m}{2}$ and $\frac{n-m}{2}$ are positive integers then $p^{2}=a^{2}+b^{2}$ where $a=\frac{n+m}{2}$ and $b=\frac{n-m}{2}$.

From above, $a=\frac{n+m}{2}$ and $b=\frac{n-m}{2}$ or $n=a+b$ and $m=a-b$.

From before, $p^{2}+4 q=n^{2}$

$$
\begin{aligned}
4 q^{2} & =n^{2}-p^{2} \\
& =(a+b)^{2}-\left(a^{2}+b^{2}\right) \\
4 q & =2 a b
\end{aligned}
$$

Therefore, $q=\frac{a b}{2}$.

Final answer: $\frac{a b}{2}$
Subfield: Geometry


Question: In the diagram, $A B=21$ and $B C=16$. Also, $\angle A B C=60^{\circ}, \angle C A D=30^{\circ}$, and $\angle A C D=45^{\circ}$. Determine the length of $C D$, to the nearest tenth.

![](https://cdn.mathpix.com/cropped/2023_12_21_e753dfdb77a4e9b70a3dg-1.jpg?height=322&width=447&top_left_y=1655&top_left_x=1311)

Solution:
By the cosine law in $\triangle C B A$,

$$
\begin{aligned}
C A^{2} & =C B^{2}+B A^{2}-2(C B)(B A) \cos (\angle C B A) \\
C A^{2} & =16^{2}+21^{2}-2(16)(21) \cos \left(60^{\circ}\right) \\
C A^{2} & =256+441-2(16)(21)\left(\frac{1}{2}\right) \\
C A^{2} & =256+441-(16)(21) \\
C A^{2} & =361 \\
C A & =\sqrt{361}=19 \quad(\text { since } C A>0)
\end{aligned}
$$

In $\triangle C A D, \angle C D A=180^{\circ}-\angle D C A-\angle D A C=180^{\circ}-45^{\circ}-30^{\circ}=105^{\circ}$.

By the sine law in $\triangle C D A$,

$$
\begin{aligned}
\frac{C D}{\sin (\angle D A C)} & =\frac{C A}{\sin (\angle C D A)} \\
C D & =\frac{19 \sin \left(30^{\circ}\right)}{\sin \left(105^{\circ}\right)} \\
C D & =\frac{19\left(\frac{1}{2}\right)}{\sin \left(105^{\circ}\right)} \\
C D & =\frac{19}{2 \sin \left(105^{\circ}\right)} \\
C D & \approx 9.835
\end{aligned}
$$

so, to the nearest tenth, $C D$ equals 9.8 .

(Note that we could have used

$$
\begin{aligned}
\sin \left(105^{\circ}\right) & =\sin \left(60^{\circ}+45^{\circ}\right)=\sin \left(60^{\circ}\right) \cos \left(45^{\circ}\right)+\cos \left(60^{\circ}\right) \sin \left(45^{\circ}\right) \\
& =\frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}}+\frac{1}{2} \cdot \frac{1}{\sqrt{2}}=\frac{\sqrt{3}+1}{2 \sqrt{2}}
\end{aligned}
$$

to say that $C D=\frac{19}{2\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right)}=\frac{19 \sqrt{2}}{\sqrt{3}+1}$ exactly, and then evaluated this expression.)

Final answer: $\frac{19 \sqrt{2}}{\sqrt{3}+1}$
Subfield: Geometry


Question: In the diagram, $A B=21$ and $B C=16$. Also, $\angle A B C=60^{\circ}, \angle C A D=30^{\circ}$, and $\angle A C D=45^{\circ}$. Determine the length of $C D$, to the nearest tenth.

<img_3209>

Solution:
By the cosine law in $\triangle C B A$,

$$
\begin{aligned}
C A^{2} & =C B^{2}+B A^{2}-2(C B)(B A) \cos (\angle C B A) \\
C A^{2} & =16^{2}+21^{2}-2(16)(21) \cos \left(60^{\circ}\right) \\
C A^{2} & =256+441-2(16)(21)\left(\frac{1}{2}\right) \\
C A^{2} & =256+441-(16)(21) \\
C A^{2} & =361 \\
C A & =\sqrt{361}=19 \quad(\text { since } C A>0)
\end{aligned}
$$

In $\triangle C A D, \angle C D A=180^{\circ}-\angle D C A-\angle D A C=180^{\circ}-45^{\circ}-30^{\circ}=105^{\circ}$.

By the sine law in $\triangle C D A$,

$$
\begin{aligned}
\frac{C D}{\sin (\angle D A C)} & =\frac{C A}{\sin (\angle C D A)} \\
C D & =\frac{19 \sin \left(30^{\circ}\right)}{\sin \left(105^{\circ}\right)} \\
C D & =\frac{19\left(\frac{1}{2}\right)}{\sin \left(105^{\circ}\right)} \\
C D & =\frac{19}{2 \sin \left(105^{\circ}\right)} \\
C D & \approx 9.835
\end{aligned}
$$

so, to the nearest tenth, $C D$ equals 9.8 .

(Note that we could have used

$$
\begin{aligned}
\sin \left(105^{\circ}\right) & =\sin \left(60^{\circ}+45^{\circ}\right)=\sin \left(60^{\circ}\right) \cos \left(45^{\circ}\right)+\cos \left(60^{\circ}\right) \sin \left(45^{\circ}\right) \\
& =\frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}}+\frac{1}{2} \cdot \frac{1}{\sqrt{2}}=\frac{\sqrt{3}+1}{2 \sqrt{2}}
\end{aligned}
$$

to say that $C D=\frac{19}{2\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right)}=\frac{19 \sqrt{2}}{\sqrt{3}+1}$ exactly, and then evaluated this expression.)

Final answer: $\frac{19 \sqrt{2}}{\sqrt{3}+1}$
Subfield: Geometry


Question: In the diagram, the large circle has radius 9 and centre $C(15,0)$. The small circles have radius 4 and centres $A$ and $B$ on the horizontal line $y=12$. Each of the two small circles is tangent to the large circle. It takes a bug 5 seconds to walk at a constant speed from $A$ to $B$ along the line $y=12$. How far does the bug walk in 1 second?

<img_3594>

Solution:
Consider $P$ on $A B$ with $C P$ perpendicular to $A B$. Note that $C P=12$.

Since the small circle with centre $A$ is tangent to the large circle with centre $C$, then $A C$ equals the sum of the radii of these circles, or $A C=4+9=13$. Similarly, $B C=13$.

This tells us that $\triangle A P C$ is congruent to $\triangle B P C$ (they have equal hypotenuses and each is right-angled and has a common side), so $B P=A P$.

By the Pythagorean Theorem in $\triangle A P C$,

$$
A P^{2}=A C^{2}-P C^{2}=13^{2}-12^{2}=169-144=25
$$

so $A P=5$ (since $A P>0)$.

Therefore, $B P=A P=5$ and so $A B=10$.

Since it takes the bug 5 seconds to walk this distance, then in 1 second, the bug walks a distance of $\frac{10}{5}=2$.

Final answer: 2
Subfield: Geometry


Question: In the diagram, the large circle has radius 9 and centre $C(15,0)$. The small circles have radius 4 and centres $A$ and $B$ on the horizontal line $y=12$. Each of the two small circles is tangent to the large circle. It takes a bug 5 seconds to walk at a constant speed from $A$ to $B$ along the line $y=12$. How far does the bug walk in 1 second?

![](https://cdn.mathpix.com/cropped/2023_12_21_e753dfdb77a4e9b70a3dg-1.jpg?height=445&width=542&top_left_y=2000&top_left_x=1255)

Solution:
Consider $P$ on $A B$ with $C P$ perpendicular to $A B$. Note that $C P=12$.

Since the small circle with centre $A$ is tangent to the large circle with centre $C$, then $A C$ equals the sum of the radii of these circles, or $A C=4+9=13$. Similarly, $B C=13$.

This tells us that $\triangle A P C$ is congruent to $\triangle B P C$ (they have equal hypotenuses and each is right-angled and has a common side), so $B P=A P$.

By the Pythagorean Theorem in $\triangle A P C$,

$$
A P^{2}=A C^{2}-P C^{2}=13^{2}-12^{2}=169-144=25
$$

so $A P=5$ (since $A P>0)$.

Therefore, $B P=A P=5$ and so $A B=10$.

Since it takes the bug 5 seconds to walk this distance, then in 1 second, the bug walks a distance of $\frac{10}{5}=2$.

Final answer: 2
Subfield: Algebra


Question: Determine all values of $k$, with $k \neq 0$, for which the parabola

$$
y=k x^{2}+(5 k+3) x+(6 k+5)
$$

has its vertex on the $x$-axis.

Solution:
For the parabola to have its vertex on the $x$-axis, the equation

$$
y=k x^{2}+(5 k+3) x+(6 k+5)=0
$$

must have two equal real roots.

That is, its discriminant must equal 0 , and so

$$
\begin{aligned}
(5 k+3)^{2}-4 k(6 k+5) & =0 \\
25 k^{2}+30 k+9-24 k^{2}-20 k & =0 \\
k^{2}+10 k+9 & =0 \\
(k+1)(k+9) & =0
\end{aligned}
$$

Therefore, $k=-1$ or $k=-9$.

Final answer: $-1,-9$