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


Question: In the diagram, line segment $F C G$ passes through vertex $C$ of square $A B C D$, with $F$ lying on $A B$ extended and $G$ lying on $A D$ extended. Prove that $\frac{1}{A B}=\frac{1}{A F}+\frac{1}{A G}$.

<img_3786>

Solution:
Without loss of generality, suppose that square $A B C D$ has side length 1 .

Suppose next that $B F=a$ and $\angle C F B=\theta$.

Since $\triangle C B F$ is right-angled at $B$, then $\angle B C F=90^{\circ}-\theta$.

Since $G C F$ is a straight line, then $\angle G C D=180^{\circ}-90^{\circ}-\left(90^{\circ}-\theta\right)=\theta$.

Therefore, $\triangle G D C$ is similar to $\triangle C B F$, since $\triangle G D C$ is right-angled at $D$.

Thus, $\frac{G D}{D C}=\frac{B C}{B F}$ or $\frac{G D}{1}=\frac{1}{a}$ or $G D=\frac{1}{a}$.

So $A F=A B+B F=1+a$ and $A G=A D+D G=1+\frac{1}{a}=\frac{a+1}{a}$.

Thus, $\frac{1}{A F}+\frac{1}{A G}=\frac{1}{1+a}+\frac{a}{a+1}=\frac{a+1}{a+1}=1=\frac{1}{A B}$, as required.
We attach a set of coordinate axes to the diagram, with $A$ at the origin, $A G$ lying along the positive $y$-axis and $A F$ lying along the positive $x$-axis.

Without loss of generality, suppose that square $A B C D$ has side length 1 , so that $C$ has coordinates $(1,1)$. (We can make this assumption without loss of generality, because if the square had a different side length, then each of the lengths in the problem would be scaled by the same factor.)



Suppose that the line through $G$ and $F$ has slope $m$.

Since this line passes through $(1,1)$, its equation is $y-1=m(x-1)$ or $y=m x+(1-m)$. The $y$-intercept of this line is $1-m$, so $G$ has coordinates $(0,1-m)$.

The $x$-intercept of this line is $\frac{m-1}{m}$, so $F$ has coordinates $\left(\frac{m-1}{m}, 0\right)$. (Note that $m \neq 0$ as the line cannot be horizontal.)

Therefore,

$$
\frac{1}{A F}+\frac{1}{A G}=\frac{m}{m-1}+\frac{1}{1-m}=\frac{m}{m-1}+\frac{-1}{m-1}=\frac{m-1}{m-1}=1=\frac{1}{A B}
$$

as required.
Join $A$ to $C$.

We know that the sum of the areas of $\triangle G C A$ and $\triangle F C A$ equals the area of $\triangle G A F$.

The area of $\triangle G C A$ (thinking of $A G$ as the base) is $\frac{1}{2}(A G)(D C)$, since $D C$ is perpendicular to $A G$.

Similarly, the area of $\triangle F C A$ is $\frac{1}{2}(A F)(C B)$.

Also, the area of $\triangle G A F$ is $\frac{1}{2}(A G)(A F)$.

Therefore,

$$
\begin{aligned}
\frac{1}{2}(A G)(D C)+\frac{1}{2}(A F)(C B) & =\frac{1}{2}(A G)(A F) \\
\frac{(A G)(D C)}{(A G)(A F)(A B)}+\frac{(A F)(C B)}{(A G)(A F)(A B)} & =\frac{(A G)(A F)}{(A G)(A F)(A B)} \\
\frac{1}{A F}+\frac{1}{A G} & =\frac{1}{A B}
\end{aligned}
$$

as required, since $A B=D C=C B$.
Subfield: Geometry


Question: In the $4 \times 4$ grid shown, three coins are randomly placed in different squares. Determine the probability that no two coins lie in the same row or column.

![](https://cdn.mathpix.com/cropped/2023_12_21_1ec73aaee00c97b3df16g-1.jpg?height=274&width=279&top_left_y=1728&top_left_x=1259)

Solution:
We consider placing the three coins individually.

Place one coin randomly on the grid.

When the second coin is placed (in any one of 15 squares), 6 of the 15 squares will leave two coins in the same row or column and 9 of the 15 squares will leave the two coins in different rows and different columns.

![](https://cdn.mathpix.com/cropped/2023_12_21_75254b466303852df435g-1.jpg?height=277&width=284&top_left_y=2130&top_left_x=1018)

Therefore, the probability that the two coins are in different rows and different columns is $\frac{9}{15}=\frac{3}{5}$.

There are 14 possible squares in which the third coin can be placed.



Of these 14 squares, 6 lie in the same row or column as the first coin and an additional 4 lie the same row or column as the second coin. Therefore, the probability that the third coin is placed in a different row and a different column than each of the first two coins is $\frac{4}{14}=\frac{2}{7}$.

Therefore, the probability that all three coins are placed in different rows and different columns is $\frac{3}{5} \times \frac{2}{7}=\frac{6}{35}$.

Final answer: $\frac{6}{35}$
Subfield: Geometry


Question: In the $4 \times 4$ grid shown, three coins are randomly placed in different squares. Determine the probability that no two coins lie in the same row or column.

<img_3676>

Solution:
We consider placing the three coins individually.

Place one coin randomly on the grid.

When the second coin is placed (in any one of 15 squares), 6 of the 15 squares will leave two coins in the same row or column and 9 of the 15 squares will leave the two coins in different rows and different columns.

<img_3640>

Therefore, the probability that the two coins are in different rows and different columns is $\frac{9}{15}=\frac{3}{5}$.

There are 14 possible squares in which the third coin can be placed.



Of these 14 squares, 6 lie in the same row or column as the first coin and an additional 4 lie the same row or column as the second coin. Therefore, the probability that the third coin is placed in a different row and a different column than each of the first two coins is $\frac{4}{14}=\frac{2}{7}$.

Therefore, the probability that all three coins are placed in different rows and different columns is $\frac{3}{5} \times \frac{2}{7}=\frac{6}{35}$.

Final answer: $\frac{6}{35}$
Subfield: Geometry


Question: In the diagram, the area of $\triangle A B C$ is 1 . Trapezoid $D E F G$ is constructed so that $G$ is to the left of $F, D E$ is parallel to $B C$, $E F$ is parallel to $A B$ and $D G$ is parallel to $A C$. Determine the maximum possible area of trapezoid $D E F G$.

![](https://cdn.mathpix.com/cropped/2023_12_21_1ec73aaee00c97b3df16g-1.jpg?height=360&width=499&top_left_y=2048&top_left_x=1255)

Solution:
Suppose that $A B=c, A C=b$ and $B C=a$.

Since $D G$ is parallel to $A C, \angle B D G=\angle B A C$ and $\angle D G B=\angle A C B$, so $\triangle D G B$ is similar to $\triangle A C B$.

(Similarly, $\triangle A E D$ and $\triangle E C F$ are also both similar to $\triangle A B C$.)

Suppose next that $D B=k c$, with $0<k<1$.

Then the ratio of the side lengths of $\triangle D G B$ to those of $\triangle A C B$ will be $k: 1$, so $B G=k a$ and $D G=k b$.

Since the ratio of the side lengths of $\triangle D G B$ to $\triangle A C B$ is $k: 1$, then the ratio of their areas will be $k^{2}: 1$, so the area of $\triangle D G B$ is $k^{2}$ (since the area of $\triangle A C B$ is 1 ).

Since $A B=c$ and $D B=k c$, then $A D=(1-k) c$, so using similar triangles as before, $D E=(1-k) a$ and $A E=(1-k) b$. Also, the area of $\triangle A D E$ is $(1-k)^{2}$.

Since $A C=b$ and $A E=(1-k) b$, then $E C=k b$, so again using similar triangles, $E F=k c$, $F C=k a$ and the area of $\triangle E C F$ is $k^{2}$.

Now the area of trapezoid $D E F G$ is the area of the large triangle minus the combined areas of the small triangles, or $1-k^{2}-k^{2}-(1-k)^{2}=2 k-3 k^{2}$.

We know that $k \geq 0$ by its definition. Also, since $G$ is to the left of $F$, then $B G+F C \leq B C$ or $k a+k a \leq a$ or $2 k a \leq a$ or $k \leq \frac{1}{2}$.

Let $f(k)=2 k-3 k^{2}$.

Since $f(k)=-3 k^{2}+2 k+0$ is a parabola opening downwards, its maximum occurs at its vertex, whose $k$-coordinate is $k=-\frac{2}{2(-3)}=\frac{1}{3}$ (which lies in the admissible range for $k$ ).

Note that $f\left(\frac{1}{3}\right)=\frac{2}{3}-3\left(\frac{1}{9}\right)=\frac{1}{3}$.

Therefore, the maximum area of the trapezoid is $\frac{1}{3}$.

Final answer: $\frac{1}{3}$
Subfield: Geometry


Question: In the diagram, the area of $\triangle A B C$ is 1 . Trapezoid $D E F G$ is constructed so that $G$ is to the left of $F, D E$ is parallel to $B C$, $E F$ is parallel to $A B$ and $D G$ is parallel to $A C$. Determine the maximum possible area of trapezoid $D E F G$.

<img_3976>

Solution:
Suppose that $A B=c, A C=b$ and $B C=a$.

Since $D G$ is parallel to $A C, \angle B D G=\angle B A C$ and $\angle D G B=\angle A C B$, so $\triangle D G B$ is similar to $\triangle A C B$.

(Similarly, $\triangle A E D$ and $\triangle E C F$ are also both similar to $\triangle A B C$.)

Suppose next that $D B=k c$, with $0<k<1$.

Then the ratio of the side lengths of $\triangle D G B$ to those of $\triangle A C B$ will be $k: 1$, so $B G=k a$ and $D G=k b$.

Since the ratio of the side lengths of $\triangle D G B$ to $\triangle A C B$ is $k: 1$, then the ratio of their areas will be $k^{2}: 1$, so the area of $\triangle D G B$ is $k^{2}$ (since the area of $\triangle A C B$ is 1 ).

Since $A B=c$ and $D B=k c$, then $A D=(1-k) c$, so using similar triangles as before, $D E=(1-k) a$ and $A E=(1-k) b$. Also, the area of $\triangle A D E$ is $(1-k)^{2}$.

Since $A C=b$ and $A E=(1-k) b$, then $E C=k b$, so again using similar triangles, $E F=k c$, $F C=k a$ and the area of $\triangle E C F$ is $k^{2}$.

Now the area of trapezoid $D E F G$ is the area of the large triangle minus the combined areas of the small triangles, or $1-k^{2}-k^{2}-(1-k)^{2}=2 k-3 k^{2}$.

We know that $k \geq 0$ by its definition. Also, since $G$ is to the left of $F$, then $B G+F C \leq B C$ or $k a+k a \leq a$ or $2 k a \leq a$ or $k \leq \frac{1}{2}$.

Let $f(k)=2 k-3 k^{2}$.

Since $f(k)=-3 k^{2}+2 k+0$ is a parabola opening downwards, its maximum occurs at its vertex, whose $k$-coordinate is $k=-\frac{2}{2(-3)}=\frac{1}{3}$ (which lies in the admissible range for $k$ ).

Note that $f\left(\frac{1}{3}\right)=\frac{2}{3}-3\left(\frac{1}{9}\right)=\frac{1}{3}$.

Therefore, the maximum area of the trapezoid is $\frac{1}{3}$.

Final answer: $\frac{1}{3}$
Subfield: Geometry


Question: The parabola $y=f(x)=x^{2}+b x+c$ has vertex $P$ and the parabola $y=g(x)=-x^{2}+d x+e$ has vertex $Q$, where $P$ and $Q$ are distinct points. The two parabolas also intersect at $P$ and $Q$.
Prove that $2(e-c)=b d$.

Solution:
The vertex of the first parabola has $x$-coordinate $x=-\frac{1}{2} b$.

Since each parabola passes through $P$, then

$$
\begin{aligned}
f\left(-\frac{1}{2} b\right) & =g\left(-\frac{1}{2} b\right) \\
\frac{1}{4} b^{2}+b\left(-\frac{1}{2} b\right)+c & =-\frac{1}{4} b^{2}+d\left(-\frac{1}{2} b\right)+e \\
\frac{1}{4} b^{2}-\frac{1}{2} b^{2}+c & =-\frac{1}{4} b^{2}-\frac{1}{2} b d+e \\
\frac{1}{2} b d & =e-c \\
b d & =2(e-c)
\end{aligned}
$$

as required. (The same result can be obtained by using the vertex of the second parabola.)
Subfield: Geometry


Question: The parabola $y=f(x)=x^{2}+b x+c$ has vertex $P$ and the parabola $y=g(x)=-x^{2}+d x+e$ has vertex $Q$, where $P$ and $Q$ are distinct points. The two parabolas also intersect at $P$ and $Q$.
Prove that the line through $P$ and $Q$ has slope $\frac{1}{2}(b+d)$ and $y$-intercept $\frac{1}{2}(c+e)$.

Solution:
First, I prove that $bd = 2(e-c)$:

The vertex of the first parabola has $x$-coordinate $x=-\frac{1}{2} b$.

Since each parabola passes through $P$, then

$$
\begin{aligned}
f\left(-\frac{1}{2} b\right) & =g\left(-\frac{1}{2} b\right) \\
\frac{1}{4} b^{2}+b\left(-\frac{1}{2} b\right)+c & =-\frac{1}{4} b^{2}+d\left(-\frac{1}{2} b\right)+e \\
\frac{1}{4} b^{2}-\frac{1}{2} b^{2}+c & =-\frac{1}{4} b^{2}-\frac{1}{2} b d+e \\
\frac{1}{2} b d & =e-c \\
b d & =2(e-c)
\end{aligned}
$$

as required. (The same result can be obtained by using the vertex of the second parabola.)

The vertex, $P$, of the first parabola has $x$-coordinate $x=-\frac{1}{2} b$ so has $y$-coordinate $f\left(-\frac{1}{2} b\right)=\frac{1}{4} b^{2}-\frac{1}{2} b^{2}+c=-\frac{1}{4} b^{2}+c$.

The vertex, $Q$, of the second parabola has $x$-coordinate $x=\frac{1}{2} d$ so has $y$-coordinate $g\left(\frac{1}{2} d\right)=-\frac{1}{4} d^{2}+\frac{1}{2} d^{2}+c=\frac{1}{4} d^{2}+e$.

Therefore, the slope of the line through $P$ and $Q$ is

$$
\begin{aligned}
\frac{\left(-\frac{1}{4} b^{2}+c\right)-\left(\frac{1}{4} d^{2}+e\right)}{-\frac{1}{2} b-\frac{1}{2} d} & =\frac{-\frac{1}{4}\left(b^{2}+d^{2}\right)-(e-c)}{-\frac{1}{2} b-\frac{1}{2} d} \\
& =\frac{-\frac{1}{4}\left(b^{2}+d^{2}\right)-\frac{1}{2} b d}{-\frac{1}{2} b-\frac{1}{2} d} \\
& =\frac{-\frac{1}{4}\left(b^{2}+2 b d+d^{2}\right)}{-\frac{1}{2}(b+d)} \\
& =\frac{1}{2}(b+d)
\end{aligned}
$$

Using the point-slope form of the line, the line thus has equation

$$
\begin{aligned}
y & =\frac{1}{2}(b+d)\left(x-\left(-\frac{1}{2} b\right)\right)+\left(-\frac{1}{4} b^{2}+c\right) \\
& =\frac{1}{2}(b+d) x+\frac{1}{4} b^{2}+\frac{1}{4} b d-\frac{1}{4} b^{2}+c \\
& =\frac{1}{2}(b+d) x+\frac{1}{4} b d+c \\
& =\frac{1}{2}(b+d) x+\frac{1}{2}(e-c)+c \\
& =\frac{1}{2}(b+d) x+\frac{1}{2}(e+c)
\end{aligned}
$$

so the $y$-intercept of the line is $\frac{1}{2}(e+c)$.
The equations of the two parabolas are $y=x^{2}+b x+c$ and $y=-x^{2}+d x+e$.

Adding the two equations, we obtain $2 y=(b+d) x+(c+e)$ or $y=\frac{1}{2}(b+d) x+\frac{1}{2}(c+e)$. This last equation is the equation of a line.

Points $P$ and $Q$, whose coordinates satisfy the equation of each parabola, must satisfy the equation of the line, and so lie on the line.

But the line through $P$ and $Q$ is unique, so this is the equation of the line through $P$ and $Q$.

Therefore, the line through $P$ and $Q$ has slope $\frac{1}{2}(b+d)$ and $y$-intercept $\frac{1}{2}(c+e)$.
Subfield: Algebra


Question: Suppose that $f(x)=x^{2}+(2 n-1) x+\left(n^{2}-22\right)$ for some integer $n$. What is the smallest positive integer $n$ for which $f(x)$ has no real roots?

Solution:
The quadratic function $f(x)=x^{2}+(2 n-1) x+\left(n^{2}-22\right)$ has no real roots exactly when its discriminant, $\Delta$, is negative.

The discriminant of this function is

$$
\begin{aligned}
\Delta & =(2 n-1)^{2}-4(1)\left(n^{2}-22\right) \\
& =\left(4 n^{2}-4 n+1\right)-\left(4 n^{2}-88\right) \\
& =-4 n+89
\end{aligned}
$$

We have $\Delta<0$ exactly when $-4 n+89<0$ or $4 n>89$.

This final inequality is equivalent to $n>\frac{89}{4}=22 \frac{1}{4}$.

Therefore, the smallest positive integer that satisfies this inequality, and hence for which $f(x)$ has no real roots, is $n=23$.

Final answer: 23
Subfield: Geometry


Question: In the diagram, $\triangle P Q R$ has $P Q=a, Q R=b, P R=21$, and $\angle P Q R=60^{\circ}$. Also, $\triangle S T U$ has $S T=a, T U=b, \angle T S U=30^{\circ}$, and $\sin (\angle T U S)=\frac{4}{5}$. Determine the values of $a$ and $b$.
![](https://cdn.mathpix.com/cropped/2023_12_21_4165463fb3d29f83c565g-1.jpg?height=228&width=792&top_left_y=1225&top_left_x=730)

Solution:
Using the cosine law in $\triangle P Q R$,

$$
\begin{aligned}
P R^{2} & =P Q^{2}+Q R^{2}-2 \cdot P Q \cdot Q R \cdot \cos (\angle P Q R) \\
21^{2} & =a^{2}+b^{2}-2 a b \cos \left(60^{\circ}\right) \\
441 & =a^{2}+b^{2}-2 a b \cdot \frac{1}{2} \\
441 & =a^{2}+b^{2}-a b
\end{aligned}
$$

Using the sine law in $\triangle S T U$, we obtain $\frac{S T}{\sin (\angle T U S)}=\frac{T U}{\sin (\angle T S U)}$ and so $\frac{a}{4 / 5}=\frac{b}{\sin \left(30^{\circ}\right)}$. Therefore, $\frac{a}{4 / 5}=\frac{b}{1 / 2}$ and so $a=\frac{4}{5} \cdot 2 b=\frac{8}{5} b$.

Substituting into the previous equation,

$$
\begin{aligned}
& 441=\left(\frac{8}{5} b\right)^{2}+b^{2}-\left(\frac{8}{5} b\right) b \\
& 441=\frac{64}{25} b^{2}+b^{2}-\frac{8}{5} b^{2} \\
& 441=\frac{64}{25} b^{2}+\frac{25}{25} b^{2}-\frac{40}{25} b^{2} \\
& 441=\frac{49}{25} b^{2} \\
& 225=b^{2}
\end{aligned}
$$

Since $b>0$, then $b=15$ and so $a=\frac{8}{5} b=\frac{8}{5} \cdot 15=24$.

Final answer: $24,15$
Subfield: Geometry


Question: In the diagram, $\triangle P Q R$ has $P Q=a, Q R=b, P R=21$, and $\angle P Q R=60^{\circ}$. Also, $\triangle S T U$ has $S T=a, T U=b, \angle T S U=30^{\circ}$, and $\sin (\angle T U S)=\frac{4}{5}$. Determine the values of $a$ and $b$.
<img_3194>

Solution:
Using the cosine law in $\triangle P Q R$,

$$
\begin{aligned}
P R^{2} & =P Q^{2}+Q R^{2}-2 \cdot P Q \cdot Q R \cdot \cos (\angle P Q R) \\
21^{2} & =a^{2}+b^{2}-2 a b \cos \left(60^{\circ}\right) \\
441 & =a^{2}+b^{2}-2 a b \cdot \frac{1}{2} \\
441 & =a^{2}+b^{2}-a b
\end{aligned}
$$

Using the sine law in $\triangle S T U$, we obtain $\frac{S T}{\sin (\angle T U S)}=\frac{T U}{\sin (\angle T S U)}$ and so $\frac{a}{4 / 5}=\frac{b}{\sin \left(30^{\circ}\right)}$. Therefore, $\frac{a}{4 / 5}=\frac{b}{1 / 2}$ and so $a=\frac{4}{5} \cdot 2 b=\frac{8}{5} b$.

Substituting into the previous equation,

$$
\begin{aligned}
& 441=\left(\frac{8}{5} b\right)^{2}+b^{2}-\left(\frac{8}{5} b\right) b \\
& 441=\frac{64}{25} b^{2}+b^{2}-\frac{8}{5} b^{2} \\
& 441=\frac{64}{25} b^{2}+\frac{25}{25} b^{2}-\frac{40}{25} b^{2} \\
& 441=\frac{49}{25} b^{2} \\
& 225=b^{2}
\end{aligned}
$$

Since $b>0$, then $b=15$ and so $a=\frac{8}{5} b=\frac{8}{5} \cdot 15=24$.

Final answer: $24,15$
Subfield: Geometry


Question: A square lattice of 16 points is constructed such that the horizontal and vertical distances between adjacent points are all exactly 1 unit. Each of four pairs of points are connected by a line segment, as shown. The intersections of these line segments are the vertices of square $A B C D$. Determine the area of square $A B C D$.

<img_3852>

Solution:
We label five additional points in the diagram:

<img_3827>

Since $P Q=Q R=R S=1$, then $P S=3$ and $P R=2$.

Since $\angle P S T=90^{\circ}$, then $P T=\sqrt{P S^{2}+S T^{2}}=\sqrt{3^{2}+1^{2}}=\sqrt{10}$ by the Pythagorean Theorem.

We are told that $A B C D$ is a square.

Thus, $P T$ is perpendicular to $Q C$ and to $R B$.

Thus, $\triangle P D Q$ is right-angled at $D$ and $\triangle P A R$ is right-angled at $A$.

Since $\triangle P D Q, \triangle P A R$ and $\triangle P S T$ are all right-angled and all share an angle at $P$, then these three triangles are similar.

This tells us that $\frac{P A}{P S}=\frac{P R}{P T}$ and so $P A=\frac{3 \cdot 2}{\sqrt{10}}$. Also, $\frac{P D}{P S}=\frac{P Q}{P T}$ and so $P D=\frac{1 \cdot 3}{\sqrt{10}}$.

Therefore,

$$
D A=P A-P D=\frac{6}{\sqrt{10}}-\frac{3}{\sqrt{10}}=\frac{3}{\sqrt{10}}
$$

This means that the area of square $A B C D$ is equal to $D A^{2}=\left(\frac{3}{\sqrt{10}}\right)^{2}=\frac{9}{10}$.
We add coordinates to the diagram as shown:

<img_3409>

We determine the side length of square $A B C D$ by determining the coordinates of $D$ and $A$ and then calculating the distance between these points.

The slope of the line through $(0,3)$ and $(3,2)$ is $\frac{3-2}{0-3}=-\frac{1}{3}$.

This equation of this line can be written as $y=-\frac{1}{3} x+3$.

The slope of the line through $(0,0)$ and $(1,3)$ is 3.

The equation of this line can be written as $y=3 x$.

The slope of the line through $(1,0)$ and $(2,3)$ is also 3.

The equation of this line can be written as $y=3(x-1)=3 x-3$.

Point $D$ is the intersection point of the lines with equations $y=-\frac{1}{3} x+3$ and $y=3 x$.

Equating expressions for $y$, we obtain $-\frac{1}{3} x+3=3 x$ and so $\frac{10}{3} x=3$ which gives $x=\frac{9}{10}$.

Since $y=3 x$, we get $y=\frac{27}{10}$ and so the coordinates of $D$ are $\left(\frac{9}{10}, \frac{27}{10}\right)$.

Point $A$ is the intersection point of the lines with equations $y=-\frac{1}{3} x+3$ and $y=3 x-3$.

Equating expressions for $y$, we obtain $-\frac{1}{3} x+3=3 x-3$ and so $\frac{10}{3} x=6$ which gives $x=\frac{18}{10}$.

Since $y=3 x-3$, we get $y=\frac{24}{10}$ and so the coordinates of $A$ are $\left(\frac{18}{10}, \frac{24}{10}\right)$. (It is easier to not reduce these fractions.)

Therefore,

$$
D A=\sqrt{\left(\frac{9}{10}-\frac{18}{10}\right)^{2}+\left(\frac{27}{10}-\frac{24}{10}\right)^{2}}=\sqrt{\left(-\frac{9}{10}\right)^{2}+\left(\frac{3}{10}\right)^{2}}=\sqrt{\frac{90}{100}}=\sqrt{\frac{9}{10}}
$$

This means that the area of square $A B C D$ is equal to $D A^{2}=\left(\sqrt{\frac{9}{10}}\right)^{2}=\frac{9}{10}$.

Final answer: $\frac{9}{10}$
Subfield: Combinatorics


Question: A bag contains 3 red marbles and 6 blue marbles. Akshan removes one marble at a time until the bag is empty. Each marble that they remove is chosen randomly from the remaining marbles. Given that the first marble that Akshan removes is red and the third marble that they remove is blue, what is the probability that the last two marbles that Akshan removes are both blue?

Solution:
Each possible order in which Akshan removes the marbles corresponds to a sequence of 9 colours, 3 of which are red and 6 of which are blue.

We write these as sequences of 3 R's and 6 B's.

Since are told that the first marble is red and the third is blue, we would like to consider all sequences of the form

$$
R \_B\_\_\_\_\_\_
$$

The 7 blanks must be filled with the remaining 2 R's and 5 B's.

There are $\left(\begin{array}{l}7 \\ 2\end{array}\right)=\frac{7 \cdot 6}{2}=21$ ways of doing this, because 2 of the 7 blanks must be chosen in which to place the R's. (We could count these 21 ways directly by working systematically through the possible pairs of blanks.)

Of these 21 ways, some have the last two marbles being blue.

These correspond to the sequences of the form

$$
R \_B \_\_\_\_ B B
$$

In these sequences, the 5 blanks must be filled with the remaining $2 \mathrm{R}$ 's and 3 B's.

There are $\left(\begin{array}{l}5 \\ 2\end{array}\right)=\frac{5 \cdot 4}{2}=10$ ways of doing this, because 2 of the 5 blanks must be chosen in which to place the R's.

Therefore, 10 of the 21 possible sequences end in two B's, and so the probability that the last two marbles removed are blue is $\frac{10}{21}$.

Final answer: $\frac{10}{21}$
Subfield: Number Theory


Question: Determine the number of quadruples of positive integers $(a, b, c, d)$ with $a<b<c<d$ that satisfy both of the following system of equations:

$$
\begin{aligned}
a c+a d+b c+b d & =2023 \\
a+b+c+d & =296
\end{aligned}
$$

Solution:
Factoring the first equation, we obtain

$$
a c+a d+b c+b d=a(c+d)+b(c+d)=(a+b)(c+d)
$$

We now have the equations

$$
\begin{aligned}
(a+b)(c+d) & =2023 \\
(a+b)+(c+d) & =296
\end{aligned}
$$

If we let $s=a+b$ and $t=c+d$, we obtain the equations

$$
\begin{aligned}
s t & =2023 \\
s+t & =296
\end{aligned}
$$

Noting that $s$ and $t$ are integers since $a, b, c$, and $d$ are integers, we look for divisor pairs of 2023 whose sum is 296 .

To find the divisors of 2023 , we first find its prime factorization:

$$
2023=7 \cdot 289=7 \cdot 17^{2}
$$

Therefore, the divisors of 2023 are 1, 7, 17, 119, 289, 2023.

This means that the divisor pairs of 2023 are

$$
2023=1 \cdot 2023=7 \cdot 289=17 \cdot 119
$$

The one divisor pair with a sum of 296 is 7 and 289. (Alternatively, we could have found these by substituting $t=206-s$ into $s t=2023$ and using the quadratic formula.)



Since $a<b<c<d$, then $a+b<c+d$ and so $s=a+b=7$ and $t=c+d=289$.

Since $a$ and $b$ are positive integers with $a<b$ and $a+b=7$, then the possible pairs $(a, b)$ are

$$
(a, b)=(1,6),(2,5),(3,4)
$$

We know that $c$ and $d$ are positive integers with $c<d$ and $c+d=289$, but also with $b<c<d$.

When $(a, b)=(1,6)$, this means that the possibilities are

$$
(c, d)=(7,282),(8,281),(9,280), \ldots,(143,146),(144,145)
$$

There are $144-7+1=138$ such pairs.

When $(a, b)=(2,5)$, the possibilities are

$$
(c, d)=(6,283),(7,282),(8,281),(9,280), \ldots,(143,146),(144,145)
$$

There are $138+1=139$ such pairs.

When $(a, b)=(3,4)$, the possibilities are

$$
(c, d)=(5,284),(6,283),(7,282),(8,281),(9,280), \ldots,(143,146),(144,145)
$$

There are $139+1=140$ such pairs.

In total, there are $138+139+140=417$ possible quadruples $(a, b, c, d)$.

Final answer: 417