di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Number Theory


Question: For how many integers $k$ with $0<k<18$ is $\frac{5 \sin \left(10 k^{\circ}\right)-2}{\sin ^{2}\left(10 k^{\circ}\right)} \geq 2$ ?

Solution:
Let $\theta=10 k^{\circ}$.

The given inequalities become $0^{\circ}<\theta<180^{\circ}$ and $\frac{5 \sin \theta-2}{\sin ^{2} \theta} \geq 2$.

When $0^{\circ}<\theta<180^{\circ}, \sin \theta \neq 0$.

This means that we can can multiply both sides by $\sin ^{2} \theta>0$ and obtain the equivalent inequalities:

$$
\begin{aligned}
\frac{5 \sin \theta-2}{\sin ^{2} \theta} & \geq 2 \\
5 \sin \theta-2 & \geq 2 \sin ^{2} \theta \\
0 & \geq 2 \sin ^{2} \theta-5 \sin \theta+2 \\
0 & \geq(2 \sin \theta-1)(\sin \theta-2)
\end{aligned}
$$

Since $\sin \theta \leq 1$, then $\sin \theta-2 \leq-1<0$ for all $\theta$.

Therefore, $(2 \sin \theta-1)(\sin \theta-2) \leq 0$ exactly when $2 \sin \theta-1 \geq 0$.

Note that $2 \sin \theta-1 \geq 0$ exactly when $\sin \theta \geq \frac{1}{2}$.

Therefore, the original inequality is true exactly when $\frac{1}{2} \leq \sin \theta \leq 1$.

Note that $\sin 30^{\circ}=\sin 150^{\circ}=\frac{1}{2}$ and $0^{\circ}<\theta<180^{\circ}$.

When $\theta=0^{\circ}, \sin \theta=0$.

From $\theta=0^{\circ}$ to $\theta=30^{\circ}, \sin \theta$ increases from 0 to $\frac{1}{2}$.

From $\theta=30^{\circ}$ to $\theta=150^{\circ}, \sin \theta$ increases from $\frac{1}{2}$ to 1 and then decreases to $\frac{1}{2}$.

From $\theta=150^{\circ}$ to $\theta=180^{\circ}, \sin \theta$ decreases from $\frac{1}{2}$ to 0 .

Therefore, the original inequality is true exactly when $30^{\circ} \leq \theta \leq 150^{\circ}$ which is equivalent to $30^{\circ} \leq 10 k^{\circ} \leq 150^{\circ}$ and to $3 \leq k \leq 15$.

The integers $k$ in this range are $k=3,4,5,6, \ldots, 12,13,14,15$, of which there are 13 .

Final answer: 13
Subfield: Algebra


Question: Determine all values of $x$ such that $\log _{2 x}(48 \sqrt[3]{3})=\log _{3 x}(162 \sqrt[3]{2})$.

Solution:
Since the base of a logarithm must be positive and cannot equal 1 , then $x>0$ and $x \neq \frac{1}{2}$ and $x \neq \frac{1}{3}$.

This tells us that $\log 2 x$ and $\log 3 x$ exist and do not equal 0 , which we will need shortly when we apply the change of base formula.

We note further that $48=2^{4} \cdot 3$ and $162=3^{4} \cdot 2$ and $\sqrt[3]{3}=3^{1 / 3}$ and $\sqrt[3]{2}=2^{1 / 3}$. Using logarithm rules, the following equations are equivalent:

$$
\begin{aligned}
\log _{2 x}(48 \sqrt[3]{3}) & =\log _{3 x}(162 \sqrt[3]{2}) \\
\frac{\log \left(2^{4} \cdot 3 \cdot 3^{1 / 3}\right)}{\log 2 x} & =\frac{\log \left(3^{4} \cdot 2 \cdot 2^{1 / 3}\right)}{\log 3 x} \quad \text { (change of base formula) } \\
\frac{\log \left(2^{4} \cdot 3^{4 / 3}\right)}{\log 2+\log x} & =\frac{\log \left(3^{4} \cdot 2^{4 / 3}\right)}{\log 3+\log x} \quad(\log a b=\log a+\log b) \\
\frac{\log \left(2^{4}\right)+\log \left(3^{4 / 3}\right)}{\log 2+\log x} & =\frac{\log \left(3^{4}\right)+\log \left(2^{4 / 3}\right)}{\log 3+\log x} \quad(\log a b=\log a+\log b) \\
\frac{4 \log 2+\frac{4}{3} \log 3}{\log 2+\log x} & =\frac{4 \log 3+\frac{4}{3} \log 2}{\log 3+\log x} \quad\left(\log \left(a^{c}\right)=c \log a\right)
\end{aligned}
$$

Cross-multiplying, we obtain

$$
\left(4 \log 2+\frac{4}{3} \log 3\right)(\log 3+\log x)=\left(4 \log 3+\frac{4}{3} \log 2\right)(\log 2+\log x)
$$

Expanding the left side, we obtain

$$
4 \log 2 \log 3+\frac{4}{3}(\log 3)^{2}+\left(4 \log 2+\frac{4}{3} \log 3\right) \log x
$$

Expanding the right side, we obtain

$$
4 \log 3 \log 2+\frac{4}{3}(\log 2)^{2}+\left(4 \log 3+\frac{4}{3} \log 2\right) \log x
$$

Simplifying and factoring, we obtain the following equivalent equations:

$$
\begin{aligned}
\frac{4}{3}(\log 3)^{2}-\frac{4}{3}(\log 2)^{2} & =\log x\left(4 \log 3+\frac{4}{3} \log 2-4 \log 2-\frac{4}{3} \log 3\right) \\
\frac{4}{3}(\log 3)^{2}-\frac{4}{3}(\log 2)^{2} & =\log x\left(\frac{8}{3} \log 3-\frac{8}{3} \log 2\right) \\
(\log 3)^{2}-(\log 2)^{2} & =2 \log x(\log 3-\log 2) \\
\log x & =\frac{(\log 3)^{2}-(\log 2)^{2}}{2(\log 3-\log 2)} \\
\log x & =\frac{(\log 3-\log 2)(\log 3+\log 2)}{2(\log 3-\log 2)} \\
\log x & =\frac{\log 3+\log 2}{2} \\
\log x & =\frac{1}{2} \log 6 \\
\log x & =\log (\sqrt{6})
\end{aligned}
$$

and so $x=\sqrt{6}$.

Final answer: $\sqrt{6}$
Subfield: Geometry


Question: In the diagram, rectangle $P Q R S$ is placed inside rectangle $A B C D$ in two different ways: first, with $Q$ at $B$ and $R$ at $C$; second, with $P$ on $A B, Q$ on $B C, R$ on $C D$, and $S$ on $D A$.
![](https://cdn.mathpix.com/cropped/2023_12_21_96cde5457e2677371bf4g-1.jpg?height=448&width=874&top_left_y=909&top_left_x=690)

If $A B=718$ and $P Q=250$, determine the length of $B C$.

Solution:
Let $B C=x, P B=b$, and $B Q=a$.

Since $B C=x$, then $A D=P S=Q R=x$.

Since $B C=x$ and $B Q=a$, then $Q C=x-a$.

Since $A B=718$ and $P B=b$, then $A P=718-b$.

Note that $P Q=S R=250$.

Let $\angle B Q P=\theta$.

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

Since $B Q C$ is a straight angle and $\angle P Q R=90^{\circ}$, then $\angle R Q C=180^{\circ}-90^{\circ}-\theta=90^{\circ}-\theta$.

Since $A P B$ is a straight angle and $\angle S P Q=90^{\circ}$, then $\angle A P S=180^{\circ}-90^{\circ}-\left(90^{\circ}-\theta\right)=\theta$.

Since $\triangle S A P$ and $\triangle Q C R$ are each right-angled and have another angle in common with $\triangle P B Q$, then these

![](https://cdn.mathpix.com/cropped/2023_12_21_4b861614b923867311b5g-1.jpg?height=634&width=659&top_left_y=206&top_left_x=1362)
three triangles are similar.

Continuing in the same way, we can show that $\triangle R D S$ is also similar to these three triangles.

Since $R S=P Q$, then $\triangle R D S$ is actually congruent to $\triangle P B Q$ (angle-side-angle).

Similarly, $\triangle S A P$ is congruent to $\triangle Q C R$.

In particular, this means that $A S=x-a, S D=a, D R=b$, and $R C=718-b$.

Since $\triangle S A P$ and $\triangle P B Q$ are similar, then $\frac{S A}{P B}=\frac{A P}{B Q}=\frac{S P}{P Q}$.

Thus, $\frac{x-a}{b}=\frac{718-b}{a}=\frac{x}{250}$.

Also, by the Pythagorean Theorem in $\triangle P B Q$, we obtain $a^{2}+b^{2}=250^{2}$.

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

$$
\begin{aligned}
x^{2} & =(x-a)^{2}+(718-b)^{2} \\
x^{2} & =x^{2}-2 a x+a^{2}+(718-b)^{2} \\
0 & =-2 a x+a^{2}+(718-b)^{2}
\end{aligned}
$$

Since $a^{2}+b^{2}=250^{2}$, then $a^{2}=250^{2}-b^{2}$.

Since $\frac{718-b}{a}=\frac{x}{250}$, then $a x=250(718-b)$.

Therefore, substituting into $(*)$, we obtain

$$
\begin{aligned}
0 & =-2(250)(718-b)+250^{2}-b^{2}+(718-b)^{2} \\
b^{2} & =250^{2}-2(250)(718-b)+(718-b)^{2} \\
b^{2} & =((718-b)-250)^{2} \quad\left(\text { since } y^{2}-2 y z+z^{2}=(y-z)^{2}\right) \\
b^{2} & =(468-b)^{2} \\
b & =468-b \quad(\text { since } b \neq b-468) \\
2 b & =468 \\
b & =234
\end{aligned}
$$

Therefore, $a^{2}=250^{2}-b^{2}=250^{2}-234^{2}=(250+234)(250-234)=484 \cdot 16=22^{2} \cdot 4^{2}=88^{2}$ and so $a=88$.

Finally, $x=\frac{250(718-b)}{a}=\frac{250 \cdot 484}{88}=1375$. Therefore, $B C=1375$.

Final answer: 1375
Subfield: Geometry


Question: In the diagram, rectangle $P Q R S$ is placed inside rectangle $A B C D$ in two different ways: first, with $Q$ at $B$ and $R$ at $C$; second, with $P$ on $A B, Q$ on $B C, R$ on $C D$, and $S$ on $D A$.
<img_3269>

If $A B=718$ and $P Q=250$, determine the length of $B C$.

Solution:
Let $B C=x, P B=b$, and $B Q=a$.

Since $B C=x$, then $A D=P S=Q R=x$.

Since $B C=x$ and $B Q=a$, then $Q C=x-a$.

Since $A B=718$ and $P B=b$, then $A P=718-b$.

Note that $P Q=S R=250$.

Let $\angle B Q P=\theta$.

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

Since $B Q C$ is a straight angle and $\angle P Q R=90^{\circ}$, then $\angle R Q C=180^{\circ}-90^{\circ}-\theta=90^{\circ}-\theta$.

Since $A P B$ is a straight angle and $\angle S P Q=90^{\circ}$, then $\angle A P S=180^{\circ}-90^{\circ}-\left(90^{\circ}-\theta\right)=\theta$.

Since $\triangle S A P$ and $\triangle Q C R$ are each right-angled and have another angle in common with $\triangle P B Q$, then these

<img_3870>
three triangles are similar.

Continuing in the same way, we can show that $\triangle R D S$ is also similar to these three triangles.

Since $R S=P Q$, then $\triangle R D S$ is actually congruent to $\triangle P B Q$ (angle-side-angle).

Similarly, $\triangle S A P$ is congruent to $\triangle Q C R$.

In particular, this means that $A S=x-a, S D=a, D R=b$, and $R C=718-b$.

Since $\triangle S A P$ and $\triangle P B Q$ are similar, then $\frac{S A}{P B}=\frac{A P}{B Q}=\frac{S P}{P Q}$.

Thus, $\frac{x-a}{b}=\frac{718-b}{a}=\frac{x}{250}$.

Also, by the Pythagorean Theorem in $\triangle P B Q$, we obtain $a^{2}+b^{2}=250^{2}$.

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

$$
\begin{aligned}
x^{2} & =(x-a)^{2}+(718-b)^{2} \\
x^{2} & =x^{2}-2 a x+a^{2}+(718-b)^{2} \\
0 & =-2 a x+a^{2}+(718-b)^{2}
\end{aligned}
$$

Since $a^{2}+b^{2}=250^{2}$, then $a^{2}=250^{2}-b^{2}$.

Since $\frac{718-b}{a}=\frac{x}{250}$, then $a x=250(718-b)$.

Therefore, substituting into $(*)$, we obtain

$$
\begin{aligned}
0 & =-2(250)(718-b)+250^{2}-b^{2}+(718-b)^{2} \\
b^{2} & =250^{2}-2(250)(718-b)+(718-b)^{2} \\
b^{2} & =((718-b)-250)^{2} \quad\left(\text { since } y^{2}-2 y z+z^{2}=(y-z)^{2}\right) \\
b^{2} & =(468-b)^{2} \\
b & =468-b \quad(\text { since } b \neq b-468) \\
2 b & =468 \\
b & =234
\end{aligned}
$$

Therefore, $a^{2}=250^{2}-b^{2}=250^{2}-234^{2}=(250+234)(250-234)=484 \cdot 16=22^{2} \cdot 4^{2}=88^{2}$ and so $a=88$.

Finally, $x=\frac{250(718-b)}{a}=\frac{250 \cdot 484}{88}=1375$. Therefore, $B C=1375$.

Final answer: 1375
Subfield: Number Theory


Question: An L-shaped triomino is composed of three unit squares, as shown:

![](https://cdn.mathpix.com/cropped/2023_12_21_96cde5457e2677371bf4g-1.jpg?height=130&width=146&top_left_y=1553&top_left_x=1518)

Suppose that $H$ and $W$ are positive integers. An $H \times W$ rectangle can be tiled if the rectangle can be completely covered with non-overlapping copies of this triomino (each of which can be rotated and/or translated) and the sum of the areas of these non-overlapping triominos equals the area of the rectangle (that is, no triomino is partly outside the rectangle). If such a rectangle can be tiled, a tiling is a specific configuration of triominos that tile the rectangle.
Determine, with justification, all integers $W$ for which a $6 \times W$ rectangle can be tiled.

Solution:
First, we note that it is possible to tile each of a $3 \times 2$ and a $2 \times 3$ rectangle:

![](https://cdn.mathpix.com/cropped/2023_12_21_3d1f6c0f030c1ddc8175g-1.jpg?height=208&width=349&top_left_y=590&top_left_x=991)

Next, we note that it is not possible to tile a $6 \times 1$ rectangle because each of the triominos needs a width of at least 2 to be placed.

Finally, we show that it is possible to tile a $6 \times W$ rectangle for every integer $W \geq 2$.

To do this, we show that such a $6 \times W$ rectangle can be made up from $3 \times 2$ and $2 \times 3$ rectangles. Since each of these types of rectangles can be tiled with triominos, then the larger rectangle can be tiled with triominos by combining these tilings.

Case 1: $W$ is even

Suppose that $W=2 k$ for some positive integer $k$.

We build a $6 \times 2 k$ rectangle by placing $k 6 \times 2$ rectangles side by side.

Each $6 \times 2$ rectangle is built by stacking two $3 \times 2$ rectangles on top of each other.

![](https://cdn.mathpix.com/cropped/2023_12_21_3d1f6c0f030c1ddc8175g-1.jpg?height=409&width=525&top_left_y=1357&top_left_x=903)

Therefore, each such rectangle can be tiled.

Case 2: $W$ is odd, $W \geq 3$

Suppose that $W=2 k+1$ for some positive integer $k$.

We build a $6 \times(2 k+1)$ rectangle by building a $6 \times 3$ rectangle and then putting $k-1$ $6 \times 2$ rectangles next to it. Note that $k-1 \geq 0$ since $k \geq 1$ and that $2 k+1=3+2(k-1)$. The $6 \times 3$ rectangle is built by stacking three $2 \times 3$ rectangles on top of each other. Each $6 \times 2$ rectangle is built by stacking two $3 \times 2$ rectangles on top of each other.

![](https://cdn.mathpix.com/cropped/2023_12_21_3d1f6c0f030c1ddc8175g-1.jpg?height=415&width=697&top_left_y=2216&top_left_x=817)



Therefore, each such rectangle can be tiled.

Thus, a $6 \times W$ rectangle can be tiled with triominos exactly when $W \geq 2$.

Final answer: integer W when $W \geq 2$
Subfield: Algebra


Question: In an infinite array with two rows, the numbers in the top row are denoted $\ldots, A_{-2}, A_{-1}, A_{0}, A_{1}, A_{2}, \ldots$ and the numbers in the bottom row are denoted $\ldots, B_{-2}, B_{-1}, B_{0}, B_{1}, B_{2}, \ldots$ For each integer $k$, the entry $A_{k}$ is directly above the entry $B_{k}$ in the array, as shown:

| $\ldots$ | $A_{-2}$ | $A_{-1}$ | $A_{0}$ | $A_{1}$ | $A_{2}$ | $\ldots$ |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| $\ldots$ | $B_{-2}$ | $B_{-1}$ | $B_{0}$ | $B_{1}$ | $B_{2}$ | $\ldots$ |

For each integer $k, A_{k}$ is the average of the entry to its left, the entry to its right, and the entry below it; similarly, each entry $B_{k}$ is the average of the entry to its left, the entry to its right, and the entry above it.
In one such array, $A_{0}=A_{1}=A_{2}=0$ and $A_{3}=1$.

Determine the value of $A_{4}$.

Solution:
We draw part of the array using the information that $A_{0}=A_{1}=A_{2}=0$ and $A_{3}=1$ :

$$
\begin{array}{l|l|l|l|l|l|l|lll|c|c|c|c|c|c}
\cdots & A_{0} & A_{1} & A_{2} & A_{3} & A_{4} & A_{5} & \cdots & \cdots & 0 & 0 & 0 & 1 & A_{4} & A_{5} & \cdots \\
\hline \cdots & B_{0} & B_{1} & B_{2} & B_{3} & B_{4} & B_{5} & \cdots & \cdots & B_{0} & B_{1} & B_{2} & B_{3} & B_{4} & B_{5} & \cdots
\end{array}
$$

Since $A_{1}$ is the average of $A_{0}, B_{1}$ and $A_{2}$, then $A_{1}=\frac{A_{0}+B_{1}+A_{2}}{3}$ or $3 A_{1}=A_{0}+B_{1}+A_{2}$. Thus, $3(0)=0+B_{1}+0$ and so $B_{1}=0$.

Since $A_{2}$ is the average of $A_{1}, B_{2}$ and $A_{3}$, then $3 A_{2}=A_{1}+B_{2}+A_{3}$ and so $3(0)=0+B_{2}+1$ which gives $B_{2}=-1$.

Since $B_{2}$ is the average of $B_{1}, A_{2}$ and $B_{3}$, then $3 B_{2}=B_{1}+A_{2}+B_{3}$ and so $3(-1)=0+0+B_{3}$ which gives $B_{3}=-3$.

So far, this gives

$$
\begin{array}{l|c|c|c|c|c|c|l}
\cdots & 0 & 0 & 0 & 1 & A_{4} & A_{5} & \cdots \\
\hline \cdots & B_{0} & 0 & -1 & -3 & B_{4} & B_{5} & \cdots
\end{array}
$$

Since $A_{3}$ is the average of $A_{2}, B_{3}$ and $A_{4}$, then $3 A_{3}=A_{2}+B_{3}+A_{4}$ and so $3(1)=$ $0+(-3)+A_{4}$ which gives $A_{4}=6$.

Final answer: 6
Subfield: Algebra


Question: In an infinite array with two rows, the numbers in the top row are denoted $\ldots, A_{-2}, A_{-1}, A_{0}, A_{1}, A_{2}, \ldots$ and the numbers in the bottom row are denoted $\ldots, B_{-2}, B_{-1}, B_{0}, B_{1}, B_{2}, \ldots$ For each integer $k$, the entry $A_{k}$ is directly above the entry $B_{k}$ in the array, as shown:

| $\ldots$ | $A_{-2}$ | $A_{-1}$ | $A_{0}$ | $A_{1}$ | $A_{2}$ | $\ldots$ |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| $\ldots$ | $B_{-2}$ | $B_{-1}$ | $B_{0}$ | $B_{1}$ | $B_{2}$ | $\ldots$ |

For each integer $k, A_{k}$ is the average of the entry to its left, the entry to its right, and the entry below it; similarly, each entry $B_{k}$ is the average of the entry to its left, the entry to its right, and the entry above it.
In another such array, we define $S_{k}=A_{k}+B_{k}$ for each integer $k$.

Prove that $S_{k+1}=2 S_{k}-S_{k-1}$ for each integer $k$.

Solution:
We draw part of the array:

| $\cdots$ | $A_{k-1}$ | $A_{k}$ | $A_{k+1}$ | $\cdots$ |
| :--- | :--- | :--- | :--- | :--- |
| $\cdots$ | $B_{k-1}$ | $B_{k}$ | $B_{k+1}$ | $\cdots$ |

Then

$$
\begin{aligned}
3 S_{k} & =3 A_{k}+3 B_{k} \\
& =3\left(\frac{A_{k-1}+B_{k}+A_{k+1}}{3}\right)+3\left(\frac{B_{k-1}+A_{k}+B_{k+1}}{3}\right) \\
& =A_{k-1}+B_{k}+A_{k+1}+B_{k-1}+A_{k}+B_{k+1} \\
& =\left(A_{k-1}+B_{k-1}\right)+\left(A_{k}+B_{k}\right)+\left(A_{k+1}+B_{k+1}\right) \\
& =S_{k-1}+S_{k}+S_{k+1}
\end{aligned}
$$

Since $3 S_{k}=S_{k-1}+S_{k}+S_{k+1}$, then $S_{k+1}=2 S_{k}-S_{k-1}$.
Subfield: Geometry


Question: How many equilateral triangles of side $1 \mathrm{~cm}$, placed as shown in the diagram, are needed to completely cover the interior of an equilateral triangle of side $10 \mathrm{~cm}$ ?

<img_3923>

Solution:
If we proceed by pattern recognition, we find after row 1 we have a total of 1 triangle, after two rows we have $2^{2}$ or 4 triangles. After ten rows we have $10^{2}$ or 100 triangles.

<img_3802>
This solution is based on the fact that the ratio of areas for similar triangles is the square of the ratio of corresponding sides. Thus the big triangle with side length ten times that of the smaller triangle has 100 times the area.

Final answer: 100
Subfield: Geometry


Question: How many equilateral triangles of side $1 \mathrm{~cm}$, placed as shown in the diagram, are needed to completely cover the interior of an equilateral triangle of side $10 \mathrm{~cm}$ ?

![](https://cdn.mathpix.com/cropped/2023_12_21_2778c1b7df4d6803a6f0g-1.jpg?height=317&width=331&top_left_y=1988&top_left_x=1363)

Solution:
If we proceed by pattern recognition, we find after row 1 we have a total of 1 triangle, after two rows we have $2^{2}$ or 4 triangles. After ten rows we have $10^{2}$ or 100 triangles.

![](https://cdn.mathpix.com/cropped/2023_12_21_7c9cbbc4456b37ad4aa9g-1.jpg?height=358&width=377&top_left_y=217&top_left_x=1297)
This solution is based on the fact that the ratio of areas for similar triangles is the square of the ratio of corresponding sides. Thus the big triangle with side length ten times that of the smaller triangle has 100 times the area.

Final answer: 100
Subfield: Algebra


Question: The populations of Alphaville and Betaville were equal at the end of 1995. The population of Alphaville decreased by $2.9 \%$ during 1996, then increased by $8.9 \%$ during 1997 , and then increased by $6.9 \%$ during 1998 . The population of Betaville increased by $r \%$ in each of the three years. If the populations of the towns are equal at the end of 1998, determine the value of $r$ correct to one decimal place.

Solution:
If $P$ is the original population of Alphaville and Betaville,

$$
\begin{aligned}
P(.971)(1.089)(1.069) & =P\left(1+\frac{r}{100}\right)^{3} \\
1.1303 & =\left(1+\frac{r}{100}\right)^{3}
\end{aligned}
$$

From here,

Possibility 1

$$
\begin{aligned}
1+\frac{r}{100} & =(1.1303)^{\frac{1}{3}} \\
1+\frac{r}{100} & =1.0416 \\
r & \doteq 4.2 \%
\end{aligned}
$$

Or, Possibility 2

$$
\begin{aligned}
3 \log \left(1+\frac{r}{100}\right) & =\log 1.1303 \\
\log \left(1+\frac{r}{100}\right) & =.01773 \\
1+\frac{r}{100} & =1.0416 \\
r & \doteq 4.2 \%
\end{aligned}
$$

Final answer: 4.2 %
Subfield: Geometry


Question: A rectangle PQRS has side PQ on the x-axis and touches the graph of $y=k \cos x$ at the points $S$ and $R$ as shown. If the length of $P Q$ is $\frac{\pi}{3}$ and the area of the rectangle is $\frac{5 \pi}{3}$, what is the value of $k ?$

![](https://cdn.mathpix.com/cropped/2023_12_21_609033623119f30122a6g-1.jpg?height=474&width=526&top_left_y=782&top_left_x=1293)

Solution:
If $P Q=\frac{\pi}{3}$, then by symmetry the coordinates of $R$

are $\left(\frac{\pi}{6}, k \cos \frac{\pi}{6}\right)$.

Area of rectangle $P Q R S=\frac{\pi}{3}\left(k \cos \frac{\pi}{6}\right)=\frac{\pi}{3}(k)\left(\frac{\sqrt{3}}{2}\right)$

But $\frac{\sqrt{3} k \pi}{6}=\frac{5 \pi}{3} \quad \therefore k=\frac{10}{\sqrt{3}}$ or $\frac{10}{3} \sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2023_12_21_703c4ffbe2eb07053abcg-1.jpg?height=545&width=591&top_left_y=199&top_left_x=1298)

Final answer: $\frac{10}{\sqrt{3}}$,$\frac{10}{3} \sqrt{3}$
Subfield: Geometry


Question: A rectangle PQRS has side PQ on the x-axis and touches the graph of $y=k \cos x$ at the points $S$ and $R$ as shown. If the length of $P Q$ is $\frac{\pi}{3}$ and the area of the rectangle is $\frac{5 \pi}{3}$, what is the value of $k ?$

<img_3538>

Solution:
If $P Q=\frac{\pi}{3}$, then by symmetry the coordinates of $R$

are $\left(\frac{\pi}{6}, k \cos \frac{\pi}{6}\right)$.

Area of rectangle $P Q R S=\frac{\pi}{3}\left(k \cos \frac{\pi}{6}\right)=\frac{\pi}{3}(k)\left(\frac{\sqrt{3}}{2}\right)$

But $\frac{\sqrt{3} k \pi}{6}=\frac{5 \pi}{3} \quad \therefore k=\frac{10}{\sqrt{3}}$ or $\frac{10}{3} \sqrt{3}$.

<img_3853>

Final answer: $\frac{10}{\sqrt{3}}$,$\frac{10}{3} \sqrt{3}$
Subfield: Geometry


Question: In determining the height, $M N$, of a tower on an island, two points $A$ and $B, 100 \mathrm{~m}$ apart, are chosen on the same horizontal plane as $N$. If $\angle N A B=108^{\circ}$, $\angle A B N=47^{\circ}$ and $\angle M B N=32^{\circ}$, determine the height of the tower to the nearest metre.

![](https://cdn.mathpix.com/cropped/2023_12_21_609033623119f30122a6g-1.jpg?height=385&width=548&top_left_y=1274&top_left_x=1271)

Solution:
In $\triangle B A N, \angle B N A=25^{\circ}$

Using the Sine Law in $\triangle B A N$,

$\frac{N B}{\sin 108^{\circ}}=\frac{100}{\sin 25^{\circ}}$

Therefore $N B=\frac{100 \sin 108^{\circ}}{\sin 25^{\circ}} \approx 225.04$,

![](https://cdn.mathpix.com/cropped/2023_12_21_703c4ffbe2eb07053abcg-1.jpg?height=431&width=635&top_left_y=1308&top_left_x=1054)

Now in $\triangle M N B, \frac{M N}{N B}=\tan 32^{\circ}$

$$
M N=\frac{100 \sin 108^{\circ}}{\sin 25^{\circ}} \times \tan 32^{\circ} \doteq 140.6
$$

The tower is approximately $141 \mathrm{~m}$ high.

Final answer: 141 m
Subfield: Geometry


Question: In determining the height, $M N$, of a tower on an island, two points $A$ and $B, 100 \mathrm{~m}$ apart, are chosen on the same horizontal plane as $N$. If $\angle N A B=108^{\circ}$, $\angle A B N=47^{\circ}$ and $\angle M B N=32^{\circ}$, determine the height of the tower to the nearest metre.

<img_3998>

Solution:
In $\triangle B A N, \angle B N A=25^{\circ}$

Using the Sine Law in $\triangle B A N$,

$\frac{N B}{\sin 108^{\circ}}=\frac{100}{\sin 25^{\circ}}$

Therefore $N B=\frac{100 \sin 108^{\circ}}{\sin 25^{\circ}} \approx 225.04$,

<img_3946>

Now in $\triangle M N B, \frac{M N}{N B}=\tan 32^{\circ}$

$$
M N=\frac{100 \sin 108^{\circ}}{\sin 25^{\circ}} \times \tan 32^{\circ} \doteq 140.6
$$

The tower is approximately $141 \mathrm{~m}$ high.

Final answer: 141 m
Subfield: Geometry


Question: The points $A, P$ and a third point $Q$ (not shown) are the vertices of a triangle which is similar to triangle $A B C$. What are the coordinates of all possible positions for $Q$ ?

![](https://cdn.mathpix.com/cropped/2023_12_21_609033623119f30122a6g-1.jpg?height=642&width=677&top_left_y=1674&top_left_x=1144)

Solution:
$Q(4,0), Q(0,4)$

$Q(2,0), Q(0,2)$

$Q(-2,2), Q(2,-2)$

![](https://cdn.mathpix.com/cropped/2023_12_21_648ed28d6b9965167750g-1.jpg?height=737&width=781&top_left_y=1212&top_left_x=366)

Final answer: $(4,0),(0,4),(2,0),(0,2),(-2,2),(2,-2)$
Subfield: Geometry


Question: The points $A, P$ and a third point $Q$ (not shown) are the vertices of a triangle which is similar to triangle $A B C$. What are the coordinates of all possible positions for $Q$ ?

<img_3242>

Solution:
$Q(4,0), Q(0,4)$

$Q(2,0), Q(0,2)$

$Q(-2,2), Q(2,-2)$

<img_3757>

Final answer: $(4,0),(0,4),(2,0),(0,2),(-2,2),(2,-2)$
Subfield: Geometry


Question: Determine the coordinates of the points of intersection of the graphs of $y=\log _{10}(x-2)$ and $y=1-\log _{10}(x+1)$.

Solution:
The intersection takes place where,

$$
\begin{aligned}
& \log _{10}(x-2)=1-\log _{10}(x+1) \\
& \log _{10}(x-2)+\log _{10}(x+1)=1 \\
& \log _{10}\left(x^{2}-x-2\right)=1
\end{aligned}
$$



$$
\begin{aligned}
& x^{2}-x-2=10 \\
& x^{2}-x-12=0 \\
& (x-4)(x+3)=0 \\
& x=4 \text { or }-3
\end{aligned}
$$

For $x=-3, y$ is not defined.

For $x=4, y=\log _{10} 2 \doteq 0.3$.

The graphs therefore intersect at $\left(4, \log _{10} 2\right)$.

Final answer: $(4, \log _{10} 2)$