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


Question: Determine all real values of $x$ for which $\log _{2}\left(2^{x-1}+3^{x+1}\right)=2 x-\log _{2}\left(3^{x}\right)$.

Solution:
We successively manipulate the given equation to produce equivalent equations:

$$
\begin{aligned}
\log _{2}\left(2^{x-1}+3^{x+1}\right) & =2 x-\log _{2}\left(3^{x}\right) \\
\log _{2}\left(2^{x-1}+3^{x+1}\right)+\log _{2}\left(3^{x}\right) & =2 x \\
\log _{2}\left(\left(2^{x-1}+3^{x+1}\right) 3^{x}\right) & =2 x \quad\left(\text { using } \log _{2} A+\log _{2} B=\log _{2} A B\right) \\
\left(2^{x-1}+3^{x+1}\right) 3^{x} & =2^{2 x} \quad \text { (exponentiating both sides) } \\
2^{-1} 2^{x} 3^{x}+3^{1} 3^{x} 3^{x} & =2^{2 x} \\
\frac{1}{2} \cdot 2^{x} 3^{x}+3 \cdot 3^{2 x} & =2^{2 x} \\
2^{x} 3^{x}+6 \cdot 3^{2 x} & \left.=2 \cdot 2^{2 x} \quad \text { (multiplying by } 2\right) \\
2^{x} 3^{x}+6 \cdot\left(3^{x}\right)^{2} & =2 \cdot\left(2^{x}\right)^{2}
\end{aligned}
$$

Next, we make the substitution $a=2^{x}$ and $b=3^{x}$.

This gives $a b+6 b^{2}=2 a^{2}$ or $2 a^{2}-a b-6 b^{2}=0$.

Factoring, we obtain $(a-2 b)(2 a+3 b)=0$.

Therefore, $a=2 b$ or $2 a=-3 b$.

Since $a>0$ and $b>0$, then $a=2 b$ which gives $2^{x}=2 \cdot 3^{x}$.

Taking $\log$ of both sides, we obtain $x \log 2=\log 2+x \log 3$ and so $x(\log 2-\log 3)=\log 2$ or $x=\frac{\log 2}{\log 2-\log 3}$.
We successively manipulate the given equation to produce equivalent equations:

$$
\begin{aligned}
\log _{2}\left(2^{x-1}+3^{x+1}\right) & =2 x-\log _{2}\left(3^{x}\right) \\
\log _{2}\left(2^{x-1}+3^{x+1}\right)+\log _{2}\left(3^{x}\right) & =2 x \\
\log _{2}\left(\left(2^{x-1}+3^{x+1}\right) 3^{x}\right) & =2 x \quad\left(\text { using } \log _{2} A+\log _{2} B=\log _{2} A B\right) \\
\left(2^{x-1}+3^{x+1}\right) 3^{x} & =2^{2 x} \quad \text { (exponentiating both sides) } \\
2^{-1} 2^{x} 3^{x}+3^{1} 3^{x} 3^{x} & =2^{2 x} \\
\frac{1}{2} \cdot 2^{x} 3^{x}+3 \cdot 3^{2 x} & =2^{2 x} \\
2^{x} 3^{x}+6 \cdot 3^{2 x} & \left.=2 \cdot 2^{2 x} \quad \text { (multiplying by } 2\right) \\
2^{x} 3^{x} 2^{-2 x}+6 \cdot 3^{2 x} 2^{-2 x} & \left.=2 \quad \text { (dividing both sides by } 2^{2 x} \neq 0\right) \\
2^{-x} 3^{x}+6 \cdot 3^{2 x} 2^{-2 x} & =2 \\
\left(\frac{3}{2}\right)^{x}+6\left(\frac{3}{2}\right)^{2 x} & =2
\end{aligned}
$$

Next, we make the substitution $t=\left(\frac{3}{2}\right)^{x}$, noting that $\left(\frac{3}{2}\right)^{2 x}=\left(\left(\frac{3}{2}\right)^{x}\right)^{2}=t^{2}$.

Thus, we obtain the equivalent equations

$$
\begin{aligned}
t+6 t^{2} & =2 \\
6 t^{2}+t-2 & =0 \\
(3 t+2)(2 t-1) & =0
\end{aligned}
$$

Therefore, $t=-\frac{2}{3}$ or $t=\frac{1}{2}$.

Since $t=\left(\frac{3}{2}\right)^{x}>0$, then we must have $t=\left(\frac{3}{2}\right)^{x}=\frac{1}{2}$.

Thus,

$$
x=\log _{3 / 2}(1 / 2)=\frac{\log (1 / 2)}{\log (3 / 2)}=\frac{\log 1-\log 2}{\log 3-\log 2}=\frac{-\log 2}{\log 3-\log 2}=\frac{\log 2}{\log 2-\log 3}
$$

Final answer: $\frac{\log 2}{\log 2-\log 3}$
Subfield: Combinatorics


Question: A multiplicative partition of a positive integer $n \geq 2$ is a way of writing $n$ as a product of one or more integers, each greater than 1. Note that we consider a positive integer to be a multiplicative partition of itself. Also, the order of the factors in a partition does not matter; for example, $2 \times 3 \times 5$ and $2 \times 5 \times 3$ are considered to be the same partition of 30 . For each positive integer $n \geq 2$, define $P(n)$ to be the number of multiplicative partitions of $n$. We also define $P(1)=1$. Note that $P(40)=7$, since the multiplicative partitions of 40 are $40,2 \times 20,4 \times 10$, $5 \times 8,2 \times 2 \times 10,2 \times 4 \times 5$, and $2 \times 2 \times 2 \times 5$.

(In each part, we use "partition" to mean "multiplicative partition". We also call the numbers being multiplied together in a given partition the "parts" of the partition.)
Determine the value of $P(64)$.

Solution:
We determine the multiplicative partitions of 64 by considering the number of parts in the various partitions. Note that 64 is a power of 2 so any divisor of 64 is also a power of 2 . In each partition, since the order of parts is not important, we list the parts in increasing order to make it easier to systematically find all of these.

* One part. There is one possibility: 64.
* Two parts. There are three possibilities: $64=2 \times 32=4 \times 16=8 \times 8$.
* Three parts. We start with the smallest possible first and second parts. We keep the first part fixed while adjusting the second and third parts. We then increase the first part and repeat.

We get: $64=2 \times 2 \times 16=2 \times 4 \times 8=4 \times 4 \times 4$.

* Four parts. A partition of 64 with four parts must include at least two $2 \mathrm{~s}$, since if it didn't, it would include at least three parts that are at least 4 , and so would be too large. With two $2 \mathrm{~s}$, the remaining two parts have a product of 16 .

We get: $64=2 \times 2 \times 2 \times 8=2 \times 2 \times 4 \times 4$.

* Five parts. A partition of 64 with five parts must include at least three $2 \mathrm{~s}$, since if it didn't, it would include at least three parts that are at least 4 , and so would be too large. With three $2 \mathrm{~s}$, the remaining two parts have a product of 8 .

We get: $64=2 \times 2 \times 2 \times 2 \times 4$.

$*$ Six parts. Since $64=2^{6}$, there is only one possibility: $64=2 \times 2 \times 2 \times 2 \times 2 \times 2$.

Therefore, $P(64)=1+3+3+2+1+1=11$.

Final answer: 11
Subfield: Geometry


Question: In the diagram, triangle ABC is right-angled at B. MT is the perpendicular bisector of $B C$ with $M$ on $B C$ and $T$ on $A C$. If $A T=A B$, what is the size of $\angle A C B$ ?

<img_3341>

Solution:
Since $M T$ is the perpendicular bisector of $B C$, then

$B M=M C$, and $T M$ is perpendicular to $B C$.

Therefore, $\triangle C M T$ is similar to $\triangle C B A$, since they share a common angle and each have a right angle.

<img_3335>

But $\frac{C M}{C B}=\frac{1}{2}$ so $\frac{C T}{C A}=\frac{C M}{C B}=\frac{1}{2}$, and thus $C T=A T=A B$, ie. $\frac{A B}{A C}=\frac{1}{2}$ or $\sin (\angle A C B)=\frac{1}{2}$.

Therefore, $\angle A C B=30^{\circ}$.
Since $T M \| A B$, and $C M=M B$, then $C T=T A=A B$.

Join $T$ to $B$.

Since $\angle A B C=90^{\circ}$, then $A C$ is the diameter of a circle passing through $A, C$ and $B$, with $T$ as its centre.

<img_3228>

Thus, $T A=A B=B T$ (all radii), and so $\triangle A B T$ is equilateral. Therefore, $\angle B A C=60^{\circ}$, and so $\angle A C B=30^{\circ}$.

Final answer: $30^{\circ}$
Subfield: Geometry


Question: In the diagram, triangle ABC is right-angled at B. MT is the perpendicular bisector of $B C$ with $M$ on $B C$ and $T$ on $A C$. If $A T=A B$, what is the size of $\angle A C B$ ?

![](https://cdn.mathpix.com/cropped/2023_12_21_d7fe03f50cac4fc44fe1g-1.jpg?height=227&width=415&top_left_y=209&top_left_x=1386)

Solution:
Since $M T$ is the perpendicular bisector of $B C$, then

$B M=M C$, and $T M$ is perpendicular to $B C$.

Therefore, $\triangle C M T$ is similar to $\triangle C B A$, since they share a common angle and each have a right angle.

![](https://cdn.mathpix.com/cropped/2023_12_21_0c20f2050e7a1226112ag-1.jpg?height=253&width=488&top_left_y=833&top_left_x=1407)

But $\frac{C M}{C B}=\frac{1}{2}$ so $\frac{C T}{C A}=\frac{C M}{C B}=\frac{1}{2}$, and thus $C T=A T=A B$, ie. $\frac{A B}{A C}=\frac{1}{2}$ or $\sin (\angle A C B)=\frac{1}{2}$.

Therefore, $\angle A C B=30^{\circ}$.
Since $T M \| A B$, and $C M=M B$, then $C T=T A=A B$.

Join $T$ to $B$.

Since $\angle A B C=90^{\circ}$, then $A C$ is the diameter of a circle passing through $A, C$ and $B$, with $T$ as its centre.

![](https://cdn.mathpix.com/cropped/2023_12_21_0c20f2050e7a1226112ag-1.jpg?height=260&width=488&top_left_y=1431&top_left_x=1407)

Thus, $T A=A B=B T$ (all radii), and so $\triangle A B T$ is equilateral. Therefore, $\angle B A C=60^{\circ}$, and so $\angle A C B=30^{\circ}$.

Final answer: $30^{\circ}$
Subfield: Algebra


Question: What are all values of $x$ such that

$$
\log _{5}(x+3)+\log _{5}(x-1)=1 ?
$$

Solution:
Combining the logarithms,

$$
\begin{aligned}
\log _{5}(x+3)+\log _{5}(x-1) & =1 \\
\log _{5}((x+3)(x-1)) & =1 \\
\log _{5}\left(x^{2}+2 x-3\right) & =1 \\
x^{2}+2 x-3 & =5 \\
x^{2}+2 x-8 & =0 \\
(x+4)(x-2) & =0
\end{aligned}
$$

Therefore, $x=-4$ or $x=2$. Substituting the two values for $x$ back into the original equation, we see that $x=2$ works, but that $x=-4$ does not, since we cannot take the logarithm of a negative number.

Final answer: 2
Subfield: Algebra


Question: A chef aboard a luxury liner wants to cook a goose. The time $t$ in hours to cook a goose at $180^{\circ} \mathrm{C}$ depends on the mass of the goose $m$ in kilograms according to the formula

$$
t=a m^{b}
$$

where $a$ and $b$ are constants. The table below gives the times observed to cook a goose at $180^{\circ} \mathrm{C}$.

| Mass, $m(\mathrm{~kg})$ | Time, $t(\mathrm{~h})$ |
| :---: | :---: |
| 3.00 | 2.75 |
| 6.00 | 3.75 |
Using the data in the table, determine both $a$ and $b$ to two decimal places.

Solution:
From the table we have two pieces of information, so we substitute both of these into the given formula.

$$
\begin{aligned}
& 2.75=a(3.00)^{b} \\
& 3.75=a(6.00)^{b}
\end{aligned}
$$

We can now proceed in either of two ways to solve for $b$.

Method 1 to find $b$

Dividing the second equation by the first, we obtain

$$
\frac{3.75}{2.75}=\frac{a(6.00)^{b}}{a(3.00)^{b}}=\frac{(6.00)^{b}}{(3.00)^{b}}=\left(\frac{6.00}{3.00}\right)^{b}=2^{b}
$$

or

$$
2^{b} \approx 1.363636
$$

Taking logarithms of both sides,



$$
\begin{aligned}
\log \left(2^{b}\right) & \approx \log (1.363636) \\
b \log (2) & \approx \log (1.363636) \\
b & \approx \frac{\log (1.363636)}{\log (2)} \\
b & \approx 0.4475
\end{aligned}
$$

Method 2 to find $b$ 

Taking logarithms of both sides of the above equations, we obtain

$$
\begin{aligned}
\log (2.75) & =\log \left(a(3.00)^{b}\right) \\
& =\log (a)+\log \left((3.00)^{b}\right) \\
& =\log (a)+b \log (3.00)
\end{aligned}
$$

Similarly,

$$
\log (3.75)=\log (a)+b \log (6.00)
$$

Subtracting the first equation from the second, we obtain

$$
\begin{aligned}
\log (3.75)-\log (2.75) & =b(\log (6.00)-\log (3.00)) \\
b & =\frac{\log (3.75)-\log (2.75)}{\log (6.00)-\log (3.00)} \\
b & \approx 0.4475
\end{aligned}
$$

We now continue in the same way for both methods.

Substituting this value for $b$ back into the first equation above,

$$
\begin{aligned}
2.75 & \approx a(3.00)^{0.4475} \\
a & \approx \frac{2.75}{(3.00)^{0.4475}} \\
a & \approx 1.6820
\end{aligned}
$$

Therefore, to two decimal places, $a=1.68$ and $b=0.45$.

Final answer: $1.68,0.45$
Subfield: Geometry


Question: In the diagram, $A B C D E F$ is a regular hexagon with a side
length of 10 . If $X, Y$ and $Z$ are the midpoints of $A B, C D$ and $E F$, respectively, what is the length of $X Z$ ?

![](https://cdn.mathpix.com/cropped/2023_12_21_1d3f148ee017cf69bf7eg-1.jpg?height=260&width=301&top_left_y=214&top_left_x=1432)

Solution:
Extend $X A$ and $Z F$ to meet at point $T$.

By symmetry, $\angle A X Z=\angle F Z X=60^{\circ}$ and $\angle T A F=\angle T F A=60^{\circ}$, and so $\triangle T A F$ and $\triangle T X Z$ are both equilateral triangles.

Since $A F=10$, then $T A=10$, which means

$T X=10+5=15$, and so $X Z=T X=15$.

![](https://cdn.mathpix.com/cropped/2023_12_21_a74c1425f4891f5bece8g-1.jpg?height=365&width=455&top_left_y=365&top_left_x=1407)
We look at the quadrilateral $A X Z F$.

Since $A B C D E F$ is a regular hexagon, then $\angle F A X=\angle A F Z=120^{\circ}$.

Note that $A F=10$, and also $A X=F Z=5$ since $X$ and $Z$ are midpoints of their respective sides.

![](https://cdn.mathpix.com/cropped/2023_12_21_a74c1425f4891f5bece8g-1.jpg?height=238&width=510&top_left_y=846&top_left_x=1320)

By symmetry, $\angle A X Z=\angle F Z X=60^{\circ}$, and so $A X Z F$ is a trapezoid.

Drop perpendiculars from $A$ and $F$ to $P$ and $Q$, respectively, on $X Z$.

By symmetry again, $P X=Q Z$. Now, $P X=A X \cos 60^{\circ}=5\left(\frac{1}{2}\right)=\frac{5}{2}$.

Since $A P Q F$ is a rectangle, then $P Q=10$.

Therefore, $X Z=X P+P Q+Q Z=\frac{5}{2}+10+\frac{5}{2}=15$.

Final answer: 15
Subfield: Geometry


Question: In the diagram, $A B C D E F$ is a regular hexagon with a side
length of 10 . If $X, Y$ and $Z$ are the midpoints of $A B, C D$ and $E F$, respectively, what is the length of $X Z$ ?

<img_3462>

Solution:
Extend $X A$ and $Z F$ to meet at point $T$.

By symmetry, $\angle A X Z=\angle F Z X=60^{\circ}$ and $\angle T A F=\angle T F A=60^{\circ}$, and so $\triangle T A F$ and $\triangle T X Z$ are both equilateral triangles.

Since $A F=10$, then $T A=10$, which means

$T X=10+5=15$, and so $X Z=T X=15$.

<img_3285>
We look at the quadrilateral $A X Z F$.

Since $A B C D E F$ is a regular hexagon, then $\angle F A X=\angle A F Z=120^{\circ}$.

Note that $A F=10$, and also $A X=F Z=5$ since $X$ and $Z$ are midpoints of their respective sides.

<img_3509>

By symmetry, $\angle A X Z=\angle F Z X=60^{\circ}$, and so $A X Z F$ is a trapezoid.

Drop perpendiculars from $A$ and $F$ to $P$ and $Q$, respectively, on $X Z$.

By symmetry again, $P X=Q Z$. Now, $P X=A X \cos 60^{\circ}=5\left(\frac{1}{2}\right)=\frac{5}{2}$.

Since $A P Q F$ is a rectangle, then $P Q=10$.

Therefore, $X Z=X P+P Q+Q Z=\frac{5}{2}+10+\frac{5}{2}=15$.

Final answer: 15
Subfield: Geometry


Question: A circle passes through the origin and the points of intersection of the parabolas $y=x^{2}-3$ and $y=-x^{2}-2 x+9$. Determine the coordinates of the centre of this circle.

Solution:
We first determine the three points through which the circle passes.

The first point is the origin $(0,0)$.

The second and third points are found by determining the points of intersection of the two parabolas $y=x^{2}-3$ and $y=-x^{2}-2 x+9$. We do this by setting the $y$ values equal.

$$
x^{2}-3=-x^{2}-2 x+9
$$

$2 x^{2}+2 x-12=0$

$x^{2}+x-6=0$

$(x+3)(x-2)=0$

so $x=-3$ or $x=2$.



We determine the points of intersection by substituting into the first parabola.

If $x=2, y=2^{2}-3=1$, so the point of intersection is $(2,1)$.

If $x=-3, y=(-3)^{2}-3=6$, so the point of intersection is $(-3,6)$.

Therefore, the circle passes through the three points $A(0,0), B(2,1)$ and $C(-3,6)$.

Let the centre of the circle be the point $Q(a, b)$.

<img_4046>

Finding the centre of the circle can be done in a variety of ways.

We use the fact $Q$ is of equal distance from each of the points $A, B$ and $C$. In particular $Q A^{2}=Q B^{2}=Q C^{2}$ or $x^{2}+y^{2}=(x-2)^{2}+(y-1)^{2}=(x+3)^{2}+(y-6)^{2}$

From the first equality,

$$
\begin{aligned}
& x^{2}+y^{2}=(x-2)^{2}+(y-1)^{2} \\
& 4 x+2 y=5
\end{aligned}
$$

<img_4000>



From the second equality,

$$
\begin{aligned}
(x-2)^{2}+(y-1)^{2} & =(x+3)^{2}+(y-6)^{2} \\
-10 x+10 y & =40 \\
y & =x+4
\end{aligned}
$$

Substituting the equation above into into $4 x+2 y=5$, we obtain $4 x+2(x+4)=5$ or $6 x=-3$ or $x=-\frac{1}{2}$. Thus, $y=-\frac{1}{2}+4=\frac{7}{2}$, and so the centre of the circle is $\left(-\frac{1}{2}, \frac{7}{2}\right)$.

Final answer: $(-\frac{1}{2}, \frac{7}{2})$
Subfield: Geometry


Question: In the diagram, $A C=2 x, B C=2 x+1$ and $\angle A C B=30^{\circ}$. If the area of $\triangle A B C$ is 18 , what is the value of $x$ ?

![](https://cdn.mathpix.com/cropped/2023_12_21_1d3f148ee017cf69bf7eg-1.jpg?height=239&width=314&top_left_y=691&top_left_x=1407)

Solution:
Using a known formula for the area of a triangle, $A=\frac{1}{2} a b \sin C$,

$$
\begin{aligned}
18 & =\frac{1}{2}(2 x+1)(2 x) \sin 30^{\circ} \\
36 & =(2 x+1)(2 x)\left(\frac{1}{2}\right) \\
0 & =2 x^{2}+x-36 \\
0 & =(2 x+9)(x-4)
\end{aligned}
$$

and so $x=4$ or $x=-\frac{9}{2}$. Since $x$ is positive, then $x=4$.
Draw a perpendicular from $A$ to $P$ on $B C$.

![](https://cdn.mathpix.com/cropped/2023_12_21_86d8b0958ad843aa94fbg-1.jpg?height=298&width=369&top_left_y=369&top_left_x=1363)

Using $\triangle A P C, A P=A C \sin 30^{\circ}=2 x\left(\frac{1}{2}\right)=x$.

Now $A P$ is the height of $\triangle A B C$, so Area $=\frac{1}{2}(B C)(A P)$.

Then

$$
\begin{aligned}
18 & =\frac{1}{2}(2 x+1)(x) \\
0 & =2 x^{2}+x-36 \\
0 & =(2 x+9)(x-4)
\end{aligned}
$$

and so $x=4$ or $x=-\frac{9}{2}$.

Since $x$ is positive, then $x=4$.

Final answer: 4
Subfield: Geometry


Question: In the diagram, $A C=2 x, B C=2 x+1$ and $\angle A C B=30^{\circ}$. If the area of $\triangle A B C$ is 18 , what is the value of $x$ ?

<img_3200>

Solution:
Using a known formula for the area of a triangle, $A=\frac{1}{2} a b \sin C$,

$$
\begin{aligned}
18 & =\frac{1}{2}(2 x+1)(2 x) \sin 30^{\circ} \\
36 & =(2 x+1)(2 x)\left(\frac{1}{2}\right) \\
0 & =2 x^{2}+x-36 \\
0 & =(2 x+9)(x-4)
\end{aligned}
$$

and so $x=4$ or $x=-\frac{9}{2}$. Since $x$ is positive, then $x=4$.
Draw a perpendicular from $A$ to $P$ on $B C$.

<img_3390>

Using $\triangle A P C, A P=A C \sin 30^{\circ}=2 x\left(\frac{1}{2}\right)=x$.

Now $A P$ is the height of $\triangle A B C$, so Area $=\frac{1}{2}(B C)(A P)$.

Then

$$
\begin{aligned}
18 & =\frac{1}{2}(2 x+1)(x) \\
0 & =2 x^{2}+x-36 \\
0 & =(2 x+9)(x-4)
\end{aligned}
$$

and so $x=4$ or $x=-\frac{9}{2}$.

Since $x$ is positive, then $x=4$.

Final answer: 4
Subfield: Geometry


Question: A ladder, $A B$, is positioned so that its bottom sits on horizontal ground and its top rests against a vertical wall, as shown. In this initial position, the ladder makes an angle of $70^{\circ}$ with the horizontal. The bottom of the ladder is then pushed $0.5 \mathrm{~m}$ away from the wall, moving the ladder to position $A^{\prime} B^{\prime}$. In this new position, the ladder makes an angle of $55^{\circ}$ with the horizontal. Calculate, to the nearest centimetre, the distance that the ladder slides down the wall (that is, the length of $B B^{\prime}$ ).

![](https://cdn.mathpix.com/cropped/2023_12_21_1d3f148ee017cf69bf7eg-1.jpg?height=499&width=331&top_left_y=992&top_left_x=1401)

Solution:
Let the length of the ladder be $L$.

Then $A C=L \cos 70^{\circ}$ and $B C=L \sin 70^{\circ}$. Also, $A^{\prime} C=L \cos 55^{\circ}$ and $B^{\prime} C=L \sin 55^{\circ}$.

Since $A^{\prime} A=0.5$, then

$$
0.5=L \cos 55^{\circ}-L \cos 70^{\circ}
$$

$$
L=\frac{0.5}{\cos 55^{\circ}-\cos 70^{\circ}}
$$

Therefore,

$$
\begin{aligned}
B B^{\prime} & =B C-B^{\prime} C \\
& =L \sin 70^{\circ}-L \sin 55^{\circ} \\
& =L\left(\sin 70^{\circ}-\sin 55^{\circ}\right) \\
& =\frac{(0.5)\left(\sin 70^{\circ}-\sin 55^{\circ}\right)}{\left(\cos 55^{\circ}-\cos 70^{\circ}\right)} \quad(\text { from }(*)) \\
& \approx 0.2603 \mathrm{~m}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2023_12_21_86d8b0958ad843aa94fbg-1.jpg?height=686&width=634&top_left_y=1039&top_left_x=1231)

Therefore, to the nearest centimetre, the distance that the ladder slides down the wall is $26 \mathrm{~cm}$.

Final answer: 26 cm
Subfield: Geometry


Question: A ladder, $A B$, is positioned so that its bottom sits on horizontal ground and its top rests against a vertical wall, as shown. In this initial position, the ladder makes an angle of $70^{\circ}$ with the horizontal. The bottom of the ladder is then pushed $0.5 \mathrm{~m}$ away from the wall, moving the ladder to position $A^{\prime} B^{\prime}$. In this new position, the ladder makes an angle of $55^{\circ}$ with the horizontal. Calculate, to the nearest centimetre, the distance that the ladder slides down the wall (that is, the length of $B B^{\prime}$ ).

<img_3632>

Solution:
Let the length of the ladder be $L$.

Then $A C=L \cos 70^{\circ}$ and $B C=L \sin 70^{\circ}$. Also, $A^{\prime} C=L \cos 55^{\circ}$ and $B^{\prime} C=L \sin 55^{\circ}$.

Since $A^{\prime} A=0.5$, then

$$
0.5=L \cos 55^{\circ}-L \cos 70^{\circ}
$$

$$
L=\frac{0.5}{\cos 55^{\circ}-\cos 70^{\circ}}
$$

Therefore,

$$
\begin{aligned}
B B^{\prime} & =B C-B^{\prime} C \\
& =L \sin 70^{\circ}-L \sin 55^{\circ} \\
& =L\left(\sin 70^{\circ}-\sin 55^{\circ}\right) \\
& =\frac{(0.5)\left(\sin 70^{\circ}-\sin 55^{\circ}\right)}{\left(\cos 55^{\circ}-\cos 70^{\circ}\right)} \quad(\text { from }(*)) \\
& \approx 0.2603 \mathrm{~m}
\end{aligned}
$$

<img_3214>

Therefore, to the nearest centimetre, the distance that the ladder slides down the wall is $26 \mathrm{~cm}$.

Final answer: 26 cm
Subfield: Algebra


Question: In a soccer league with 5 teams, each team plays 20 games(that is, 5 games with each of the other 4 teams). For each team, every game ends in a win (W), a loss (L), or a tie (T). The numbers of wins, losses and ties for each team at the end of the season are shown in the table. Determine the values of $x, y$ and $z$.

| Team | W | L | T |
| :---: | ---: | ---: | ---: |
| A | 2 | 15 | 3 |
| B | 7 | 9 | 4 |
| C | 6 | 12 | 2 |
| D | 10 | 8 | 2 |
| E | $x$ | $y$ | $z$ |

Solution:
In total, there are $\frac{1}{2} \times 5 \times 20=50$ games played, since each of 5 teams plays 20 games (we divide by 2 since each game is double-counted).

In each game, there is either a loss or a tie.

The number of games with a loss is $44+y$ from the second column, and the number of games with a tie is $\frac{1}{2}(11+z)$ (since any game ending in a tie has 2 ties).



So

$$
\begin{aligned}
50 & =44+y+\frac{1}{2}(11+z) \\
100 & =88+2 y+11+z \\
1 & =2 y+z
\end{aligned}
$$

Since $y$ and $z$ are non-negative integers, $z=1$ and $y=0$. So $x=19$ since Team E plays 20 games.
In any game played, the final result is either both teams earning a tie, or one team earning a win, and the other getting a loss. Therefore, the total number of wins among all teams equals the total number of losses, ie.

$$
\begin{aligned}
25+x & =44+y \\
x-y & =19
\end{aligned}
$$

Also, since team E plays 20 games, then

$$
x+y+z=20
$$

So from (1), $x$ must be at least 19, and from (2), $x$ can be at most 20.

Lastly, we know that the total of all of the teams numbers of ties must be even, ie. $11+z$ is even, ie. $z$ is odd.

Since $x$ is at least 19, then $z$ can be at most 1 by (2).

Therefore, $z=1$. Thus, $x=19$ and $y=0$.
In any game played, the final result is either both teams earning a tie, or one team earning a win, and the other getting a loss. Therefore, the total number of wins among all teams equals the total number of losses, ie.

$$
\begin{aligned}
25+x & =44+y \\
x-y & =19
\end{aligned}
\tag{1}
$$

Also, since team E plays 20 games, then

$$
x+y+z=20
\tag{2}
$$

So from (1), $x$ must be at least 19, and from (2), $x$ can be at most 20.

Consider the possibility that $x=20$. From (2), then $y=z=0$, which does not agree with (1).

Thus, the only possibility is $x=19$. From (1), $y=0$, and so $z=1$ from (2). (These three values agree with both equations (1) and (2).)

Final answer: 19,0,1