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


Question: The geometric sequence with $n$ terms $t_{1}, t_{2}, \ldots, t_{n-1}, t_{n}$ has $t_{1} t_{n}=3$. Also, the product of all $n$ terms equals 59049 (that is, $t_{1} t_{2} \cdots t_{n-1} t_{n}=59049$ ). Determine the value of $n$.

(A geometric sequence is a sequence in which each term after the first is obtained from the previous term by multiplying it by a constant. For example, $3,6,12$ is a geometric sequence with three terms.)

Solution:
Suppose that the first term in the geometric sequence is $t_{1}=a$ and the common ratio in the sequence is $r$.

Then the sequence, which has $n$ terms, is $a, a r, a r^{2}, a r^{3}, \ldots, a r^{n-1}$.

In general, the $k$ th term is $t_{k}=a r^{k-1}$; in particular, the $n$th term is $t_{n}=a r^{n-1}$.

Since $t_{1} t_{n}=3$, then $a \cdot a r^{n-1}=3$ or $a^{2} r^{n-1}=3$.

Since $t_{1} t_{2} \cdots t_{n-1} t_{n}=59049$, then

$$
\begin{aligned}
(a)(a r) \cdots\left(a r^{n-2}\right)\left(a r^{n-1}\right) & =59049 \\
a^{n} r r^{2} \cdots r^{n-2} r^{n-1} & =59049 \\
a^{n} r^{1+2+\cdots+(n-2)+(n-1)} & =59049 \\
a^{n} r^{\frac{1}{2}(n-1)(n)} & =59049
\end{aligned}
$$

$$
a^{n} r r^{2} \cdots r^{n-2} r^{n-1}=59049 \quad \text { (since there are } n \text { factors of } a \text { on the left side) }
$$

since $1+2+\cdots+(n-2)+(n-1)=\frac{1}{2}(n-1)(n)$.

Since $a^{2} r^{n-1}=3$, then $\left(a^{2} r^{n-1}\right)^{n}=3^{n}$ or $a^{2 n} r^{(n-1)(n)}=3^{n}$.

Since $a^{n} r^{\frac{1}{2}(n-1)(n)}=59049$, then $\left(a^{n} r^{\frac{1}{2}(n-1)(n)}\right)^{2}=59049^{2}$ or $a^{2 n} r^{(n-1)(n)}=59049^{2}$.

Since the left sides of these equations are the same, then $3^{n}=59049^{2}$.

Now

$$
59049=3(19683)=3^{2}(6561)=3^{3}(2187)=3^{4}(729)=3^{5}(243)=3^{6}(81)=3^{6} 3^{4}=3^{10}
$$

Since $59049=3^{10}$, then $59049^{2}=3^{20}$ and so $3^{n}=3^{20}$, which gives $n=20$.

Final answer: 20
Subfield: Algebra


Question: If $\frac{(x-2013)(y-2014)}{(x-2013)^{2}+(y-2014)^{2}}=-\frac{1}{2}$, what is the value of $x+y$ ?

Solution:
Let $a=x-2013$ and let $b=y-2014$.

The given equation becomes $\frac{a b}{a^{2}+b^{2}}=-\frac{1}{2}$, which is equivalent to $2 a b=-a^{2}-b^{2}$ and $a^{2}+2 a b+b^{2}=0$.

This is equivalent to $(a+b)^{2}=0$ which is equivalent to $a+b=0$.

Since $a=x-2013$ and $b=y-2014$, then $x-2013+y-2014=0$ or $x+y=4027$.

Final answer: 4027
Subfield: Algebra


Question: Determine all real numbers $x$ for which

$$
\left(\log _{10} x\right)^{\log _{10}\left(\log _{10} x\right)}=10000
$$

Solution:
Let $a=\log _{10} x$.

Then $\left(\log _{10} x\right)^{\log _{10}\left(\log _{10} x\right)}=10000$ becomes $a^{\log _{10} a}=10^{4}$.

Taking the base 10 logarithm of both sides and using the fact that $\log _{10}\left(a^{b}\right)=b \log _{10} a$, we obtain $\left(\log _{10} a\right)\left(\log _{10} a\right)=4$ or $\left(\log _{10} a\right)^{2}=4$.

Therefore, $\log _{10} a= \pm 2$ and so $\log _{10}\left(\log _{10} x\right)= \pm 2$.

If $\log _{10}\left(\log _{10} x\right)=2$, then $\log _{10} x=10^{2}=100$ and so $x=10^{100}$.

If $\log _{10}\left(\log _{10} x\right)=-2$, then $\log _{10} x=10^{-2}=\frac{1}{100}$ and so $x=10^{1 / 100}$.

Therefore, $x=10^{100}$ or $x=10^{1 / 100}$.

We check these answers in the original equation.

If $x=10^{100}$, then $\log _{10} x=100$.

Thus, $\left(\log _{10} x\right)^{\log _{10}\left(\log _{10} x\right)}=100^{\log _{10} 100}=100^{2}=10000$.

If $x=10^{1 / 100}$, then $\log _{10} x=1 / 100=10^{-2}$.

Thus, $\left(\log _{10} x\right)^{\log _{10}\left(\log _{10} x\right)}=\left(10^{-2}\right)^{\log _{10}\left(10^{-2}\right)}=\left(10^{-2}\right)^{-2}=10^{4}=10000$.

Final answer: $10^{100}$,$10^{1 / 100}$
Subfield: Geometry


Question: In the diagram, $\angle A C B=\angle A D E=90^{\circ}$. If $A B=75, B C=21, A D=20$, and $C E=47$, determine the exact length of $B D$.

<img_3927>

Solution:
We use the cosine law in $\triangle A B D$ to determine the length of $B D$ :

$$
B D^{2}=A B^{2}+A D^{2}-2(A B)(A D) \cos (\angle B A D)
$$

We are given that $A B=75$ and $A D=20$, so we need to determine $\cos (\angle B A D)$.

Now

$$
\begin{aligned}
\cos (\angle B A D) & =\cos (\angle B A C+\angle E A D) \\
& =\cos (\angle B A C) \cos (\angle E A D)-\sin (\angle B A C) \sin (\angle E A D) \\
& =\frac{A C}{A B} \frac{A D}{A E}-\frac{B C}{A B} \frac{E D}{A E}
\end{aligned}
$$

since $\triangle A B C$ and $\triangle A D E$ are right-angled.

Since $A B=75$ and $B C=21$, then by the Pythagorean Theorem,

$$
A C=\sqrt{A B^{2}-B C^{2}}=\sqrt{75^{2}-21^{2}}=\sqrt{5625-441}=\sqrt{5184}=72
$$

since $A C>0$.

Since $A C=72$ and $C E=47$, then $A E=A C-C E=25$.

Since $A E=25$ and $A D=20$, then by the Pythagorean Theorem,

$$
E D=\sqrt{A E^{2}-A D^{2}}=\sqrt{25^{2}-20^{2}}=\sqrt{625-400}=\sqrt{225}=15
$$

since $E D>0$.

Therefore,

$$
\cos (\angle B A D)=\frac{A C}{A B} \frac{A D}{A E}-\frac{B C}{A B} \frac{E D}{A E}=\frac{72}{75} \frac{20}{25}-\frac{21}{75} \frac{15}{25}=\frac{1440-315}{75(25)}=\frac{1125}{75(25)}=\frac{45}{75}=\frac{3}{5}
$$



Finally,

$$
\begin{aligned}
B D^{2} & =A B^{2}+A D^{2}-2(A B)(A D) \cos (\angle B A D) \\
& =75^{2}+20^{2}-2(75)(20)\left(\frac{3}{5}\right) \\
& =5625+400-1800 \\
& =4225
\end{aligned}
$$

Since $B D>0$, then $B D=\sqrt{4225}=65$, as required.

Final answer: 65
Subfield: Geometry


Question: In the diagram, $\angle A C B=\angle A D E=90^{\circ}$. If $A B=75, B C=21, A D=20$, and $C E=47$, determine the exact length of $B D$.

![](https://cdn.mathpix.com/cropped/2023_12_21_09b264570ad909ff6decg-1.jpg?height=380&width=379&top_left_y=588&top_left_x=1250)

Solution:
We use the cosine law in $\triangle A B D$ to determine the length of $B D$ :

$$
B D^{2}=A B^{2}+A D^{2}-2(A B)(A D) \cos (\angle B A D)
$$

We are given that $A B=75$ and $A D=20$, so we need to determine $\cos (\angle B A D)$.

Now

$$
\begin{aligned}
\cos (\angle B A D) & =\cos (\angle B A C+\angle E A D) \\
& =\cos (\angle B A C) \cos (\angle E A D)-\sin (\angle B A C) \sin (\angle E A D) \\
& =\frac{A C}{A B} \frac{A D}{A E}-\frac{B C}{A B} \frac{E D}{A E}
\end{aligned}
$$

since $\triangle A B C$ and $\triangle A D E$ are right-angled.

Since $A B=75$ and $B C=21$, then by the Pythagorean Theorem,

$$
A C=\sqrt{A B^{2}-B C^{2}}=\sqrt{75^{2}-21^{2}}=\sqrt{5625-441}=\sqrt{5184}=72
$$

since $A C>0$.

Since $A C=72$ and $C E=47$, then $A E=A C-C E=25$.

Since $A E=25$ and $A D=20$, then by the Pythagorean Theorem,

$$
E D=\sqrt{A E^{2}-A D^{2}}=\sqrt{25^{2}-20^{2}}=\sqrt{625-400}=\sqrt{225}=15
$$

since $E D>0$.

Therefore,

$$
\cos (\angle B A D)=\frac{A C}{A B} \frac{A D}{A E}-\frac{B C}{A B} \frac{E D}{A E}=\frac{72}{75} \frac{20}{25}-\frac{21}{75} \frac{15}{25}=\frac{1440-315}{75(25)}=\frac{1125}{75(25)}=\frac{45}{75}=\frac{3}{5}
$$



Finally,

$$
\begin{aligned}
B D^{2} & =A B^{2}+A D^{2}-2(A B)(A D) \cos (\angle B A D) \\
& =75^{2}+20^{2}-2(75)(20)\left(\frac{3}{5}\right) \\
& =5625+400-1800 \\
& =4225
\end{aligned}
$$

Since $B D>0$, then $B D=\sqrt{4225}=65$, as required.

Final answer: 65
Subfield: Algebra


Question: If $\log _{10} x=3+\log _{10} y$, what is the value of $\frac{x}{y}$ ?

Solution:
$$
\begin{gathered}
\log _{10} x-\log _{10} y=3 \\
\Leftrightarrow \log _{10}\left(\frac{x}{y}\right)=3 \\
\Leftrightarrow \frac{x}{y}=10^{3}=1000
\end{gathered}
$$

Final answer: 1000
Subfield: Algebra


Question: If $x+\frac{1}{x}=\frac{13}{6}$, determine all values of $x^{2}+\frac{1}{x^{2}}$.

Solution:
$\left(x+\frac{1}{x}\right)^{2}=\left(\frac{13}{6}\right)^{2}$; squaring

$x^{2}+2+\frac{1}{x^{2}}=\frac{169}{36}$

$x^{2}+\frac{1}{x^{2}}=\frac{169}{32}-2$

$x^{2}+\frac{1}{x^{2}}=\frac{169}{36}-\frac{72}{36}=\frac{97}{36}$
$6 x\left(x+\frac{1}{x}\right)=6 x\left(\frac{13}{6}\right)$

$6 x^{2}+6=13 x$

$6 x^{2}-13 x+6=0$

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



$x=\frac{2}{3}$ or $x=\frac{3}{2}$

For $x=\frac{2}{3}, x^{2}+\frac{1}{x^{2}}$

$=\left(\frac{2}{3}\right)^{2}+\frac{1}{\left(\frac{2}{3}\right)^{2}}$

$=\frac{4}{9}+\frac{9}{4}$

For $x=\frac{3}{2},\left(\frac{3}{2}\right)^{2}+\frac{1}{\left(\frac{3}{2}\right)^{2}}$

$=\frac{9}{4}+\frac{4}{9}$

$=\frac{97}{36}$

$=\frac{81+16}{36}$

$=\frac{97}{36}$

Final answer: $\frac{97}{36}$
Subfield: Geometry


Question: A circle, with diameter $A B$ as shown, intersects the positive $y$-axis at point $D(0, d)$. Determine $d$.

<img_3467>

Solution:
The centre of the circle is $(3,0)$ and the circle has a radius of 5.

Thus $\sqrt{d^{2}+3^{2}}=5$

$$
\begin{aligned}
& d^{2}=5^{2}-3^{2} \\
& d^{2}=16
\end{aligned}
$$

Therefore $d=4$, since $d>0$.
Since $A B$ is a diameter of the circle, $\angle A D B=90^{\circ}$ and $\angle A O D=90^{\circ}$.

$\triangle A D O \sim \triangle D B O$

Therefore, $\frac{O D}{A O}=\frac{B O}{O D}$

and $d^{2}=2(8)$

$$
\begin{aligned}
d^{2} & =16 \\
d & =4, \text { since } d>0 .
\end{aligned}
$$
$\angle A D B=\angle A O D=\angle B O D=90^{\circ}$

In $\triangle A O D, A D^{2}=4+d^{2}$.

In $\triangle B O D, D B^{2}=64+d^{2}$.

In $\Delta A D B,\left(4+d^{2}\right)+\left(64+d^{2}\right)=100$

$$
\begin{aligned}
2 d^{2} & =32 \\
d & =4, d>0
\end{aligned}
$$

Final answer: 4
Subfield: Geometry


Question: A circle, with diameter $A B$ as shown, intersects the positive $y$-axis at point $D(0, d)$. Determine $d$.

![](https://cdn.mathpix.com/cropped/2023_12_21_47e3f6341f2fa24a72f4g-1.jpg?height=474&width=618&top_left_y=210&top_left_x=1190)

Solution:
The centre of the circle is $(3,0)$ and the circle has a radius of 5.

Thus $\sqrt{d^{2}+3^{2}}=5$

$$
\begin{aligned}
& d^{2}=5^{2}-3^{2} \\
& d^{2}=16
\end{aligned}
$$

Therefore $d=4$, since $d>0$.
Since $A B$ is a diameter of the circle, $\angle A D B=90^{\circ}$ and $\angle A O D=90^{\circ}$.

$\triangle A D O \sim \triangle D B O$

Therefore, $\frac{O D}{A O}=\frac{B O}{O D}$

and $d^{2}=2(8)$

$$
\begin{aligned}
d^{2} & =16 \\
d & =4, \text { since } d>0 .
\end{aligned}
$$
$\angle A D B=\angle A O D=\angle B O D=90^{\circ}$

In $\triangle A O D, A D^{2}=4+d^{2}$.

In $\triangle B O D, D B^{2}=64+d^{2}$.

In $\Delta A D B,\left(4+d^{2}\right)+\left(64+d^{2}\right)=100$

$$
\begin{aligned}
2 d^{2} & =32 \\
d & =4, d>0
\end{aligned}
$$

Final answer: 4
Subfield: Geometry


Question: A square $P Q R S$ with side of length $x$ is subdivided into four triangular regions as shown so that area (A) + area $(B)=\text{area}(C)$. If $P T=3$ and $R U=5$, determine the value of $x$.

![](https://cdn.mathpix.com/cropped/2023_12_21_47e3f6341f2fa24a72f4g-1.jpg?height=376&width=374&top_left_y=701&top_left_x=1361)

Solution:
Since the side length of the square is $x, T S=x-3$ and $V S=x-5$

Area of triangle $A=\frac{1}{2}(3)(x)$.

Area of triangle $B=\frac{1}{2}(5)(x)$

Area of triangle $C=\frac{1}{2}(x-5)(x-3)$.

From the given information, $\frac{1}{2}(3 x)+\frac{1}{2}(5 x)=\frac{1}{2}(x-5)(x-3)$. Labelled diagram

$3 x+5 x=x^{2}-8 x+15$

$x^{2}-16 x+15=0$

![](https://cdn.mathpix.com/cropped/2023_12_21_ccfc71aa0aef1ed6786cg-1.jpg?height=440&width=484&top_left_y=1755&top_left_x=1388)

Thus $x=15$ or $x=1$.

Therefore $x=15$ since $x=1$ is inadmissible.

Final answer: 15
Subfield: Geometry


Question: A square $P Q R S$ with side of length $x$ is subdivided into four triangular regions as shown so that area (A) + area $(B)=\text{area}(C)$. If $P T=3$ and $R U=5$, determine the value of $x$.

<img_3567>

Solution:
Since the side length of the square is $x, T S=x-3$ and $V S=x-5$

Area of triangle $A=\frac{1}{2}(3)(x)$.

Area of triangle $B=\frac{1}{2}(5)(x)$

Area of triangle $C=\frac{1}{2}(x-5)(x-3)$.

From the given information, $\frac{1}{2}(3 x)+\frac{1}{2}(5 x)=\frac{1}{2}(x-5)(x-3)$. Labelled diagram

$3 x+5 x=x^{2}-8 x+15$

$x^{2}-16 x+15=0$

<img_3393>

Thus $x=15$ or $x=1$.

Therefore $x=15$ since $x=1$ is inadmissible.

Final answer: 15
Subfield: Combinatorics


Question: A die, with the numbers $1,2,3,4,6$, and 8 on its six faces, is rolled. After this roll, if an odd number appears on the top face, all odd numbers on the die are doubled. If an even number appears on the top face, all the even numbers are halved. If the given die changes in this way, what is the probability that a 2 will appear on the second roll of the die?

Solution:
There are only two possibilities on the first roll - it can either be even or odd.

Possibility 1 'The first roll is odd'

The probability of an odd outcome on the first roll is $\frac{1}{3}$.

After doubling all the numbers, the possible outcomes on the second roll would now be 2, 2, 6, $4,6,8$ with the probability of a 2 being $\frac{1}{3}$.

Thus the probability of a 2 on the second roll would be $\frac{1}{3} \times \frac{1}{3}=\frac{1}{9}$.

Possibility 2 'The first is even'

The probability of an even outcome on the first roll is $\frac{2}{3}$.

After halving all the numbers, the possible outcomes on the second roll would be 1, 1, 3, 2, 3, 8 .

The probability of a 2 on the second die would now be $\frac{1}{6}$.

Thus the probability of a 2 on the second roll is $\frac{2}{3} \times \frac{1}{6}=\frac{1}{9}$.

The probability of a 2 appear on the top face is $\frac{1}{9}+\frac{1}{9}=\frac{2}{9}$.

Final answer: $\frac{2}{9}$
Subfield: Algebra


Question: The table below gives the final standings for seven of the teams in the English Cricket League in 1998. At the end of the year, each team had played 17 matches and had obtained the total number of points shown in the last column. Each win $W$, each draw $D$, each bonus bowling point $A$, and each bonus batting point $B$ received $w, d, a$ and $b$ points respectively, where $w, d, a$ and $b$ are positive integers. No points are given for a loss. Determine the values of $w, d, a$ and $b$ if total points awarded are given by the formula: Points $=w \times W+d \times D+a \times A+b \times B$.



Final Standings

|  | $W$ | Losses | $D$ | $A$ | $B$ | Points |
| :--- | :---: | :---: | :---: | :---: | :---: | :---: |
| Sussex | 6 | 7 | 4 | 30 | 63 | 201 |
| Warks | 6 | 8 | 3 | 35 | 60 | 200 |
| Som | 6 | 7 | 4 | 30 | 54 | 192 |
| Derbys | 6 | 7 | 4 | 28 | 55 | 191 |
| Kent | 5 | 5 | 7 | 18 | 59 | 178 |
| Worcs | 4 | 6 | 7 | 32 | 59 | 176 |
| Glam | 4 | 6 | 7 | 36 | 55 | 176 |

Solution:
There are a variety of ways to find the unknowns.

The most efficient way is to choose equations that have like coefficients. Here is one way to solve the problem using this method.

For Sussex: $\quad 6 w+4 d+30 a+63 b=201$

For Som: $\quad 6 w+4 d+30 a+54 b=192$

Subtracting, $\quad 9 b=9 b=1$

If $b=1$

For Derbys: $\quad 6 w+4 d+28 a+55=191$ 

$$
6 w+4 d+28 a=136 \tag{1}
$$

For Sussex: $\quad 6 w+4 d+30 a+63=201$

$$
6 w+4 d+30 a=138 \tag{2}
$$

Subtracting, (2) - (1)

$$
2 a=2
$$

$$
a=1 \text {. }
$$

We can now calculate $d$ and $w$ by substituting $a=1, b=1$ into a pair of equations.

An efficient way of doing this is by substituting $a=1, b=1$ into Som and Worcs.

For Som: $\quad 6 w+4 d+84=192$

$$
6 w+4 d=108 \tag{3}
$$

For Worcs: $\quad 6 w+3 d+85=200$

$$
6 w+3 d=105 \tag{4}
$$

Subtracting, (3) - (4) $\quad d=3$.

Substituting $d=3$ in either (3) or (4), $6 w+4(3)=108$ (substituting in (3))

$$
\begin{aligned}
6 w & =96 \\
w & =16 .
\end{aligned}
$$

Therefore $w=16, d=3, a=b=1$.

Final answer: 16,3,1,1
Subfield: Geometry


Question: In the diagram, $A D=D C, \sin \angle D B C=0.6$ and $\angle A C B=90^{\circ}$. What is the value of $\tan \angle A B C$ ?

<img_3406>

Solution:
Let $D B=10$.

Therefore, $D C=A D=6$.

By the theorem of Pythagoras, $B C^{2}=10^{2}-6^{2}=64$.

Therefore, $B C=8$.



Thus, $\tan \angle A B C=\frac{12}{8}=\frac{3}{2}$.

Final answer: $\frac{3}{2}$
Subfield: Geometry


Question: In the diagram, $A D=D C, \sin \angle D B C=0.6$ and $\angle A C B=90^{\circ}$. What is the value of $\tan \angle A B C$ ?

![](https://cdn.mathpix.com/cropped/2023_12_21_b1dc13dda57202a677cdg-1.jpg?height=366&width=357&top_left_y=207&top_left_x=1342)

Solution:
Let $D B=10$.

Therefore, $D C=A D=6$.

By the theorem of Pythagoras, $B C^{2}=10^{2}-6^{2}=64$.

Therefore, $B C=8$.



Thus, $\tan \angle A B C=\frac{12}{8}=\frac{3}{2}$.

Final answer: $\frac{3}{2}$
Subfield: Geometry


Question: On a cross-sectional diagram of the Earth, the $x$ and $y$-axes are placed so that $O(0,0)$ is the centre of the Earth and $C(6.40,0.00)$ is the location of Cape Canaveral. A space shuttle is forced to land on an island at $A(5.43,3.39)$, as shown. Each unit represents $1000 \mathrm{~km}$.

Determine the distance from Cape Canaveral to the island, measured on the surface of the earth, to the nearest $10 \mathrm{~km}$.

<img_3336>

Solution:
$\tan \angle A O C=\frac{3.39}{5.43}$

$\angle A O C=\tan ^{-1}\left(\frac{3.39}{5.43}\right)=31.97^{\circ}$

The arc length $\overparen{A C}=\frac{31.97}{360^{\circ}}[(2 \pi)(6.40)]=3.57$ units

The distance is approximately $3570 \mathrm{~km}$.

Final answer: 3570 km
Subfield: Geometry


Question: On a cross-sectional diagram of the Earth, the $x$ and $y$-axes are placed so that $O(0,0)$ is the centre of the Earth and $C(6.40,0.00)$ is the location of Cape Canaveral. A space shuttle is forced to land on an island at $A(5.43,3.39)$, as shown. Each unit represents $1000 \mathrm{~km}$.

Determine the distance from Cape Canaveral to the island, measured on the surface of the earth, to the nearest $10 \mathrm{~km}$.

![](https://cdn.mathpix.com/cropped/2023_12_21_b1dc13dda57202a677cdg-1.jpg?height=315&width=493&top_left_y=634&top_left_x=1315)

Solution:
$\tan \angle A O C=\frac{3.39}{5.43}$

$\angle A O C=\tan ^{-1}\left(\frac{3.39}{5.43}\right)=31.97^{\circ}$

The arc length $\overparen{A C}=\frac{31.97}{360^{\circ}}[(2 \pi)(6.40)]=3.57$ units

The distance is approximately $3570 \mathrm{~km}$.

Final answer: 3570 km
Subfield: Algebra


Question: Let $\lfloor x\rfloor$ represent the greatest integer which is less than or equal to $x$. For example, $\lfloor 3\rfloor=3,\lfloor 2.6\rfloor=2$. If $x$ is positive and $x\lfloor x\rfloor=17$, what is the value of $x$ ?

Solution:
We deduce that $4<x<5$.

Otherwise, if $x \leq 4, x\lfloor x\rfloor \leq 16$, and if $x \geq 5, x\lfloor x\rfloor \geq 25$.

Therefore $\lfloor x\rfloor=4$

Since $x\lfloor x\rfloor=17$

$$
\begin{aligned}
4 x & =17 \\
x & =4.25
\end{aligned}
$$

Final answer: 4.25
Subfield: Geometry


Question: The parabola $y=-x^{2}+4$ has vertex $P$ and intersects the $x$-axis at $A$ and $B$. The parabola is translated from its original position so that its vertex moves along the line $y=x+4$ to the point $Q$. In this position, the parabola intersects the $x$-axis at $B$ and $C$. Determine the coordinates of $C$.

<img_3623>

Solution:
The parabola $y=-x^{2}+4$ has vertex $P(0,4)$ and intersects the $x$-axis at $A(-2,0)$ and $B(2,0)$. The intercept $B(2,0)$ has its pre-image, $B^{\prime}$ on the parabola $y=-x^{2}+4$. To find $B^{\prime}$, we find the point of intersection of the line passing through $B(2,0)$, with slope 1 , and the parabola $y=-x^{2}+4$.

The equation of the line is $y=x-2$.

Intersection points, $x-2=-x^{2}+4$

$$
\begin{array}{r}
x^{2}+x-6=0 \\
(x+3)(x-2)=0 .
\end{array}
$$

Therefore, $x=-3$ or $x=2$.

For $x=-3, y=-3-2=-5$. Thus $B^{\prime}$ has coordinates $(-3,-5)$.

If $(-3,-5) \rightarrow(2,0)$ then the required general translation mapping $y=-x^{2}+4$ onto the parabola with vertex $Q$ is $(x, y) \rightarrow(x+5, y+5)$.

Possibility 1

Using the general translation, we find the coordinates of $Q$ to be, $P(0,4) \rightarrow Q(0+5,4+5)=Q(5,9)$.

If $C$ is the reflection of $B$ in the axis of symmetry of the parabola, i.e. $x=5, C$ has coordinates $(8,0)$.

Possibility 2

If $B^{\prime}$ has coordinates $(-3,-5)$ then $C^{\prime}$ is the reflection of $B^{\prime}$ in the $y$-axis. Thus $C^{\prime}$ has coordinates $(3,-5)$.

If we apply the general translation then $C$ has coordinates $(3+5,-5+5)$ or $(8,0)$.

Thus $C$ has coordinates $(8,0)$.

Possibility 3

Using the general translation, we find the coordinates of $Q$ to be, $P(0,4) \rightarrow Q(0+5,4+5)=Q(5,9)$.

The equation of the image parabola is $y=-(x-5)^{2}+9$.



To find its intercepts, $-(x-5)^{2}+9=0$

$$
\begin{aligned}
(x-5)^{2} & =9 \\
x-5 & = \pm 3 .
\end{aligned}
$$

Therefore $x=8$ or $x=2$.

Thus $C$ has coordinates $(8,0)$.
The translation moving the parabola with equation $y=-x^{2}+4$ onto the parabola with vertex $Q$ is $T(t, t)$ because the slope of the line $y=x+4$ is 1 .

The pre-image of $B^{\prime}$ is $(2-t,-t)$.

Since $B^{\prime}$ is on the parabola with vertex $P$, we have

$$
\begin{aligned}
-t & =-(2-t)^{2}+4 \\
-t & =-4+4 t-t^{2}+4 \\
t^{2}-5 t & =0 \\
t(t-5) & =0
\end{aligned}
$$

Therefore, $t=0$ or $t=5$.

Thus $B^{\prime}$ is $(-3,-5)$.

Let $C$ have coordinates $(c, 0)$.

The pre-image of $C$ is $(c-5,-5)$.

Therefore, $-5=-(c-5)^{2}+4$.

Or, $(c-5)^{2}=9$.

Therefore $c-5=3$ or $c-5=-3$.

$$
c=8 \text { or } \quad c=2
$$

Thus $C$ has coordinates $(8,0)$.

Final answer: $(8,0)$
Subfield: Geometry


Question: The parabola $y=-x^{2}+4$ has vertex $P$ and intersects the $x$-axis at $A$ and $B$. The parabola is translated from its original position so that its vertex moves along the line $y=x+4$ to the point $Q$. In this position, the parabola intersects the $x$-axis at $B$ and $C$. Determine the coordinates of $C$.

![](https://cdn.mathpix.com/cropped/2023_12_21_b1dc13dda57202a677cdg-1.jpg?height=540&width=596&top_left_y=1340&top_left_x=1212)

Solution:
The parabola $y=-x^{2}+4$ has vertex $P(0,4)$ and intersects the $x$-axis at $A(-2,0)$ and $B(2,0)$. The intercept $B(2,0)$ has its pre-image, $B^{\prime}$ on the parabola $y=-x^{2}+4$. To find $B^{\prime}$, we find the point of intersection of the line passing through $B(2,0)$, with slope 1 , and the parabola $y=-x^{2}+4$.

The equation of the line is $y=x-2$.

Intersection points, $x-2=-x^{2}+4$

$$
\begin{array}{r}
x^{2}+x-6=0 \\
(x+3)(x-2)=0 .
\end{array}
$$

Therefore, $x=-3$ or $x=2$.

For $x=-3, y=-3-2=-5$. Thus $B^{\prime}$ has coordinates $(-3,-5)$.

If $(-3,-5) \rightarrow(2,0)$ then the required general translation mapping $y=-x^{2}+4$ onto the parabola with vertex $Q$ is $(x, y) \rightarrow(x+5, y+5)$.

Possibility 1

Using the general translation, we find the coordinates of $Q$ to be, $P(0,4) \rightarrow Q(0+5,4+5)=Q(5,9)$.

If $C$ is the reflection of $B$ in the axis of symmetry of the parabola, i.e. $x=5, C$ has coordinates $(8,0)$.

Possibility 2

If $B^{\prime}$ has coordinates $(-3,-5)$ then $C^{\prime}$ is the reflection of $B^{\prime}$ in the $y$-axis. Thus $C^{\prime}$ has coordinates $(3,-5)$.

If we apply the general translation then $C$ has coordinates $(3+5,-5+5)$ or $(8,0)$.

Thus $C$ has coordinates $(8,0)$.

Possibility 3

Using the general translation, we find the coordinates of $Q$ to be, $P(0,4) \rightarrow Q(0+5,4+5)=Q(5,9)$.

The equation of the image parabola is $y=-(x-5)^{2}+9$.



To find its intercepts, $-(x-5)^{2}+9=0$

$$
\begin{aligned}
(x-5)^{2} & =9 \\
x-5 & = \pm 3 .
\end{aligned}
$$

Therefore $x=8$ or $x=2$.

Thus $C$ has coordinates $(8,0)$.
The translation moving the parabola with equation $y=-x^{2}+4$ onto the parabola with vertex $Q$ is $T(t, t)$ because the slope of the line $y=x+4$ is 1 .

The pre-image of $B^{\prime}$ is $(2-t,-t)$.

Since $B^{\prime}$ is on the parabola with vertex $P$, we have

$$
\begin{aligned}
-t & =-(2-t)^{2}+4 \\
-t & =-4+4 t-t^{2}+4 \\
t^{2}-5 t & =0 \\
t(t-5) & =0
\end{aligned}
$$

Therefore, $t=0$ or $t=5$.

Thus $B^{\prime}$ is $(-3,-5)$.

Let $C$ have coordinates $(c, 0)$.

The pre-image of $C$ is $(c-5,-5)$.

Therefore, $-5=-(c-5)^{2}+4$.

Or, $(c-5)^{2}=9$.

Therefore $c-5=3$ or $c-5=-3$.

$$
c=8 \text { or } \quad c=2
$$

Thus $C$ has coordinates $(8,0)$.

Final answer: $(8,0)$
Subfield: Geometry


Question: A cube has edges of length $n$, where $n$ is an integer. Three faces, meeting at a corner, are painted red. The cube is then cut into $n^{3}$ smaller cubes of unit length. If exactly 125 of these cubes have no faces painted red, determine the value of $n$.

Solution:
If we remove the cubes which have red paint, we are left with a smaller cube with measurements, $(n-1) \times(n-1) \times(n-1)$

Thus, $(n-1)^{3}=125$

$$
n=6 \text {. }
$$

Final answer: 6