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


Question: The Sieve of Sundaram uses the following infinite table of positive integers:

| 4 | 7 | 10 | 13 | $\cdots$ |
| :---: | :---: | :---: | :---: | :---: |
| 7 | 12 | 17 | 22 | $\cdots$ |
| 10 | 17 | 24 | 31 | $\cdots$ |
| 13 | 22 | 31 | 40 | $\cdots$ |
| $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ |  |

The numbers in each row in the table form an arithmetic sequence. The numbers in each column in the table form an arithmetic sequence. The first four entries in each of the first four rows and columns are shown.
Prove that if $N$ is an entry in the table, then $2 N+1$ is composite.

Solution:
First, we prove that the number in the $R$ th row and $C$ th column equals $2RC+R+C$.
Proof. First, we determine the first entry in the $R$ th row.

Since the first column is an arithmetic sequence with common difference 3 , then the $R$ th entry in the first column (that is, the first entry in the $R$ th row) is $4+(R-1)(3)$ or $4+3 R-3=3 R+1$.

Second, we determine the common difference in the $R$ th row by determining the second entry in the $R$ th row.

Since the second column is an arithmetic sequence with common difference 5 , then the $R$ th entry in the second column (that is, the second entry in the $R$ th row) is $7+(R-1)(5)$ or $7+5 R-5=5 R+2$.

Therefore, the common difference in the $R$ th row must be $(5 R+2)-(3 R+1)=2 R+1$. Thus, the $C$ th entry in the $R$ th row (that is, the number in the $R$ th row and the $C$ th column) is

$$
3 R+1+(C-1)(2 R+1)=3 R+1+2 R C+C-2 R-1=2 R C+R+C
$$

Suppose that $N$ is an entry in the table, say in the $R$ th row and $C$ th column.

From the proof above, then $N=2 R C+R+C$ and so $2 N+1=4 R C+2 R+2 C+1$.

Now $4 R C+2 R+2 C+1=2 R(2 C+1)+2 C+1=(2 R+1)(2 C+1)$.

Since $R$ and $C$ are integers with $R \geq 1$ and $C \geq 1$, then $2 R+1$ and $2 C+1$ are each integers that are at least 3 .

Therefore, $2 N+1=(2 R+1)(2 C+1)$ must be composite, since it is the product of two integers that are each greater than 1.
Subfield: Number Theory


Question: Let $\lfloor x\rfloor$ denote the greatest integer less than or equal to $x$. For example, $\lfloor 3.1\rfloor=3$ and $\lfloor-1.4\rfloor=-2$.

Suppose that $f(n)=2 n-\left\lfloor\frac{1+\sqrt{8 n-7}}{2}\right\rfloor$ and $g(n)=2 n+\left\lfloor\frac{1+\sqrt{8 n-7}}{2}\right\rfloor$ for each positive integer $n$.
Determine the value of $g(2011)$.

Solution:
If $n=2011$, then $8 n-7=16081$ and so $\sqrt{8 n-7} \approx 126.81$.

Thus, $\frac{1+\sqrt{8 n-7}}{2} \approx \frac{1+126.81}{2} \approx 63.9$.

Therefore, $g(2011)=2(2011)+\left\lfloor\frac{1+\sqrt{8(2011)-7}}{2}\right\rfloor=4022+\lfloor 63.9\rfloor=4022+63=4085$.

Final answer: 4085
Subfield: Number Theory


Question: Let $\lfloor x\rfloor$ denote the greatest integer less than or equal to $x$. For example, $\lfloor 3.1\rfloor=3$ and $\lfloor-1.4\rfloor=-2$.

Suppose that $f(n)=2 n-\left\lfloor\frac{1+\sqrt{8 n-7}}{2}\right\rfloor$ and $g(n)=2 n+\left\lfloor\frac{1+\sqrt{8 n-7}}{2}\right\rfloor$ for each positive integer $n$.
Determine a value of $n$ for which $f(n)=100$.

Solution:
To determine a value of $n$ for which $f(n)=100$, we need to solve the equation

$$
2 n-\left\lfloor\frac{1+\sqrt{8 n-7}}{2}\right\rfloor=100
$$

We first solve the equation

$$
2 x-\frac{1+\sqrt{8 x-7}}{2}=100 \quad(* *)
$$

because the left sides of $(*)$ and $(* *)$ do not differ by much and so the solutions are likely close together. We will try integers $n$ in $(*)$ that are close to the solutions to $(* *)$.

Manipulating $(* *)$, we obtain

$$
\begin{aligned}
4 x-(1+\sqrt{8 x-7}) & =200 \\
4 x-201 & =\sqrt{8 x-7} \\
(4 x-201)^{2} & =8 x-7 \\
16 x^{2}-1608 x+40401 & =8 x-7 \\
16 x^{2}-1616 x+40408 & =0 \\
2 x^{2}-202 x+5051 & =0
\end{aligned}
$$

By the quadratic formula,

$$
x=\frac{202 \pm \sqrt{202^{2}-4(2)(5051)}}{2(2)}=\frac{202 \pm \sqrt{396}}{4}=\frac{101 \pm \sqrt{99}}{2}
$$

and so $x \approx 55.47$ or $x \approx 45.53$.

We try $n=55$, which is close to 55.47 :

$$
f(55)=2(55)-\left\lfloor\frac{1+\sqrt{8(55)-7}}{2}\right\rfloor=110-\left\lfloor\frac{1+\sqrt{433}}{2}\right\rfloor
$$

Since $\sqrt{433} \approx 20.8$, then $\frac{1+\sqrt{433}}{2} \approx 10.9$, which gives $\left\lfloor\frac{1+\sqrt{433}}{2}\right\rfloor=10$.

Thus, $f(55)=110-10=100$.

Therefore, a value of $n$ for which $f(n)=100$ is $n=55$.

Final answer: 55
Subfield: Combinatorics


Question: Six tickets numbered 1 through 6 are placed in a box. Two tickets are randomly selected and removed together. What is the probability that the smaller of the two numbers on the tickets selected is less than or equal to 4 ?

Solution:
The possible pairs of numbers on the tickets are (listed as ordered pairs): (1,2), (1,3), $(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6)$, and $(5,6)$.

There are fifteen such pairs. (We treat the pair of tickets numbered 2 and 4 as being the same as the pair numbered 4 and 2.)

The pairs for which the smaller of the two numbers is less than or equal to 4 are $(1,2)$, $(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5)$, and $(4,6)$. There are fourteen such pairs.

Therefore, the probability of selecting such a pair of tickets is $\frac{14}{15}$.
We find the probability that the smaller number on the two tickets is NOT less than or equal to 4.

Therefore, the smaller number on the two tickets is at least 5.

Thus, the pair of numbers must be 5 and 6 , since two distinct numbers less than or equal to 6 are being chosen.

The possible pairs of numbers on the tickets are (listed as ordered pairs): (1,2), (1,3), $(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6)$, and $(5,6)$.

There are fifteen such pairs. (We treat the pair of tickets numbered 2 and 4 as being the same as the pair numbered 4 and 2.)

Therefore, the probability that the smaller number on the two tickets is NOT less than or equal to 4 is $\frac{1}{15}$, so the probability that the smaller number on the two tickets IS less than or equal to 4 is $1-\frac{1}{15}=\frac{14}{15}$.

Final answer: $\frac{14}{15}$
Subfield: Geometry


Question: A helicopter hovers at point $H$, directly above point $P$ on level ground. Lloyd sits on the ground at a point $L$ where $\angle H L P=60^{\circ}$. A ball is droppped from the helicopter. When the ball is at point $B, 400 \mathrm{~m}$ directly below the helicopter, $\angle B L P=30^{\circ}$. What is the distance between $L$ and $P$ ?

![](https://cdn.mathpix.com/cropped/2023_12_21_dea74421c52c577440bcg-1.jpg?height=374&width=441&top_left_y=434&top_left_x=1254)

Solution:
Since $\angle H L P=60^{\circ}$ and $\angle B L P=30^{\circ}$, then $\angle H L B=\angle H L P-\angle B L P=30^{\circ}$.

Also, since $\angle H L P=60^{\circ}$ and $\angle H P L=90^{\circ}$, then $\angle L H P=180^{\circ}-90^{\circ}-60^{\circ}=30^{\circ}$.

![](https://cdn.mathpix.com/cropped/2023_12_21_79f76cd9df8987116379g-1.jpg?height=512&width=609&top_left_y=945&top_left_x=861)

Therefore, $\triangle H B L$ is isosceles and $B L=H B=400 \mathrm{~m}$.

In $\triangle B L P, B L=400 \mathrm{~m}$ and $\angle B L P=30^{\circ}$, so $L P=B L \cos \left(30^{\circ}\right)=400\left(\frac{\sqrt{3}}{2}\right)=200 \sqrt{3}$ m.

Therefore, the distance between $L$ and $P$ is $200 \sqrt{3} \mathrm{~m}$.
Since $\angle H L P=60^{\circ}$ and $\angle B L P=30^{\circ}$, then $\angle H L B=\angle H L P-\angle B L P=30^{\circ}$.

Also, since $\angle H L P=60^{\circ}$ and $\angle H P L=90^{\circ}$, then $\angle L H P=180^{\circ}-90^{\circ}-60^{\circ}=30^{\circ}$. Also, $\angle L B P=60^{\circ}$.

Let $L P=x$.

![](https://cdn.mathpix.com/cropped/2023_12_21_79f76cd9df8987116379g-1.jpg?height=518&width=613&top_left_y=2018&top_left_x=859)

Since $\triangle B L P$ is $30^{\circ}-60^{\circ}-90^{\circ}$, then $B P: L P=1: \sqrt{3}$, so $B P=\frac{1}{\sqrt{3}} L P=\frac{1}{\sqrt{3}} x$.



Since $\triangle H L P$ is $30^{\circ}-60^{\circ}-90^{\circ}$, then $H P: L P=\sqrt{3}: 1$, so $H P=\sqrt{3} L P=\sqrt{3} x$.

But $H P=H B+B P$ so

$$
\begin{aligned}
\sqrt{3} x & =400+\frac{1}{\sqrt{3}} x \\
3 x & =400 \sqrt{3}+x \\
2 x & =400 \sqrt{3} \\
x & =200 \sqrt{3}
\end{aligned}
$$

Therefore, the distance from $L$ to $P$ is $200 \sqrt{3} \mathrm{~m}$.

Final answer: $200 \sqrt{3}$ m
Subfield: Geometry


Question: A helicopter hovers at point $H$, directly above point $P$ on level ground. Lloyd sits on the ground at a point $L$ where $\angle H L P=60^{\circ}$. A ball is droppped from the helicopter. When the ball is at point $B, 400 \mathrm{~m}$ directly below the helicopter, $\angle B L P=30^{\circ}$. What is the distance between $L$ and $P$ ?

<img_3984>

Solution:
Since $\angle H L P=60^{\circ}$ and $\angle B L P=30^{\circ}$, then $\angle H L B=\angle H L P-\angle B L P=30^{\circ}$.

Also, since $\angle H L P=60^{\circ}$ and $\angle H P L=90^{\circ}$, then $\angle L H P=180^{\circ}-90^{\circ}-60^{\circ}=30^{\circ}$.

<img_3808>

Therefore, $\triangle H B L$ is isosceles and $B L=H B=400 \mathrm{~m}$.

In $\triangle B L P, B L=400 \mathrm{~m}$ and $\angle B L P=30^{\circ}$, so $L P=B L \cos \left(30^{\circ}\right)=400\left(\frac{\sqrt{3}}{2}\right)=200 \sqrt{3}$ m.

Therefore, the distance between $L$ and $P$ is $200 \sqrt{3} \mathrm{~m}$.
Since $\angle H L P=60^{\circ}$ and $\angle B L P=30^{\circ}$, then $\angle H L B=\angle H L P-\angle B L P=30^{\circ}$.

Also, since $\angle H L P=60^{\circ}$ and $\angle H P L=90^{\circ}$, then $\angle L H P=180^{\circ}-90^{\circ}-60^{\circ}=30^{\circ}$. Also, $\angle L B P=60^{\circ}$.

Let $L P=x$.

<img_3794>

Since $\triangle B L P$ is $30^{\circ}-60^{\circ}-90^{\circ}$, then $B P: L P=1: \sqrt{3}$, so $B P=\frac{1}{\sqrt{3}} L P=\frac{1}{\sqrt{3}} x$.



Since $\triangle H L P$ is $30^{\circ}-60^{\circ}-90^{\circ}$, then $H P: L P=\sqrt{3}: 1$, so $H P=\sqrt{3} L P=\sqrt{3} x$.

But $H P=H B+B P$ so

$$
\begin{aligned}
\sqrt{3} x & =400+\frac{1}{\sqrt{3}} x \\
3 x & =400 \sqrt{3}+x \\
2 x & =400 \sqrt{3} \\
x & =200 \sqrt{3}
\end{aligned}
$$

Therefore, the distance from $L$ to $P$ is $200 \sqrt{3} \mathrm{~m}$.

Final answer: $200 \sqrt{3}$ m
Subfield: Geometry


Question: A goat starts at the origin $(0,0)$ and then makes several moves. On move 1 , it travels 1 unit up to $(0,1)$. On move 2 , it travels 2 units right to $(2,1)$. On move 3 , it travels 3 units down to $(2,-2)$. On move 4 , it travels 4 units to $(-2,-2)$. It continues in this fashion, so that on move $n$, it turns $90^{\circ}$ in a clockwise direction from its previous heading and travels $n$ units in this new direction. After $n$ moves, the goat has travelled a total of 55 units. Determine the coordinates of its position at this time.

Solution:
After 2 moves, the goat has travelled $1+2=3$ units.

After 3 moves, the goat has travelled $1+2+3=6$ units.

Similarly, after $n$ moves, the goat has travelled a total of $1+2+3+\cdots+n$ units.

For what value of $n$ is $1+2+3+\cdots+n$ equal to 55 ?

The fastest way to determine the value of $n$ is by adding the first few integers until we obtain a sum of 55 . This will be $n=10$.

(We could also do this by remembering that $1+2+3+\cdots+n=\frac{1}{2} n(n+1)$ and solving for $n$ this way.)

So we must determine the coordinates of the goat after 10 moves.

We consider first the $x$-coordinate.

Since starting at $(0,0)$ the goat has moved 2 units in the positive $x$ direction, 4 units in the negative $x$ direction, 6 units in the positive $x$ direction, 8 units in the negative $x$ direction and 10 units in the positive $x$ direction, so its $x$ coordinate should be $2-4+6-8+10=6$. Similarly, its $y$-coordinate should be $1-3+5-7+9=5$.

Therefore, after having travelled a distance of 55 units, the goat is at the point $(6,5)$.

Final answer: (6,5)
Subfield: Algebra


Question: Determine all possible values of $r$ such that the three term geometric sequence 4, $4 r, 4 r^{2}$ is also an arithmetic sequence.

(An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant. For example, 3, 5, 7, 9, 11 is an arithmetic sequence.)

Solution:
Since the sequence $4,4 r, 4 r^{2}$ is also arithmetic, then the difference between $4 r^{2}$ and $4 r$ equals the difference between $4 r$ and 4 , or

$$
\begin{aligned}
4 r^{2}-4 r & =4 r-4 \\
4 r^{2}-8 r+4 & =0 \\
r^{2}-2 r+1 & =0 \\
(r-1)^{2} & =0
\end{aligned}
$$

Therefore, the only value of $r$ is $r=1$.
Since the sequence $4,4 r, 4 r^{2}$ is also arithmetic, then we can write $4 r=4+d$ and $4 r^{2}=4+2 d$ for some real number $d$. (Here, $d$ is the common difference in this arithmetic sequence.)

Then $d=4 r-4$ and $2 d=4 r^{2}-4$ or $d=2 r^{2}-2$.

Therefore, equating the two expressions for $d$, we obtain $2 r^{2}-2=4 r-4$ or $2 r^{2}-4 r+2=0$ or $r^{2}-2 r+1=0$ or $(r-1)^{2}=0$.

Therefore, the only value of $r$ is $r=1$.

Final answer: 1
Subfield: Geometry


Question: In the diagram, $A B C D$ is a quadrilateral in which $\angle A+\angle C=180^{\circ}$. What is the length of $C D$ ?

<img_3146>

Solution:
In order to determine $C D$, we must determine one of the angles (or at least some information about one of the angles) in $\triangle B C D$.

To do this, we look at $\angle A$ use the fact that $\angle A+\angle C=180^{\circ}$.



<img_3524>

Using the cosine law in $\triangle A B D$, we obtain

$$
\begin{aligned}
7^{2} & =5^{2}+6^{2}-2(5)(6) \cos (\angle A) \\
49 & =61-60 \cos (\angle A) \\
\cos (\angle A) & =\frac{1}{5}
\end{aligned}
$$

Since $\cos (\angle A)=\frac{1}{5}$ and $\angle A+\angle C=180^{\circ}$, then $\cos (\angle C)=-\cos \left(180^{\circ}-\angle A\right)=-\frac{1}{5}$.

(We could have calculated the actual size of $\angle A$ using $\cos (\angle A)=\frac{1}{5}$ and then used this to calculate the size of $\angle C$, but we would introduce the possibility of rounding error by doing this.)

Then, using the cosine law in $\triangle B C D$, we obtain

$$
\begin{aligned}
7^{2} & =4^{2}+C D^{2}-2(4)(C D) \cos (\angle C) \\
49 & =16+C D^{2}-8(C D)\left(-\frac{1}{5}\right) \\
0 & =5 C D^{2}+8 C D-165 \\
0 & =(5 C D+33)(C D-5)
\end{aligned}
$$

So $C D=-\frac{33}{5}$ or $C D=5$. (We could have also determined these roots using the quadratic formula.)

Since $C D$ is a length, it must be positive, so $C D=5$.

(We could have also proceeded by using the sine law in $\triangle B C D$ to determine $\angle B D C$ and then found the size of $\angle D B C$, which would have allowed us to calculate $C D$ using the sine law. However, this would again introduce the potential of rounding error.)

Final answer: 5
Subfield: Geometry


Question: In the diagram, $A B C D$ is a quadrilateral in which $\angle A+\angle C=180^{\circ}$. What is the length of $C D$ ?

![](https://cdn.mathpix.com/cropped/2023_12_21_9babb378c9215bc0c2fdg-1.jpg?height=444&width=437&top_left_y=735&top_left_x=1256)

Solution:
In order to determine $C D$, we must determine one of the angles (or at least some information about one of the angles) in $\triangle B C D$.

To do this, we look at $\angle A$ use the fact that $\angle A+\angle C=180^{\circ}$.



![](https://cdn.mathpix.com/cropped/2023_12_21_72eb55817192542b5c6cg-1.jpg?height=442&width=439&top_left_y=153&top_left_x=946)

Using the cosine law in $\triangle A B D$, we obtain

$$
\begin{aligned}
7^{2} & =5^{2}+6^{2}-2(5)(6) \cos (\angle A) \\
49 & =61-60 \cos (\angle A) \\
\cos (\angle A) & =\frac{1}{5}
\end{aligned}
$$

Since $\cos (\angle A)=\frac{1}{5}$ and $\angle A+\angle C=180^{\circ}$, then $\cos (\angle C)=-\cos \left(180^{\circ}-\angle A\right)=-\frac{1}{5}$.

(We could have calculated the actual size of $\angle A$ using $\cos (\angle A)=\frac{1}{5}$ and then used this to calculate the size of $\angle C$, but we would introduce the possibility of rounding error by doing this.)

Then, using the cosine law in $\triangle B C D$, we obtain

$$
\begin{aligned}
7^{2} & =4^{2}+C D^{2}-2(4)(C D) \cos (\angle C) \\
49 & =16+C D^{2}-8(C D)\left(-\frac{1}{5}\right) \\
0 & =5 C D^{2}+8 C D-165 \\
0 & =(5 C D+33)(C D-5)
\end{aligned}
$$

So $C D=-\frac{33}{5}$ or $C D=5$. (We could have also determined these roots using the quadratic formula.)

Since $C D$ is a length, it must be positive, so $C D=5$.

(We could have also proceeded by using the sine law in $\triangle B C D$ to determine $\angle B D C$ and then found the size of $\angle D B C$, which would have allowed us to calculate $C D$ using the sine law. However, this would again introduce the potential of rounding error.)

Final answer: 5
Subfield: Algebra


Question: If $f(x)=\sin ^{2} x-2 \sin x+2$, what are the minimum and maximum values of $f(x)$ ?

Solution:
We rewrite by completing the square as $f(x)=\sin ^{2} x-2 \sin x+2=(\sin x-1)^{2}+1$.

Therefore, since $(\sin x-1)^{2} \geq 0$, then $f(x) \geq 1$, and in fact $f(x)=1$ when $\sin x=1$ (which occurs for instance when $x=90^{\circ}$ ).

Thus, the minimum value of $f(x)$ is 1 .

To maximize $f(x)$, we must maximize $(\sin x-1)^{2}$.

Since $-1 \leq \sin x \leq 1$, then $(\sin x-1)^{2}$ is maximized when $\sin x=-1$ (for instance, when $\left.x=270^{\circ}\right)$. In this case, $(\sin x-1)^{2}=4$, so $f(x)=5$.

Thus, the maximum value of $f(x)$ is 5 .

Final answer: 5,1
Subfield: Geometry


Question: In the diagram, the parabola

$$
y=-\frac{1}{4}(x-r)(x-s)
$$

intersects the axes at three points. The vertex of this parabola is the point $V$. Determine the value of $k$ and the coordinates of $V$.

![](https://cdn.mathpix.com/cropped/2023_12_21_9babb378c9215bc0c2fdg-1.jpg?height=550&width=566&top_left_y=1346&top_left_x=1251)

Solution:
From the diagram, the $x$-intercepts of the parabola are $x=-k$ and $x=3 k$.



![](https://cdn.mathpix.com/cropped/2023_12_21_f83bb2c63ced8ec40925g-1.jpg?height=544&width=571&top_left_y=167&top_left_x=880)

Since we are given that $y=-\frac{1}{4}(x-r)(x-s)$, then the $x$-intercepts are $r$ and $s$, so $r$ and $s$ equal $-k$ and $3 k$ in some order.

Therefore, we can rewrite the parabola as $y=-\frac{1}{4}(x-(-k))(x-3 k)$.

Since the point $(0,3 k)$ lies on the parabola, then $3 k=-\frac{1}{4}(0+k)(0-3 k)$ or $12 k=3 k^{2}$ or $k^{2}-4 k=0$ or $k(k-4)=0$.

Thus, $k=0$ or $k=4$.

Since the two roots are distinct, then we cannot have $k=0$ (otherwise both $x$-intercepts would be 0 ).

Thus, $k=4$.

This tells us that the equation of the parabola is $y=-\frac{1}{4}(x+4)(x-12)$ or $y=-\frac{1}{4} x^{2}+$ $2 x+12$.

We still have to determine the coordinates of the vertex, $V$.

Since the $x$-intercepts of the parabola are -4 and 12 , then the $x$-coordinate of the vertex is the average of these intercepts, or 4.

(We could have also used the fact that the $x$-coordinate is $-\frac{b}{2 a}=-\frac{2}{2\left(-\frac{1}{4}\right)}$.)

Therefore, the $y$-coordinate of the vertex is $y=-\frac{1}{4}\left(4^{2}\right)+2(4)+12=16$.

Thus, the coordinates of the vertex are $(4,16)$.

Final answer: $4,(4,16)$
Subfield: Geometry


Question: In the diagram, the parabola

$$
y=-\frac{1}{4}(x-r)(x-s)
$$

intersects the axes at three points. The vertex of this parabola is the point $V$. Determine the value of $k$ and the coordinates of $V$.

<img_3872>

Solution:
From the diagram, the $x$-intercepts of the parabola are $x=-k$ and $x=3 k$.



<img_3883>

Since we are given that $y=-\frac{1}{4}(x-r)(x-s)$, then the $x$-intercepts are $r$ and $s$, so $r$ and $s$ equal $-k$ and $3 k$ in some order.

Therefore, we can rewrite the parabola as $y=-\frac{1}{4}(x-(-k))(x-3 k)$.

Since the point $(0,3 k)$ lies on the parabola, then $3 k=-\frac{1}{4}(0+k)(0-3 k)$ or $12 k=3 k^{2}$ or $k^{2}-4 k=0$ or $k(k-4)=0$.

Thus, $k=0$ or $k=4$.

Since the two roots are distinct, then we cannot have $k=0$ (otherwise both $x$-intercepts would be 0 ).

Thus, $k=4$.

This tells us that the equation of the parabola is $y=-\frac{1}{4}(x+4)(x-12)$ or $y=-\frac{1}{4} x^{2}+$ $2 x+12$.

We still have to determine the coordinates of the vertex, $V$.

Since the $x$-intercepts of the parabola are -4 and 12 , then the $x$-coordinate of the vertex is the average of these intercepts, or 4.

(We could have also used the fact that the $x$-coordinate is $-\frac{b}{2 a}=-\frac{2}{2\left(-\frac{1}{4}\right)}$.)

Therefore, the $y$-coordinate of the vertex is $y=-\frac{1}{4}\left(4^{2}\right)+2(4)+12=16$.

Thus, the coordinates of the vertex are $(4,16)$.

Final answer: $4,(4,16)$
Subfield: Geometry


Question: In the diagram, two circles are tangent to each other at point $B$. A straight line is drawn through $B$ cutting the two circles at $A$ and $C$, as shown. Tangent lines are drawn to the circles at $A$ and $C$. Prove that these two tangent lines are parallel.

<img_3533>

Solution:
Let the centres of the two circles be $O_{1}$ and $O_{2}$.

Join $A$ and $B$ to $O_{1}$ and $B$ and $C$ to $O_{2}$.

Designate two points $W$ and $X$ on either side of $A$ on one tangent line, and two points $Y$ and $Z$ on either side of $C$ on the other tangent line.

<img_3553>

Let $\angle X A B=\theta$.

Since $W X$ is tangent to the circle with centre $O_{1}$ at $A$, then $O_{1} A$ is perpendicular to $W X$, so $\angle O_{1} A B=90^{\circ}-\theta$.

Since $O_{1} A=O_{1} B$ because both are radii, then $\triangle A O_{1} B$ is isosceles, so $\angle O_{1} B A=$ $\angle O_{1} A B=90^{\circ}-\theta$.

Since the two circles are tangent at $B$, then the line segment joining $O_{1}$ and $O_{2}$ passes through $B$, ie. $O_{1} B O_{2}$ is a straight line segment.

Thus, $\angle O_{2} B C=\angle O_{1} B A=90^{\circ}-\theta$, by opposite angles.

Since $O_{2} B=O_{2} C$, then similarly to above, $\angle O_{2} C B=\angle O_{2} B C=90^{\circ}-\theta$.

Since $Y Z$ is tangent to the circle with centre $O_{2}$ at $C$, then $O_{2} C$ is perpendicular to $Y Z$. Thus, $\angle Y C B=90^{\circ}-\angle O_{2} C B=\theta$.

Since $\angle X A B=\angle Y C B$, then $W X$ is parallel to $Y Z$, by alternate angles, as required.
Let the centres of the two circles be $O_{1}$ and $O_{2}$.

Join $A$ and $B$ to $O_{1}$ and $B$ and $C$ to $O_{2}$.

Since $A O_{1}$ and $B O_{1}$ are radii of the same circle, $A O_{1}=B O_{1}$ so $\triangle A O_{1} B$ is isosceles, so $\angle O_{1} A B=\angle O_{1} B A$.

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Since $\mathrm{BO}_{2}$ and $\mathrm{CO}_{2}$ are radii of the same circle, $B O_{2}=C_{2}$ so $\triangle B O_{2} C$ is isosceles, so $\angle O_{2} B C=\angle O_{2} C B$.

Since the two circles are tangent at $B$, then $O_{1} B O_{2}$ is a line segment (ie. the line segment joining $O_{1}$ and $O_{2}$ passes through the point of tangency of the two circles).

Since $O_{1} B O_{2}$ is straight, then $\angle O_{1} B A=\angle O_{2} B C$, by opposite angles.

Thus, $\angle O_{1} A B=\angle O_{1} B A=\angle O_{2} B C=\angle O_{2} C B$.

This tells us that $\triangle A O_{1} B$ is similar to $\triangle B O_{2} C$, so $\angle A O_{1} B=\angle B O_{2} C$ or $\angle A O_{1} O_{2}=$ $\angle C O_{2} O_{1}$.

Therefore, $A O_{1}$ is parallel to $C_{2}$, by alternate angles.

But $A$ and $C$ are points of tangency, $A O_{1}$ is perpendicular to the tangent line at $A$ and $\mathrm{CO}_{2}$ is perpendicular to the tangent line at $C$.

Since $A O_{1}$ and $C O_{2}$ are parallel, then the two tangent lines must be parallel.