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


Question: Prove that it is not possible to create a sequence of 4 numbers $a, b, c, d$, such that the sum of any two consecutive terms is positive, and the sum of any three consecutive terms is negative.

Solution:
Assume such a sequence $a, b, c, d$ exists. (We proceed by contradiction.)

Since the sum of any two consecutive terms is positive, $a+b>0, b+c>0$, and

$c+d>0$. Adding these three inequalities, $(a+b)+(b+c)+(c+d)>0$ or

$a+2 b+2 c+d>0$.

We are going to show that this statement contradicts the facts that are known about the sequence. We are told that the sum of any three consecutive terms is negative, ie.

$a+b+c<0$ and $b+c+d<0$. Adding these two inequalities, $(a+b+c)+(b+c+d)<0$ or $a+2 b+2 c+d<0$.

This is a contradiction, since the two conditions $a+2 b+2 c+d>0$ and $a+2 b+2 c+d<0$ cannot occur simultaneously.

Therefore, our original assumption is false, and so no such sequence exists.
Assume such a sequence $a, b, c, d$ exists. (We proceed by contradiction.)

We consider two cases.

Case 1: $a \leq 0$

In this case, $b>0$ since $a+b>0$

Then, since $a+b+c<0$, we must have that $c<0$.

But $c+d>0$, so $d>0$.

This means that we have $b>0$ and $c+d>0$, ie. $b+c+d>0$.

But from the conditions on the sequence, $b+c+d<0$, a contradiction.

Therefore, no such sequence exists with $a \leq 0$.

Case 2: $a>0$

In this case, it is not immediately clear whether $b$ has to be positive or negative.

However, we do know that $a+b>0$ and $a+b+c<0$, so it must be true that $c<0$.

Then since $b+c>0$ and $c+d>0$, we must have both $b>0$ and $d>0$. But then $b+c+d=b+(c+d)>0$ since $c+d>0$ and $b>0$.

This is again a contradiction.

Therefore, no such sequence exists with $a>0$.
Subfield: Geometry


Question: In triangle $A B C, \angle A B C=90^{\circ}$. Rectangle $D E F G$ is inscribed in $\triangle A B C$, as shown. Squares $J K G H$ and $M L F N$ are inscribed in $\triangle A G D$ and $\triangle C F E$, respectively. If the side length of $J H G K$ is $v$, the side length of $M L F N$ is $w$, and $D G=u$, prove that $u=v+w$.

<img_3545>

Solution:
Let $\angle B A C=\theta$. Then by parallel lines, $\angle D J H=\angle B D E=\theta$.

Thus, $\angle B E D=90^{\circ}-\theta$ and so

$\angle N E M=\theta$ since

$\angle D E F=90^{\circ}$.

Since $D G=u$ and $H G=v$,

then $D H=u-v$.

<img_3789>

Similarly, $E N=u-w$.

Looking at $\triangle D H J$ and $\triangle M N E$, we see that $\tan \theta=\frac{u-v}{v}$ and $\tan \theta=\frac{w}{u-w}$.

Therefore,

$$
\begin{aligned}
\frac{u-v}{v} & =\frac{w}{u-w} \\
(u-v)(u-w) & =v w \\
u^{2}-u v-u w+v w & =v w \\
u(u-v-w) & =0
\end{aligned}
$$

and since $u \neq 0$, we must have $u-v-w=0$ or $u=v+w$.

[Note: If $u=0$, then the height of rectangle $D E F G$ is 0 , ie. $D$ coincides with point $A$ and $E$ coincides with point $C$, which says that we must also have $v=w=0$, ie. the squares have no place to go!]
Subfield: Geometry


Question: In triangle $A B C, \angle A B C=90^{\circ}$. Rectangle $D E F G$ is inscribed in $\triangle A B C$, as shown. Squares $J K G H$ and $M L F N$ are inscribed in $\triangle A G D$ and $\triangle C F E$, respectively. If the side length of $J H G K$ is $v$, the side length of $M L F N$ is $w$, and $D G=u$, prove that $u=v+w$.

![](https://cdn.mathpix.com/cropped/2023_12_21_02b1e9456f98fdbe5dd9g-1.jpg?height=274&width=477&top_left_y=199&top_left_x=1298)

Solution:
Let $\angle B A C=\theta$. Then by parallel lines, $\angle D J H=\angle B D E=\theta$.

Thus, $\angle B E D=90^{\circ}-\theta$ and so

$\angle N E M=\theta$ since

$\angle D E F=90^{\circ}$.

Since $D G=u$ and $H G=v$,

then $D H=u-v$.

![](https://cdn.mathpix.com/cropped/2023_12_21_98602fddd2dfab207e62g-1.jpg?height=450&width=856&top_left_y=306&top_left_x=930)

Similarly, $E N=u-w$.

Looking at $\triangle D H J$ and $\triangle M N E$, we see that $\tan \theta=\frac{u-v}{v}$ and $\tan \theta=\frac{w}{u-w}$.

Therefore,

$$
\begin{aligned}
\frac{u-v}{v} & =\frac{w}{u-w} \\
(u-v)(u-w) & =v w \\
u^{2}-u v-u w+v w & =v w \\
u(u-v-w) & =0
\end{aligned}
$$

and since $u \neq 0$, we must have $u-v-w=0$ or $u=v+w$.

[Note: If $u=0$, then the height of rectangle $D E F G$ is 0 , ie. $D$ coincides with point $A$ and $E$ coincides with point $C$, which says that we must also have $v=w=0$, ie. the squares have no place to go!]

Final answer: 证明题，略
Subfield: Geometry


Question: Three thin metal rods of lengths 9,12 and 15 are welded together to form a right-angled triangle, which is held in a horizontal position. A solid sphere of radius 5 rests in the triangle so that it is tangent to each of the three sides. Assuming that the thickness of the rods can be neglected, how high above the plane of the triangle is the top of the sphere?

Solution:
Consider the cross-section of the sphere in the plane defined by the triangle. This crosssection will be a circle, since any cross-section of a sphere is a circle. This circle will be tangent to the three sides of the triangle, ie. will be the inscribed circle (or incircle) of the triangle. Let the centre of this circle be $O$, and its radius be $r$. We calculate the value of $r$.

<img_3449>

Join $O$ to the three points of tangency, $P, Q, R$, and to the three vertices $A, B, C$. Then $O P, O Q$ and $O R$ (radii) will form right angles with the three sides of the triangle. Consider the three triangles $\triangle A O B$, $\triangle B O C$ and $\triangle C O A$. Each of these triangles has a height of $r$ and they have bases 15, 9 and 12, respectively. Since the area of $\triangle A B C$ is equal to the sum of the areas of $\triangle A O B, \triangle B O C$, and $\triangle C O A$, So comparing areas,

$$
\begin{aligned}
\frac{1}{2}(9)(12) & =\frac{1}{2}(9)(r)+\frac{1}{2}(12)(r)+\frac{1}{2}(15)(r) \\
54 & =\frac{1}{2} r(9+12+15) \\
r & =3
\end{aligned}
$$

<img_3572>

Now join the centre of the cross-sectional circle to the centre of the sphere and let this distance be $h$. Now, the line joining the centre of the circle to the centre of the sphere will be perpendicular to the plane of the triangle, so we can form a right-angled triangle by joining the centre of the sphere to any point on the circumference of the cross-sectional circle. By Pythagoras,

$$
\begin{aligned}
h^{2}+r^{2} & =25 \\
h & =4
\end{aligned}
$$

This tells us that the top of the sphere is 9 units above the plane of the triangle, since the top of the sphere is 5 units above the centre of the sphere.

Final answer: 5
Subfield: Geometry


Question: A triangle is called Heronian if each of its side lengths is an integer and its area is also an integer. A triangle is called Pythagorean if it is right-angled and each of its side lengths is an integer.
Show that every Pythagorean triangle is Heronian.

Solution:
Consider a Pythagorean triangle with integer side lengths $a, b, c$ satisfying $a^{2}+b^{2}=c^{2}$. To show that this triangle is Heronian, we must show that it has an integer area. Now we know that the area is equal to $\frac{1}{2} a b$, so we must show that either $a$ or $b$ is an even integer.

<img_3317>

Suppose that both $a$ and $b$ are odd. (We proceed by contradiction.)

In this case, let $a=2 k+1$ and $b=2 l+1$. Then both $a^{2}$ and $b^{2}$ are odd, and so $c^{2}$ is even since $a^{2}+b^{2}=c^{2}$. Therefore, $c$ itself must be even, so let $c=2 m$.

Therefore,

$$
\begin{aligned}
(2 k+1)^{2}+(2 l+1)^{2} & =(2 m)^{2} \\
4 k^{2}+4 k+1+4 l^{2}+4 l+1 & =4 m^{2} \\
4\left(k^{2}+k+l^{2}+l\right)+2 & =4\left(m^{2}\right)
\end{aligned}
$$

But the right side is a multiple of 4 , and the left side is not a multiple of 4 . This is a contradiction.

Therefore, one of $a$ or $b$ must be even, and so the area of the triangle is an integer. Thus, any Pythagorean triangle is Heronian.
Subfield: Geometry


Question: A triangle is called Heronian if each of its side lengths is an integer and its area is also an integer. A triangle is called Pythagorean if it is right-angled and each of its side lengths is an integer.
Show that every odd integer greater than 1 is a side length of some Pythagorean triangle.

Solution:
We examine the first few smallest Pythagorean triples:

$$
345 \quad\left(3^{2}=4+5\right)
$$



$$
\begin{array}{llll}
5 & 12 & 13 & \left(5^{2}=12+13\right) \\
6 & 8 & 10 & \text { (Does not fit pattern) } \\
7 & 24 & 25 & \left(7^{2}=24+25\right)
\end{array}
$$

It appears from the first few examples that perhaps we can form a Pythagorean triple by using any odd number greater than 1 as its shortest leg.

Next, we notice from the pattern that the sum of the second leg and the hypotenuse is the square of the shortest leg, and that these two side lengths differ by 1 .

Will this pattern always hold?

Let $a=2 k+1$ with $k \geq 1$. (This formula will generate all odd integers greater than or equal to 3.) Can we always find $b$ so that $c=b+1$ and $a^{2}+b^{2}=c^{2}$ ?

Consider the equation

$$
\begin{aligned}
(2 k+1)^{2}+b^{2} & =(b+1)^{2} \\
4 k^{2}+4 k+1+b^{2} & =b^{2}+2 b+1 \\
4 k^{2}+4 k & =2 b \\
b & =2 k^{2}+2 k
\end{aligned}
$$

So we can always find a $b$ to make the equation true. Therefore, since $a$ can be any odd integer greater than or equal to 3 , then we can make any odd number the shortest leg of a Pythagorean triangle, namely the Pythagorean triangle $a=2 k+1, b=2 k^{2}+2 k$, $c=2 k^{2}+2 k+1$. (Check that $a^{2}+b^{2}=c^{2}$ does indeed hold here!)
Subfield: Geometry


Question: A triangle is called Heronian if each of its side lengths is an integer and its area is also an integer. A triangle is called Pythagorean if it is right-angled and each of its side lengths is an integer.
Find a Heronian triangle which has all side lengths different, and no side length divisible by $3,5,7$ or 11 .

Solution:
We consider forming a triangle by joining two Pythagorean triangles along a common side. Since any Pythagorean triangle is Heronian, then the triangle that is formed by joining two Pythagorean triangles in the manner shown will have integer side lengths and will have integer area, thus making it Heronian.

![](https://cdn.mathpix.com/cropped/2023_12_21_d5600eed21b7775d92fdg-1.jpg?height=184&width=328&top_left_y=1529&top_left_x=1489)

So again, we make a list of Pythagorean triples

| 3 | 4 | 5 |
| :--- | :--- | :--- |
| 5 | 12 | 13 |
| 6 | 8 | 10 |
| 7 | 24 | 25 |
| 8 | 15 | 17 |
| 9 | 40 | 41 |
| 10 | 24 | 26 |
| 11 | 60 | 61 |



We notice that we can scale any Pythagorean triangle by an integer factor and obtain another Pythagorean triangle. This will enable us to create two Pythagorean triangles with a common side length.

Also, we note that when joining two Pythagorean triangles, the hypotenuse of each triangle becomes a side length in the new triangle. Since we cannot have a side length divisible by $3,5,7$ or 11 , this eliminates the $3-4-5,6-8-10$, and $7-24-25$ triangles from the list above.

Suppose we scale the 8-15-17 triangle by a factor of 4 to obtain $32-60-68$ and join to the 11-60-61 triangle in the manner shown.

Thus we obtain a 43-61-68 triangle, which has integer area because its height is an even integer.

Therefore, a 43-61-68 triangle is Heronian.

Final answer: a 43-61-68 triangle
Subfield: Geometry


Question: Triangle $A B C$ has vertices $A(0,5), B(3,0)$ and $C(8,3)$. Determine the measure of $\angle A C B$.

Solution:
First, we calculate the side lengths of $\triangle A B C$ :

$$
\begin{aligned}
& A B=\sqrt{(0-3)^{2}+(5-0)^{2}}=\sqrt{34} \\
& B C=\sqrt{(3-8)^{2}+(0-3)^{2}}=\sqrt{34} \\
& A C=\sqrt{(0-8)^{2}+(5-3)^{2}}=\sqrt{68}
\end{aligned}
$$

Since $A B=B C$ and $A C=\sqrt{2} A B=\sqrt{2} B C$, then $\triangle A B C$ is an isosceles right-angled triangle, with the

<img_3758>
right angle at $B$.

Therefore, $\angle A C B=45^{\circ}$.
First, we calculate the side lengths of $\triangle A B C$ :

$$
\begin{aligned}
& A B=\sqrt{(0-3)^{2}+(5-0)^{2}}=\sqrt{34} \\
& B C=\sqrt{(3-8)^{2}+(0-3)^{2}}=\sqrt{34} \\
& A C=\sqrt{(0-8)^{2}+(5-3)^{2}}=\sqrt{68}
\end{aligned}
$$

Line segment $A B$ has slope $\frac{5-0}{0-3}=-\frac{5}{3}$.

Line segment $B C$ has slope $\frac{0-3}{3-8}=\frac{3}{5}$.

Since the product of these two slopes is -1 , then $A B$ and $B C$ are perpendicular.

Therefore, $\triangle A B C$ is right-angled at $B$.

Since $A B=B C$, then $\triangle A B C$ is an isosceles right-angled triangle, so $\angle A C B=45^{\circ}$.
First, we calculate the side lengths of $\triangle A B C$ :

$$
\begin{aligned}
& A B=\sqrt{(0-3)^{2}+(5-0)^{2}}=\sqrt{34} \\
& B C=\sqrt{(3-8)^{2}+(0-3)^{2}}=\sqrt{34} \\
& A C=\sqrt{(0-8)^{2}+(5-3)^{2}}=\sqrt{68}
\end{aligned}
$$

Using the cosine law,

$$
\begin{aligned}
A B^{2} & =A C^{2}+B C^{2}-2(A C)(B C) \cos (\angle A C B) \\
34 & =68+34-2(\sqrt{68})(\sqrt{34}) \cos (\angle A C B) \\
0 & =68-2(\sqrt{2} \sqrt{34})(\sqrt{34}) \cos (\angle A C B) \\
0 & =68-68 \sqrt{2} \cos (\angle A C B) \\
68 \sqrt{2} \cos (\angle A C B) & =68 \\
\cos (\angle A C B) & =\frac{1}{\sqrt{2}}
\end{aligned}
$$

Since $\cos (\angle A C B)=\frac{1}{\sqrt{2}}$ and $0^{\circ}<\angle A C B<180^{\circ}$, then $\angle A C B=45^{\circ}$.

Final answer: $45^{\circ}$
Subfield: Geometry


Question: In the diagram, $P Q R S$ is an isosceles trapezoid with $P Q=7, P S=Q R=8$, and $S R=15$. Determine the length of the diagonal $P R$.

![](https://cdn.mathpix.com/cropped/2023_12_21_040d04737e70e698092cg-1.jpg?height=287&width=607&top_left_y=770&top_left_x=1212)

Solution:
Draw perpendiculars from $P$ and $Q$ to $X$ and $Y$, respectively, on $S R$.

![](https://cdn.mathpix.com/cropped/2023_12_21_8bce42a3784597e137dag-1.jpg?height=298&width=609&top_left_y=241&top_left_x=861)

Since $P Q$ is parallel to $S R$ (because $P Q R S$ is a trapezoid) and $P X$ and $Q Y$ are perpendicular to $S R$, then $P Q Y X$ is a rectangle.

Thus, $X Y=P Q=7$ and $P X=Q Y$.

Since $\triangle P X S$ and $\triangle Q Y R$ are right-angled with $P S=Q R$ and $P X=Q Y$, then these triangles are congruent, and so $S X=Y R$.

Since $X Y=7$ and $S R=15$, then $S X+7+Y R=15$ or $2 \times S X=8$ and so $S X=4$.

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

$$
P X^{2}=P S^{2}-S X^{2}=8^{2}-4^{2}=64-16=48
$$

Now $P R$ is the hypotenuse of right-angled $\triangle P X R$.

Since $P R>0$, then by the Pythagorean Theorem,

$$
P R=\sqrt{P X^{2}+X R^{2}}=\sqrt{48+(7+4)^{2}}=\sqrt{48+11^{2}}=\sqrt{48+121}=\sqrt{169}=13
$$

Therefore, $P R=13$.

Final answer: 13
Subfield: Geometry


Question: In the diagram, $P Q R S$ is an isosceles trapezoid with $P Q=7, P S=Q R=8$, and $S R=15$. Determine the length of the diagonal $P R$.

<img_3239>

Solution:
Draw perpendiculars from $P$ and $Q$ to $X$ and $Y$, respectively, on $S R$.

<img_3755>

Since $P Q$ is parallel to $S R$ (because $P Q R S$ is a trapezoid) and $P X$ and $Q Y$ are perpendicular to $S R$, then $P Q Y X$ is a rectangle.

Thus, $X Y=P Q=7$ and $P X=Q Y$.

Since $\triangle P X S$ and $\triangle Q Y R$ are right-angled with $P S=Q R$ and $P X=Q Y$, then these triangles are congruent, and so $S X=Y R$.

Since $X Y=7$ and $S R=15$, then $S X+7+Y R=15$ or $2 \times S X=8$ and so $S X=4$.

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

$$
P X^{2}=P S^{2}-S X^{2}=8^{2}-4^{2}=64-16=48
$$

Now $P R$ is the hypotenuse of right-angled $\triangle P X R$.

Since $P R>0$, then by the Pythagorean Theorem,

$$
P R=\sqrt{P X^{2}+X R^{2}}=\sqrt{48+(7+4)^{2}}=\sqrt{48+11^{2}}=\sqrt{48+121}=\sqrt{169}=13
$$

Therefore, $P R=13$.

Final answer: 13
Subfield: Combinatorics


Question: Blaise and Pierre will play 6 games of squash. Since they are equally skilled, each is equally likely to win any given game. (In squash, there are no ties.) The probability that each of them will win 3 of the 6 games is $\frac{5}{16}$. What is the probability that Blaise will win more games than Pierre?

Solution:
There are two possibilities: either each player wins three games or one player wins more games than the other.

Since the probability that each player wins three games is $\frac{5}{16}$, then the probability that any one player wins more games than the other is $1-\frac{5}{16}=\frac{11}{16}$.

Since each of Blaise and Pierre is equally likely to win any given game, then each must be equally likely to win more games than the other.

Therefore, the probability that Blaise wins more games than Pierre is $\frac{1}{2} \times \frac{11}{16}=\frac{11}{32}$.
We consider the results of the 6 games as a sequence of 6 Bs or Ps, with each letter a B if Blaise wins the corresponding game or $\mathrm{P}$ if Pierre wins.

Since the two players are equally skilled, then the probability that each wins a given game is $\frac{1}{2}$. This means that the probability of each letter being a $B$ is $\frac{1}{2}$ and the probability of each letter being a $\mathrm{P}$ is also $\frac{1}{2}$.

Since each sequence consists of 6 letters, then the probability of a particular sequence occurring is $\left(\frac{1}{2}\right)^{6}=\frac{1}{64}$, because each of the letters is specified.

Since they play 6 games in total, then the probability that Blaise wins more games than Pierre is the sum of the probabilities that Blaise wins 4 games, that Blaise wins 5 games, and that Blaise wins 6 games.

If Blaise wins 6 games, then the sequence consists of 6 Bs. The probability of this is $\frac{1}{64}$, since there is only one way to arrange $6 \mathrm{Bs}$.

If Blaise wins 5 games, then the sequence consists of $5 \mathrm{Bs}$ and $1 \mathrm{P}$. The probability of this is $6 \times \frac{1}{64}=\frac{6}{64}$, since there are 6 possible positions in the list for the $1 \mathrm{P}$ (eg. PBBBBB,BPBBBB, BBPBBB, BBBPBB, BBBBPB, BBBBBP).

The probability that Blaise wins 4 games is $\left(\begin{array}{l}6 \\ 2\end{array}\right) \times \frac{1}{64}=\frac{15}{64}$, since there are $\left(\begin{array}{l}6 \\ 2\end{array}\right)=15$ ways for 4 Bs and 2 Ps to be arranged.

Therefore, the probability that Blaise wins more games than Pierre is $\frac{1}{64}+\frac{6}{64}+\frac{15}{64}=\frac{22}{64}=\frac{11}{32}$.

Final answer: $\frac{11}{32}$
Subfield: Algebra


Question: Determine all real values of $x$ for which

$$
3^{x+2}+2^{x+2}+2^{x}=2^{x+5}+3^{x}
$$

Solution:
Using exponent rules and arithmetic, we manipulate the given equation:

$$
\begin{aligned}
3^{x+2}+2^{x+2}+2^{x} & =2^{x+5}+3^{x} \\
3^{x} 3^{2}+2^{x} 2^{2}+2^{x} & =2^{x} 2^{5}+3^{x} \\
9\left(3^{x}\right)+4\left(2^{x}\right)+2^{x} & =32\left(2^{x}\right)+3^{x} \\
8\left(3^{x}\right) & =27\left(2^{x}\right) \\
\frac{3^{x}}{2^{x}} & =\frac{27}{8} \\
\left(\frac{3}{2}\right)^{x} & =\left(\frac{3}{2}\right)^{3}
\end{aligned}
$$

Since the two expressions are equal and the bases are equal, then the exponents must be equal, so $x=3$.

Final answer: 3
Subfield: Geometry


Question: In the diagram, $\triangle A B C$ has $A B=A C$ and $\angle B A C<60^{\circ}$. Point $D$ is on $A C$ with $B C=B D$. Point $E$ is on $A B$ with $B E=E D$. If $\angle B A C=\theta$, determine $\angle B E D$ in terms of $\theta$.

<img_3901>

Solution:
Since $A B=A C$, then $\triangle A B C$ is isosceles and $\angle A B C=\angle A C B$. Note that $\angle B A C=\theta$.

<img_3938>

The angles in $\triangle A B C$ add to $180^{\circ}$, so $\angle A B C+\angle A C B+\angle B A C=180^{\circ}$.

Thus, $2 \angle A C B+\theta=180^{\circ}$ or $\angle A B C=\angle A C B=\frac{1}{2}\left(180^{\circ}-\theta\right)=90^{\circ}-\frac{1}{2} \theta$.

Now $\triangle B C D$ is isosceles as well with $B C=B D$ and so $\angle C D B=\angle D C B=90^{\circ}-\frac{1}{2} \theta$.

Since the angles in $\triangle B C D$ add to $180^{\circ}$, then

$$
\angle C B D=180^{\circ}-\angle D C B-\angle C D B=180^{\circ}-\left(90^{\circ}-\frac{1}{2} \theta\right)-\left(90^{\circ}-\frac{1}{2} \theta\right)=\theta
$$

Now $\angle E B D=\angle A B C-\angle D B C=\left(90^{\circ}-\frac{1}{2} \theta\right)-\theta=90^{\circ}-\frac{3}{2} \theta$.

Since $B E=E D$, then $\angle E D B=\angle E B D=90^{\circ}-\frac{3}{2} \theta$.

Therefore, $\angle B E D=180^{\circ}-\angle E B D-\angle E D B=180^{\circ}-\left(90^{\circ}-\frac{3}{2} \theta\right)-\left(90^{\circ}-\frac{3}{2} \theta\right)=3 \theta$.

Final answer: $3 \theta$
Subfield: Geometry


Question: In the diagram, $\triangle A B C$ has $A B=A C$ and $\angle B A C<60^{\circ}$. Point $D$ is on $A C$ with $B C=B D$. Point $E$ is on $A B$ with $B E=E D$. If $\angle B A C=\theta$, determine $\angle B E D$ in terms of $\theta$.

![](https://cdn.mathpix.com/cropped/2023_12_21_040d04737e70e698092cg-1.jpg?height=496&width=455&top_left_y=1628&top_left_x=1320)

Solution:
Since $A B=A C$, then $\triangle A B C$ is isosceles and $\angle A B C=\angle A C B$. Note that $\angle B A C=\theta$.

![](https://cdn.mathpix.com/cropped/2023_12_21_920fb565baecc5ceb3fcg-1.jpg?height=494&width=460&top_left_y=1260&top_left_x=930)

The angles in $\triangle A B C$ add to $180^{\circ}$, so $\angle A B C+\angle A C B+\angle B A C=180^{\circ}$.

Thus, $2 \angle A C B+\theta=180^{\circ}$ or $\angle A B C=\angle A C B=\frac{1}{2}\left(180^{\circ}-\theta\right)=90^{\circ}-\frac{1}{2} \theta$.

Now $\triangle B C D$ is isosceles as well with $B C=B D$ and so $\angle C D B=\angle D C B=90^{\circ}-\frac{1}{2} \theta$.

Since the angles in $\triangle B C D$ add to $180^{\circ}$, then

$$
\angle C B D=180^{\circ}-\angle D C B-\angle C D B=180^{\circ}-\left(90^{\circ}-\frac{1}{2} \theta\right)-\left(90^{\circ}-\frac{1}{2} \theta\right)=\theta
$$

Now $\angle E B D=\angle A B C-\angle D B C=\left(90^{\circ}-\frac{1}{2} \theta\right)-\theta=90^{\circ}-\frac{3}{2} \theta$.

Since $B E=E D$, then $\angle E D B=\angle E B D=90^{\circ}-\frac{3}{2} \theta$.

Therefore, $\angle B E D=180^{\circ}-\angle E B D-\angle E D B=180^{\circ}-\left(90^{\circ}-\frac{3}{2} \theta\right)-\left(90^{\circ}-\frac{3}{2} \theta\right)=3 \theta$.

Final answer: $3 \theta$
Subfield: Geometry


Question: In the diagram, the ferris wheel has a diameter of $18 \mathrm{~m}$ and rotates at a constant rate. When Kolapo rides the ferris wheel and is at its lowest point, he is $1 \mathrm{~m}$ above the ground. When Kolapo is at point $P$ that is $16 \mathrm{~m}$ above the ground and is rising, it takes him 4 seconds to reach the highest point, $T$. He continues to travel for another 8 seconds reaching point $Q$. Determine Kolapo's height above the ground when he reaches point $Q$.

<img_3685>

Solution:
Let $O$ be the centre of the ferris wheel and $B$ the lowest point on the wheel.

Since the radius of the ferris wheel is $9 \mathrm{~m}$ (half of the diameter of $18 \mathrm{~m}$ ) and $B$ is $1 \mathrm{~m}$ above the ground, then $O$ is $9+1=10 \mathrm{~m}$ above the ground.

Let $\angle T O P=\theta$.

<img_3230>

Since the ferris wheel rotates at a constant speed, then in 8 seconds, the angle through which the wheel rotates is twice the angle through which it rotates in 4 seconds. In other words, $\angle T O Q=2 \theta$.

Draw a perpendicular from $P$ to $R$ on $T B$ and from $Q$ to $G$ on $T B$.

Since $P$ is $16 \mathrm{~m}$ above the ground and $O$ is $10 \mathrm{~m}$ above the ground, then $O R=6 \mathrm{~m}$.

Since $O P$ is a radius of the circle, then $O P=9 \mathrm{~m}$.

Looking at right-angled $\triangle O R P$, we see that $\cos \theta=\frac{O R}{O P}=\frac{6}{9}=\frac{2}{3}$.

Since $\cos \theta=\frac{2}{3}<\frac{1}{\sqrt{2}}=\cos \left(45^{\circ}\right)$, then $\theta>45^{\circ}$.

This means that $2 \theta>90^{\circ}$, which means that $Q$ is below the horizontal diameter through $O$ and so $G$ is below $O$.

Since $\angle T O Q=2 \theta$, then $\angle Q O G=180^{\circ}-2 \theta$.

Kolapo's height above the ground at $Q$ equals $1 \mathrm{~m}$ plus the length of $B G$.

Now $B G=O B-O G$. We know that $O B=9 \mathrm{~m}$.

Also, considering right-angled $\triangle Q O G$, we have

$$
O G=O Q \cos (\angle Q O G)=9 \cos \left(180^{\circ}-2 \theta\right)=-9 \cos (2 \theta)=-9\left(2 \cos ^{2} \theta-1\right)
$$

Since $\cos \theta=\frac{2}{3}$, then $O G=-9\left(2\left(\frac{2}{3}\right)^{2}-1\right)=-9\left(\frac{8}{9}-1\right)=1 \mathrm{~m}$.

Therefore, $B G=9-1=8 \mathrm{~m}$ and so $Q$ is $1+8=9 \mathrm{~m}$ above the ground.

Final answer: 9 m