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


Question: In the diagram, $\triangle A B C$ is right-angled at $B$ and $A B=10$. If $\cos (\angle B A C)=\frac{5}{13}$, what is the value of $\tan (\angle A C B)$ ?

![](https://cdn.mathpix.com/cropped/2023_12_21_e7d09dafb0c4acfee33dg-1.jpg?height=385&width=284&top_left_y=1678&top_left_x=1251)

Solution:
Since $\cos (\angle B A C)=\frac{A B}{A C}$ and $\cos (\angle B A C)=\frac{5}{13}$ and $A B=10$, then $A C=\frac{13}{5} A B=26$.

Since $\triangle A B C$ is right-angled at $B$, then by the Pythagorean Theorem, $B C^{2}=A C^{2}-A B^{2}=26^{2}-10^{2}=576$ so $B C=24$ since $B C>0$.

Therefore, $\tan (\angle A C B)=\frac{A B}{B C}=\frac{10}{24}=\frac{5}{12}$.

Final answer: $\frac{5}{12}$
Subfield: Geometry


Question: In the diagram, $\triangle A B C$ is right-angled at $B$ and $A B=10$. If $\cos (\angle B A C)=\frac{5}{13}$, what is the value of $\tan (\angle A C B)$ ?

<img_4032>

Solution:
Since $\cos (\angle B A C)=\frac{A B}{A C}$ and $\cos (\angle B A C)=\frac{5}{13}$ and $A B=10$, then $A C=\frac{13}{5} A B=26$.

Since $\triangle A B C$ is right-angled at $B$, then by the Pythagorean Theorem, $B C^{2}=A C^{2}-A B^{2}=26^{2}-10^{2}=576$ so $B C=24$ since $B C>0$.

Therefore, $\tan (\angle A C B)=\frac{A B}{B C}=\frac{10}{24}=\frac{5}{12}$.

Final answer: $\frac{5}{12}$
Subfield: Geometry


Question: Suppose $0^{\circ}<x<90^{\circ}$ and $2 \sin ^{2} x+\cos ^{2} x=\frac{25}{16}$. What is the value of $\sin x$ ?

Solution:
Since $2 \sin ^{2} x+\cos ^{2} x=\frac{25}{16}$ and $\sin ^{2} x+\cos ^{2} x=1\left(\right.$ so $\left.\cos ^{2} x=1-\sin ^{2} x\right)$, then we get

$$
\begin{aligned}
2 \sin ^{2} x+\left(1-\sin ^{2} x\right) & =\frac{25}{16} \\
\sin ^{2} x & =\frac{25}{16}-1 \\
\sin ^{2} x & =\frac{9}{16} \\
\sin x & = \pm \frac{3}{4}
\end{aligned}
$$

so $\sin x=\frac{3}{4}$ since $\sin x>0$ because $0^{\circ}<x<90^{\circ}$.

Final answer: $\frac{3}{4}$
Subfield: Geometry


Question: In the diagram, $A B=B C=2 \sqrt{2}, C D=D E$, $\angle C D E=60^{\circ}$, and $\angle E A B=75^{\circ}$. Determine the perimeter of figure $A B C D E$. Explain how you got your answer.

<img_3224>

Solution:
Since $\triangle A B C$ is isosceles and right-angled, then $\angle B A C=45^{\circ}$.

Also, $A C=\sqrt{2} A B=\sqrt{2}(2 \sqrt{2})=4$.

Since $\angle E A B=75^{\circ}$ and $\angle B A C=45^{\circ}$, then $\angle C A E=\angle E A B-\angle B A C=30^{\circ}$.

Since $\triangle A E C$ is right-angled and has a $30^{\circ}$ angle, then $\triangle A E C$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.

Thus, $E C=\frac{1}{2} A C=2$ (since $E C$ is opposite the $30^{\circ}$ angle) and $A E=\frac{\sqrt{3}}{2} A C=2 \sqrt{3}$ (since $A E$ is opposite the $60^{\circ}$ angle).

In $\triangle C D E, E D=D C$ and $\angle E D C=60^{\circ}$, so $\triangle C D E$ is equilateral.

Therefore, $E D=C D=E C=2$.

Overall, the perimeter of $A B C D E$ is

$$
A B+B C+C D+D E+E A=2 \sqrt{2}+2 \sqrt{2}+2+2+2 \sqrt{3}=4+4 \sqrt{2}+2 \sqrt{3}
$$

Final answer: $4+4 \sqrt{2}+2 \sqrt{3}$
Subfield: Geometry


Question: In the diagram, $A B=B C=2 \sqrt{2}, C D=D E$, $\angle C D E=60^{\circ}$, and $\angle E A B=75^{\circ}$. Determine the perimeter of figure $A B C D E$. Explain how you got your answer.

![](https://cdn.mathpix.com/cropped/2023_12_21_e7d09dafb0c4acfee33dg-1.jpg?height=333&width=464&top_left_y=2208&top_left_x=1313)

Solution:
Since $\triangle A B C$ is isosceles and right-angled, then $\angle B A C=45^{\circ}$.

Also, $A C=\sqrt{2} A B=\sqrt{2}(2 \sqrt{2})=4$.

Since $\angle E A B=75^{\circ}$ and $\angle B A C=45^{\circ}$, then $\angle C A E=\angle E A B-\angle B A C=30^{\circ}$.

Since $\triangle A E C$ is right-angled and has a $30^{\circ}$ angle, then $\triangle A E C$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.

Thus, $E C=\frac{1}{2} A C=2$ (since $E C$ is opposite the $30^{\circ}$ angle) and $A E=\frac{\sqrt{3}}{2} A C=2 \sqrt{3}$ (since $A E$ is opposite the $60^{\circ}$ angle).

In $\triangle C D E, E D=D C$ and $\angle E D C=60^{\circ}$, so $\triangle C D E$ is equilateral.

Therefore, $E D=C D=E C=2$.

Overall, the perimeter of $A B C D E$ is

$$
A B+B C+C D+D E+E A=2 \sqrt{2}+2 \sqrt{2}+2+2+2 \sqrt{3}=4+4 \sqrt{2}+2 \sqrt{3}
$$

Final answer: $4+4 \sqrt{2}+2 \sqrt{3}$
Subfield: Algebra


Question: The first term of a sequence is 2007. Each term, starting with the second, is the sum of the cubes of the digits of the previous term. What is the 2007th term?

Solution:
From the given information, the first term in the sequence is 2007 and each term starting with the second can be determined from the previous term.

The second term is $2^{3}+0^{3}+0^{3}+7^{3}=8+0+0+343=351$.

The third term is $3^{3}+5^{3}+1^{3}=27+125+1=153$.

The fourth term is $1^{3}+5^{3}+3^{3}=27+125+1=153$.

Since two consecutive terms are equal, then every term thereafter will be equal, because each term depends only on the previous term and a term of 153 always makes the next term 153.

Thus, the 2007th term will be 153 .

Final answer: 153
Subfield: Algebra


Question: Sequence A has $n$th term $n^{2}-10 n+70$.

(The first three terms of sequence $\mathrm{A}$ are $61,54,49$. )

Sequence B is an arithmetic sequence with first term 5 and common difference 10. (The first three terms of sequence $\mathrm{B}$ are $5,15,25$.)

Determine all $n$ for which the $n$th term of sequence $\mathrm{A}$ is equal to the $n$th term of sequence B. Explain how you got your answer.

Solution:
The $n$th term of sequence $\mathrm{A}$ is $n^{2}-10 n+70$.

Since sequence B is arithmetic with first term 5 and common difference 10 , then the $n$th term of sequence $\mathrm{B}$ is equal to $5+10(n-1)=10 n-5$. (Note that this formula agrees with the first few terms.)

For the $n$th term of sequence $\mathrm{A}$ to be equal to the $n$th term of sequence $\mathrm{B}$, we must have

$$
\begin{aligned}
n^{2}-10 n+70 & =10 n-5 \\
n^{2}-20 n+75 & =0 \\
(n-5)(n-15) & =0
\end{aligned}
$$

Therefore, $n=5$ or $n=15$. That is, 5 th and 15 th terms of sequence $\mathrm{A}$ and sequence $\mathrm{B}$ are equal to each other.

Final answer: 5,15
Subfield: Algebra


Question: Determine all values of $x$ for which $2+\sqrt{x-2}=x-2$.

Solution:
Rearranging and then squaring both sides,

$$
\begin{aligned}
2+\sqrt{x-2} & =x-2 \\
\sqrt{x-2} & =x-4 \\
x-2 & =(x-4)^{2} \\
x-2 & =x^{2}-8 x+16 \\
0 & =x^{2}-9 x+18 \\
0 & =(x-3)(x-6)
\end{aligned}
$$

so $x=3$ or $x=6$.

We should check both solutions, because we may have introduced extraneous solutions by squaring.

If $x=3$, the left side equals $2+\sqrt{1}=3$ and the right side equals 1 , so $x=3$ must be rejected.

If $x=6$, the left side equals $2+\sqrt{4}=4$ and the right side equals 4 , so $x=6$ is the only solution.
Suppose $u=\sqrt{x-2}$.

The equation becomes $2+u=u^{2}$ or $u^{2}-u-2=0$ or $(u-2)(u+1)=0$.

Therefore, $u=2$ or $u=-1$.

But we cannot have $\sqrt{x-2}=-1$ (as square roots are always non-negative).

Therefore, $\sqrt{x-2}=2$ or $x-2=4$ or $x=6$.

Final answer: 6
Subfield: Geometry


Question: In the diagram, the parabola intersects the $x$-axis at $A(-3,0)$ and $B(3,0)$ and has its vertex at $C$ below the $x$-axis. The area of $\triangle A B C$ is 54 . Determine the equation of the parabola. Explain how you got your answer.

<img_4015>

Solution:
From the diagram, the parabola has $x$-intercepts $x=3$ and $x=-3$.

Therefore, the equation of the parabola is of the form $y=a(x-3)(x+3)$ for some real number $a$.

Triangle $A B C$ can be considered as having base $A B$ (of length $3-(-3)=6$ ) and height $O C$ (where $O$ is the origin).

Suppose $C$ has coordinates $(0,-c)$. Then $O C=c$.

Thus, the area of $\triangle A B C$ is $\frac{1}{2}(A B)(O C)=3 c$. But we know that the area of $\triangle A B C$ is 54 , so $3 c=54$ or $c=18$.

Since the parabola passes through $C(0,-18)$, then this point must satisfy the equation of the parabola.

Therefore, $-18=a(0-3)(0+3)$ or $-18=-9 a$ or $a=2$.

Thus, the equation of the parabola is $y=2(x-3)(x+3)=2 x^{2}-18$.
Triangle $A B C$ can be considered as having base $A B$ (of length $3-(-3)=6$ ) and height $O C$ (where $O$ is the origin).

Suppose $C$ has coordinates $(0,-c)$. Then $O C=c$.

Thus, the area of $\triangle A B C$ is $\frac{1}{2}(A B)(O C)=3 c$. But we know that the area of $\triangle A B C$ is 54 , so $3 c=54$ or $c=18$.

Therefore, the parabola has vertex $C(0,-18)$, so has equation $y=a(x-0)^{2}-18$.

(The vertex of the parabola must lie on the $y$-axis since its roots are equally distant from the $y$-axis, so $C$ must be the vertex.)

Since the parabola passes through $B(3,0)$, then these coordinates satisfy the equation, so $0=3^{2} a-18$ or $9 a=18$ or $a=2$.

Therefore, the equation of the parabola is $y=2 x^{2}-18$.

Final answer: $y=2 x^{2}-18$
Subfield: Geometry


Question: In the diagram, the parabola intersects the $x$-axis at $A(-3,0)$ and $B(3,0)$ and has its vertex at $C$ below the $x$-axis. The area of $\triangle A B C$ is 54 . Determine the equation of the parabola. Explain how you got your answer.

![](https://cdn.mathpix.com/cropped/2023_12_21_84558404cbbb20558c95g-1.jpg?height=406&width=553&top_left_y=797&top_left_x=1255)

Solution:
From the diagram, the parabola has $x$-intercepts $x=3$ and $x=-3$.

Therefore, the equation of the parabola is of the form $y=a(x-3)(x+3)$ for some real number $a$.

Triangle $A B C$ can be considered as having base $A B$ (of length $3-(-3)=6$ ) and height $O C$ (where $O$ is the origin).

Suppose $C$ has coordinates $(0,-c)$. Then $O C=c$.

Thus, the area of $\triangle A B C$ is $\frac{1}{2}(A B)(O C)=3 c$. But we know that the area of $\triangle A B C$ is 54 , so $3 c=54$ or $c=18$.

Since the parabola passes through $C(0,-18)$, then this point must satisfy the equation of the parabola.

Therefore, $-18=a(0-3)(0+3)$ or $-18=-9 a$ or $a=2$.

Thus, the equation of the parabola is $y=2(x-3)(x+3)=2 x^{2}-18$.
Triangle $A B C$ can be considered as having base $A B$ (of length $3-(-3)=6$ ) and height $O C$ (where $O$ is the origin).

Suppose $C$ has coordinates $(0,-c)$. Then $O C=c$.

Thus, the area of $\triangle A B C$ is $\frac{1}{2}(A B)(O C)=3 c$. But we know that the area of $\triangle A B C$ is 54 , so $3 c=54$ or $c=18$.

Therefore, the parabola has vertex $C(0,-18)$, so has equation $y=a(x-0)^{2}-18$.

(The vertex of the parabola must lie on the $y$-axis since its roots are equally distant from the $y$-axis, so $C$ must be the vertex.)

Since the parabola passes through $B(3,0)$, then these coordinates satisfy the equation, so $0=3^{2} a-18$ or $9 a=18$ or $a=2$.

Therefore, the equation of the parabola is $y=2 x^{2}-18$.

Final answer: $y=2 x^{2}-18$
Subfield: Geometry


Question: In the diagram, $A(0, a)$ lies on the $y$-axis above $D$. If the triangles $A O B$ and $B C D$ have the same area, determine the value of $a$. Explain how you got your answer.

<img_3690>

Solution:
$\triangle A O B$ is right-angled at $O$, so has area $\frac{1}{2}(A O)(O B)=\frac{1}{2} a(1)=\frac{1}{2} a$.

We next need to calculate the area of $\triangle B C D$.

Method 1: Completing the trapezoid

Drop a perpendicular from $C$ to $P(3,0)$ on the $x$-axis.

<img_3403>

Then $D O P C$ is a trapezoid with parallel sides $D O$ of length 1 and $P C$ of length 2 and height $O P$ (which is indeed perpendicular to the parallel sides) of length 3.

The area of the trapezoid is thus $\frac{1}{2}(D O+P C)(O P)=\frac{1}{2}(1+2)(3)=\frac{9}{2}$.

But the area of $\triangle B C D$ equals the area of trapezoid $D O P C$ minus the areas of $\triangle D O B$ and $\triangle B P C$.

$\triangle D O B$ is right-angled at $O$, so has area $\frac{1}{2}(D O)(O B)=\frac{1}{2}(1)(1)=\frac{1}{2}$.

$\triangle B P C$ is right-angled at $P$, so has area $\frac{1}{2}(B P)(P C)=\frac{1}{2}(2)(2)=2$.

Thus, the area of $\triangle D B C$ is $\frac{9}{2}-\frac{1}{2}-2=2$.



(A similar method for calculating the area of $\triangle D B C$ would be to drop a perpendicular to $Q$ on the $y$-axis, creating a rectangle $Q O P C$.)

Method 2: $\triangle D B C$ is right-angled

The slope of line segment $D B$ is $\frac{1-0}{0-1}=-1$.

The slope of line segment $B C$ is $\frac{2-0}{3-1}=1$.

Since the product of these slopes is -1 (that is, their slopes are negative reciprocals), then $D B$ and $B C$ are perpendicular.

Therefore, the area of $\triangle D B C$ is $\frac{1}{2}(D B)(B C)$.

Now $D B=\sqrt{(1-0)^{2}+(0-1)^{2}}=\sqrt{2}$ and $B C=\sqrt{(3-1)^{2}+(2-0)^{2}}=\sqrt{8}$.

Thus, the area of $\triangle D B C$ is $\frac{1}{2} \sqrt{2} \sqrt{8}=2$.

Since the area of $\triangle A O B$ equals the area of $\triangle D B C$, then $\frac{1}{2} a=2$ or $a=4$.

Final answer: 4
Subfield: Geometry


Question: In the diagram, $A(0, a)$ lies on the $y$-axis above $D$. If the triangles $A O B$ and $B C D$ have the same area, determine the value of $a$. Explain how you got your answer.

![](https://cdn.mathpix.com/cropped/2023_12_21_84558404cbbb20558c95g-1.jpg?height=434&width=485&top_left_y=1621&top_left_x=1259)

Solution:
$\triangle A O B$ is right-angled at $O$, so has area $\frac{1}{2}(A O)(O B)=\frac{1}{2} a(1)=\frac{1}{2} a$.

We next need to calculate the area of $\triangle B C D$.

Method 1: Completing the trapezoid

Drop a perpendicular from $C$ to $P(3,0)$ on the $x$-axis.

![](https://cdn.mathpix.com/cropped/2023_12_21_99bcabad849ac4365f35g-1.jpg?height=423&width=502&top_left_y=1664&top_left_x=909)

Then $D O P C$ is a trapezoid with parallel sides $D O$ of length 1 and $P C$ of length 2 and height $O P$ (which is indeed perpendicular to the parallel sides) of length 3.

The area of the trapezoid is thus $\frac{1}{2}(D O+P C)(O P)=\frac{1}{2}(1+2)(3)=\frac{9}{2}$.

But the area of $\triangle B C D$ equals the area of trapezoid $D O P C$ minus the areas of $\triangle D O B$ and $\triangle B P C$.

$\triangle D O B$ is right-angled at $O$, so has area $\frac{1}{2}(D O)(O B)=\frac{1}{2}(1)(1)=\frac{1}{2}$.

$\triangle B P C$ is right-angled at $P$, so has area $\frac{1}{2}(B P)(P C)=\frac{1}{2}(2)(2)=2$.

Thus, the area of $\triangle D B C$ is $\frac{9}{2}-\frac{1}{2}-2=2$.



(A similar method for calculating the area of $\triangle D B C$ would be to drop a perpendicular to $Q$ on the $y$-axis, creating a rectangle $Q O P C$.)

Method 2: $\triangle D B C$ is right-angled

The slope of line segment $D B$ is $\frac{1-0}{0-1}=-1$.

The slope of line segment $B C$ is $\frac{2-0}{3-1}=1$.

Since the product of these slopes is -1 (that is, their slopes are negative reciprocals), then $D B$ and $B C$ are perpendicular.

Therefore, the area of $\triangle D B C$ is $\frac{1}{2}(D B)(B C)$.

Now $D B=\sqrt{(1-0)^{2}+(0-1)^{2}}=\sqrt{2}$ and $B C=\sqrt{(3-1)^{2}+(2-0)^{2}}=\sqrt{8}$.

Thus, the area of $\triangle D B C$ is $\frac{1}{2} \sqrt{2} \sqrt{8}=2$.

Since the area of $\triangle A O B$ equals the area of $\triangle D B C$, then $\frac{1}{2} a=2$ or $a=4$.

Final answer: 4
Subfield: Geometry


Question: The Little Prince lives on a spherical planet which has a radius of $24 \mathrm{~km}$ and centre $O$. He hovers in a helicopter $(H)$ at a height of $2 \mathrm{~km}$ above the surface of the planet. From his position in the helicopter, what is the distance, in kilometres, to the furthest point on the surface of the planet that he can see?

![](https://cdn.mathpix.com/cropped/2023_12_21_1ec73aaee00c97b3df16g-1.jpg?height=363&width=290&top_left_y=274&top_left_x=1248)

Solution:
Suppose that $O$ is the centre of the planet, $H$ is the place where His Highness hovers in the helicopter, and $P$ is the furthest point on the surface of the planet that he can see.

![](https://cdn.mathpix.com/cropped/2023_12_21_86b73a5af3c2f1283f30g-1.jpg?height=366&width=307&top_left_y=1216&top_left_x=1012)

Then $H P$ must be a tangent to the surface of the planet (otherwise he could see further), so $O P$ (a radius) is perpendicular to $H P$ (a tangent).

We are told that $O P=24 \mathrm{~km}$.

Since the helicopter hovers at a height of $2 \mathrm{~km}$, then $O H=24+2=26 \mathrm{~km}$.

Therefore, $H P^{2}=O H^{2}-O P^{2}=26^{2}-24^{2}=100$, so $H P=10 \mathrm{~km}$.

Therefore, the distance to the furthest point that he can see is $10 \mathrm{~km}$.

Final answer: 10 km
Subfield: Geometry


Question: The Little Prince lives on a spherical planet which has a radius of $24 \mathrm{~km}$ and centre $O$. He hovers in a helicopter $(H)$ at a height of $2 \mathrm{~km}$ above the surface of the planet. From his position in the helicopter, what is the distance, in kilometres, to the furthest point on the surface of the planet that he can see?

<img_3900>

Solution:
Suppose that $O$ is the centre of the planet, $H$ is the place where His Highness hovers in the helicopter, and $P$ is the furthest point on the surface of the planet that he can see.

<img_3899>

Then $H P$ must be a tangent to the surface of the planet (otherwise he could see further), so $O P$ (a radius) is perpendicular to $H P$ (a tangent).

We are told that $O P=24 \mathrm{~km}$.

Since the helicopter hovers at a height of $2 \mathrm{~km}$, then $O H=24+2=26 \mathrm{~km}$.

Therefore, $H P^{2}=O H^{2}-O P^{2}=26^{2}-24^{2}=100$, so $H P=10 \mathrm{~km}$.

Therefore, the distance to the furthest point that he can see is $10 \mathrm{~km}$.

Final answer: 10 km
Subfield: Geometry


Question: In the diagram, points $A$ and $B$ are located on islands in a river full of rabid aquatic goats. Determine the distance from $A$ to $B$, to the nearest metre. (Luckily, someone has measured the angles shown in the diagram as well as the distances $C D$ and $D E$.)

<img_3143>

Solution:
Since we know the measure of $\angle A D B$, then to find the distance $A B$, it is enough to find the distances $A D$ and $B D$ and then apply the cosine law.

In $\triangle D B E$, we have $\angle D B E=180^{\circ}-20^{\circ}-70^{\circ}=90^{\circ}$, so $\triangle D B E$ is right-angled, giving $B D=100 \cos \left(20^{\circ}\right) \approx 93.969$.

In $\triangle D A C$, we have $\angle D A C=180^{\circ}-50^{\circ}-45^{\circ}=85^{\circ}$.

Using the sine law, $\frac{A D}{\sin \left(50^{\circ}\right)}=\frac{C D}{\sin \left(85^{\circ}\right)}$, so $A D=\frac{150 \sin \left(50^{\circ}\right)}{\sin \left(85^{\circ}\right)} \approx 115.346$.



Finally, using the cosine law in $\triangle A B D$, we get

$$
\begin{aligned}
A B^{2} & =A D^{2}+B D^{2}-2(A D)(B D) \cos (\angle A D B) \\
A B^{2} & \approx(115.346)^{2}+(93.969)^{2}-2(115.346)(93.969) \cos \left(35^{\circ}\right) \\
A B^{2} & \approx 4377.379 \\
A B & \approx 66.16
\end{aligned}
$$

Therefore, the distance from $A$ to $B$ is approximately $66 \mathrm{~m}$.

Final answer: 66 m
Subfield: Geometry


Question: In the diagram, points $A$ and $B$ are located on islands in a river full of rabid aquatic goats. Determine the distance from $A$ to $B$, to the nearest metre. (Luckily, someone has measured the angles shown in the diagram as well as the distances $C D$ and $D E$.)

![](https://cdn.mathpix.com/cropped/2023_12_21_1ec73aaee00c97b3df16g-1.jpg?height=490&width=754&top_left_y=682&top_left_x=1076)

Solution:
Since we know the measure of $\angle A D B$, then to find the distance $A B$, it is enough to find the distances $A D$ and $B D$ and then apply the cosine law.

In $\triangle D B E$, we have $\angle D B E=180^{\circ}-20^{\circ}-70^{\circ}=90^{\circ}$, so $\triangle D B E$ is right-angled, giving $B D=100 \cos \left(20^{\circ}\right) \approx 93.969$.

In $\triangle D A C$, we have $\angle D A C=180^{\circ}-50^{\circ}-45^{\circ}=85^{\circ}$.

Using the sine law, $\frac{A D}{\sin \left(50^{\circ}\right)}=\frac{C D}{\sin \left(85^{\circ}\right)}$, so $A D=\frac{150 \sin \left(50^{\circ}\right)}{\sin \left(85^{\circ}\right)} \approx 115.346$.



Finally, using the cosine law in $\triangle A B D$, we get

$$
\begin{aligned}
A B^{2} & =A D^{2}+B D^{2}-2(A D)(B D) \cos (\angle A D B) \\
A B^{2} & \approx(115.346)^{2}+(93.969)^{2}-2(115.346)(93.969) \cos \left(35^{\circ}\right) \\
A B^{2} & \approx 4377.379 \\
A B & \approx 66.16
\end{aligned}
$$

Therefore, the distance from $A$ to $B$ is approximately $66 \mathrm{~m}$.

Final answer: 66 m
Subfield: Geometry


Question: Determine all values of $x$ for which $(\sqrt{x})^{\log _{10} x}=100$.

Solution:
Using rules for manipulating logarithms,

$$
\begin{aligned}
(\sqrt{x})^{\log _{10} x} & =100 \\
\log _{10}\left((\sqrt{x})^{\log _{10} x}\right) & =\log _{10} 100 \\
\left(\log _{10} x\right)\left(\log _{10} \sqrt{x}\right) & =2 \\
\left(\log _{10} x\right)\left(\log _{10} x^{\frac{1}{2}}\right) & =2 \\
\left(\log _{10} x\right)\left(\frac{1}{2} \log _{10} x\right) & =2 \\
\left(\log _{10} x\right)^{2} & =4 \\
\log _{10} x & = \pm 2 \\
x & =10^{ \pm 2}
\end{aligned}
$$

Therefore, $x=100$ or $x=\frac{1}{100}$.

(We can check by substitution that each is indeed a solution.)

Final answer: $100,\frac{1}{100}$
Subfield: Geometry


Question: In the diagram, line segment $F C G$ passes through vertex $C$ of square $A B C D$, with $F$ lying on $A B$ extended and $G$ lying on $A D$ extended. Prove that $\frac{1}{A B}=\frac{1}{A F}+\frac{1}{A G}$.

![](https://cdn.mathpix.com/cropped/2023_12_21_1ec73aaee00c97b3df16g-1.jpg?height=344&width=545&top_left_y=1319&top_left_x=1256)

Solution:
Without loss of generality, suppose that square $A B C D$ has side length 1 .

Suppose next that $B F=a$ and $\angle C F B=\theta$.

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

Since $G C F$ is a straight line, then $\angle G C D=180^{\circ}-90^{\circ}-\left(90^{\circ}-\theta\right)=\theta$.

Therefore, $\triangle G D C$ is similar to $\triangle C B F$, since $\triangle G D C$ is right-angled at $D$.

Thus, $\frac{G D}{D C}=\frac{B C}{B F}$ or $\frac{G D}{1}=\frac{1}{a}$ or $G D=\frac{1}{a}$.

So $A F=A B+B F=1+a$ and $A G=A D+D G=1+\frac{1}{a}=\frac{a+1}{a}$.

Thus, $\frac{1}{A F}+\frac{1}{A G}=\frac{1}{1+a}+\frac{a}{a+1}=\frac{a+1}{a+1}=1=\frac{1}{A B}$, as required.
We attach a set of coordinate axes to the diagram, with $A$ at the origin, $A G$ lying along the positive $y$-axis and $A F$ lying along the positive $x$-axis.

Without loss of generality, suppose that square $A B C D$ has side length 1 , so that $C$ has coordinates $(1,1)$. (We can make this assumption without loss of generality, because if the square had a different side length, then each of the lengths in the problem would be scaled by the same factor.)



Suppose that the line through $G$ and $F$ has slope $m$.

Since this line passes through $(1,1)$, its equation is $y-1=m(x-1)$ or $y=m x+(1-m)$. The $y$-intercept of this line is $1-m$, so $G$ has coordinates $(0,1-m)$.

The $x$-intercept of this line is $\frac{m-1}{m}$, so $F$ has coordinates $\left(\frac{m-1}{m}, 0\right)$. (Note that $m \neq 0$ as the line cannot be horizontal.)

Therefore,

$$
\frac{1}{A F}+\frac{1}{A G}=\frac{m}{m-1}+\frac{1}{1-m}=\frac{m}{m-1}+\frac{-1}{m-1}=\frac{m-1}{m-1}=1=\frac{1}{A B}
$$

as required.
Join $A$ to $C$.

We know that the sum of the areas of $\triangle G C A$ and $\triangle F C A$ equals the area of $\triangle G A F$.

The area of $\triangle G C A$ (thinking of $A G$ as the base) is $\frac{1}{2}(A G)(D C)$, since $D C$ is perpendicular to $A G$.

Similarly, the area of $\triangle F C A$ is $\frac{1}{2}(A F)(C B)$.

Also, the area of $\triangle G A F$ is $\frac{1}{2}(A G)(A F)$.

Therefore,

$$
\begin{aligned}
\frac{1}{2}(A G)(D C)+\frac{1}{2}(A F)(C B) & =\frac{1}{2}(A G)(A F) \\
\frac{(A G)(D C)}{(A G)(A F)(A B)}+\frac{(A F)(C B)}{(A G)(A F)(A B)} & =\frac{(A G)(A F)}{(A G)(A F)(A B)} \\
\frac{1}{A F}+\frac{1}{A G} & =\frac{1}{A B}
\end{aligned}
$$

as required, since $A B=D C=C B$.

Final answer: 证明题，略