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


Question: A helicopter is flying due west over level ground at a constant altitude of $222 \mathrm{~m}$ and at a constant speed. A lazy, stationary goat, which is due west of the helicopter, takes two measurements of the angle between the ground and the helicopter. The first measurement the goat makes is $6^{\circ}$ and the second measurement, which he makes 1 minute later, is $75^{\circ}$. If the helicopter has not yet passed over the goat, as shown, how fast is the helicopter travelling to the nearest kilometre per hour?

![](https://cdn.mathpix.com/cropped/2023_12_21_1f6e0e123b672977527ag-1.jpg?height=379&width=881&top_left_y=2182&top_left_x=641)

Solution:
Let $G$ be the point where the goat is standing, $H$ the position of the helicopter when the goat first measures the angle, $P$ the point directly below the helicopter at this time, $J$ the position of the helicopter one minute later, and $Q$ the point directly below the helicopter at this time.

![](https://cdn.mathpix.com/cropped/2023_12_21_94ca7e940013ecf0c964g-1.jpg?height=498&width=808&top_left_y=1426&top_left_x=1103)

Using the initial position of the helicopter, $\tan \left(6^{\circ}\right)=\frac{H P}{P G}$ or $P G=\frac{222}{\tan \left(6^{\circ}\right)} \approx 2112.19 \mathrm{~m}$.

Using the second position of the helicopter, $\tan \left(75^{\circ}\right)=\frac{J Q}{Q G}$ or $Q G=\frac{222}{\tan \left(75^{\circ}\right)} \approx 59.48 \mathrm{~m}$.

So in the one minute that has elapsed, the helicopter has travelled

$2112.19 \mathrm{~m}-59.48 \mathrm{~m}=2052.71 \mathrm{~m}$ or $2.0527 \mathrm{~km}$.

Therefore, in one hour, the helicopter will travel $60(2.0527)=123.162 \mathrm{~km}$.

Thus, the helicopter is travelling $123 \mathrm{~km} / \mathrm{h}$.

Final answer: 123 km/h
Subfield: Algebra


Question: The function $f(x)$ has the property that $f(2 x+3)=2 f(x)+3$ for all $x$. If $f(0)=6$, what is the value of $f(9)$ ?

Solution:
Since we are looking for the value of $f(9)$, then it makes sense to use the given equation and to set $x=3$ in order to obtain $f(9)=2 f(3)+3$.

So we need to determine the value of $f(3)$. We use the equation again and set $x=0$ since we will then get $f(3)$ on the left side and $f(0)$ (whose value we already know) on the right side, ie.

$$
f(3)=2 f(0)+3=2(6)+3=15
$$

Thus, $f(9)=2(15)+3=33$.

Final answer: 33
Subfield: Algebra


Question: Suppose that the functions $f(x)$ and $g(x)$ satisfy the system of equations

$$
\begin{aligned}
f(x)+3 g(x) & =x^{2}+x+6 \\
2 f(x)+4 g(x) & =2 x^{2}+4
\end{aligned}
$$

for all $x$. Determine the values of $x$ for which $f(x)=g(x)$.

Solution:
We solve the system of equations for $f(x)$ and $g(x)$.

Dividing out the common factor of 2 from the second equation, we get

$f(x)+2 g(x)=x^{2}+2$.

Subtracting from the first equation, we get $g(x)=x+4$.

Thus, $f(x)=x^{2}+2-2 g(x)=x^{2}+2-2(x+4)=x^{2}-2 x-6$.

Equating $f(x)$ and $g(x)$, we obtain

$$
\begin{aligned}
x^{2}-2 x-6 & =x+4 \\
x^{2}-3 x-10 & =0 \\
(x-5)(x+2) & =0
\end{aligned}
$$

Therefore, $x=5$ or $x=-2$.
Instead of considering the equation $f(x)=g(x)$, we consider the equation $f(x)-g(x)=0$, and we try to obtain an expression for $f(x)-g(x)$ by manipulating the two given equations.

In fact, after some experimentation, we can see that

$$
\begin{aligned}
f(x)-g(x) & =2(2 f(x)+4 g(x))-3(f(x)+3 g(x)) \\
& =2\left(2 x^{2}+4\right)-3\left(x^{2}+x+6\right) \\
& =x^{2}-3 x-10
\end{aligned}
$$

So to solve $f(x)-g(x)=0$, we solve $x^{2}-3 x-10=0$ or $(x-5)(x+2)=0$. Therefore, $x=5$ or $x=-2$.

Final answer: $5,-2$
Subfield: Combinatorics


Question: In a short-track speed skating event, there are five finalists including two Canadians. The first three skaters to finish the race win a medal. If all finalists have the same chance of finishing in any position, what is the probability that neither Canadian wins a medal?

Solution:
We label the 5 skaters A, B, C, D, and E, where D and E are the two Canadians.

There are then $5 !=5 \times 4 \times 3 \times 2 \times 1=120$ ways of arranging these skaters in their order of finish (for example, $\mathrm{ADBCE}$ indicates that A finished first, $\mathrm{D}$ second, etc.), because there are 5 choices for the winner, 4 choices for the second place finisher, 3 choices for the third place finisher, etc.



If the two Canadians finish without winning medals, then they must finish fourth and fifth. So the $\mathrm{D}$ and $\mathrm{E}$ are in the final two positions, and $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ in the first three. There are $3 !=6$ ways of arranging the $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$, and $2 !=2$ ways to arrange the $\mathrm{D}$ and E. Thus, there are $6 \times 2=12$ ways or arranging the skaters so that neither Canadian wins a medal.

Therefore, the probability that neither Canadian wins a medal is

$$
\frac{\# \text { of ways where Canadians don't win medals }}{\text { Total } \# \text { of arrangements }}=\frac{12}{120}=\frac{1}{10}
$$
We label the 5 skaters as A, B, C, D, and E, where D and E are the two Canadians. In any race, two of the skaters finish fourth and fifth. Also, any pair of skaters are equally as likely to finish fourth and fifth, since the probability of every skater is equally likely to finish in a given position.

How many pairs of 2 skaters can we form from the 5 skaters? There are ten such pairs:

$$
\{A, B\},\{A, C\},\{A, D\},\{A, E\},\{B, C\},\{B, D\},\{B, E\},\{C, D\},\{C, E\},\{D, E\}
$$

Only one of these ten pairs is made up of the two Canadians. Therefore, the probability is $\frac{1}{10}$, since one out of ten choices gives the desired result.

Final answer: $\frac{1}{10}$
Subfield: Number Theory


Question: Determine the number of positive integers less than or equal to 300 that are multiples of 3 or 5 , but are not multiples of 10 or 15 .

Solution:
Since the least common multiple of $3,5,10$ and 15 is 30 , then we can count the number of positive integers less than or equal to 30 satisfying these conditions, and multiply the total by 10 to obtain the number less than 300. (This is because each group of 30 consecutive integers starting with 1 more than a multiple of 30 will have the same number of integers having these properties, because we can subtract 30 from each one and not change these properties.)

So from 1 to 30, we have:

$$
3,5,6,9,12,18,21,24,25,27
$$

Thus there are 10 less than or equal to 30 , and so 100 such positive integers less than or equal to 300 .
We proceed by doing a (careful!) count.

The number of positive multiples of 3 less than or equal to 300 is 100.

The number of positive multiples of 5 less than or equal to 300 is 60 .

Thus, we have 160 candidates, but have included multiples of 15 twice (since 15 is a multiple of each of 3 and 5), and have also included multiples of 10.

The number of multiples of 15 less than or equal to 300 is 20 , so to remove the multiples of 15 , we must remove 40 from 160 to get 120 positive integers less than or equal to 300 which are multiples of 3 or 5 but not of 15 .



This total still included some multiples of 10 that are less or equal to 300 (but not all, since we have already removed 30 , for instance).

In fact, there are 30 multiples of 10 less than or equal 300,10 of which are multiples of 15 as well (that is, the multiples of 30). So we must remove 20 from the total of 120. We then obtain that there are 100 positive integers less than or equal to 300 which are multiples of 3 or 5 , but not of 10 or 15 .

Final answer: 100
Subfield: Algebra


Question: In the series of odd numbers $1+3+5-7-9-11+13+15+17-19-21-23 \ldots$ the signs alternate every three terms, as shown. What is the sum of the first 300 terms of the series?

Solution:
Since the signs alternate every three terms, it makes sense to look at the terms in groups of 6 .

The sum of the first 6 terms is $1+3+5-7-9-11=-18$.

The sum of the next 6 terms is $13+15+17-19-21-23=-18$.

In fact, the sum of each group of 6 terms will be the same, since in each group, 12 has been added to the numerical value of each term when compared to the previous group of 6 , so overall 12 has been added three times and subtracted three times.

Since we are looking for the sum of the first 300 terms, then we are looking at 50 groups of 6 terms, so the sum must be $50(-18)=-900$.

Final answer: -900
Subfield: Algebra


Question: A two-digit number has the property that the square of its tens digit plus ten times its units digit equals the square of its units digit plus ten times its tens digit. Determine all two-digit numbers which have this property, and are prime numbers.

Solution:
Let the two digit integer have tens digit $a$ and units digit $b$. Then the given information tells us

$$
\begin{aligned}
a^{2}+10 b & =b^{2}+10 a \\
a^{2}-b^{2}-10 a+10 b & =0 \\
(a+b)(a-b)-10(a-b) & =0 \\
(a-b)(a+b-10) & =0
\end{aligned}
$$

and so $a=b$ or $a+b=10$.

So the possibilities for the integer are 11, 22, 33, 44, 55, 66, 77, 88, 99, 19, 28, 37, 46, 55, $64,73,82,91$. We now must determine which integers in this list are prime.

We can quickly reject all multiples of 11 bigger than 11 and all of the even integers, to reduce the list to $11,19,37,73,91$.

All of these are prime, except for $91=13 \times 7$.

Therefore, the required integers are 11, 19, 37, and 73 .

Final answer: 11,19,37,73
Subfield: Algebra


Question: A lead box contains samples of two radioactive isotopes of iron. Isotope A decays so that after every 6 minutes, the number of atoms remaining is halved. Initially, there are twice as many atoms of isotope $\mathrm{A}$ as of isotope $\mathrm{B}$, and after 24 minutes there are the same number of atoms of each isotope. How long does it take the number of atoms of isotope B to halve?

Solution:
In 24 minutes, the number of atoms of isotope $\mathrm{A}$ has halved 4 times, so the initial number of atoms is $2^{4}=16$ times the number of atoms of isotope $\mathrm{A}$ at time 24 minutes.

But there were initially half as many atoms of isotope B as of isotope B, so there was 8 times the final number of atoms. Therefore, the number of atoms of isotope B halves 3 times in the 24 minutes, so it takes 8 minutes for the number of atoms of isotope B to halve.
Initially, there is twice as many atoms of isotope A as of isotope B, so let the original numbers of atoms of each be $2 x$ and $x$, respectively.

Considering isotope A, after 24 minutes, if it loses half of its atoms every 6 minutes, there will be $2 x\left(\frac{1}{2}\right)^{\frac{24}{6}}$ atoms remaining.

Similarly for isotope B, after 24 minutes, there will be $x\left(\frac{1}{2}\right)^{\frac{24}{T}}$ atoms remaining, where $T$ is the length of time (in minutes) that it takes for the number of atoms to halve.

From the given information,

$$
\begin{aligned}
2 x\left(\frac{1}{2}\right)^{\frac{24}{6}} & =x\left(\frac{1}{2}\right)^{\frac{24}{T}} \\
2\left(\frac{1}{2}\right)^{4} & =\left(\frac{1}{2}\right)^{\frac{24}{T}} \\
\left(\frac{1}{2}\right)^{3} & =\left(\frac{1}{2}\right)^{\frac{24}{T}} \\
\frac{24}{T} & =3 \\
T & =8
\end{aligned}
$$

Therefore, it takes 8 minutes for the number of atoms of isotope B to halve.

Final answer: 8 min
Subfield: Algebra


Question: Solve the system of equations:

$$
\begin{aligned}
& \log _{10}\left(x^{3}\right)+\log _{10}\left(y^{2}\right)=11 \\
& \log _{10}\left(x^{2}\right)-\log _{10}\left(y^{3}\right)=3
\end{aligned}
$$

Solution:
Using the facts that $\log _{10} A+\log _{10} B=\log _{10} A B$ and that $\log _{10} A-\log _{10} B=\log _{10} \frac{A}{B}$, then we can convert the two equations to

$$
\begin{aligned}
\log _{10}\left(x^{3} y^{2}\right) & =11 \\
\log _{10}\left(\frac{x^{2}}{y^{3}}\right) & =3
\end{aligned}
$$

Raising both sides to the power of 10 , we obtain

$$
\begin{aligned}
x^{3} y^{2} & =10^{11} \\
\frac{x^{2}}{y^{3}} & =10^{3}
\end{aligned}
$$

To eliminate the $y$ 's, we raise the first equation to the power 3 and the second to the power 2 to obtain

$$
\begin{aligned}
x^{9} y^{6} & =10^{33} \\
\frac{x^{4}}{y^{6}} & =10^{6}
\end{aligned}
$$

and multiply to obtain $x^{9} x^{4}=x^{13}=10^{39}=10^{33} 10^{6}$.

Therefore, since $x^{13}=10^{39}$, then $x=10^{3}$.



Substituting back into $x^{3} y^{2}=10^{11}$, we get $y^{2}=10^{2}$, and so $y= \pm 10$. However, substituting into $\frac{x^{2}}{y^{3}}=10^{3}$ we see that $y$ must be positive, so $y=10$.

Therefore, the solution to the system of equation is $x=10^{3}$ and $y=10$.
Since the domain of the logarithm is the positive real numbers, then the quantities $\log _{10}\left(x^{3}\right)$ and $\log _{10}\left(y^{3}\right)$ tell us that $x$ and $y$ are positive.

Using the fact that $\log _{10}\left(a^{b}\right)=b \log _{10}(a)$, we rewrite the equations as

$$
\begin{aligned}
& 3 \log _{10} x+2 \log _{10} y=11 \\
& 2 \log _{10} x-3 \log _{10} y=3
\end{aligned}
$$

We solve the system of equations for $\log _{10} x$ and $\log _{10} y$ by multiplying the first equation by 3 and adding two times the second equation in order to eliminate $\log _{10} y$. Thus we obtain $13 \log _{10} x=39$ or $\log _{10} x=3$.

Substituting back into the first equation, we obtain $\log _{10} y=1$.

Therefore, $x=10^{3}$ and $y=10$.

Final answer: $10^{3},10$
Subfield: Geometry


Question: A regular hexagon is a six-sided figure which has all of its angles equal and all of its side lengths equal. In the diagram, $A B C D E F$ is a regular hexagon with an area of 36. The region common to the equilateral triangles $A C E$ and $B D F$ is a hexagon, which is shaded as shown. What is the area of the shaded hexagon?

![](https://cdn.mathpix.com/cropped/2023_12_21_fe07897436972c8371a1g-1.jpg?height=331&width=352&top_left_y=1992&top_left_x=1385)

Solution:
We label the vertices of the shaded hexagon $U, V, W, X$, $Y$, and $Z$.

By symmetry, all of the six triangles with two vertices on the inner hexagon and one on the outer hexagon (eg. triangle $U V A$ ) are congruent equilateral triangles. In order to determine the area of the inner hexagon, we determine the ratio of the side lengths of the two hexagons.

![](https://cdn.mathpix.com/cropped/2023_12_21_31cf9a291a17adc4339dg-1.jpg?height=439&width=485&top_left_y=1298&top_left_x=1362)

Let the side length of the inner hexagon be $x$. Then $A U=U F=x$. Then triangle $A U F$ has a $120^{\circ}$ between the two sides of length $x$. If we draw a perpendicular from $U$ to point $P$ on side $A F$, then $U P$ divides

$\triangle A U F$ into two $30^{\circ}-60^{\circ}-90^{\circ}$ triangles. Thus, $F P=P A=\frac{\sqrt{3}}{2} x$ and so $A F=\sqrt{3} x$.

So the ratio of the side lengths of the hexagons is $\sqrt{3}: 1$, and so the ratio of

![](https://cdn.mathpix.com/cropped/2023_12_21_31cf9a291a17adc4339dg-1.jpg?height=401&width=810&top_left_y=1865&top_left_x=1075)
their areas is $(\sqrt{3})^{2}: 1=3: 1$.

Since the area of the larger hexagon is 36 , then the area of the inner hexagon is 12.
We label the vertices of the hexagon $U, V, W, X, Y$, and $Z$. By symmetry, all of the six triangles with two vertices on the inner hexagon and one on the outer hexagon (eg. triangle $U V A$ ) are congruent equilateral triangles.

We also join the opposite vertices of the inner hexagon, ie. we join $U$ to $X, V$ to $Y$, and $W$ to $Z$. (These 3 line segments all meet at a single point, say $O$.) This divides the inner hexagon into 6 small equilateral triangles identical to the

![](https://cdn.mathpix.com/cropped/2023_12_21_78c42ff33d676905b437g-1.jpg?height=439&width=485&top_left_y=301&top_left_x=1400)
six earlier mentioned equilateral triangles.

Let the area of one of these triangles be $a$. Then we can label the 12 small equilateral triangles as all having area $a$.

But triangle $A U F$ also has area $a$, because if we consider triangle $A F V$, then $A U$ is a median (since $F U=A U=U V$ by symmetry) and so divides triangle $A F V$ into two triangles of equal area. Since the area of

![](https://cdn.mathpix.com/cropped/2023_12_21_78c42ff33d676905b437g-1.jpg?height=230&width=369&top_left_y=915&top_left_x=1363)
triangle $A U V$ is $a$, then the area of triangle $A U F$ is also $a$.

Therefore, hexagon $A B C D E F$ is divided into 18 equal areas. Thus, $a=2$ since the area of the large hexagon is 36.

Since the area of $U V W X Y Z$ is $6 a$, then its area is 12 .

Final answer: 12
Subfield: Geometry


Question: A regular hexagon is a six-sided figure which has all of its angles equal and all of its side lengths equal. In the diagram, $A B C D E F$ is a regular hexagon with an area of 36. The region common to the equilateral triangles $A C E$ and $B D F$ is a hexagon, which is shaded as shown. What is the area of the shaded hexagon?

<img_3159>

Solution:
We label the vertices of the shaded hexagon $U, V, W, X$, $Y$, and $Z$.

By symmetry, all of the six triangles with two vertices on the inner hexagon and one on the outer hexagon (eg. triangle $U V A$ ) are congruent equilateral triangles. In order to determine the area of the inner hexagon, we determine the ratio of the side lengths of the two hexagons.

<img_3663>

Let the side length of the inner hexagon be $x$. Then $A U=U F=x$. Then triangle $A U F$ has a $120^{\circ}$ between the two sides of length $x$. If we draw a perpendicular from $U$ to point $P$ on side $A F$, then $U P$ divides

$\triangle A U F$ into two $30^{\circ}-60^{\circ}-90^{\circ}$ triangles. Thus, $F P=P A=\frac{\sqrt{3}}{2} x$ and so $A F=\sqrt{3} x$.

So the ratio of the side lengths of the hexagons is $\sqrt{3}: 1$, and so the ratio of

<img_3205>
their areas is $(\sqrt{3})^{2}: 1=3: 1$.

Since the area of the larger hexagon is 36 , then the area of the inner hexagon is 12.
We label the vertices of the hexagon $U, V, W, X, Y$, and $Z$. By symmetry, all of the six triangles with two vertices on the inner hexagon and one on the outer hexagon (eg. triangle $U V A$ ) are congruent equilateral triangles.

We also join the opposite vertices of the inner hexagon, ie. we join $U$ to $X, V$ to $Y$, and $W$ to $Z$. (These 3 line segments all meet at a single point, say $O$.) This divides the inner hexagon into 6 small equilateral triangles identical to the

<img_3385>
six earlier mentioned equilateral triangles.

Let the area of one of these triangles be $a$. Then we can label the 12 small equilateral triangles as all having area $a$.

But triangle $A U F$ also has area $a$, because if we consider triangle $A F V$, then $A U$ is a median (since $F U=A U=U V$ by symmetry) and so divides triangle $A F V$ into two triangles of equal area. Since the area of

<img_3160>
triangle $A U V$ is $a$, then the area of triangle $A U F$ is also $a$.

Therefore, hexagon $A B C D E F$ is divided into 18 equal areas. Thus, $a=2$ since the area of the large hexagon is 36.

Since the area of $U V W X Y Z$ is $6 a$, then its area is 12 .

Final answer: 12
Subfield: Geometry


Question: At the Big Top Circus, Herc the Human Cannonball is fired out of the cannon at ground level. (For the safety of the spectators, the cannon is partially buried in the sand floor.) Herc's trajectory is a parabola until he catches the vertical safety net, on his way down, at point $B$. Point $B$ is $64 \mathrm{~m}$ directly above point $C$ on the floor of the tent. If Herc reaches a maximum height of $100 \mathrm{~m}$, directly above a point $30 \mathrm{~m}$ from the cannon, determine the horizontal distance from the cannon to the net.

<img_3392>

Solution:
We assign coordinates to the diagram, with the mouth of the cannon at the point $(0,0)$, with the positive $x$-axis in the horizontal direction towards the safety net from the cannon, and the positive $y$ axis upwards from $(0,0)$.

Since Herc reaches his maximum height when his horizontal distance is $30 \mathrm{~m}$, then the axis of symmetry of the parabola is the line $x=30$. Since the parabola has a root at $x=0$, then the other root must be at $x=60$.

Therefore, the parabola has the form $y=\operatorname{ax}(x-60)$.

In order to determine the value of $a$, we note that Herc passes through the point $(30,100)$, and so

<img_3831>



$$
\begin{aligned}
100 & =30 a(-30) \\
a & =-\frac{1}{9}
\end{aligned}
$$

Thus, the equation of the parabola is $y=-\frac{1}{9} x(x-60)$.

(Alternatively, we could say that since the parabola has its maximum point at $(30,100)$, then it must be of the form $y=a(x-30)^{2}+100$.

Since the parabola passes through $(0,0)$, then we have

$$
\begin{aligned}
& 0=a(0-30)^{2}+100 \\
& 0=900 a+100 \\
& a=-\frac{1}{9}
\end{aligned}
$$

Thus, the parabola has the equation $y=-\frac{1}{9}(x-30)^{2}+100$.)

We would like to find the points on the parabola which have $y$-coordinate 64 , so we solve

$$
\begin{aligned}
64 & =-\frac{1}{9} x(x-60) \\
0 & =x^{2}-60 x+576 \\
0 & =(x-12)(x-48)
\end{aligned}
$$

Since we want a point after Herc has passed his highest point, then $x=48$, ie. the horizontal distance from the cannon to the safety net is $48 \mathrm{~m}$.

Final answer: 48 m
Subfield: Geometry


Question: At the Big Top Circus, Herc the Human Cannonball is fired out of the cannon at ground level. (For the safety of the spectators, the cannon is partially buried in the sand floor.) Herc's trajectory is a parabola until he catches the vertical safety net, on his way down, at point $B$. Point $B$ is $64 \mathrm{~m}$ directly above point $C$ on the floor of the tent. If Herc reaches a maximum height of $100 \mathrm{~m}$, directly above a point $30 \mathrm{~m}$ from the cannon, determine the horizontal distance from the cannon to the net.

![](https://cdn.mathpix.com/cropped/2023_12_21_8c2daaa43e7e62803ac0g-1.jpg?height=794&width=683&top_left_y=202&top_left_x=1122)

Solution:
We assign coordinates to the diagram, with the mouth of the cannon at the point $(0,0)$, with the positive $x$-axis in the horizontal direction towards the safety net from the cannon, and the positive $y$ axis upwards from $(0,0)$.

Since Herc reaches his maximum height when his horizontal distance is $30 \mathrm{~m}$, then the axis of symmetry of the parabola is the line $x=30$. Since the parabola has a root at $x=0$, then the other root must be at $x=60$.

Therefore, the parabola has the form $y=\operatorname{ax}(x-60)$.

In order to determine the value of $a$, we note that Herc passes through the point $(30,100)$, and so

![](https://cdn.mathpix.com/cropped/2023_12_21_78c42ff33d676905b437g-1.jpg?height=870&width=651&top_left_y=1468&top_left_x=1233)



$$
\begin{aligned}
100 & =30 a(-30) \\
a & =-\frac{1}{9}
\end{aligned}
$$

Thus, the equation of the parabola is $y=-\frac{1}{9} x(x-60)$.

(Alternatively, we could say that since the parabola has its maximum point at $(30,100)$, then it must be of the form $y=a(x-30)^{2}+100$.

Since the parabola passes through $(0,0)$, then we have

$$
\begin{aligned}
& 0=a(0-30)^{2}+100 \\
& 0=900 a+100 \\
& a=-\frac{1}{9}
\end{aligned}
$$

Thus, the parabola has the equation $y=-\frac{1}{9}(x-30)^{2}+100$.)

We would like to find the points on the parabola which have $y$-coordinate 64 , so we solve

$$
\begin{aligned}
64 & =-\frac{1}{9} x(x-60) \\
0 & =x^{2}-60 x+576 \\
0 & =(x-12)(x-48)
\end{aligned}
$$

Since we want a point after Herc has passed his highest point, then $x=48$, ie. the horizontal distance from the cannon to the safety net is $48 \mathrm{~m}$.

Final answer: 48 m