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


Question: Digital images consist of a very large number of equally spaced dots called pixels The resolution of an image is the number of pixels/cm in each of the horizontal and vertical directions.

Thus, an image with dimensions $10 \mathrm{~cm}$ by $15 \mathrm{~cm}$ and a resolution of 75 pixels/cm has a total of $(10 \times 75) \times(15 \times 75)=843750$ pixels.

If each of these dimensions was increased by $n \%$ and the resolution was decreased by $n \%$, the image would have 345600 pixels.

Determine the value of $n$.

Solution:
When the dimensions were increased by $n \%$ from 10 by 15 , the new dimensions were $10\left(1+\frac{n}{100}\right)$ by $15\left(1+\frac{n}{100}\right)$.

When the resolution was decreased by $n$ percent, the new resolution was $75\left(1-\frac{n}{100}\right)$.

(Note that $n$ cannot be larger than 100, since the resolution cannot be decreased by more than $100 \%$.)

Therefore, the number of pixels in the new image is

$$
\left[10\left(1+\frac{n}{100}\right) \times 75\left(1-\frac{n}{100}\right)\right] \times\left[15\left(1+\frac{n}{100}\right) \times 75\left(1-\frac{n}{100}\right)\right]
$$

Since we know that the number of pixels in the new image is 345600 , then

$$
\begin{aligned}
{\left[10\left(1+\frac{n}{100}\right) \times 75\left(1-\frac{n}{100}\right)\right] \times\left[15\left(1+\frac{n}{100}\right) \times 75\left(1-\frac{n}{100}\right)\right] } & =345600 \\
{[10 \times 75] \times[15 \times 75] \times\left(1+\frac{n}{100}\right)^{2} \times\left(1-\frac{n}{100}\right)^{2} } & =345600 \\
843750\left(1+\frac{n}{100}\right)^{2}\left(1-\frac{n}{100}\right)^{2} & =345600 \\
\left(1-\frac{n^{2}}{100^{2}}\right)^{2} & =0.4096 \\
1-\frac{n^{2}}{100^{2}} & = \pm 0.64 \\
1-\frac{n^{2}}{100^{2}} & =0.64 \\
\frac{n^{2}}{100^{2}} & =0.36 \\
\frac{n}{100} & =0.6 \\
n & =60
\end{aligned}
$$

$$
\begin{array}{rlrl}
1-\frac{n^{2}}{100^{2}} & =0.64 & & (n \text { cannot be larger than } 100) \\
\frac{n^{2}}{100^{2}} & =0.36 & \\
\frac{n}{100} & =0.6 & & (\text { since } n \text { must be positive })
\end{array}
$$

Thus, $n=60$.

Final answer: 60
Subfield: Geometry


Question: In the diagram, $A C=B C, A D=7, D C=8$, and $\angle A D C=120^{\circ}$. What is the value of $x$ ?

<img_3788>

Solution:
We first calculate the length of $A C$ using the cosine law:

$$
\begin{aligned}
A C^{2} & =7^{2}+8^{2}-2(7)(8) \cos \left(120^{\circ}\right) \\
A C^{2} & =49+64-112\left(-\frac{1}{2}\right) \\
A C^{2} & =169 \\
A C & =13
\end{aligned}
$$

Since triangle $A B C$ is right-angled and isosceles, then $x=A B=\sqrt{2}(A C)=13 \sqrt{2}$.

Final answer: $13 \sqrt{2}$
Subfield: Geometry


Question: In the diagram, $A C=B C, A D=7, D C=8$, and $\angle A D C=120^{\circ}$. What is the value of $x$ ?

![](https://cdn.mathpix.com/cropped/2023_12_21_bfecee96824679158031g-1.jpg?height=456&width=366&top_left_y=905&top_left_x=1389)

Solution:
We first calculate the length of $A C$ using the cosine law:

$$
\begin{aligned}
A C^{2} & =7^{2}+8^{2}-2(7)(8) \cos \left(120^{\circ}\right) \\
A C^{2} & =49+64-112\left(-\frac{1}{2}\right) \\
A C^{2} & =169 \\
A C & =13
\end{aligned}
$$

Since triangle $A B C$ is right-angled and isosceles, then $x=A B=\sqrt{2}(A C)=13 \sqrt{2}$.

Final answer: $13 \sqrt{2}$
Subfield: Algebra


Question: If $T=x^{2}+\frac{1}{x^{2}}$, determine the values of $b$ and $c$ so that $x^{6}+\frac{1}{x^{6}}=T^{3}+b T+c$ for all non-zero real numbers $x$.

Solution:
Consider the right side of the given equation:

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

For this expression to be equal to $x^{6}+\frac{1}{x^{6}}$ for all values of $x$, we want $b+3=0$ or $b=-3$ and $c=0$.

Final answer: -3,0
Subfield: Combinatorics


Question: A Skolem sequence of order $n$ is a sequence $\left(s_{1}, s_{2}, \ldots, s_{2 n}\right)$ of $2 n$ integers satisfying the conditions:

i) for every $k$ in $\{1,2,3, \ldots, n\}$, there exist exactly two elements $s_{i}$ and $s_{j}$ with $s_{i}=s_{j}=k$, and

ii) if $s_{i}=s_{j}=k$ with $i<j$, then $j-i=k$.


For example, $(4,2,3,2,4,3,1,1)$ is a Skolem sequence of order 4.
List all Skolem sequences of order 4.

Solution:
We start by placing the two 4's. We systematically try each pair of possible positions from positions 1 and 5 to positions 4 and 8 . For each of these positions, we try placing


the two 3's in each pair of possible positions, and then see if the two 2's and two 1's will fit.

(We can reduce our work by noticing that if a Skolem sequence has the two 4's in positions 1 and 5 , then reversing the sequence will give a Skolem sequence with the two 4 's in positions 4 and 8 . So we only need to consider putting the two 4's in positions 1 and 5 , and in positions 2 and 6 . The remaining possibilities can be dealt with by reversing.)

Thus, the six possible Skolem sequences of order 4 are:

$(4,2,3,2,4,3,1,1)$ and its reverse, $(1,1,3,4,2,3,2,4)$

$(4,1,1,3,4,2,3,2)$ and its reverse, $(2,3,2,4,3,1,1,4)$

$(3,4,2,3,2,4,1,1)$ and its reverse, $(1,1,4,2,3,2,4,3)$

Final answer: (4,2,3,2,4,3,1,1),(1,1,3,4,2,3,2,4),(4,1,1,3,4,2,3,2),(2,3,2,4,3,1,1,4),(3,4,2,3,2,4,1,1),(1,1,4,2,3,2,4,3)
Subfield: Combinatorics


Question: A Skolem sequence of order $n$ is a sequence $\left(s_{1}, s_{2}, \ldots, s_{2 n}\right)$ of $2 n$ integers satisfying the conditions:

i) for every $k$ in $\{1,2,3, \ldots, n\}$, there exist exactly two elements $s_{i}$ and $s_{j}$ with $s_{i}=s_{j}=k$, and

ii) if $s_{i}=s_{j}=k$ with $i<j$, then $j-i=k$.


For example, $(4,2,3,2,4,3,1,1)$ is a Skolem sequence of order 4.
Determine, with justification, all Skolem sequences of order 9 which satisfy all of the following three conditions:
I) $s_{3}=1$,

II) $s_{18}=8$, and

III) between any two equal even integers, there is exactly one odd integer.

Solution:
Since we are trying to create a Skolem sequence of order 9 , then there are 18 positions to fill with 10 odd numbers and 8 even numbers.

We are told that $s_{18}=8$, so we must have $s_{10}=8$, since the 8 's must be 8 positions apart. By condition III, between the two 8's, there can be only one odd integer. But there are 7 positions between the two 8 's and only 6 remaining even numbers to place. Thus, all 6 remaining even numbers are placed between the two 8's. The only way in which this is possible is with the two 6's next to the two 8's, then the two 4's, then the two 2's. (The two 8 's are 8 positions apart, and the two 6 's must be 6 positions apart.)

Thus, the sequence so far is:
$( \ldots,\ldots,1,\ldots,\ldots,\ldots,\ldots,\ldots,\ldots,8,6,4,2, \ldots, 2,4,6,8)$

The numbers that we have left to place are $1,3,3,5,5,7,7,9,9$, and empty positions are $1,2,4,5,6,7,8,9,14$.

Since the 9's must be 9 positions apart, they must be placed in positions 5 and 14 .

Thus, we have

$$
(\ldots, \ldots, 1, \ldots, 9, \ldots, \ldots, \ldots, \ldots, 8,6,4,2,9,2,4,6,8)
$$

The remaining 1 must be placed in position 2 or 4 . If it is placed in position 2 , then the 7 's can only go in positions 1 and 8 , giving

$$
(7,1,1, \ldots, 9, \ldots, \ldots, 7, \ldots, 8,6,4,2,9,2,4,6,8)
$$

But we now cannot place both the two 3's and the two 5's. (The placing of one of these pairs means that the other pair cannot be placed.)

We conclude that the only possibility is that the remaining 1 must be placed in position 4 . This gives

$$
(\ldots, \ldots, 1,1,9, \ldots, \ldots, \ldots, 8,6,4,2,9,2,4,6,8)
$$

with 3, 3, 5, 5, 7, 7 left to be placed in positions $1,2,6,7,8,9$.

Now the two 3's must be placed in positions 6 and 9, so the 7's must be placed in positions 1 and 8 , and finally the 5's must be placed in positions 2 and 7 .

Therefore, the only Skolem sequence satisfying the given conditions is $(7,5,1,1,9,3,5,7,3,8,6,4,2,9,2,4,6,8)$

Final answer: (7,5,1,1,9,3,5,7,3,8,6,4,2,9,2,4,6,8)
Subfield: Number Theory


Question: The three-digit positive integer $m$ is odd and has three distinct digits. If the hundreds digit of $m$ equals the product of the tens digit and ones (units) digit of $m$, what is $m$ ?

Solution:
Suppose that $m$ has hundreds digit $a$, tens digit $b$, and ones (units) digit $c$.

From the given information, $a, b$ and $c$ are distinct, each of $a, b$ and $c$ is less than 10, $a=b c$, and $c$ is odd (since $m$ is odd).

The integer $m=623$ satisfies all of these conditions. Since we are told there is only one such number, then 623 must be the only answer.

Why is this the only possible value of $m$ ?

We note that we cannot have $b=1$ or $c=1$, otherwise $a=c$ or $a=b$.

Thus, $b \geq 2$ and $c \geq 2$.

Since $c \geq 2$ and $c$ is odd, then $c$ can equal $3,5,7$, or 9 .

Since $b \geq 2$ and $a=b c$, then if $c$ equals 5,7 or $9, a$ would be larger than 10 , which is not possible.

Thus, $c=3$.

Since $b \geq 2$ and $b \neq c$, then $b=2$ or $b \geq 4$.

If $b \geq 4$ and $c=3$, then $a>10$, which is not possible.

Therefore, we must have $c=3$ and $b=2$, which gives $a=6$.

Final answer: 623
Subfield: Combinatorics


Question: Eleanor has 100 marbles, each of which is black or gold. The ratio of the number of black marbles to the number of gold marbles is $1: 4$. How many gold marbles should she add to change this ratio to $1: 6$ ?

Solution:
Since Eleanor has 100 marbles which are black and gold in the ratio $1: 4$, then $\frac{1}{5}$ of her marbles are black, which means that she has $\frac{1}{5} \cdot 100=20$ black marbles.

When more gold marbles are added, the ratio of black to gold is $1: 6$, which means that she has $6 \cdot 20=120$ gold marbles.

Eleanor now has $20+120=140$ marbles, which means that she added $140-100=40$ gold marbles.

Final answer: 40
Subfield: Number Theory


Question: Suppose that $n$ is a positive integer and that the value of $\frac{n^{2}+n+15}{n}$ is an integer. Determine all possible values of $n$.

Solution:
First, we see that $\frac{n^{2}+n+15}{n}=\frac{n^{2}}{n}+\frac{n}{n}+\frac{15}{n}=n+1+\frac{15}{n}$.

This means that $\frac{n^{2}+n+15}{n}$ is an integer exactly when $n+1+\frac{15}{n}$ is an integer.

Since $n+1$ is an integer, then $\frac{n^{2}+n+15}{n}$ is an integer exactly when $\frac{15}{n}$ is an integer.

The expression $\frac{15}{n}$ is an integer exactly when $n$ is a divisor of 15 .

Since $n$ is a positive integer, then the possible values of $n$ are 1, 3, 5, and 15 .

Final answer: 1, 3, 5, 15
Subfield: Geometry


Question: Donna has a laser at $C$. She points the laser beam at the point $E$. The beam reflects off of $D F$ at $E$ and then off of $F H$ at $G$, as shown, arriving at point $B$ on $A D$. If $D E=E F=1 \mathrm{~m}$, what is the length of $B D$, in metres?

<img_3944>

Solution:
First, we note that a triangle with one right angle and one angle with measure $45^{\circ}$ is isosceles.

This is because the measure of the third angle equals $180^{\circ}-90^{\circ}-45^{\circ}=45^{\circ}$ which means that the triangle has two equal angles.

In particular, $\triangle C D E$ is isosceles with $C D=D E$ and $\triangle E F G$ is isosceles with $E F=F G$. Since $D E=E F=1 \mathrm{~m}$, then $C D=F G=1 \mathrm{~m}$.

Join $C$ to $G$.

<img_3379>

Consider quadrilateral $C D F G$. Since the angles at $D$ and $F$ are right angles and since $C D=G F$, it must be the case that $C D F G$ is a rectangle.

This means that $C G=D F=2 \mathrm{~m}$ and that the angles at $C$ and $G$ are right angles.

Since $\angle C G F=90^{\circ}$ and $\angle D C G=90^{\circ}$, then $\angle B G C=180^{\circ}-90^{\circ}-45^{\circ}=45^{\circ}$ and $\angle B C G=90^{\circ}$.

This means that $\triangle B C G$ is also isosceles with $B C=C G=2 \mathrm{~m}$.

Finally, $B D=B C+C D=2 \mathrm{~m}+1 \mathrm{~m}=3 \mathrm{~m}$.

Final answer: 3
Subfield: Geometry


Question: Donna has a laser at $C$. She points the laser beam at the point $E$. The beam reflects off of $D F$ at $E$ and then off of $F H$ at $G$, as shown, arriving at point $B$ on $A D$. If $D E=E F=1 \mathrm{~m}$, what is the length of $B D$, in metres?

![](https://cdn.mathpix.com/cropped/2023_12_21_55f6cf88d5d7c500e366g-1.jpg?height=456&width=317&top_left_y=276&top_left_x=1354)

Solution:
First, we note that a triangle with one right angle and one angle with measure $45^{\circ}$ is isosceles.

This is because the measure of the third angle equals $180^{\circ}-90^{\circ}-45^{\circ}=45^{\circ}$ which means that the triangle has two equal angles.

In particular, $\triangle C D E$ is isosceles with $C D=D E$ and $\triangle E F G$ is isosceles with $E F=F G$. Since $D E=E F=1 \mathrm{~m}$, then $C D=F G=1 \mathrm{~m}$.

Join $C$ to $G$.

![](https://cdn.mathpix.com/cropped/2023_12_21_efd50db100807eeb16e0g-1.jpg?height=463&width=358&top_left_y=1384&top_left_x=989)

Consider quadrilateral $C D F G$. Since the angles at $D$ and $F$ are right angles and since $C D=G F$, it must be the case that $C D F G$ is a rectangle.

This means that $C G=D F=2 \mathrm{~m}$ and that the angles at $C$ and $G$ are right angles.

Since $\angle C G F=90^{\circ}$ and $\angle D C G=90^{\circ}$, then $\angle B G C=180^{\circ}-90^{\circ}-45^{\circ}=45^{\circ}$ and $\angle B C G=90^{\circ}$.

This means that $\triangle B C G$ is also isosceles with $B C=C G=2 \mathrm{~m}$.

Finally, $B D=B C+C D=2 \mathrm{~m}+1 \mathrm{~m}=3 \mathrm{~m}$.

Final answer: 3
Subfield: Combinatorics


Question: Ada starts with $x=10$ and $y=2$, and applies the following process:

Step 1: Add $x$ and $y$. Let $x$ equal the result. The value of $y$ does not change. Step 2: Multiply $x$ and $y$. Let $x$ equal the result. The value of $y$ does not change.

Step 3: Add $y$ and 1. Let $y$ equal the result. The value of $x$ does not change.

Ada keeps track of the values of $x$ and $y$ :

|  | $x$ | $y$ |
| :---: | :---: | :---: |
| Before Step 1 | 10 | 2 |
| After Step 1 | 12 | 2 |
| After Step 2 | 24 | 2 |
| After Step 3 | 24 | 3 |

Continuing now with $x=24$ and $y=3$, Ada applies the process two more times. What is the final value of $x$ ?

Solution:
We apply the process two more times:

|  | $x$ | $y$ |
| :---: | :---: | :---: |
| Before Step 1 | 24 | 3 |
| After Step 1 | 27 | 3 |
| After Step 2 | 81 | 3 |
| After Step 3 | 81 | 4 |


|  | $x$ | $y$ |
| :---: | :---: | :---: |
| Before Step 1 | 81 | 4 |
| After Step 1 | 85 | 4 |
| After Step 2 | 340 | 4 |
| After Step 3 | 340 | 5 |

Therefore, the final value of $x$ is 340 .

Final answer: 340
Subfield: Number Theory


Question: Determine all integers $k$, with $k \neq 0$, for which the parabola with equation $y=k x^{2}+6 x+k$ has two distinct $x$-intercepts.

Solution:
The parabola with equation $y=k x^{2}+6 x+k$ has two distinct $x$-intercepts exactly when the discriminant of the quadratic equation $k x^{2}+6 x+k=0$ is positive.

Here, the disciminant equals $\Delta=6^{2}-4 \cdot k \cdot k=36-4 k^{2}$.

The inequality $36-4 k^{2}>0$ is equivalent to $k^{2}<9$.

Since $k$ is an integer and $k \neq 0$, then $k$ can equal $-2,-1,1,2$.

(If $k \geq 3$ or $k \leq-3$, we get $k^{2} \geq 9$ so no values of $k$ in these ranges give the desired result.)

Final answer: -2,-1,1,2
Subfield: Number Theory


Question: The positive integers $a$ and $b$ have no common divisor larger than 1 . If the difference between $b$ and $a$ is 15 and $\frac{5}{9}<\frac{a}{b}<\frac{4}{7}$, what is the value of $\frac{a}{b}$ ?

Solution:
Since $\frac{a}{b}<\frac{4}{7}$ and $\frac{4}{7}<1$, then $\frac{a}{b}<1$.

Since $a$ and $b$ are positive integers, then $a<b$.

Since the difference between $a$ and $b$ is 15 and $a<b$, then $b=a+15$.

Therefore, we have $\frac{5}{9}<\frac{a}{a+15}<\frac{4}{7}$.

We multiply both sides of the left inequality by $9(a+15)$ (which is positive) to obtain $5(a+15)<9 a$ from which we get $5 a+75<9 a$ and so $4 a>75$.

From this, we see that $a>\frac{75}{4}=18.75$.

Since $a$ is an integer, then $a \geq 19$.

We multiply both sides of the right inequality by $7(a+15)$ (which is positive) to obtain $7 a<4(a+15)$ from which we get $7 a<4 a+60$ and so $3 a<60$.

From this, we see that $a<20$.

Since $a$ is an integer, then $a \leq 19$.

Since $a \geq 19$ and $a \leq 19$, then $a=19$, which means that $\frac{a}{b}=\frac{19}{34}$.

Final answer: $\frac{19}{34}$
Subfield: Algebra


Question: A geometric sequence has first term 10 and common ratio $\frac{1}{2}$.

An arithmetic sequence has first term 10 and common difference $d$.

The ratio of the 6th term in the geometric sequence to the 4th term in the geometric sequence equals the ratio of the 6th term in the arithmetic sequence to the 4 th term in the arithmetic sequence.

Determine all possible values of $d$.

(An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant, called the common difference. For example, 3, 5, 7, 9 are the first four terms of an arithmetic sequence.

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

Solution:
The first 6 terms of a geometric sequence with first term 10 and common ratio $\frac{1}{2}$ are $10,5, \frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \frac{5}{16}$.

Here, the ratio of its 6 th term to its 4 th term is $\frac{5 / 16}{5 / 4}$ which equals $\frac{1}{4}$. (We could have determined this without writing out the sequence, since moving from the 4th term to the 6th involves multiplying by $\frac{1}{2}$ twice.)

The first 6 terms of an arithmetic sequence with first term 10 and common difference $d$ are $10,10+d, 10+2 d, 10+3 d, 10+4 d, 10+5 d$.

Here, the ratio of the 6 th term to the 4 th term is $\frac{10+5 d}{10+3 d}$.

Since these ratios are equal, then $\frac{10+5 d}{10+3 d}=\frac{1}{4}$, which gives $4(10+5 d)=10+3 d$ and so $40+20 d=10+3 d$ or $17 d=-30$ and so $d=-\frac{30}{17}$.

Final answer: $-\frac{30}{17}$
Subfield: Algebra


Question: For each positive real number $x$, define $f(x)$ to be the number of prime numbers $p$ that satisfy $x \leq p \leq x+10$. What is the value of $f(f(20))$ ?

Solution:
Let $a=f(20)$. Then $f(f(20))=f(a)$.

To calculate $f(f(20))$, we determine the value of $a$ and then the value of $f(a)$.

By definition, $a=f(20)$ is the number of prime numbers $p$ that satisfy $20 \leq p \leq 30$.

The prime numbers between 20 and 30, inclusive, are 23 and 29 , so $a=f(20)=2$.

Thus, $f(f(20))=f(a)=f(2)$.

By definition, $f(2)$ is the number of prime numbers $p$ that satisfy $2 \leq p \leq 12$.

The prime numbers between 2 and 12, inclusive, are $2,3,5,7,11$, of which there are 5 .

Therefore, $f(f(20))=5$.

Final answer: 5
Subfield: Algebra


Question: Determine all triples $(x, y, z)$ of real numbers that satisfy the following system of equations:

$$
\begin{aligned}
(x-1)(y-2) & =0 \\
(x-3)(z+2) & =0 \\
x+y z & =9
\end{aligned}
$$

Solution:
Since $(x-1)(y-2)=0$, then $x=1$ or $y=2$.

Suppose that $x=1$. In this case, the remaining equations become:

$$
\begin{aligned}
(1-3)(z+2) & =0 \\
1+y z & =9
\end{aligned}
$$

or

$$
\begin{array}{r}
-2(z+2)=0 \\
y z=8
\end{array}
$$

From the first of these equations, $z=-2$.

From the second of these equations, $y(-2)=8$ and so $y=-4$.

Therefore, if $x=1$, the only solution is $(x, y, z)=(1,-4,-2)$.

Suppose that $y=2$. In this case, the remaining equations become:

$$
\begin{aligned}
(x-3)(z+2) & =0 \\
x+2 z & =9
\end{aligned}
$$

From the first equation $x=3$ or $z=-2$.

If $x=3$, then $3+2 z=9$ and so $z=3$.

If $z=-2$, then $x+2(-2)=9$ and so $x=13$.

Therefore, if $y=2$, the solutions are $(x, y, z)=(3,2,3)$ and $(x, y, z)=(13,2,-2)$.

In summary, the solutions to the system of equations are

$$
(x, y, z)=(1,-4,-2),(3,2,3),(13,2,-2)
$$

We can check by substitution that each of these triples does indeed satisfy each of the equations.

Final answer: (1,-4,-2),(3,2,3),(13,2,-2)
Subfield: Algebra


Question: Suppose that the function $g$ satisfies $g(x)=2 x-4$ for all real numbers $x$ and that $g^{-1}$ is the inverse function of $g$. Suppose that the function $f$ satisfies $g\left(f\left(g^{-1}(x)\right)\right)=2 x^{2}+16 x+26$ for all real numbers $x$. What is the value of $f(\pi)$ ?

Solution:
Since the function $g$ is linear and has positive slope, then it is one-to-one and so invertible. This means that $g^{-1}(g(a))=a$ for every real number $a$ and $g\left(g^{-1}(b)\right)=b$ for every real number $b$.

Therefore, $g\left(f\left(g^{-1}(g(a))\right)\right)=g(f(a))$ for every real number $a$.

This means that

$$
\begin{aligned}
g(f(a)) & =g\left(f\left(g^{-1}(g(a))\right)\right) \\
& =2(g(a))^{2}+16 g(a)+26 \\
& =2(2 a-4)^{2}+16(2 a-4)+26 \\
& =2\left(4 a^{2}-16 a+16\right)+32 a-64+26 \\
& =8 a^{2}-6
\end{aligned}
$$

Furthermore, if $b=f(a)$, then $g^{-1}(g(f(a)))=g^{-1}(g(b))=b=f(a)$.

Therefore,

$$
f(a)=g^{-1}(g(f(a)))=g^{-1}\left(8 a^{2}-6\right)
$$

Since $g(x)=2 x-4$, then $y=2 g^{-1}(y)-4$ and so $g^{-1}(y)=\frac{1}{2} y+2$.

Therefore,

$$
f(a)=\frac{1}{2}\left(8 a^{2}-6\right)+2=4 a^{2}-1
$$

and so $f(\pi)=4 \pi^{2}-1$.
Since the function $g$ is linear and has positive slope, then it is one-to-one and so invertible. To find a formula for $g^{-1}(y)$, we start with the equation $g(x)=2 x-4$, convert to $y=2 g^{-1}(y)-4$ and then solve for $g^{-1}(y)$ to obtain $2 g^{-1}(y)=y+4$ and so $g^{-1}(y)=\frac{y+4}{2}$. We are given that $g\left(f\left(g^{-1}(x)\right)\right)=2 x^{2}+16 x+26$.

We can apply the function $g^{-1}$ to both sides to obtain successively:

$$
\begin{aligned}
f\left(g^{-1}(x)\right) & =g^{-1}\left(2 x^{2}+16 x+26\right) \\
f\left(g^{-1}(x)\right) & \left.=\frac{\left(2 x^{2}+16 x+26\right)+4}{2} \quad \text { (knowing a formula for } g^{-1}\right) \\
f\left(g^{-1}(x)\right) & =x^{2}+8 x+15 \\
f\left(\frac{x+4}{2}\right) & \left.=x^{2}+8 x+15 \quad \text { (knowing a formula for } g^{-1}\right) \\
f\left(\frac{x+4}{2}\right) & =x^{2}+8 x+16-1 \\
f\left(\frac{x+4}{2}\right) & =(x+4)^{2}-1
\end{aligned}
$$

We want to determine the value of $f(\pi)$.

Thus, we can replace $\frac{x+4}{2}$ with $\pi$, which is equivalent to replacing $x+4$ with $2 \pi$.

Thus, $f(\pi)=(2 \pi)^{2}-1=4 \pi^{2}-1$.

Final answer: $4 \pi^{2}-1$