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


Question: An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:

$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$

Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write

$$
S_{3}[263415]=(132,312,231,213)
$$

More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.

In this power question, you will be asked to analyze some of the properties of labels and signatures.


We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :

![](https://cdn.mathpix.com/cropped/2023_12_21_793ce3b512229d118445g-1.jpg?height=236&width=442&top_left_y=2213&top_left_x=836)



A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:

![](https://cdn.mathpix.com/cropped/2023_12_21_54786ae6b210123098f4g-1.jpg?height=257&width=464&top_left_y=419&top_left_x=825)
Show that $(231,213,123,132)$ is a unique 3 -signature.

Solution:
Let $L=a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$. We have $a_{4}<a_{6}<a_{5}$ (from window \#4), $a_{3}<a_{1}<a_{2}$ (from window \#1), and $a_{2}<a_{4}$ (from window \#2). Linking these inequalities, we get

$$
a_{3}<a_{1}<a_{2}<a_{4}<a_{6}<a_{5} \quad \Rightarrow \quad L=231465
$$

so $S_{3}[L]$ is unique.

Final answer: 证明题，略
Subfield: Combinatorics


Question: An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:

$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$

Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write

$$
S_{3}[263415]=(132,312,231,213)
$$

More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.

In this power question, you will be asked to analyze some of the properties of labels and signatures.


We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :

<img_3144>



A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:

<img_3777>
Show that $(231,213,123,132)$ is a unique 3 -signature.

Solution:
Let $L=a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$. We have $a_{4}<a_{6}<a_{5}$ (from window \#4), $a_{3}<a_{1}<a_{2}$ (from window \#1), and $a_{2}<a_{4}$ (from window \#2). Linking these inequalities, we get

$$
a_{3}<a_{1}<a_{2}<a_{4}<a_{6}<a_{5} \quad \Rightarrow \quad L=231465
$$

so $S_{3}[L]$ is unique.
Subfield: Combinatorics


Question: An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:

$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$

Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write

$$
S_{3}[263415]=(132,312,231,213)
$$

More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.

In this power question, you will be asked to analyze some of the properties of labels and signatures.


We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :

![](https://cdn.mathpix.com/cropped/2023_12_21_793ce3b512229d118445g-1.jpg?height=236&width=442&top_left_y=2213&top_left_x=836)



A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:

![](https://cdn.mathpix.com/cropped/2023_12_21_54786ae6b210123098f4g-1.jpg?height=257&width=464&top_left_y=419&top_left_x=825)
Find two 5-labels with unique 2-signatures.

Solution:
12345 and 54321 are the only ones.

Final answer: 12345, 54321
Subfield: Combinatorics


Question: An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:

$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$

Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write

$$
S_{3}[263415]=(132,312,231,213)
$$

More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.

In this power question, you will be asked to analyze some of the properties of labels and signatures.


We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :

<img_3144>



A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:

<img_3777>
Find two 5-labels with unique 2-signatures.

Solution:
12345 and 54321 are the only ones.

Final answer: 12345, 54321
Subfield: Combinatorics


Question: An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:

$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$

Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write

$$
S_{3}[263415]=(132,312,231,213)
$$

More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.

In this power question, you will be asked to analyze some of the properties of labels and signatures.


We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :

<img_3144>



A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:

<img_3777>
Prove that $S_{5}[495138627]$ is unique.

Solution:
Let $L=a_{1}, \ldots, a_{9}$ and suppose $S_{5}[L]=S_{5}[495138627]=\left(\omega_{1}, \ldots, \omega_{5}\right)$. Then we get the following inequalities:

| $a_{4}<a_{8}$ | $\left[\right.$ from $\left.\omega_{4}\right]$ | $a_{8}<a_{5}$ | $\left[\right.$ from $\left.\omega_{4}\right]$ |
| :--- | :--- | :--- | :--- |
| $a_{5}<a_{1}$ | $\left[\right.$ from $\left.\omega_{1}\right]$ | $a_{1}<a_{3}$ | $\left[\right.$ from $\left.\omega_{1}\right]$ |
| $a_{3}<a_{7}$ | $\left[\right.$ from $\left.\omega_{3}\right]$ | $a_{7}<a_{9}$ | $\left[\right.$ from $\left.\omega_{5}\right]$ |
| $a_{9}<a_{6}$ | $\left[\right.$ from $\left.\omega_{5}\right]$ | $a_{6}<a_{2}$ | [from $\left.\omega_{2}\right]$ |

Combining, we get $a_{4}<a_{8}<a_{5}<a_{1}<a_{3}<a_{7}<a_{9}<a_{6}<a_{2}$, which forces $a_{4}=1, a_{8}=2, \ldots, a_{2}=9$. So the label $L$ is forced and $S_{5}[495138627]$ is therefore unique.
Subfield: Geometry


Question: In $\triangle A B C, D$ is on $\overline{A C}$ so that $\overline{B D}$ is the angle bisector of $\angle B$. Point $E$ is on $\overline{A B}$ and $\overline{C E}$ intersects $\overline{B D}$ at $P$. Quadrilateral $B C D E$ is cyclic, $B P=12$ and $P E=4$. Compute the ratio $\frac{A C}{A E}$.

Solution:
Let $\omega$ denote the circle that circumscribes quadrilateral $B C D E$. Draw in line segment $\overline{D E}$. Note that $\angle D P E$ and $\angle C P B$ are congruent, and $\angle D E C$ and $\angle D B C$ are congruent, since they cut off the same arc of $\omega$. Therefore, $\triangle B C P$ and $\triangle E D P$ are similar. Thus $\frac{B C}{D E}=\frac{B P}{E P}=$ $\frac{12}{4}=3$.

Because $\angle B C E$ and $\angle B D E$ cut off the same arc of $\omega$, these angles are congruent. Let $\alpha$ be the measure of these angles. Similarly, $\angle D C E$ and $\angle D B E$ cut off the same arc of $\omega$. Let $\beta$ be the measure of these angles. Since $B D$ is an angle bisector, $\mathrm{m} \angle C B D=\beta$.

Note that $\mathrm{m} \angle A D E=180^{\circ}-\mathrm{m} \angle B D E-\mathrm{m} \angle B D C$. It follows that

$$
\begin{aligned}
\mathrm{m} \angle A D E & =180^{\circ}-\mathrm{m} \angle B D E-\left(180^{\circ}-\mathrm{m} \angle C B D-\mathrm{m} \angle B C D\right) \\
\Rightarrow \mathrm{m} \angle A D E & =180^{\circ}-\mathrm{m} \angle B D E-\left(180^{\circ}-\mathrm{m} \angle C B D-\mathrm{m} \angle B C E-\mathrm{m} \angle D C E\right) \\
\Rightarrow \mathrm{m} \angle A D E & =180^{\circ}-\alpha-\left(180^{\circ}-\beta-\alpha-\beta\right) \\
\Rightarrow \mathrm{m} \angle A D E & =2 \beta=\mathrm{m} \angle C B D .
\end{aligned}
$$

Thus $\angle A D E$ is congruent to $\angle C B D$, and it follows that $\triangle A D E$ is similar to $\triangle A B C$. Hence $\frac{B C}{D E}=\frac{A C}{A E}$, and by substituting in given values, we have $\frac{A C}{A E}=\mathbf{3}$.

Final answer: 3
Subfield: Combinatorics


Question: Complete the following "cross-number puzzle", where each "Across" answer represents a fourdigit number, and each "Down" answer represents a three-digit number. No answer begins with the digit 0.

## Across:

1. $\underline{A} \underline{B} \underline{C} \underline{D}$ is the cube of the sum of the digits in the answer to 1 Down.
2. From left to right, the digits in $E \underline{F} \underline{G} \underline{H}$ are strictly decreasing.
3. From left to right, the digits in $\underline{I} \underline{J} K \underline{L}$ are strictly decreasing.

## Down:

1. $\underline{A} \underline{E}$ I is a perfect fourth power.
2. $\underline{B} \underline{F} \underline{J}$ is a perfect square.
3. The digits in $\underline{C} G \underline{K}$ form a geometric progression.
4. $\underline{D} \underline{H} \underline{L}$ has a two-digit prime factor.

![](https://cdn.mathpix.com/cropped/2023_12_21_94af3fb67559aed23a01g-1.jpg?height=480&width=545&top_left_y=1454&top_left_x=839)

Solution:
From 1 Down, $\underline{A} \underline{E} \underline{I}=256$ or 625 , either of which make $\underline{A} \underline{B} \underline{C} \underline{D}=2197$, so $\underline{A} \underline{E} \underline{I}=256$.

From $\underline{B}=1$ together with 2 Down, $\underline{B} \underline{F} \underline{J}=121$ or 144 . But $\underline{J}=1$ does not work because then 6 Across could not be satisfied. Therefore $\underline{B} \underline{F} \underline{J}=144$.

From $\underline{C}=9$ together with 5 Across and 3 Down, we have $\underline{C} \underline{G} \underline{K}=931$.

From $\underline{D}=7$ together with 5 and 6 Across, we get $\underline{D} \underline{H} \underline{L}=720$ or 710 , but only 710 has a two-digit prime factor.

Final answer: | ${ }^{1} 2$ | ${ }^{2} 1$ | 9 | 7 |
| :--- | :--- | :--- | :--- |
| ${ }^{5} 5$ | 4 | 3 | 1 |
| ${ }^{6} 6$ | 4 | 1 | 0 |
Subfield: Geometry


Question: In rectangle $M N P Q$, point $A$ lies on $\overline{Q N}$. Segments parallel to the rectangle's sides are drawn through point $A$, dividing the rectangle into four regions. The areas of regions I, II, and III are integers in geometric progression. If the area of $M N P Q$ is 2009 , compute the maximum possible area of region I.

<img_3600>

Solution:
Because $A$ is on diagonal $\overline{N Q}$, rectangles $N X A B$ and $A C Q Y$ are similar. Thus $\frac{A B}{A X}=\frac{Q Y}{Q C}=$ $\frac{A C}{A Y} \Rightarrow A B \cdot A Y=A C \cdot A X$. Therefore, we have $2009=[\mathrm{I}]+2[\mathrm{II}]+[\mathrm{III}]$.

Let the common ratio of the geometric progression be $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers ( $q$ may equal 1 ). Then [I] must be some integer multiple of $q^{2}$, which we will call $a q^{2}$. This gives $[\mathrm{II}]=a p q$ and [III $=a p^{2}$. By factoring, we get

$$
2009=a q^{2}+2 a p q+a p^{2} \Rightarrow 7^{2} \cdot 41=a(p+q)^{2}
$$

Thus we must have $p+q=7$ and $a=41$. Since $[\mathrm{I}]=a q^{2}$ and $p, q>0$, the area is maximized when $\frac{p}{q}=\frac{1}{6}$, giving $[\mathrm{I}]=41 \cdot 36=\mathbf{1 4 7 6}$. The areas of the other regions are 246,246, and 41 .

Final answer: 1476
Subfield: Geometry


Question: In rectangle $M N P Q$, point $A$ lies on $\overline{Q N}$. Segments parallel to the rectangle's sides are drawn through point $A$, dividing the rectangle into four regions. The areas of regions I, II, and III are integers in geometric progression. If the area of $M N P Q$ is 2009 , compute the maximum possible area of region I.

![](https://cdn.mathpix.com/cropped/2023_12_21_5bb283f04ad556d7c2dcg-1.jpg?height=398&width=624&top_left_y=492&top_left_x=794)

Solution:
Because $A$ is on diagonal $\overline{N Q}$, rectangles $N X A B$ and $A C Q Y$ are similar. Thus $\frac{A B}{A X}=\frac{Q Y}{Q C}=$ $\frac{A C}{A Y} \Rightarrow A B \cdot A Y=A C \cdot A X$. Therefore, we have $2009=[\mathrm{I}]+2[\mathrm{II}]+[\mathrm{III}]$.

Let the common ratio of the geometric progression be $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers ( $q$ may equal 1 ). Then [I] must be some integer multiple of $q^{2}$, which we will call $a q^{2}$. This gives $[\mathrm{II}]=a p q$ and [III $=a p^{2}$. By factoring, we get

$$
2009=a q^{2}+2 a p q+a p^{2} \Rightarrow 7^{2} \cdot 41=a(p+q)^{2}
$$

Thus we must have $p+q=7$ and $a=41$. Since $[\mathrm{I}]=a q^{2}$ and $p, q>0$, the area is maximized when $\frac{p}{q}=\frac{1}{6}$, giving $[\mathrm{I}]=41 \cdot 36=\mathbf{1 4 7 6}$. The areas of the other regions are 246,246, and 41 .

Final answer: 1476
Subfield: Combinatorics


Question: Let $N$ be a six-digit number formed by an arrangement of the digits $1,2,3,3,4,5$. Compute the smallest value of $N$ that is divisible by 264 .

Solution:
Note that $264=3 \cdot 8 \cdot 11$, so we will need to address all these factors. Because the sum of the digits is 18 , it follows that 3 divides $N$, regardless of how we order the digits of $N$. In order for 8 to divide $N$, we need $N$ to end in $\underline{O} 12, \underline{O} 52, \underline{E} 32$, or $\underline{E} 24$, where $O$ and $E$ denote odd and even digits. Now write $N=\underline{U} \underline{V} \underline{W} \underline{X} \underline{Y} \underline{Z}$. Note that $N$ is divisible by 11 if and only if $(U+W+Y)-(V+X+Z)$ is divisible by 11. Because the sum of the three largest digits is only 12 , we must have $U+W+Y=V+X+Z=9$.

Because $Z$ must be even, this implies that $V, X, Z$ are $2,3,4$ (in some order). This means $Y \neq 2$, and so we must have $Z \neq 4 \Rightarrow Z=2$. Of the three remaining possibilities, $\underline{E} 32$ gives the smallest solution, 135432.

Final answer: 135432
Subfield: Geometry


Question: In triangle $A B C, A B=4, B C=6$, and $A C=8$. Squares $A B Q R$ and $B C S T$ are drawn external to and lie in the same plane as $\triangle A B C$. Compute $Q T$.

Solution:
Set $\mathrm{m} \angle A B C=x$ and $\mathrm{m} \angle T B Q=y$. Then $x+y=180^{\circ}$ and so $\cos x+\cos y=0$. Applying the Law of Cosines to triangles $A B C$ and $T B Q$ gives $A C^{2}=A B^{2}+B C^{2}-2 A B \cdot B C \cos x$ and $Q T^{2}=B T^{2}+B Q^{2}-2 B T \cdot B Q \cos y$, which, after substituting values, become $8^{2}=$ $4^{2}+6^{2}-48 \cos x$ and $Q T^{2}=4^{2}+6^{2}-48 \cos y$.

Adding the last two equations yields $Q T^{2}+8^{2}=2\left(4^{2}+6^{2}\right)$ or $Q T=\mathbf{2} \sqrt{\mathbf{1 0}}$.

Final answer: $2 \sqrt{10}$
Subfield: Geometry


Question: An ellipse in the first quadrant is tangent to both the $x$-axis and $y$-axis. One focus is at $(3,7)$, and the other focus is at $(d, 7)$. Compute $d$.

Solution:
See the diagram below. The center of the ellipse is $C=\left(\frac{d+3}{2}, 7\right)$. The major axis of the ellipse is the line $y=7$, and the minor axis is the line $x=\frac{d+3}{2}$. The ellipse is tangent to the coordinate axes at $T_{x}=\left(\frac{d+3}{2}, 0\right)$ and $T_{y}=(0,7)$. Let $F_{1}=(3,7)$ and $F_{2}=(d, 7)$. Using the locus definition of an ellipse, we have $F_{1} T_{x}+F_{2} T_{x}=F_{1} T_{y}+F_{2} T_{y}$; that is,

$$
2 \sqrt{\left(\frac{d-3}{2}\right)^{2}+7^{2}}=d+3 \quad \text { or } \quad \sqrt{(d-3)^{2}+14^{2}}=d+3
$$

Squaring both sides of the last equation gives $d^{2}-6 d+205=d^{2}+6 d+9$ or $196=12 d$, so $d=\frac{49}{3}$.

<img_4034>

Final answer: $\frac{49}{3}$
Subfield: Geometry


Question: Let $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6} A_{7} A_{8}$ be a regular octagon. Let $\mathbf{u}$ be the vector from $A_{1}$ to $A_{2}$ and let $\mathbf{v}$ be the vector from $A_{1}$ to $A_{8}$. The vector from $A_{1}$ to $A_{4}$ can be written as $a \mathbf{u}+b \mathbf{v}$ for a unique ordered pair of real numbers $(a, b)$. Compute $(a, b)$.

Solution:
We can scale the octagon so that $A_{1} A_{2}=\sqrt{2}$. Because the exterior angle of the octagon is $45^{\circ}$, we can place the octagon in the coordinate plane with $A_{1}$ being the origin, $A_{2}=(\sqrt{2}, 0)$, and $A_{8}=(1,1)$.

<img_3693>

Then $A_{3}=(1+\sqrt{2}, 1)$ and $A_{4}=(1+\sqrt{2}, 1+\sqrt{2})$. It follows that $\mathbf{u}=\langle\sqrt{2}, 0\rangle, \mathbf{v}=\langle-1,1\rangle$, and

$$
\overrightarrow{A_{1} A_{4}}=\langle 1+\sqrt{2}, 1+\sqrt{2}\rangle=a\langle\sqrt{2}, 0\rangle+b\langle-1,1\rangle=\langle a \sqrt{2}-b, b\rangle .
$$

Thus $b=\sqrt{2}+1$ and $a \sqrt{2}-b=\sqrt{2}+1$, or $a=2+\sqrt{2}$, so $(a, b)=(2+\sqrt{2}, \sqrt{2}+1)$.
Extend $\overline{A_{1} A_{2}}$ and $\overline{A_{5} A_{4}}$ to meet at point $Q$; let $P$ be the intersection of $\widehat{A_{1} Q}$ and $\overleftrightarrow{A_{6} A_{3}}$. Then $A_{1} A_{2}=\|\mathbf{u}\|, A_{2} P=\|\mathbf{u}\| \sqrt{2}$, and $P Q=\|\mathbf{u}\|$, so $A_{1} Q=(2+\sqrt{2})\|\mathbf{u}\|$.



Because $A_{1} Q A_{4}$ is a $45^{\circ}-45^{\circ}-90^{\circ}$ right triangle, $A_{4} Q=\frac{A_{1} Q}{\sqrt{2}}=(\sqrt{2}+1)\|\mathbf{u}\|$. Thus $\overrightarrow{A_{1} A_{4}}=\overrightarrow{A_{1} Q}+\overrightarrow{Q A_{4}}$, and because $\|\mathbf{u}\|=\|\mathbf{v}\|$, we have $(a, b)=(2+\sqrt{2}, \sqrt{2}+\mathbf{1})$.

Final answer: $\quad(2+\sqrt{2}, 1+\sqrt{2})$