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1735

Geometry

Three circular arcs $\gamma_{1}, \gamma_{2}$, and $\gamma_{3}$ connect the points $A$ and $C$. These arcs lie in the same half-plane defined by line $A C$ in such a way that $\operatorname{arc} \gamma_{2}$ lies between the $\operatorname{arcs} \gamma_{1}$ and $\gamma_{3}$. Point $B$ lies on the segment $A C$. Let $h_{1}, h_{2}$, and $h_{3}$ be three rays starting at $B$, lying in the same half-plane, $h_{2}$ being between $h_{1}$ and $h_{3}$. For $i, j=1,2,3$, denote by $V_{i j}$ the point of intersection of $h_{i}$ and $\gamma_{j}$ (see the Figure below). Denote by $\overparen{V_{i j} V_{k j}} \overparen{k_{k \ell} V_{i \ell}}$ the curved quadrilateral, whose sides are the segments $V_{i j} V_{i \ell}, V_{k j} V_{k \ell}$ and $\operatorname{arcs} V_{i j} V_{k j}$ and $V_{i \ell} V_{k \ell}$. We say that this quadrilateral is circumscribed if there exists a circle touching these two segments and two arcs. Prove that if the curved quadrilaterals $\sqrt{V_{11} V_{21}} \sqrt{V_{22} V_{12}}, \sqrt{V_{12} V_{22}} \sqrt{V_{23} V_{13}}, \sqrt{V_{21} V_{31}} \sqrt{V_{32} V_{22}}$ are circumscribed, then the curved quadrilateral $\overparen{V_{22} V_{32}} \overparen{V_{33} V_{23}}$ is circumscribed, too. Fig. 1

TP_MM_maths_en_COMP

1975

Combinatorics

Construct a tetromino by attaching two $2 \times 1$ dominoes along their longer sides such that the midpoint of the longer side of one domino is a corner of the other domino. This construction yields two kinds of tetrominoes with opposite orientations. Let us call them Sand Z-tetrominoes, respectively. S-tetrominoes Z-tetrominoes Assume that a lattice polygon $P$ can be tiled with $S$-tetrominoes. Prove than no matter how we tile $P$ using only $\mathrm{S}$ - and Z-tetrominoes, we always use an even number of Z-tetrominoes.

TP_MM_maths_en_COMP

2039

Combinatorics

An anti-Pascal pyramid is a finite set of numbers, placed in a triangle-shaped array so that the first row of the array contains one number, the second row contains two numbers, the third row contains three numbers and so on; and, except for the numbers in the bottom row, each number equals the absolute value of the difference of the two numbers below it. For instance, the triangle below is an anti-Pascal pyramid with four rows, in which every integer from 1 to $1+2+3+4=10$ occurs exactly once: Prove that it is impossible to form an anti-Pascal pyramid with 2018 rows, using every integer from 1 to $1+2+\cdots+2018$ exactly once.

TP_MM_maths_en_COMP

2184

Geometry

Let $A B C D$ be a cyclic quadrilateral, and let diagonals $A C$ and $B D$ intersect at $X$. Let $C_{1}, D_{1}$ and $M$ be the midpoints of segments $C X$, $D X$ and $C D$, respectively. Lines $A D_{1}$ and $B C_{1}$ intersect at $Y$, and line $M Y$ intersects diagonals $A C$ and $B D$ at different points $E$ and $F$, respectively. Prove that line $X Y$ is tangent to the circle through $E, F$ and $X$.

TP_MM_maths_en_COMP

2227

Geometry

Let $H$ be the orthocenter and $G$ be the centroid of acute-angled triangle $\triangle A B C$ with $A B \neq A C$. The line $A G$ intersects the circumcircle of $\triangle A B C$ at $A$ and $P$. Let $P^{\prime}$ be the reflection of $P$ in the line $B C$. Prove that $\angle C A B=60^{\circ}$ if and only if $H G=G P^{\prime}$.

TP_MM_maths_en_COMP

2270

Geometry

In the diagram, two circles are tangent to each other at point $B$. A straight line is drawn through $B$ cutting the two circles at $A$ and $C$, as shown. Tangent lines are drawn to the circles at $A$ and $C$. Prove that these two tangent lines are parallel.

TP_MM_maths_en_COMP

2274

Combinatorics

A school has a row of $n$ open lockers, numbered 1 through $n$. After arriving at school one day, Josephine starts at the beginning of the row and closes every second locker until reaching the end of the row, as shown in the example below. Then on her way back, she closes every second locker that is still open. She continues in this manner along the row, until only one locker remains open. Define $f(n)$ to be the number of the last open locker. For example, if there are 15 lockers, then $f(15)=11$ as shown below: Prove that there is no positive integer $n$ such that $f(n)=2005$ and there are infinitely many positive integers $n$ such that $f(n)=f(2005)$.

TP_MM_maths_en_COMP

2348

Geometry

In the diagram, $C$ lies on $B D$. Also, $\triangle A B C$ and $\triangle E C D$ are equilateral triangles. If $M$ is the midpoint of $B E$ and $N$ is the midpoint of $A D$, prove that $\triangle M N C$ is equilateral.

TP_MM_maths_en_COMP

2363

Geometry

In parallelogram $A B C D, A B=a$ and $B C=b$, where $a>b$. The points of intersection of the angle bisectors are the vertices of quadrilateral $P Q R S$. Prove that $P Q R S$ is a rectangle.

TP_MM_maths_en_COMP

2364

Geometry

In parallelogram $A B C D, A B=a$ and $B C=b$, where $a>b$. The points of intersection of the angle bisectors are the vertices of quadrilateral $P Q R S$. Prove that $P R=a-b$.

TP_MM_maths_en_COMP

2367

Geometry

An equilateral triangle $A B C$ has side length 2 . A square, $P Q R S$, is such that $P$ lies on $A B, Q$ lies on $B C$, and $R$ and $S$ lie on $A C$ as shown. The points $P, Q, R$, and $S$ move so that $P, Q$ and $R$ always remain on the sides of the triangle and $S$ moves from $A C$ to $A B$ through the interior of the triangle. If the points $P, Q, R$ and $S$ always form the vertices of a square, show that the path traced out by $S$ is a straight line parallel to $B C$.

TP_MM_maths_en_COMP

2398

Geometry

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

TP_MM_maths_en_COMP

2448

Geometry

A circle with its centre on the $y$-axis intersects the graph of $y=|x|$ at the origin, $O$, and exactly two other distinct points, $A$ and $B$, as shown. Prove that the ratio of the area of triangle $A B O$ to the area of the circle is always $1: \pi$.

TP_MM_maths_en_COMP

2449

Geometry

In the diagram, triangle $A B C$ has a right angle at $B$ and $M$ is the midpoint of $B C$. A circle is drawn using $B C$ as its diameter. $P$ is the point of intersection of the circle with $A C$. The tangent to the circle at $P$ cuts $A B$ at $Q$. Prove that $Q M$ is parallel to $A C$.

TP_MM_maths_en_COMP

2462

Geometry

A large square $A B C D$ is drawn, with a second smaller square $P Q R S$ completely inside it so that the squares do not touch. Line segments $A P, B Q, C R$, and $D S$ are drawn, dividing the region between the squares into four nonoverlapping convex quadrilaterals, as shown. If the sides of $P Q R S$ are not parallel to the sides of $A B C D$, prove that the sum of the areas of quadrilaterals $A P S D$ and $B C R Q$ equals the sum of the areas of quadrilaterals $A B Q P$ and $C D S R$. (Note: A convex quadrilateral is a quadrilateral in which the measure of each of the four interior angles is less than $180^{\circ}$.)

TP_MM_maths_en_COMP

2474

Geometry

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

TP_MM_maths_en_COMP

2488

Geometry

In the diagram, quadrilateral $A B C D$ has points $M$ and $N$ on $A B$ and $D C$, respectively, with $\frac{A M}{A B}=\frac{N C}{D C}$. Line segments $A N$ and $D M$ intersect at $P$, while $B N$ and $C M$ intersect at $Q$. Prove that the area of quadrilateral $P M Q N$ equals the sum of the areas of $\triangle A P D$ and $\triangle B Q C$.

TP_MM_maths_en_COMP

2499

Geometry

In the diagram, $A B$ and $B C$ are chords of the circle with $A B<B C$. If $D$ is the point on the circle such that $A D$ is perpendicular to $B C$ and $E$ is the point on the circle such that $D E$ is parallel to $B C$, carefully prove, explaining all steps, that $\angle E A C+\angle A B C=90^{\circ}$.

TP_MM_maths_en_COMP

2516

Number Theory

Suppose that $m$ and $n$ are positive integers with $m \geq 2$. The $(m, n)$-sawtooth sequence is a sequence of consecutive integers that starts with 1 and has $n$ teeth, where each tooth starts with 2, goes up to $m$ and back down to 1 . For example, the $(3,4)$-sawtooth sequence is The $(3,4)$-sawtooth sequence includes 17 terms and the average of these terms is $\frac{33}{17}$. Prove that, for all pairs of positive integers $(m, n)$ with $m \geq 2$, the average of the terms in the $(m, n)$-sawtooth sequence is not an integer.

TP_MM_maths_en_COMP

2526

Geometry

In the diagram, $A B C D$ is a square. Points $E$ and $F$ are chosen on $A C$ so that $\angle E D F=45^{\circ}$. If $A E=x, E F=y$, and $F C=z$, prove that $y^{2}=x^{2}+z^{2}$.

TP_MM_maths_en_COMP

2538

Geometry

In the diagram, $A B$ is tangent to the circle with centre $O$ and radius $r$. The length of $A B$ is $p$. Point $C$ is on the circle and $D$ is inside the circle so that $B C D$ is a straight line, as shown. If $B C=C D=D O=q$, prove that $q^{2}+r^{2}=p^{2}$.

TP_MM_maths_en_COMP

2542

Algebra

Suppose there are $n$ plates equally spaced around a circular table. Ross wishes to place an identical gift on each of $k$ plates, so that no two neighbouring plates have gifts. Let $f(n, k)$ represent the number of ways in which he can place the gifts. For example $f(6,3)=2$, as shown below. Throughout this problem, we represent the states of the $n$ plates as a string of 0's and 1's (called a binary string) of length $n$ of the form $p_{1} p_{2} \cdots p_{n}$, with the $r$ th digit from the left (namely $p_{r}$ ) equal to 1 if plate $r$ contains a gift and equal to 0 if plate $r$ does not. We call a binary string of length $n$ allowable if it satisfies the requirements - that is, if no two adjacent digits both equal 1. Note that digit $p_{n}$ is also "adjacent" to digit $p_{1}$, so we cannot have $p_{1}=p_{n}=1$. Prove that $f(n, k)=f(n-1, k)+f(n-2, k-1)$ for all integers $n \geq 3$ and $k \geq 2$.

TP_MM_maths_en_COMP

2547

Geometry

In trapezoid $A B C D, B C$ is parallel to $A D$ and $B C$ is perpendicular to $A B$. Also, the lengths of $A D, A B$ and $B C$, in that order, form a geometric sequence. Prove that $A C$ is perpendicular to $B D$. (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.)

TP_MM_maths_en_COMP

2561

Geometry

In the graph, the parabola $y=x^{2}$ has been translated to the position shown. Prove that $d e=f$.

TP_MM_maths_en_COMP

2562

Geometry

In quadrilateral $K W A D$, the midpoints of $K W$ and $A D$ are $M$ and $N$ respectively. If $M N=\frac{1}{2}(A W+D K)$, prove that $WA$ is parallel to $K D$.

TP_MM_maths_en_COMP

2563

Number Theory

Consider the first $2 n$ natural numbers. Pair off the numbers, as shown, and multiply the two members of each pair. Prove that there is no value of $n$ for which two of the $n$ products are equal.

TP_MM_maths_en_COMP

2803

Combinatorics

This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. Consider the hideout map $M$ below. Show that one Cop can always catch the Robber.

TP_MM_maths_en_COMP

2804

Combinatorics

This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. The map shown below is $\mathcal{C}_{6}$, the cyclic graph with six hideouts. Show that three Cops are sufficient to catch the Robber on $\mathcal{C}_{6}$, so that $C\left(\mathcal{C}_{6}\right) \leq 3$.

TP_MM_maths_en_COMP

2805

Combinatorics

This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. Show that for all maps $M, C(M)<h(M)$.

TP_MM_maths_en_COMP

2807

Combinatorics

This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. Show that $C(M) \leq 3$ for the map below.

TP_MM_maths_en_COMP

2809

Combinatorics

This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$. A path on $n$ hideouts is a map with $n$ hideouts, connected in one long string. (More formally, a map is a path if and only if two hideouts are adjacent to exactly one hideout each and all other hideouts are adjacent to exactly two hideouts each.) It is denoted by $\mathcal{P}_{n}$. The maps $\mathcal{P}_{3}$ through $\mathcal{P}_{6}$ are shown below. Show that $D\left(\mathcal{P}_{n}, 1\right) \leq 2 n$ for $n \geq 3$.

TP_MM_maths_en_COMP

2811

Combinatorics

This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$. Definition: The workday number of $M$, denoted $W(M)$, is the minimum number of Cop workdays needed to guarantee the Robber's capture. For example, a strategy that guarantees capture within three days using 17 Cops on the first day, 11 Cops on the second day, and only 6 Cops on the third day would require a total of $17+11+6=34$ Cop workdays. Let $M$ be a map with $n \geq 3$ hideouts. Prove that $2 \leq W(M) \leq n$, and that these bounds cannot be improved. In other words, prove that for each $n \geq 3$, there exist maps $M_{1}$ and $M_{2}$ such that $W\left(M_{1}\right)=2$ and $W\left(M_{2}\right)=n$.

TP_MM_maths_en_COMP

2812

Combinatorics

This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$. Definition: A map is bipartite if it can be partitioned into two sets of hideouts, $\mathcal{A}$ and $\mathcal{B}$, such that $\mathcal{A} \cap \mathcal{B}=\emptyset$, and each hideout in $\mathcal{A}$ is adjacent only to hideouts in $\mathcal{B}$, and each hideout in $\mathcal{B}$ is adjacent only to hideouts in $\mathcal{A}$. Prove that if $M$ is bipartite, then $C(M) \leq n / 2$.

TP_MM_maths_en_COMP

2813

Combinatorics

This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$. Definition: A map is bipartite if it can be partitioned into two sets of hideouts, $\mathcal{A}$ and $\mathcal{B}$, such that $\mathcal{A} \cap \mathcal{B}=\emptyset$, and each hideout in $\mathcal{A}$ is adjacent only to hideouts in $\mathcal{B}$, and each hideout in $\mathcal{B}$ is adjacent only to hideouts in $\mathcal{A}$. Prove that $C(M) \leq n / 2$ for any map $M$ with the property that, for all hideouts $H_{1}$ and $H_{2}$, either all paths from $H_{1}$ to $H_{2}$ contain an odd number of edges, or all paths from $H_{1}$ to $H_{2}$ contain an even number of edges.

TP_MM_maths_en_COMP

2814

Combinatorics

This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$. Definition: A map is bipartite if it can be partitioned into two sets of hideouts, $\mathcal{A}$ and $\mathcal{B}$, such that $\mathcal{A} \cap \mathcal{B}=\emptyset$, and each hideout in $\mathcal{A}$ is adjacent only to hideouts in $\mathcal{B}$, and each hideout in $\mathcal{B}$ is adjacent only to hideouts in $\mathcal{A}$. A map $M$ is called $k$-perfect if its hideout set $H$ can be partitioned into equal-sized subsets $\mathcal{A}_{1}, \mathcal{A}_{2}, \ldots, \mathcal{A}_{k}$ such that for any $j$, the hideouts of $\mathcal{A}_{j}$ are only adjacent to hideouts in $\mathcal{A}_{j+1}$ or $\mathcal{A}_{j-1}$. (The indices are taken modulo $k$ : the hideouts of $\mathcal{A}_{1}$ may be adjacent to the hideouts of $\mathcal{A}_{k}$.) Show that if $M$ is $k$-perfect, then $C(M) \leq \frac{2 n}{k}$.

TP_MM_maths_en_COMP

2815

Combinatorics

This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$. Definition: The workday number of $M$, denoted $W(M)$, is the minimum number of Cop workdays needed to guarantee the Robber's capture. For example, a strategy that guarantees capture within three days using 17 Cops on the first day, 11 Cops on the second day, and only 6 Cops on the third day would require a total of $17+11+6=34$ Cop workdays. Definition: A map is bipartite if it can be partitioned into two sets of hideouts, $\mathcal{A}$ and $\mathcal{B}$, such that $\mathcal{A} \cap \mathcal{B}=\emptyset$, and each hideout in $\mathcal{A}$ is adjacent only to hideouts in $\mathcal{B}$, and each hideout in $\mathcal{B}$ is adjacent only to hideouts in $\mathcal{A}$. Find an example of a map $M$ with 2012 hideouts such that $C(M)=17$ and $W(M)=34$, or prove that no such map exists.

TP_MM_maths_en_COMP

2816

Combinatorics

This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$. Definition: The workday number of $M$, denoted $W(M)$, is the minimum number of Cop workdays needed to guarantee the Robber's capture. For example, a strategy that guarantees capture within three days using 17 Cops on the first day, 11 Cops on the second day, and only 6 Cops on the third day would require a total of $17+11+6=34$ Cop workdays. Find an example of a map $M$ with 4 or more hideouts such that $W(M)=3$, or prove that no such map exists.

TP_MM_maths_en_COMP

2873

Combinatorics

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 $(12,12,12,21,12,21)$ : 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: Prove that the following signature is possible. $(123,132,213)$,

TP_MM_maths_en_COMP

2874

Combinatorics

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 $(12,12,12,21,12,21)$ : 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: Prove that the following signature is impossible: $(321,312,213)$.

TP_MM_maths_en_COMP

2877

Combinatorics

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 $(12,12,12,21,12,21)$ : 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: Show that $(312,231,312,132)$ is not a unique 3 -signature.

TP_MM_maths_en_COMP

2878

Combinatorics

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 $(12,12,12,21,12,21)$ : 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: Show that $(231,213,123,132)$ is a unique 3 -signature.

TP_MM_maths_en_COMP

2881

Combinatorics

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 $(12,12,12,21,12,21)$ : 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: Prove that $S_{5}[495138627]$ is unique.

TP_MM_maths_en_COMP

2883

Combinatorics

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 $(12,12,12,21,12,21)$ : 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: Show that for each $k \geq 2$, the number of unique $2^{k-1}$-signatures on the set of $2^{k}$-labels is at least $2^{2^{k}-3}$.

TP_MM_maths_en_COMP

2928

Geometry

A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. Circle $A$ has curvature 2; circle $B$ has curvature 1 . Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Without using Descartes' Circle Formula, Show that the circles marked off and sold on day 1 are centered at $\left(0, \pm \frac{2}{3}\right)$ with radius $\frac{1}{3}$.

TP_MM_maths_en_COMP

2934

Geometry

A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. Circle $A$ has curvature 2; circle $B$ has curvature 1 . Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Given three mutually tangent circles with curvatures $a, b, c>0$, suppose that $(a, b, c, 0)$ does not satisfy Descartes' Circle Formula. Show that there are two distinct values of $r$ such that there is a circle of radius $r$ tangent to the given circles.

TP_MM_maths_en_COMP

2935

Geometry

A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. Circle $A$ has curvature 2; circle $B$ has curvature 1 . Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Algebraically, it is possible for a quadruple $(a, b, c, 0)$ to satisfy Descartes' Circle Formula, as occurs when $a=b=1$ and $c=4$. Find a geometric interpretation for this situation.

TP_MM_maths_en_COMP

2936

Geometry

A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. Circle $A$ has curvature 2; circle $B$ has curvature 1 . Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Let $\phi=\frac{1+\sqrt{5}}{2}$, and let $\rho=\phi+\sqrt{\phi}$. Prove that $\rho^{4}=2 \rho^{3}+2 \rho^{2}+2 \rho-1$.

TP_MM_maths_en_COMP

2937

Geometry

A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. Circle $A$ has curvature 2; circle $B$ has curvature 1 . Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Let $\phi=\frac{1+\sqrt{5}}{2}$, and let $\rho=\phi+\sqrt{\phi}$. Show that four pairwise externally tangent circles with nonequal radii in geometric progression must have common ratio $\rho$.

TP_MM_maths_en_COMP

2938

Geometry

A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. Circle $A$ has curvature 2; circle $B$ has curvature 1 . Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Given $A, B, C, D$ as above with $s=a+b+c+d$, if there is a second circle $A^{\prime}$ with curvature $a^{\prime}$ also tangent to $B, C$, and $D$. We can describe $A$ and $A^{\prime}$ as conjugate circles. Use Descartes' Circle Formula to show that $a^{\prime}=2 s-3 a$ and therefore $s^{\prime}=a^{\prime}+b+c+d=$ $3 s-4 a$.

TP_MM_maths_en_COMP

2939

Geometry

A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. Circle $A$ has curvature 2; circle $B$ has curvature 1 . Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Show that by area, $12 \%$ of the kingdom is sold on day 2 .

TP_MM_maths_en_COMP

2941

Geometry

A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. Circle $A$ has curvature 2; circle $B$ has curvature 1 . Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Show that the plots sold on day 3 have mean curvature of 23.

TP_MM_maths_en_COMP

2942

Geometry

A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. Circle $A$ has curvature 2; circle $B$ has curvature 1 . Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Prove that the curvature of each circular plot is an integer.

TP_MM_maths_en_COMP

2943

Combinatorics

A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. Circle $A$ has curvature 2; circle $B$ has curvature 1 . Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Descartes' Circle Formula can be extended by interpreting the coordinates of points on the plane as complex numbers in the usual way: the point $(x, y)$ represents the complex number $x+y i$. On the complex plane, let $z_{A}, z_{B}, z_{C}, z_{D}$ be the centers of circles $A, B, C, D$ respectively; as before, $a, b, c, d$ are the curvatures of their respective circles. Then Descartes' Extended Circle Formula states $$ \left(a \cdot z_{A}+b \cdot z_{B}+c \cdot z_{C}+d \cdot z_{D}\right)^{2}=2\left(a^{2} z_{A}^{2}+b^{2} z_{B}^{2}+c^{2} z_{C}^{2}+d^{2} z_{D}^{2}\right) . $$ Prove that the center of each circular plot has coordinates $\left(\frac{u}{c}, \frac{v}{c}\right)$ where $u$ and $v$ are integers, and $c$ is the curvature of the plot.

TP_MM_maths_en_COMP

2944

Combinatorics

A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. Circle $A$ has curvature 2; circle $B$ has curvature 1 . Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Descartes' Circle Formula can be extended by interpreting the coordinates of points on the plane as complex numbers in the usual way: the point $(x, y)$ represents the complex number $x+y i$. On the complex plane, let $z_{A}, z_{B}, z_{C}, z_{D}$ be the centers of circles $A, B, C, D$ respectively; as before, $a, b, c, d$ are the curvatures of their respective circles. Then Descartes' Extended Circle Formula states $$ \left(a \cdot z_{A}+b \cdot z_{B}+c \cdot z_{C}+d \cdot z_{D}\right)^{2}=2\left(a^{2} z_{A}^{2}+b^{2} z_{B}^{2}+c^{2} z_{C}^{2}+d^{2} z_{D}^{2}\right) . $$ Given a c-triangle $T$, let $a, b$, and $c$ be the curvatures of the three $\operatorname{arcs}$ bounding $T$, with $a \leq b \leq c$, and let $d$ be the curvature of the incircle of $T$. Define the circle configuration associated with $T$ to be $\mathcal{C}(T)=(a, b, c, d)$. Define the c-triangle $T$ to be proper if $c \leq d$. For example, circles of curvatures $-1,2$, and 3 determine two c-triangles. The incircle of one has curvature 6 , so it is proper; the incircle of the other has curvature 2 , so it is not proper. Let $P$ and $Q$ be two c-triangles, with associated configurations $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=$ $(w, x, y, z)$. We say that $P$ dominates $Q$ if $a \leq w, b \leq x, c \leq y$, and $d \leq z$. (The term "dominates" refers to the fact that the radii of the arcs defining $Q$ cannot be larger than the radii of the arcs defining $P$.) Removing the incircle from $T$ gives three c-triangles, $T^{(1)}, T^{(2)}, T^{(3)}$, each bounded by the incircle of $T$ and two of the arcs that bound $T$. These triangles have associated configurations $$ \begin{aligned} \mathcal{C}\left(T^{(1)}\right) & =\left(b, c, d, a^{\prime}\right), \\ \mathcal{C}\left(T^{(2)}\right) & =\left(a, c, d, b^{\prime}\right), \\ \mathcal{C}\left(T^{(3)}\right) & =\left(a, b, d, c^{\prime}\right), \end{aligned} $$ Let $P$ and $Q$ be two proper c-triangles such that $P$ dominates $Q$. Let $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=(w, x, y, z)$. Show that $P^{(3)}$ dominates $P^{(2)}$ and that $P^{(2)}$ dominates $P^{(1)}$.

TP_MM_maths_en_COMP

2945

Combinatorics

A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. Circle $A$ has curvature 2; circle $B$ has curvature 1 . Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Descartes' Circle Formula can be extended by interpreting the coordinates of points on the plane as complex numbers in the usual way: the point $(x, y)$ represents the complex number $x+y i$. On the complex plane, let $z_{A}, z_{B}, z_{C}, z_{D}$ be the centers of circles $A, B, C, D$ respectively; as before, $a, b, c, d$ are the curvatures of their respective circles. Then Descartes' Extended Circle Formula states $$ \left(a \cdot z_{A}+b \cdot z_{B}+c \cdot z_{C}+d \cdot z_{D}\right)^{2}=2\left(a^{2} z_{A}^{2}+b^{2} z_{B}^{2}+c^{2} z_{C}^{2}+d^{2} z_{D}^{2}\right) . $$ Given a c-triangle $T$, let $a, b$, and $c$ be the curvatures of the three $\operatorname{arcs}$ bounding $T$, with $a \leq b \leq c$, and let $d$ be the curvature of the incircle of $T$. Define the circle configuration associated with $T$ to be $\mathcal{C}(T)=(a, b, c, d)$. Define the c-triangle $T$ to be proper if $c \leq d$. For example, circles of curvatures $-1,2$, and 3 determine two c-triangles. The incircle of one has curvature 6 , so it is proper; the incircle of the other has curvature 2 , so it is not proper. Let $P$ and $Q$ be two c-triangles, with associated configurations $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=$ $(w, x, y, z)$. We say that $P$ dominates $Q$ if $a \leq w, b \leq x, c \leq y$, and $d \leq z$. (The term "dominates" refers to the fact that the radii of the arcs defining $Q$ cannot be larger than the radii of the arcs defining $P$.) Removing the incircle from $T$ gives three c-triangles, $T^{(1)}, T^{(2)}, T^{(3)}$, each bounded by the incircle of $T$ and two of the arcs that bound $T$. These triangles have associated configurations $$ \begin{aligned} \mathcal{C}\left(T^{(1)}\right) & =\left(b, c, d, a^{\prime}\right), \\ \mathcal{C}\left(T^{(2)}\right) & =\left(a, c, d, b^{\prime}\right), \\ \mathcal{C}\left(T^{(3)}\right) & =\left(a, b, d, c^{\prime}\right), \end{aligned} $$ Let $P$ and $Q$ be two proper c-triangles such that $P$ dominates $Q$. Let $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=(w, x, y, z)$. Prove that $P^{(1)}$ dominates $Q^{(1)}$.

TP_MM_maths_en_COMP

2946

Combinatorics

A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. Circle $A$ has curvature 2; circle $B$ has curvature 1 . Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Descartes' Circle Formula can be extended by interpreting the coordinates of points on the plane as complex numbers in the usual way: the point $(x, y)$ represents the complex number $x+y i$. On the complex plane, let $z_{A}, z_{B}, z_{C}, z_{D}$ be the centers of circles $A, B, C, D$ respectively; as before, $a, b, c, d$ are the curvatures of their respective circles. Then Descartes' Extended Circle Formula states $$ \left(a \cdot z_{A}+b \cdot z_{B}+c \cdot z_{C}+d \cdot z_{D}\right)^{2}=2\left(a^{2} z_{A}^{2}+b^{2} z_{B}^{2}+c^{2} z_{C}^{2}+d^{2} z_{D}^{2}\right) . $$ Given a c-triangle $T$, let $a, b$, and $c$ be the curvatures of the three $\operatorname{arcs}$ bounding $T$, with $a \leq b \leq c$, and let $d$ be the curvature of the incircle of $T$. Define the circle configuration associated with $T$ to be $\mathcal{C}(T)=(a, b, c, d)$. Define the c-triangle $T$ to be proper if $c \leq d$. For example, circles of curvatures $-1,2$, and 3 determine two c-triangles. The incircle of one has curvature 6 , so it is proper; the incircle of the other has curvature 2 , so it is not proper. Let $P$ and $Q$ be two c-triangles, with associated configurations $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=$ $(w, x, y, z)$. We say that $P$ dominates $Q$ if $a \leq w, b \leq x, c \leq y$, and $d \leq z$. (The term "dominates" refers to the fact that the radii of the arcs defining $Q$ cannot be larger than the radii of the arcs defining $P$.) Removing the incircle from $T$ gives three c-triangles, $T^{(1)}, T^{(2)}, T^{(3)}$, each bounded by the incircle of $T$ and two of the arcs that bound $T$. These triangles have associated configurations $$ \begin{aligned} \mathcal{C}\left(T^{(1)}\right) & =\left(b, c, d, a^{\prime}\right), \\ \mathcal{C}\left(T^{(2)}\right) & =\left(a, c, d, b^{\prime}\right), \\ \mathcal{C}\left(T^{(3)}\right) & =\left(a, b, d, c^{\prime}\right), \end{aligned} $$ Let $P$ and $Q$ be two proper c-triangles such that $P$ dominates $Q$. Let $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=(w, x, y, z)$. Prove that $P^{(3)}$ dominates $Q^{(3)}$.

TP_MM_maths_en_COMP

3052

Combinatorics

The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or "Mountain of Gems". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle ("PT") and its variants. Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below. PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. Consider the parity of each entry: define $$ \operatorname{PaP}(n, k)= \begin{cases}1 & \text { if } \mathrm{Pa}(n, k) \text { is odd } \\ 0 & \text { if } \mathrm{Pa}(n, k) \text { is even }\end{cases} $$ Prove that $\operatorname{PaP}(n, 0)=\operatorname{PaP}(n, n)=1$ for all nonnegative integers $n$.

TP_MM_maths_en_COMP

3054

Combinatorics

The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or "Mountain of Gems". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle ("PT") and its variants. Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below. PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. Consider the parity of each entry: define $$ \operatorname{PaP}(n, k)= \begin{cases}1 & \text { if } \mathrm{Pa}(n, k) \text { is odd } \\ 0 & \text { if } \mathrm{Pa}(n, k) \text { is even }\end{cases} $$ If $n=2^{j}$ for some nonnegative integer $j$, and $0<k<n$, show that $\operatorname{PaP}(n, k)=0$.

TP_MM_maths_en_COMP

3055

Combinatorics

The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or "Mountain of Gems". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle ("PT") and its variants. Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below. PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. Consider the parity of each entry: define $$ \operatorname{PaP}(n, k)= \begin{cases}1 & \text { if } \mathrm{Pa}(n, k) \text { is odd } \\ 0 & \text { if } \mathrm{Pa}(n, k) \text { is even }\end{cases} $$ Let $j \geq 0$, and suppose $n \geq 2^{j}$. Prove that $\mathrm{Pa}(n, k)$ has the same parity as the sum $\operatorname{Pa}\left(n-2^{j}, k-2^{j}\right)+\operatorname{Pa}\left(n-2^{j}, k\right)$, i.e., either both $\operatorname{Pa}(n, k)$ and the given sum are even, or both are odd.

TP_MM_maths_en_COMP

3056

Combinatorics

The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or "Mountain of Gems". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle ("PT") and its variants. Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below. PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. Consider the parity of each entry: define $$ \operatorname{PaP}(n, k)= \begin{cases}1 & \text { if } \mathrm{Pa}(n, k) \text { is odd } \\ 0 & \text { if } \mathrm{Pa}(n, k) \text { is even }\end{cases} $$ If $j$ is an integer such that $2^{j} \leq n<2^{j+1}$, and $k<2^{j}$, prove that $$ \operatorname{PaP}(n, k)=\operatorname{PaP}\left(n-2^{j}, k\right) . $$

TP_MM_maths_en_COMP

3059

Combinatorics

The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or "Mountain of Gems". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle ("PT") and its variants. Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below. PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. Clark's Triangle: If the left side of PT is replaced with consecutive multiples of 6 , starting with 0 , but the right entries (except the first) and the generating rule are left unchanged, the result is called Clark's Triangle. If the $k^{\text {th }}$ entry of the $n^{\text {th }}$ row is denoted by $\mathrm{Cl}(n, k)$, then the formal rule is: $$ \begin{cases}\mathrm{Cl}(n, 0)=6 n & \text { for all } n \\ \mathrm{Cl}(n, n)=1 & \text { for } n \geq 1 \\ \mathrm{Cl}(n, k)=\mathrm{Cl}(n-1, k-1)+\mathrm{Cl}(n-1, k) & \text { for } n \geq 1 \text { and } 1 \leq k \leq n-1\end{cases} $$ The first four rows of Clark's Triangle are given below. Prove the formula $\mathrm{Cl}(n, 1)=3 n^{2}-3 n+1$.

TP_MM_maths_en_COMP

3072

Combinatorics

Leibniz's Harmonic Triangle: Consider the triangle formed by the rule $$ \begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases} $$ This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below. For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$. If $\sum_{i=m}^{\infty} \operatorname{Le}(i, m)=\operatorname{Le}(n, k)$, prove that $n=k=m-1$.

TP_MM_maths_en_COMP

938

Mechanics

4. A complex dance In this problem, we will solve a number of differential equations corresponding to very different physical phenomena that are unified by the idea of oscillation. Oscillations are captured elegantly by extending our notion of numbers to include the imaginary unit number $i$, strangely defined to obey $i^{2}=-1$. In other words, rather than using real numbers, it is more convenient for us to work in terms of complex numbers. Exponentials are usually associated with rapid growth or decay. However, with the inclusion of complex numbers, imaginary "growth" and "decay" can be translated into oscillations by the Euler identity: $$ e^{i \theta}=\cos \theta+i \sin \theta \tag{1} $$

(a) The usual form of Newton's second law $(\vec{F}=m \vec{a})$ breaks down when we go into a rotating frame, where both the centrifugal and Coriolis forces become important to account for. Newton's second law then takes the form $$ \vec{F}=m(\vec{a}+2 \vec{v} \times \vec{\Omega}+\vec{\Omega} \times(\vec{\Omega} \times \vec{r})) \tag{2} $$ For a particle free of forces confined to the $x-y$ plane in a frame which rotates about the $z$ axis with angular frequency $\Omega$, this becomes the complicated-looking system of differential equations, $$ \begin{aligned} & 0=\ddot{x}+2 \Omega \dot{y}-\Omega^{2} x \\ & 0=\ddot{x}-2 \Omega \dot{x}-\Omega^{2} y \end{aligned} \tag{3} $$ where dots represent time derivatives. Defining $\eta=x+i y$, show that Equations 3 are equivalent to the following single (complex) equation: $$ 0=\ddot{\eta}-2 i \Omega \dot{\eta}-\Omega^{2} \eta \tag{4} $$

TP_TO_physics_en_COMP

943

Mechanics

4. A complex dance In this problem, we will solve a number of differential equations corresponding to very different physical phenomena that are unified by the idea of oscillation. Oscillations are captured elegantly by extending our notion of numbers to include the imaginary unit number $i$, strangely defined to obey $i^{2}=-1$. In other words, rather than using real numbers, it is more convenient for us to work in terms of complex numbers. Exponentials are usually associated with rapid growth or decay. However, with the inclusion of complex numbers, imaginary "growth" and "decay" can be translated into oscillations by the Euler identity: $$ e^{i \theta}=\cos \theta+i \sin \theta \tag{1} $$ Context question: (a) The usual form of Newton's second law $(\vec{F}=m \vec{a})$ breaks down when we go into a rotating frame, where both the centrifugal and Coriolis forces become important to account for. Newton's second law then takes the form $$ \vec{F}=m(\vec{a}+2 \vec{v} \times \vec{\Omega}+\vec{\Omega} \times(\vec{\Omega} \times \vec{r})) \tag{2} $$ For a particle free of forces confined to the $x-y$ plane in a frame which rotates about the $z$ axis with angular frequency $\Omega$, this becomes the complicated-looking system of differential equations, $$ \begin{aligned} & 0=\ddot{x}+2 \Omega \dot{y}-\Omega^{2} x \\ & 0=\ddot{x}-2 \Omega \dot{x}-\Omega^{2} y \end{aligned} \tag{3} $$ where dots represent time derivatives. Defining $\eta=x+i y$, show that Equations 3 are equivalent to the following single (complex) equation: $$ 0=\ddot{\eta}-2 i \Omega \dot{\eta}-\Omega^{2} \eta \tag{4} $$ Context answer: \boxed{证明题} Context question: (b) Equation 4 is a version of the damped harmonic oscillator, and can be solved by guessing a solution $\eta=\alpha e^{\lambda t}$. Plugging in this guess, what must $\lambda$ be? Context answer: \boxed{$\lambda=i \Omega$} Context question: (c) Using your answer to part (b), and defining $\alpha=A e^{i \phi}$ where $A$ and $\phi$ are real, find $\mathbf{x}(\mathbf{t})$ and $\mathbf{y}(\mathbf{t})$. This is the trajectory for a particle which is stationary with respect to the symmetry axis. While not required for this problem, an additional guess would reveal that $\eta=\beta t e^{\lambda t}$ is also a solution. Context answer: \boxed{$x(t)=A \cos (\Omega t+\phi)$ , $y(t)=A \sin (\Omega t+\phi)$} Context question: (d) The one-dimensional diffusion equation (also called the "heat equation") is given (for a free particle) by $$ \frac{\partial \psi}{\partial t}=a \frac{\partial^{2} \psi}{\partial x^{2}} \tag{5} $$ A spatial wave can be written as $\sim e^{i k x}$ (larger $k$ 's correspond to waves oscillating on smaller length scales). Guessing a solution $\psi(x, t)=A e^{i k x-i \omega t}$, find $\omega$ in terms of k. A relationship of this time is called a "dispersion relation." Context answer: \boxed{$\omega=-i k^{2} a$} Context question: (e) The most important equation of non-relativistic quantum mechanics is the Schrödinger equation, which is given by $$ i \hbar \frac{\partial \psi}{\partial t}=-\frac{\hbar^{2}}{2 m} \frac{\partial^{2} \psi}{\partial x^{2}} \tag{6} $$ Using your answer to part (d), what is the dispersion relation of the Schrödinger equation? Context answer: \boxed{$\omega=\frac{\hbar k^{2}}{2 m}$}

(f) If the energy of a wave is $E=\hbar \omega$ and the momentum is $p=\hbar k$, show that the dispersion relation found in part (e) resembles the classical expectation for the kinetic energy of a particle, $\mathrm{E}=\mathrm{mv}^{2} / \mathbf{2}$.

TP_TO_physics_en_COMP

946

Electromagnetism

5. Polarization and Oscillation In this problem, we will understand the polarization of metallic bodies and the method of images that simplifies the math in certain geometrical configurations. Throughout the problem, suppose that metals are excellent conductors and they polarize significantly faster than the classical relaxation of the system. Context question: (a) Explain in words why there can't be a non-zero electric field in a metallic body, and why this leads to constant electric potential throughout the body. Context answer: 开放性回答

(b) Laplace's equation is a second order differential equation $$ \nabla^{2} \phi=\frac{\partial^{2} \phi}{\partial x^{2}}+\frac{\partial^{2} \phi}{\partial y^{2}}+\frac{\partial^{2} \phi}{\partial z^{2}}=0 \tag{8} $$ Solutions to this equation are called harmonic functions. One of the most important properties satisfied by these functions is the maximum principle. It states that a harmonic function attains extremes on the boundary. Using this, prove the uniqueness theorem: Solution to Laplace's equation in a volume $V$ is uniquely determined if its solution on the boundary is specified. That is, if $\nabla^{2} \phi_{1}=0$, $\nabla^{2} \phi_{2}=0$ and $\phi_{1}=\phi_{2}$ on the boundary of $V$, then $\phi_{1}=\phi_{2}$ in $V$. Hint: Consider $\phi=\phi_{1}-\phi_{2}$.

TP_TO_physics_en_COMP

963

Modern Physics

4. Lorentz Boost In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts: $$ \begin{aligned} x^{\prime} & =x-v t \\ t^{\prime} & =t \end{aligned} $$ In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts: $$ \begin{aligned} x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\ t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right) \end{aligned} $$ In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal "speed limit" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal "speed limit," which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$. The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$, $$ \begin{aligned} x^{\prime} & =X(x, t, v) \\ t^{\prime} & =T(x, t, v) \end{aligned} $$ which we will refer to as the generalized boost.

(a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.)

TP_TO_physics_en_COMP

964

Modern Physics

4. Lorentz Boost In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts: $$ \begin{aligned} x^{\prime} & =x-v t \\ t^{\prime} & =t \end{aligned} $$ In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts: $$ \begin{aligned} x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\ t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right) \end{aligned} $$ In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal "speed limit" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal "speed limit," which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$. The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$, $$ \begin{aligned} x^{\prime} & =X(x, t, v) \\ t^{\prime} & =T(x, t, v) \end{aligned} $$ which we will refer to as the generalized boost. Context question: (a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.) Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Analogously, we can conclude additionally that $\frac{\partial X}{\partial x}$ is independent of both $x$ and $t$ and $\frac{\partial T}{\partial t}$ is independent of $x$ and $t$. It can be shown that $X, T$ may be given in the form $$ \begin{aligned} X(x, t, v) & =A(v) x+B(v) t \\ T(x, t, v) & =C(v) x+D(v) t \end{aligned} $$ where $A, B, C, D$ are functions of $v$. In other words, the generalized boost is a linear transformation of coordinates.

(b) The isotropy of space implies that there is no preferred direction in the universe, i.e., that the laws of physics are the same in all directions. Use this to study the general coordinate transformations $X, T$ after setting $x \rightarrow-x$ and $x^{\prime} \rightarrow-x^{\prime}$ and conclude that $A(v), D(v)$ are even functions of $v$ and $B(v), C(v)$ are odd functions of $v$. (Hint: the relative velocity $v$ is a number which is measured by the $F$ frame using $v=\frac{d x}{d t}$.)

TP_TO_physics_en_COMP

965

Modern Physics

4. Lorentz Boost In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts: $$ \begin{aligned} x^{\prime} & =x-v t \\ t^{\prime} & =t \end{aligned} $$ In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts: $$ \begin{aligned} x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\ t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right) \end{aligned} $$ In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal "speed limit" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal "speed limit," which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$. The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$, $$ \begin{aligned} x^{\prime} & =X(x, t, v) \\ t^{\prime} & =T(x, t, v) \end{aligned} $$ which we will refer to as the generalized boost. Context question: (a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.) Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Analogously, we can conclude additionally that $\frac{\partial X}{\partial x}$ is independent of both $x$ and $t$ and $\frac{\partial T}{\partial t}$ is independent of $x$ and $t$. It can be shown that $X, T$ may be given in the form $$ \begin{aligned} X(x, t, v) & =A(v) x+B(v) t \\ T(x, t, v) & =C(v) x+D(v) t \end{aligned} $$ where $A, B, C, D$ are functions of $v$. In other words, the generalized boost is a linear transformation of coordinates. Context question: (b) The isotropy of space implies that there is no preferred direction in the universe, i.e., that the laws of physics are the same in all directions. Use this to study the general coordinate transformations $X, T$ after setting $x \rightarrow-x$ and $x^{\prime} \rightarrow-x^{\prime}$ and conclude that $A(v), D(v)$ are even functions of $v$ and $B(v), C(v)$ are odd functions of $v$. (Hint: the relative velocity $v$ is a number which is measured by the $F$ frame using $v=\frac{d x}{d t}$.) Context answer: \boxed{证明题}

(c) The principle of relativity implies that the laws of physics are agreed upon by observers in inertial frames. This implies that the general coordinate transformations $X, T$ are invertible and their inverses have the same functional form as $X, T$ after setting $v \rightarrow-v$. Use this fact to show the following system of equations hold: $$ \begin{aligned} A(v)^{2}-B(v) C(v) & =1 \\ D(v)^{2}-B(v) C(v) & =1 \\ C(v)(A(v)-D(v)) & =0 \\ B(v)(A(v)-D(v)) & =0 . \end{aligned} $$ (Hint: It's convenient to write $X, T$ as matrices and recall the definition of matrix inverses.) Physically, we must have that $B(v)$ and $C(v)$ are not both identically zero for nonzero $v$. So, we can conclude from the above that $D(v)=A(v)$ and $C(v)=\frac{A(v)^{2}-1}{B(v)}$.

TP_TO_physics_en_COMP

966

Modern Physics

4. Lorentz Boost In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts: $$ \begin{aligned} x^{\prime} & =x-v t \\ t^{\prime} & =t \end{aligned} $$ In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts: $$ \begin{aligned} x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\ t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right) \end{aligned} $$ In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal "speed limit" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal "speed limit," which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$. The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$, $$ \begin{aligned} x^{\prime} & =X(x, t, v) \\ t^{\prime} & =T(x, t, v) \end{aligned} $$ which we will refer to as the generalized boost. Context question: (a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.) Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Analogously, we can conclude additionally that $\frac{\partial X}{\partial x}$ is independent of both $x$ and $t$ and $\frac{\partial T}{\partial t}$ is independent of $x$ and $t$. It can be shown that $X, T$ may be given in the form $$ \begin{aligned} X(x, t, v) & =A(v) x+B(v) t \\ T(x, t, v) & =C(v) x+D(v) t \end{aligned} $$ where $A, B, C, D$ are functions of $v$. In other words, the generalized boost is a linear transformation of coordinates. Context question: (b) The isotropy of space implies that there is no preferred direction in the universe, i.e., that the laws of physics are the same in all directions. Use this to study the general coordinate transformations $X, T$ after setting $x \rightarrow-x$ and $x^{\prime} \rightarrow-x^{\prime}$ and conclude that $A(v), D(v)$ are even functions of $v$ and $B(v), C(v)$ are odd functions of $v$. (Hint: the relative velocity $v$ is a number which is measured by the $F$ frame using $v=\frac{d x}{d t}$.) Context answer: \boxed{证明题} Context question: (c) The principle of relativity implies that the laws of physics are agreed upon by observers in inertial frames. This implies that the general coordinate transformations $X, T$ are invertible and their inverses have the same functional form as $X, T$ after setting $v \rightarrow-v$. Use this fact to show the following system of equations hold: $$ \begin{aligned} A(v)^{2}-B(v) C(v) & =1 \\ D(v)^{2}-B(v) C(v) & =1 \\ C(v)(A(v)-D(v)) & =0 \\ B(v)(A(v)-D(v)) & =0 . \end{aligned} $$ (Hint: It's convenient to write $X, T$ as matrices and recall the definition of matrix inverses.) Physically, we must have that $B(v)$ and $C(v)$ are not both identically zero for nonzero $v$. So, we can conclude from the above that $D(v)=A(v)$ and $C(v)=\frac{A(v)^{2}-1}{B(v)}$. Context answer: \boxed{证明题}

(d) Use the previous results and the fact that the location of the $F^{\prime}$ frame may be given by $x=v t$ in the $F$ frame to conclude that the coordinate transformations have the following form: $$ \begin{aligned} x^{\prime} & =A(v) x-v A(v) t \\ t^{\prime} & =-\left(\frac{A(v)^{2}-1}{v A(v)}\right) x+A(v) t \end{aligned} $$

TP_TO_physics_en_COMP

967

Modern Physics

4. Lorentz Boost In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts: $$ \begin{aligned} x^{\prime} & =x-v t \\ t^{\prime} & =t \end{aligned} $$ In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts: $$ \begin{aligned} x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\ t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right) \end{aligned} $$ In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal "speed limit" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal "speed limit," which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$. The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$, $$ \begin{aligned} x^{\prime} & =X(x, t, v) \\ t^{\prime} & =T(x, t, v) \end{aligned} $$ which we will refer to as the generalized boost. Context question: (a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.) Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Analogously, we can conclude additionally that $\frac{\partial X}{\partial x}$ is independent of both $x$ and $t$ and $\frac{\partial T}{\partial t}$ is independent of $x$ and $t$. It can be shown that $X, T$ may be given in the form $$ \begin{aligned} X(x, t, v) & =A(v) x+B(v) t \\ T(x, t, v) & =C(v) x+D(v) t \end{aligned} $$ where $A, B, C, D$ are functions of $v$. In other words, the generalized boost is a linear transformation of coordinates. Context question: (b) The isotropy of space implies that there is no preferred direction in the universe, i.e., that the laws of physics are the same in all directions. Use this to study the general coordinate transformations $X, T$ after setting $x \rightarrow-x$ and $x^{\prime} \rightarrow-x^{\prime}$ and conclude that $A(v), D(v)$ are even functions of $v$ and $B(v), C(v)$ are odd functions of $v$. (Hint: the relative velocity $v$ is a number which is measured by the $F$ frame using $v=\frac{d x}{d t}$.) Context answer: \boxed{证明题} Context question: (c) The principle of relativity implies that the laws of physics are agreed upon by observers in inertial frames. This implies that the general coordinate transformations $X, T$ are invertible and their inverses have the same functional form as $X, T$ after setting $v \rightarrow-v$. Use this fact to show the following system of equations hold: $$ \begin{aligned} A(v)^{2}-B(v) C(v) & =1 \\ D(v)^{2}-B(v) C(v) & =1 \\ C(v)(A(v)-D(v)) & =0 \\ B(v)(A(v)-D(v)) & =0 . \end{aligned} $$ (Hint: It's convenient to write $X, T$ as matrices and recall the definition of matrix inverses.) Physically, we must have that $B(v)$ and $C(v)$ are not both identically zero for nonzero $v$. So, we can conclude from the above that $D(v)=A(v)$ and $C(v)=\frac{A(v)^{2}-1}{B(v)}$. Context answer: \boxed{证明题} Context question: (d) Use the previous results and the fact that the location of the $F^{\prime}$ frame may be given by $x=v t$ in the $F$ frame to conclude that the coordinate transformations have the following form: $$ \begin{aligned} x^{\prime} & =A(v) x-v A(v) t \\ t^{\prime} & =-\left(\frac{A(v)^{2}-1}{v A(v)}\right) x+A(v) t \end{aligned} $$ Context answer: \boxed{证明题}

(e) Assume that a composition of boosts results in a boost of the same functional form. Use this fact and all the previous results you have derived about these generalized boosts to conclude that $$ \frac{A(v)^{2}-1}{v^{2} A(v)}=\kappa . $$ for an arbitrary constant $\kappa$.

TP_TO_physics_en_COMP

968

Modern Physics

4. Lorentz Boost In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts: $$ \begin{aligned} x^{\prime} & =x-v t \\ t^{\prime} & =t \end{aligned} $$ In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts: $$ \begin{aligned} x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\ t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right) \end{aligned} $$ In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal "speed limit" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal "speed limit," which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$. The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$, $$ \begin{aligned} x^{\prime} & =X(x, t, v) \\ t^{\prime} & =T(x, t, v) \end{aligned} $$ which we will refer to as the generalized boost. Context question: (a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.) Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Analogously, we can conclude additionally that $\frac{\partial X}{\partial x}$ is independent of both $x$ and $t$ and $\frac{\partial T}{\partial t}$ is independent of $x$ and $t$. It can be shown that $X, T$ may be given in the form $$ \begin{aligned} X(x, t, v) & =A(v) x+B(v) t \\ T(x, t, v) & =C(v) x+D(v) t \end{aligned} $$ where $A, B, C, D$ are functions of $v$. In other words, the generalized boost is a linear transformation of coordinates. Context question: (b) The isotropy of space implies that there is no preferred direction in the universe, i.e., that the laws of physics are the same in all directions. Use this to study the general coordinate transformations $X, T$ after setting $x \rightarrow-x$ and $x^{\prime} \rightarrow-x^{\prime}$ and conclude that $A(v), D(v)$ are even functions of $v$ and $B(v), C(v)$ are odd functions of $v$. (Hint: the relative velocity $v$ is a number which is measured by the $F$ frame using $v=\frac{d x}{d t}$.) Context answer: \boxed{证明题} Context question: (c) The principle of relativity implies that the laws of physics are agreed upon by observers in inertial frames. This implies that the general coordinate transformations $X, T$ are invertible and their inverses have the same functional form as $X, T$ after setting $v \rightarrow-v$. Use this fact to show the following system of equations hold: $$ \begin{aligned} A(v)^{2}-B(v) C(v) & =1 \\ D(v)^{2}-B(v) C(v) & =1 \\ C(v)(A(v)-D(v)) & =0 \\ B(v)(A(v)-D(v)) & =0 . \end{aligned} $$ (Hint: It's convenient to write $X, T$ as matrices and recall the definition of matrix inverses.) Physically, we must have that $B(v)$ and $C(v)$ are not both identically zero for nonzero $v$. So, we can conclude from the above that $D(v)=A(v)$ and $C(v)=\frac{A(v)^{2}-1}{B(v)}$. Context answer: \boxed{证明题} Context question: (d) Use the previous results and the fact that the location of the $F^{\prime}$ frame may be given by $x=v t$ in the $F$ frame to conclude that the coordinate transformations have the following form: $$ \begin{aligned} x^{\prime} & =A(v) x-v A(v) t \\ t^{\prime} & =-\left(\frac{A(v)^{2}-1}{v A(v)}\right) x+A(v) t \end{aligned} $$ Context answer: \boxed{证明题} Context question: (e) Assume that a composition of boosts results in a boost of the same functional form. Use this fact and all the previous results you have derived about these generalized boosts to conclude that $$ \frac{A(v)^{2}-1}{v^{2} A(v)}=\kappa . $$ for an arbitrary constant $\kappa$. Context answer: \boxed{证明题}

(f) (1 point) Show that $\kappa$ has dimensions of (velocity $)^{-2}$, and show that the generalized boost now has the form $$ \begin{aligned} x^{\prime} & =\frac{1}{\sqrt{1-\kappa v^{2}}}(x-v t) \\ t^{\prime} & =\frac{1}{\sqrt{1-\kappa v^{2}}}(t-\kappa v x) \end{aligned} $$

TP_TO_physics_en_COMP

969

Modern Physics

4. Lorentz Boost In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts: $$ \begin{aligned} x^{\prime} & =x-v t \\ t^{\prime} & =t \end{aligned} $$ In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts: $$ \begin{aligned} x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\ t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right) \end{aligned} $$ In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal "speed limit" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal "speed limit," which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$. The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$, $$ \begin{aligned} x^{\prime} & =X(x, t, v) \\ t^{\prime} & =T(x, t, v) \end{aligned} $$ which we will refer to as the generalized boost. Context question: (a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.) Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Analogously, we can conclude additionally that $\frac{\partial X}{\partial x}$ is independent of both $x$ and $t$ and $\frac{\partial T}{\partial t}$ is independent of $x$ and $t$. It can be shown that $X, T$ may be given in the form $$ \begin{aligned} X(x, t, v) & =A(v) x+B(v) t \\ T(x, t, v) & =C(v) x+D(v) t \end{aligned} $$ where $A, B, C, D$ are functions of $v$. In other words, the generalized boost is a linear transformation of coordinates. Context question: (b) The isotropy of space implies that there is no preferred direction in the universe, i.e., that the laws of physics are the same in all directions. Use this to study the general coordinate transformations $X, T$ after setting $x \rightarrow-x$ and $x^{\prime} \rightarrow-x^{\prime}$ and conclude that $A(v), D(v)$ are even functions of $v$ and $B(v), C(v)$ are odd functions of $v$. (Hint: the relative velocity $v$ is a number which is measured by the $F$ frame using $v=\frac{d x}{d t}$.) Context answer: \boxed{证明题} Context question: (c) The principle of relativity implies that the laws of physics are agreed upon by observers in inertial frames. This implies that the general coordinate transformations $X, T$ are invertible and their inverses have the same functional form as $X, T$ after setting $v \rightarrow-v$. Use this fact to show the following system of equations hold: $$ \begin{aligned} A(v)^{2}-B(v) C(v) & =1 \\ D(v)^{2}-B(v) C(v) & =1 \\ C(v)(A(v)-D(v)) & =0 \\ B(v)(A(v)-D(v)) & =0 . \end{aligned} $$ (Hint: It's convenient to write $X, T$ as matrices and recall the definition of matrix inverses.) Physically, we must have that $B(v)$ and $C(v)$ are not both identically zero for nonzero $v$. So, we can conclude from the above that $D(v)=A(v)$ and $C(v)=\frac{A(v)^{2}-1}{B(v)}$. Context answer: \boxed{证明题} Context question: (d) Use the previous results and the fact that the location of the $F^{\prime}$ frame may be given by $x=v t$ in the $F$ frame to conclude that the coordinate transformations have the following form: $$ \begin{aligned} x^{\prime} & =A(v) x-v A(v) t \\ t^{\prime} & =-\left(\frac{A(v)^{2}-1}{v A(v)}\right) x+A(v) t \end{aligned} $$ Context answer: \boxed{证明题} Context question: (e) Assume that a composition of boosts results in a boost of the same functional form. Use this fact and all the previous results you have derived about these generalized boosts to conclude that $$ \frac{A(v)^{2}-1}{v^{2} A(v)}=\kappa . $$ for an arbitrary constant $\kappa$. Context answer: \boxed{证明题} Context question: (f) (1 point) Show that $\kappa$ has dimensions of (velocity $)^{-2}$, and show that the generalized boost now has the form $$ \begin{aligned} x^{\prime} & =\frac{1}{\sqrt{1-\kappa v^{2}}}(x-v t) \\ t^{\prime} & =\frac{1}{\sqrt{1-\kappa v^{2}}}(t-\kappa v x) \end{aligned} $$ Context answer: \boxed{证明题}

(g) Assume that $v$ may be infinite. Argue that $\kappa=0$ and show that you recover the Galilean boost. Under this assumption, explain using a Galilean boost why this implies that a particle may travel arbitrarily fast.

TP_TO_physics_en_COMP

970

Modern Physics

4. Lorentz Boost In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts: $$ \begin{aligned} x^{\prime} & =x-v t \\ t^{\prime} & =t \end{aligned} $$ In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts: $$ \begin{aligned} x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\ t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right) \end{aligned} $$ In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal "speed limit" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal "speed limit," which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$. The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$, $$ \begin{aligned} x^{\prime} & =X(x, t, v) \\ t^{\prime} & =T(x, t, v) \end{aligned} $$ which we will refer to as the generalized boost. Context question: (a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.) Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Analogously, we can conclude additionally that $\frac{\partial X}{\partial x}$ is independent of both $x$ and $t$ and $\frac{\partial T}{\partial t}$ is independent of $x$ and $t$. It can be shown that $X, T$ may be given in the form $$ \begin{aligned} X(x, t, v) & =A(v) x+B(v) t \\ T(x, t, v) & =C(v) x+D(v) t \end{aligned} $$ where $A, B, C, D$ are functions of $v$. In other words, the generalized boost is a linear transformation of coordinates. Context question: (b) The isotropy of space implies that there is no preferred direction in the universe, i.e., that the laws of physics are the same in all directions. Use this to study the general coordinate transformations $X, T$ after setting $x \rightarrow-x$ and $x^{\prime} \rightarrow-x^{\prime}$ and conclude that $A(v), D(v)$ are even functions of $v$ and $B(v), C(v)$ are odd functions of $v$. (Hint: the relative velocity $v$ is a number which is measured by the $F$ frame using $v=\frac{d x}{d t}$.) Context answer: \boxed{证明题} Context question: (c) The principle of relativity implies that the laws of physics are agreed upon by observers in inertial frames. This implies that the general coordinate transformations $X, T$ are invertible and their inverses have the same functional form as $X, T$ after setting $v \rightarrow-v$. Use this fact to show the following system of equations hold: $$ \begin{aligned} A(v)^{2}-B(v) C(v) & =1 \\ D(v)^{2}-B(v) C(v) & =1 \\ C(v)(A(v)-D(v)) & =0 \\ B(v)(A(v)-D(v)) & =0 . \end{aligned} $$ (Hint: It's convenient to write $X, T$ as matrices and recall the definition of matrix inverses.) Physically, we must have that $B(v)$ and $C(v)$ are not both identically zero for nonzero $v$. So, we can conclude from the above that $D(v)=A(v)$ and $C(v)=\frac{A(v)^{2}-1}{B(v)}$. Context answer: \boxed{证明题} Context question: (d) Use the previous results and the fact that the location of the $F^{\prime}$ frame may be given by $x=v t$ in the $F$ frame to conclude that the coordinate transformations have the following form: $$ \begin{aligned} x^{\prime} & =A(v) x-v A(v) t \\ t^{\prime} & =-\left(\frac{A(v)^{2}-1}{v A(v)}\right) x+A(v) t \end{aligned} $$ Context answer: \boxed{证明题} Context question: (e) Assume that a composition of boosts results in a boost of the same functional form. Use this fact and all the previous results you have derived about these generalized boosts to conclude that $$ \frac{A(v)^{2}-1}{v^{2} A(v)}=\kappa . $$ for an arbitrary constant $\kappa$. Context answer: \boxed{证明题} Context question: (f) (1 point) Show that $\kappa$ has dimensions of (velocity $)^{-2}$, and show that the generalized boost now has the form $$ \begin{aligned} x^{\prime} & =\frac{1}{\sqrt{1-\kappa v^{2}}}(x-v t) \\ t^{\prime} & =\frac{1}{\sqrt{1-\kappa v^{2}}}(t-\kappa v x) \end{aligned} $$ Context answer: \boxed{证明题} Context question: (g) Assume that $v$ may be infinite. Argue that $\kappa=0$ and show that you recover the Galilean boost. Under this assumption, explain using a Galilean boost why this implies that a particle may travel arbitrarily fast. Context answer: \boxed{证明题}

(h) Assume that $v$ must be smaller than a finite value. Show that $1 / \sqrt{\kappa}$ is the maximum allowable speed, and that this speed is frame invariant, i.e., $\frac{d x^{\prime}}{d t^{\prime}}=\frac{d x}{d t}$ for something moving at speed $1 / \sqrt{\kappa}$. Experiment has shown that this speed is $c$, the speed of light. Setting $\kappa=1 / c^{2}$, show that you recover the Lorentz boost.

TP_TO_physics_en_COMP

975

Electromagnetism

2. Johnson-Nyquist noise In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ : $$ \frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R \tag{2} $$ Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise. Electromagnetic modes in a resistor We first establish the properties of thermally excited electromagnetic modes $$ V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right) \tag{3} $$ in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$.

(a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$. Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$.

TP_TO_physics_en_COMP

976

Electromagnetism

2. Johnson-Nyquist noise In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ : $$ \frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R \tag{2} $$ Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise. Electromagnetic modes in a resistor We first establish the properties of thermally excited electromagnetic modes $$ V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right) \tag{3} $$ in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$. Context question: (a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$. Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$. Context answer: \boxed{证明题}

(b) Each mode $n$ in the resistor can be thought of as a species of particle, called a bosonic collective mode. This particle obeys Bose-Einstein statistics: the average number of particles $\left\langle N_{n}\right\rangle$ in the mode $n$ is $$ \left\langle N_{n}\right\rangle=\frac{1}{\exp \frac{\hbar \omega_{n}}{k T}-1} \tag{4} $$ In the low-energy limit $\hbar \omega_{n} \ll k T$, show that $$ \left\langle N_{n}\right\rangle \approx \frac{k T}{\hbar \omega_{n}} \tag{5} $$ You can use the Taylor expansion $e^{x} \approx 1+x$ for small $x$.

TP_TO_physics_en_COMP

977

Electromagnetism

2. Johnson-Nyquist noise In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ : $$ \frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R \tag{2} $$ Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise. Electromagnetic modes in a resistor We first establish the properties of thermally excited electromagnetic modes $$ V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right) \tag{3} $$ in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$. Context question: (a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$. Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$. Context answer: \boxed{证明题} Context question: (b) Each mode $n$ in the resistor can be thought of as a species of particle, called a bosonic collective mode. This particle obeys Bose-Einstein statistics: the average number of particles $\left\langle N_{n}\right\rangle$ in the mode $n$ is $$ \left\langle N_{n}\right\rangle=\frac{1}{\exp \frac{\hbar \omega_{n}}{k T}-1} \tag{4} $$ In the low-energy limit $\hbar \omega_{n} \ll k T$, show that $$ \left\langle N_{n}\right\rangle \approx \frac{k T}{\hbar \omega_{n}} \tag{5} $$ You can use the Taylor expansion $e^{x} \approx 1+x$ for small $x$. Context answer: \boxed{证明题}

(c) By analogy to the photon, explain why the energy of each particle in the mode $n$ is $\hbar \omega_{n}$.

TP_TO_physics_en_COMP

978

Electromagnetism

2. Johnson-Nyquist noise In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ : $$ \frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R \tag{2} $$ Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise. Electromagnetic modes in a resistor We first establish the properties of thermally excited electromagnetic modes $$ V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right) \tag{3} $$ in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$. Context question: (a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$. Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$. Context answer: \boxed{证明题} Context question: (b) Each mode $n$ in the resistor can be thought of as a species of particle, called a bosonic collective mode. This particle obeys Bose-Einstein statistics: the average number of particles $\left\langle N_{n}\right\rangle$ in the mode $n$ is $$ \left\langle N_{n}\right\rangle=\frac{1}{\exp \frac{\hbar \omega_{n}}{k T}-1} \tag{4} $$ In the low-energy limit $\hbar \omega_{n} \ll k T$, show that $$ \left\langle N_{n}\right\rangle \approx \frac{k T}{\hbar \omega_{n}} \tag{5} $$ You can use the Taylor expansion $e^{x} \approx 1+x$ for small $x$. Context answer: \boxed{证明题} Context question: (c) By analogy to the photon, explain why the energy of each particle in the mode $n$ is $\hbar \omega_{n}$. Context answer: \boxed{证明题}

(d) Using parts (a), (b), and (c), show that the average power delivered to the resistor (or produced by the resistor) per frequency interval is $$ P[f, f+d f] \approx k T d f . \tag{6} $$ Here, $f=\omega / 2 \pi$ is the frequency. $P[f, f+d f]$ is known as the available noise power of the resistor. (Hint: Power is delivered to the resistor when particles enter at $x=0$, with speed $c^{\prime}$, and produced by the resistor when they exit at $x=L$.)

TP_TO_physics_en_COMP

979

Electromagnetism

2. Johnson-Nyquist noise In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ : $$ \frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R \tag{2} $$ Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise. Electromagnetic modes in a resistor We first establish the properties of thermally excited electromagnetic modes $$ V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right) \tag{3} $$ in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$. Context question: (a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$. Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$. Context answer: \boxed{证明题} Context question: (b) Each mode $n$ in the resistor can be thought of as a species of particle, called a bosonic collective mode. This particle obeys Bose-Einstein statistics: the average number of particles $\left\langle N_{n}\right\rangle$ in the mode $n$ is $$ \left\langle N_{n}\right\rangle=\frac{1}{\exp \frac{\hbar \omega_{n}}{k T}-1} \tag{4} $$ In the low-energy limit $\hbar \omega_{n} \ll k T$, show that $$ \left\langle N_{n}\right\rangle \approx \frac{k T}{\hbar \omega_{n}} \tag{5} $$ You can use the Taylor expansion $e^{x} \approx 1+x$ for small $x$. Context answer: \boxed{证明题} Context question: (c) By analogy to the photon, explain why the energy of each particle in the mode $n$ is $\hbar \omega_{n}$. Context answer: \boxed{证明题} Context question: (d) Using parts (a), (b), and (c), show that the average power delivered to the resistor (or produced by the resistor) per frequency interval is $$ P[f, f+d f] \approx k T d f . \tag{6} $$ Here, $f=\omega / 2 \pi$ is the frequency. $P[f, f+d f]$ is known as the available noise power of the resistor. (Hint: Power is delivered to the resistor when particles enter at $x=0$, with speed $c^{\prime}$, and produced by the resistor when they exit at $x=L$.) Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Nyquist equivalent noisy voltage source The formula $\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R$ is the per-frequency, mean-squared value of an equivalent noisy voltage source, $V$, which would dissipate the available noise power, $\frac{d P}{d f}=k T$, from the resistor $R$ into a second resistor $r$.

(a) Assume that resistors $R$ and $r$ are in series with a voltage $V . R$ and $V$ are fixed, but $r$ can vary. Show the maximum power dissipation across $r$ is $$ P_{\max }=\frac{V^{2}}{4 R} . \tag{7} $$ Give the optimal value of $r$ in terms of $R$ and $V$.

TP_TO_physics_en_COMP