This problem is from ``Mathematics for the Liberal Arts'' (Wiley, 2012).
Leonhard Euler (1707-1783) proved the formula
\[ {1\over 1^2} + {1\over 2^2} + {1\over 3^2} + \cdots =\frac{\pi^2}{6}.\]
Can you use Euler's formula to find the exact value of
\[
\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+\frac{1}{5^2}-\frac{1}{6^2}+\cdots ?
\]

My math page: https://sites.google.com/site/martinerickson/

## Tuesday, August 14, 2012

## Sunday, August 12, 2012

### Six Mathematical Puzzles

Let $S = 1 + 11 + 111 + \cdots + 1\ldots 1$, where the last summand has $111$ decimal digits each equal to $1$. What is the sum of the digits of $S$?

Suppose that five particles travel back and forth on the unit interval $[0,1]$. At the start, all five particles move to the right with the same velocity. When a particle reaches $0$ or $1$, it reverses direction but maintains its speed. When two particles collide, they reverse direction and maintain speed. How many particle-particle collisions occur before the five particles occupy their original positions and are moving to the right?

In how many ways can a chess Queen move from one corner of the chess board to the opposite corner, moving closer to the goal square at every step?

What is the only positive integer $n$ such that there are exactly $n$ incongruent triangles with integer sides and perimeter $n$?

How many ways can you make one million dollars using any number of pennies, nickels, dimes, quarters, one-dollar bills, five-dollar bills, ten-dollar bills, twenty-dollar bills, fifty-dollar bills, and hundred-dollar bills?

What is noteworthy about the multisets $\{2,2\}$, $\{1,1,2,2,2,3,3,4\}$, and $\{1,1,1,2,2,3,3,3,4,6\}$? Can you characterize all multisets of positive integers with this property?

My math page: https://sites.google.com/site/martinerickson/

Suppose that five particles travel back and forth on the unit interval $[0,1]$. At the start, all five particles move to the right with the same velocity. When a particle reaches $0$ or $1$, it reverses direction but maintains its speed. When two particles collide, they reverse direction and maintain speed. How many particle-particle collisions occur before the five particles occupy their original positions and are moving to the right?

In how many ways can a chess Queen move from one corner of the chess board to the opposite corner, moving closer to the goal square at every step?

What is the only positive integer $n$ such that there are exactly $n$ incongruent triangles with integer sides and perimeter $n$?

How many ways can you make one million dollars using any number of pennies, nickels, dimes, quarters, one-dollar bills, five-dollar bills, ten-dollar bills, twenty-dollar bills, fifty-dollar bills, and hundred-dollar bills?

What is noteworthy about the multisets $\{2,2\}$, $\{1,1,2,2,2,3,3,4\}$, and $\{1,1,1,2,2,3,3,3,4,6\}$? Can you characterize all multisets of positive integers with this property?

My math page: https://sites.google.com/site/martinerickson/

## Monday, August 6, 2012

### Four Spheres in a Tetrahedron

Four spheres of unit radius are contained in a regular tetrahedron in such a way that each is tangent to three faces of the tetrahedron and to the other three spheres. What is the side length of the tetrahedron?
This problem is from the book ``Beautiful Mathematics," published by the Mathematical Association of America.

My math page: https://sites.google.com/site/martinerickson/

My math page: https://sites.google.com/site/martinerickson/

## Wednesday, August 1, 2012

### A Square Root Calculation

Calculate
\[
\sqrt{\phantom{XXXXXXXXXXXXX}}
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
{(\underbrace{111\ldots1}_{\mbox{$100$ $1$'s}})(1\underbrace{000\ldots0}_{\mbox{$99$ $0$'s}}5)+1} \,\,.
\]

Solution: We can write the expression inside the square root as \[ \frac{1}{9}(10^{100}-1)(10^{100}+5)+1=\frac{(10^{100}+2)^2}{9}. \] Hence the answer is $(10^{100}+2)/3=\underbrace{3\ldots3}_{\mbox{$99$ $3$'s}}4$.

My math page: https://sites.google.com/site/martinerickson/

Solution: We can write the expression inside the square root as \[ \frac{1}{9}(10^{100}-1)(10^{100}+5)+1=\frac{(10^{100}+2)^2}{9}. \] Hence the answer is $(10^{100}+2)/3=\underbrace{3\ldots3}_{\mbox{$99$ $3$'s}}4$.

My math page: https://sites.google.com/site/martinerickson/

## Sunday, July 22, 2012

### Triangular Numbers in Base 9

A triangular number is a number of the form $n(n+1)/2$, where $n$ is a positive integer. Prove that $1$, $11$, $111$, $1111$, $\ldots$ are all triangular numbers in base $9$.

Solution: The proof is by mathematical induction. The first number, $1$, is triangular: $1=\frac{1(1+1)}{2}$. Each successive number is $9$ times the previous number (which puts a $0$ at the right) plus $1$. We claim that the operation of multiplying a number by $9$ and adding $1$ transforms any triangular number into another triangular number. Let's see: \[ 9 \frac{n(n+1)}{2}+1=\frac{9n^2+9n+2}{2}=\frac{(3n+1)(3n+2)}{2}. \] This is a triangular number. And that completes the proof.

The operation ``multiply by $9$ and add $1$,'' used to produce new triangular numbers, can be generalized. If $k$ is a positive integer, multiplying any triangular number by $(2k+1)^2$ and adding $k(k+1)/2$ gives another triangular number. This is because of the identity \[ (2k+1)^2\frac{n(n+1)}{2}+\frac{k(k+1)}{2}=\frac{[(2k+1)n+k][(2k+1)n+k+1]}{2}. \] For example (with $k=2$), multiplying any triangular number by $25$ and adding $3$ gives another triangular number. It follows that $3$, $33$, $333$, $3333$, $\ldots$ are all triangular numbers in base $25$.

My math page: https://sites.google.com/site/martinerickson/

Solution: The proof is by mathematical induction. The first number, $1$, is triangular: $1=\frac{1(1+1)}{2}$. Each successive number is $9$ times the previous number (which puts a $0$ at the right) plus $1$. We claim that the operation of multiplying a number by $9$ and adding $1$ transforms any triangular number into another triangular number. Let's see: \[ 9 \frac{n(n+1)}{2}+1=\frac{9n^2+9n+2}{2}=\frac{(3n+1)(3n+2)}{2}. \] This is a triangular number. And that completes the proof.

The operation ``multiply by $9$ and add $1$,'' used to produce new triangular numbers, can be generalized. If $k$ is a positive integer, multiplying any triangular number by $(2k+1)^2$ and adding $k(k+1)/2$ gives another triangular number. This is because of the identity \[ (2k+1)^2\frac{n(n+1)}{2}+\frac{k(k+1)}{2}=\frac{[(2k+1)n+k][(2k+1)n+k+1]}{2}. \] For example (with $k=2$), multiplying any triangular number by $25$ and adding $3$ gives another triangular number. It follows that $3$, $33$, $333$, $3333$, $\ldots$ are all triangular numbers in base $25$.

My math page: https://sites.google.com/site/martinerickson/

## Tuesday, July 10, 2012

### Chess on an Infinite Board

Two players (White and Black) are playing on an infinite chess board (extending infinitely in all directions).
First, White places a certain number of queens (and no other pieces) on the board.
Black then places a king on any unoccupied, unattacked square of the board.
The players take turns moving until Black is checkmated.
What is the minimum number of queens White needs to force a checkmate?
Answer the same problem if White starts with rooks instead of queens.
Do the same for bishops and knights.
Let Q, R, B, and N be the minimum number of queens, rooks, bishops, and knights, respectively. What is the sum 1/Q + 1/R + 1/B + 1/N?

This problem appeared as the IBM Research "Ponder This" challenge of December 2011. The problem and solution may be found at: http://domino.research.ibm.com/comm/wwwr_ponder.nsf/challenges/December2011.html

My math page: https://sites.google.com/site/martinerickson/

This problem appeared as the IBM Research "Ponder This" challenge of December 2011. The problem and solution may be found at: http://domino.research.ibm.com/comm/wwwr_ponder.nsf/challenges/December2011.html

My math page: https://sites.google.com/site/martinerickson/

## Saturday, July 7, 2012

### Lines on a Hyperbolic Paraboloid

A ruled surface has the property that through every point on the surface there is a straight line lying on the surface. Cylinders and cones are ruled surfaces. Another example is a hyperbolic paraboloid (a saddle surface). In fact, a hyperbolic paraboloid contains a pair of straight lines through any given point on it. Find equations of a pair of lines through the point $(1,2,3)$ on the hyperbolic paraboloid $z=y^2-x^2$. Prove that these are the only two such lines.

Solution: Remember that a line in space can be given in parametric form $x=x_0+\alpha t$, $y=y_0+\beta t$, $z=z_0+\gamma t$, where $x_0$, $y_0$, $z_0$, $\alpha$, $\beta$, $\gamma$ are constants, and $t \in \mathbf{R}$.

Let $x=1+t$, $y=2+t$. Then \[ z=(2+t)^2-(1+t)^2=3+2t. \] So we have a line $(x,y,z)=(1+t,2+t,3+2t)$. Similarly, let $x=1+t$, $y=2-t$. Then \[ z=(2-t)^2-(1+t)^2=3-6t. \] So we have a line $(x,y,z)=(1+t,2-t,3-6t)$.

We will prove that these are the only two lines on $z=y^2-x^2$ through $(1,2,3)$. Assume that $x=1+\alpha t$, $y=2+\beta t$. Then \[ z=(2+\beta t)^2-(1+ \alpha t)^2=4+4 \beta t + \beta^2 t^2 -1- 2 \alpha t - \alpha^2 t^2. \] In order for us to have a line, the coefficient of $t^2$ must be zero. Hence $\alpha^2=\beta^2$, and $\beta= \pm \alpha$. For definiteness, assume that $\alpha=1$. Then $\beta=\pm 1$, and we have the two cases above.

We could make the same kind of argument to show that given any point on the surface $z=y^2-x^2$, there are exactly two lines through the point and lying on the surface. But there is a method that renders the computations even easier. We make the change of variables \[ x=X-Y, \quad y=X+Y, \quad z=4Z. \] Then \[ 4Z=(X+Y)^2-(X-Y)^2=X^2+2XY+Y^2-X^2-Y^2+2XY, \] and we obtain the particularly simple equation \[ Z=XY. \] The critical thing is that the change of variables is linear; lines are mapped to lines. In our problem, the point was $(1,2,3)$. Under the change of variables, this becomes $(\frac{3}{2},\frac{1}{2},\frac{3}{4})$. The lines that pass through this point and lie on the saddle surface have either $X$ or $Y$ constant: \[ (X,Y,Z)=\left(\frac{3}{2}+s,\frac{1}{2},\frac{3}{4}+\frac{1}{2}s\right) \quad \mbox{and} \quad (X,Y,Z)=\left(\frac{3}{2},\frac{1}{2}+t,\frac{3}{4}+\frac{3}{2}t\right). \] In general, the two lines passing through the point $(X_0,Y_0,Z_0)$ and lying on the surface $Z=XY$ are \[ (X,Y,Z)=(X_0+s,Y_0,Z_0+sY_0) \quad \mbox{and} \quad (X,Y,Z)=(X_0,Y_0+t,Z_0+tX_0). \] Thus we have two families of lines that rule the saddle surface. The lines in each family do not intersect because if two such lines intersected, the intersection point would be on three lines on the saddle surface. Furthermore, every pair of lines from different families intersect. For two such lines have equations \[ (X,Y,Z)=(X_1+s,Y_1,Z_1+sY_1) \quad \mbox{and} \quad (X,Y,Z)=(X_2,Y_2+t,Z_2+tX_2), \] and it is easy to see that the lines intersect at the point $(X_2,Y_1,X_2Y_1)$, when $s=X_2-X_1$ and $t=Y_1-Y_2$.

My math page: https://sites.google.com/site/martinerickson/

Solution: Remember that a line in space can be given in parametric form $x=x_0+\alpha t$, $y=y_0+\beta t$, $z=z_0+\gamma t$, where $x_0$, $y_0$, $z_0$, $\alpha$, $\beta$, $\gamma$ are constants, and $t \in \mathbf{R}$.

Let $x=1+t$, $y=2+t$. Then \[ z=(2+t)^2-(1+t)^2=3+2t. \] So we have a line $(x,y,z)=(1+t,2+t,3+2t)$. Similarly, let $x=1+t$, $y=2-t$. Then \[ z=(2-t)^2-(1+t)^2=3-6t. \] So we have a line $(x,y,z)=(1+t,2-t,3-6t)$.

We will prove that these are the only two lines on $z=y^2-x^2$ through $(1,2,3)$. Assume that $x=1+\alpha t$, $y=2+\beta t$. Then \[ z=(2+\beta t)^2-(1+ \alpha t)^2=4+4 \beta t + \beta^2 t^2 -1- 2 \alpha t - \alpha^2 t^2. \] In order for us to have a line, the coefficient of $t^2$ must be zero. Hence $\alpha^2=\beta^2$, and $\beta= \pm \alpha$. For definiteness, assume that $\alpha=1$. Then $\beta=\pm 1$, and we have the two cases above.

We could make the same kind of argument to show that given any point on the surface $z=y^2-x^2$, there are exactly two lines through the point and lying on the surface. But there is a method that renders the computations even easier. We make the change of variables \[ x=X-Y, \quad y=X+Y, \quad z=4Z. \] Then \[ 4Z=(X+Y)^2-(X-Y)^2=X^2+2XY+Y^2-X^2-Y^2+2XY, \] and we obtain the particularly simple equation \[ Z=XY. \] The critical thing is that the change of variables is linear; lines are mapped to lines. In our problem, the point was $(1,2,3)$. Under the change of variables, this becomes $(\frac{3}{2},\frac{1}{2},\frac{3}{4})$. The lines that pass through this point and lie on the saddle surface have either $X$ or $Y$ constant: \[ (X,Y,Z)=\left(\frac{3}{2}+s,\frac{1}{2},\frac{3}{4}+\frac{1}{2}s\right) \quad \mbox{and} \quad (X,Y,Z)=\left(\frac{3}{2},\frac{1}{2}+t,\frac{3}{4}+\frac{3}{2}t\right). \] In general, the two lines passing through the point $(X_0,Y_0,Z_0)$ and lying on the surface $Z=XY$ are \[ (X,Y,Z)=(X_0+s,Y_0,Z_0+sY_0) \quad \mbox{and} \quad (X,Y,Z)=(X_0,Y_0+t,Z_0+tX_0). \] Thus we have two families of lines that rule the saddle surface. The lines in each family do not intersect because if two such lines intersected, the intersection point would be on three lines on the saddle surface. Furthermore, every pair of lines from different families intersect. For two such lines have equations \[ (X,Y,Z)=(X_1+s,Y_1,Z_1+sY_1) \quad \mbox{and} \quad (X,Y,Z)=(X_2,Y_2+t,Z_2+tX_2), \] and it is easy to see that the lines intersect at the point $(X_2,Y_1,X_2Y_1)$, when $s=X_2-X_1$ and $t=Y_1-Y_2$.

My math page: https://sites.google.com/site/martinerickson/

## Friday, July 6, 2012

### The Age of Diophantus

Diophantus of Alexandria (c. 200--c. 284 CE) discovered methods for finding integer or rational solutions to certain types of algebraic equations. A riddle from the 5th century purports to express the number of years that Diophantus lived (it isn't known whether the facts in the riddle are accurate):

Diophantus lived one-sixth of his life as a child, then one-twelfth of his life later he grew a beard. After another one-seventh of his life he married, and five years after that he had a son. His son lived only half as long as he did. Four years after his son's death, Diophantus died. How many years did Diophantus live?

Solution: Since the riddle talks about one-twelfth of Diophantus' life and one-seventh of his life, we guess that the number of years he lived is a multiple of both $12$ and $7$, and hence a multiple of $84$. But the only multiple of $84$ reasonable for a human lifespan is $84$, and we see that $84$ years satisfies the conditions of the problem:

\[

\frac{84}{6}+\frac{84}{12}+\frac{84}{7}+5+\frac{84}{2}+4=14+7+12+5+42+4=84.

\]

The reasonable guess succeeded. We can also solve the problem directly. Let $x$ be the number of years Diophantus lived. Then

\[

\frac{x}{6}+\frac{x}{12}+\frac{x}{7}+5+\frac{x}{2}+4=x.

\]

Hence

\[

9=x\left(\frac{9}{84}\right),

\]

and $x=84$.

My math page: https://sites.google.com/site/martinerickson/

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