nyuuzyou/wikis huggingface dataset:
Hellinger-Toeplitz Theorem/Proof 1

== Theorem ==
== Proof ==
Let $\struct {\HH \times \HH, \norm \cdot_{\HH \times \HH} }$ be the direct product of $\HH$ with itself, with the direct product norm.
From the Closed Graph Theorem, it suffices to show that: 
:$G_T = \set {\tuple {x, T x} \in \HH \times \HH : x \in \HH}$
is closed in $\struct {\HH \times \HH, \norm \cdot_{\HH \times \HH} }$.
Let $\sequence {\tuple {x_n, T x_n} }_{n \mathop \in \N}$ be a sequence in $G_T$ converging to $\tuple {x, y}$ in $\struct {\HH \times \HH, \norm \cdot_{\HH \times \HH} }$.
From Convergence in Direct Product Norm, we have: 
:$\sequence {x_n}_{n \mathop \in \N}$ converges to $x$ in $X$
and:
:$\sequence {y_n}_{n \mathop \in \N}$ converges to $y$ in $Y$. 
We show that $y = T x$.
Let $z \in \HH$. 
Then we have: 
:$\innerprod {T x_n} z = \innerprod {x_n} {T z}$
Taking $n \to \infty$, we have: 
:$\innerprod y z = \innerprod x {T z}$ 
from Inner Product is Continuous.
Since $T$ is Hermitian, we have: 
:$\innerprod x {T z} = \innerprod {T x} z$ 
Then we have: 
:$\innerprod y z = \innerprod {T x} z$ for all $z \in \HH$.
So from Inner Product is Sesquilinear, we have: 
:$\innerprod {y - T x} z = 0$ for all $z \in \HH$.
In particular: 
:$\innerprod {y - T x} {y - T x} = 0$
so: 
:$y = T x$
So $G_T$ is closed in $\struct {\HH \times \HH, \norm \cdot_{\HH \times \HH} }$.
So, by the Closed Graph Theorem, $T$ is bounded.
Category:Hellinger-Toeplitz Theorem
Hellinger-Toeplitz Theorem/Proof 2

== Theorem ==
== Proof ==
 suppose that $T$ is not bounded.
Then there does not exist $C > 0$ such that: 
:$\norm {T x}_\HH \le C$ for each $x \in \HH$ with $\norm x_\HH = 1$.
That is, for each $n \in \N$ there exists $y_n \in \HH$ such that: 
:$\norm {T y_n}_\HH \ge n$ 
with $\norm {y_n}_\HH = 1$.
For each $n \in \N$, define the linear functional $f_n : \HH \to \HH$ by: 
:$\map {f_n} x = \innerprod x {T y_n}$
Then, from the Riesz Representation Theorem (Hilbert Spaces), we have that $f_n$ is a bounded linear functional with: 
:$\norm {f_n}_{\HH^\ast} = \norm {T y_n}_\HH \ge n$
So:
:$\ds \sup_n \norm {f_n}_{\HH^\ast} = \infty$ 
Then, from the Principle of Condensation of Singularities, there exists $x \in X$ such that: 
:$\ds \sup_n \cmod {\map {f_n} x} = \infty$
Since $T$ is Hermitian, we have: 
:$\map {f_n} x = \innerprod {T x} {y_n}$ for each $n \in \N$.
Then, we have:
But then we have: 
:$\ds \sup_n \cmod {\map {f_n} x} \le \norm {T x}_\HH < \infty$
a contradiction.
So $T$ is bounded.
Category:Hellinger-Toeplitz Theorem
Helly's Theorem

== Theorem ==
Let $A_1, A_2, \ldots, A_r \in \R^n$ be convex sets in real Euclidean $n$-space such that $r > n$.
Let $A_1, A_2, \ldots, A_r$ have the property that every collection of $n + 1$ of $A_1, A_2, \ldots, A_r$ have a point in common.
Then all of $A_1, A_2, \ldots, A_r$ have a point in common.
== Proof ==
== Historical Note ==
== Sources ==
* 
* 
Category:Helly's Theorem
Category:Convex Sets (Vector Spaces)
Category:Linear Algebra
Helmholtz's Theorem

== Theorem ==
Let $R$ be a region of ordinary space.
Let $\mathbf V$ be a vector field over $R$.
Let $\mathbf V$ be both non-conservative and non-solenoidal.
Then $\mathbf V$ can be decomposed into the sum of $2$ vector fields:
:one being conservative, with scalar potential $S$, but not solenoidal
:one being solenoidal, with vector potential $\mathbf A$, but not conservative.
Thus $\mathbf V$ satifies the partial differential equations:
where:
:$\operatorname {div}$ denotes the divergence operator
:$\grad$ denotes the gradient operator
:$\curl$ denotes the curl operator
:$\nabla^2$ denotes the Laplacian.
== Proof ==
Let us write:
:$\mathbf V = \grad S + \curl \mathbf A$
where:
:$S$ is a scalar field
:$\mathbf A$ is a vector field chosen to be solenoidal
Then:
and:
== Sources ==
* : Chapter $\text {V}$: Further Applications of the Operator $\nabla$: $7$. The Classification of Vector Fields: $\text {(iv)}$
Category:Vector Calculus
Henry Ernest Dudeney/536 Puzzles & Curious Problems/102 - An Exceptional Number

== by : $102$ ==
;An Exceptional Number
:A number is formed of $6$ successive digits (not necessarily in regular order)
::so that the number formed by the first $2$ multiplied by the $3$rd digit will produce the number expressed by the last $3$.
=== Click here for solution ===
== Historical Note ==
== Sources ==
* : Answers: $102$. An Exceptional Number
Category:Henry Ernest Dudeney/536 Puzzles & Curious Problems
Henry Ernest Dudeney/536 Puzzles & Curious Problems/102 - An Exceptional Number/Historical Note

== Historical Note on by : $102$ -An Exceptional Number ==
 includes this variant on $51$ - An Exceptional Number from 's as an aside in the Answers section of his $1968$ repackaging .
He attributes its discovery to .
== Sources ==
* : Answers: $102$. An Exceptional Number
Category:Historical Notes
Henry Ernest Dudeney/536 Puzzles & Curious Problems/102 - An Exceptional Number/Solution

== by : $102$ ==
;An Exceptional Number
== Solution ==
We have:
:$9 \ 4 \ 7 \ 6 \ 5 \ 8$
as:
:$94 \times 7 = 658$
== Historical Note ==
== Sources ==
* : Answers: $102$. An Exceptional Number
Category:Henry Ernest Dudeney/536 Puzzles & Curious Problems/Solutions
Henry Ernest Dudeney/536 Puzzles & Curious Problems/1 - Concerning a Cheque

== by : $1$ ==
;Concerning a Cheque
:A man went into a bank to cash a check.
:In handing over the money the cashier, by mistake, gave him dollars for cents and cents for dollars.
:He pocketed the money without examining it, and spent a nickel on his way home.
:He then found that he possessed exactly twice the amount of the check.
:He had no money in his pocket before going to the bank.
:What was the exact amount of that check?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $1$. Concerning a Cheque
Category:Henry Ernest Dudeney/536 Puzzles & Curious Problems
Henry Ernest Dudeney/536 Puzzles & Curious Problems/1 - Concerning a Cheque/Solution

== by : $1$ ==
;Concerning a Cheque
== Solution ==
The amount on the the check was $\$ 31.63$.
After leaving the bank, he had $\$ 63.31$.
After arriving home, he had $\$ 63.26$, which is twice $\$ 31.63$.
== Proof ==
Let $C$ be the amount written on the check.
Let $C$ consist of $x$ dollars and $y$ cents.
Let $C_1$ be the value of the check in cents.
Let $C_2$ be the money the man left the bank with in cents.
Let $C_3$ be the money the man arrived home with in cents.
We have:
:$C_1 = 100 x + y$
After coming out of the bank, the man has:
:$C_2 = 100 y + x$
After arriving home, the man has:
:$C_3 = 100 y + x - 5$
But we have:
:$C_1 \times 2 = C_3$
which leads us to:
This is a linear Diophantine equation.
Using the Euclidean Algorithm:
Thus we have that:
:$\gcd \set {199, -98} = 1$
which is a divisor of $-5$:
:$-5 = -5 \times 1$
So, from Solution of Linear Diophantine Equation, a solution exists.
Next we find a single solution to $199 x - 98 y = -5$.
Again with the Euclidean Algorithm:
and so:
is a solution.
From Solution of Linear Diophantine Equation, the general solution is:
:$\forall t \in \Z: x = x_0 + \dfrac b d t, y = y_0 - \dfrac a d t$
where $d = \gcd \set {a, b}$.
In this case:
In order to make $x$ and $y$ minimally positive, we see that $t = -2$:
Thus we arrive at:
:$x = 31$
and:
:$y = 63$
and the solution follows.
== Sources ==
* : Answers: $1$. Concerning a Cheque
Category:Henry Ernest Dudeney/536 Puzzles & Curious Problems/Solutions
Henry Ernest Dudeney/536 Puzzles & Curious Problems/3 - Loose Cash

== by : $3$ ==
;Loose Cash
:What is the largest sum of money -- all in current silver coins and no silver dollars --
::that I could have in my pocket without being able to give change for a dollar, half dollar, quarter, dime or nickel?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $3$. Loose Cash
Category:Henry Ernest Dudeney/536 Puzzles & Curious Problems
Henry Ernest Dudeney/536 Puzzles & Curious Problems/3 - Loose Cash/Solution

== by : $3$ ==
;Loose Cash
== Solution ==
The largest sum possible is $\$ 1.19$, composed of:
:$1$ half dollar
:$1$ quarter
:$4$ dimes
:$4$ cents.
== Proof ==
The coins available are:
:The half dollar, which is $50$ cents
:The quarter, which is $25$ cents
:The dime, which is $10$ cents
:The nickel, which is $5$ cents
:The cent itself.
We can have $4$ cents, but not $5$ cents as then we will have change for a nickel.
We can have $1$ nickel, but not $2$ nickels as then we will have change for a dime.
We can have $4$ dimes, but not $5$ dimes as then we will have change for a half dollar.
But if we have $2$ dimes as well as a nickel, then we will have change for a quarter.
Now $4$ dimes is more than $2$ dimes and a nickel, so for our maximum we need to lose the nickel.
We can have $1$ quarter, but not $2$ quarters as then we will have change for a half dollar.
We can have $1$ half dollar, but not $2$ half dollar as then we will have change for a dollar.
We have the maximum number of all the coins we are allowed.
Hence the result.
== Sources ==
* : Answers: $3$. Loose Cash
Category:Henry Ernest Dudeney/536 Puzzles & Curious Problems/Solutions
Henry Ernest Dudeney/Modern Puzzles

== : ==
== Arithmetical and Algebraical Problems ==
== Geometrical Problems ==
== Moving Counter Problems ==
== Unicursal and Route Problems ==
== Combination and Group Problems ==
== Magic Square Problems ==
== Magic Star Problems ==
== Measuring, Weighing, and Packing Problems ==
== Crossing River Problems ==
== Problems Concerning Games ==
== Puzzle Games ==
== Wheel Paradox Problems ==
== Unclassified Problems ==
Category:Recreational Mathematics
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/100 - Odds and Evens

== by : $100$ ==
;Odds and Evens
:Here is a little parlour trick, the explanation of which is quite easy.
:Ask a friend to take an even number of coins in one hand and an odd number in the other.
:You then undertake to tell him which hand holds the odd and which the even.
:Tell him to multiply the number in the right hand by $7$ and the number in the left by $6$,
::add the two products together, and tell you the result.
:You can then immediately give him the required answer.
:How are you to do it?
:In practice, after he had given me the result, I should say,
::"Let us add your age, now deduct the day of the month," and so on --
::a lot of absurd apparent calculations merely devised for his mystification,
::for you already know the answer.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $100$. -- Odds and Evens
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/100 - Odds and Evens/Solution

== by : $100$ ==
;Odds and Evens
== Solution ==
If the friend gives an odd total, the right hand holds the odd number.
Otherwise the left hand holds the odd number.
== Proof ==
Let $r$ be the number of coins in the right hand, and $l$ be the number of coins in the left hand.
If $r$ is odd, then $7 r$ is also odd, while $6 l$ is always even.
Hence the total $7 r + 6 l$ is in this case odd.
If $r$ is even, then $7 r$ is also even, while again $6 l$ is always even.
Hence the total $7 r + 6 l$ is in this case even.
So if the friend gives an odd total, the right hand holds the odd number.
Otherwise the left hand holds the odd number.
It is to be noted that with the low emphasis currently placed on ability to perform mental arithmetic, it is likely to be the case that a friend who is capable of multiplying the number of coins by $7$ and $6$ is also probably sufficiently mathematically sophisticated as to understand immediately how the trick works.
== Also see ==
* Henry Ernest Dudeney: Puzzles and Curious Problems: $352$ - Locating the Coins, which revisits the same idea.
== Sources ==
* : Solutions: $100$. -- Odds and Evens
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/101 - Twenty Questions

== by : $101$ ==
;Twenty Questions
:I am reminded of an interesting old game I used to play as a bachelor.
:Somebody thinks of an object -- say , or ...
:You have then to discover the object, by putting not more than $20$ questions, each of which must be answered by "yes" or "no".
:You have to word your questions discreetly, because if you ask, for example,
::"Is it animal, vegetable or mineral?" you might get the unsatisfactory answer "Yes," and so waste a question.
:We found that the expert rarely failed to get an exact solution, and I have known some most remarkably difficult cases solved by the twenty questions.
:A novel limitation of the game is suggested to me, which will call for some ingenuity, and the puzzle will doubtless be attacked in various ways by different persons.
:It is simply this.
:I think of a number containing six figures.
:Can you discover what it is by putting to me twenty questions,
::each of which can only be answered by "yes" or "no"?
:After the twentieth question you must give the number.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $101$. -- Twenty Questions
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $182$. Twenty Questions
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/101 - Twenty Questions/Solution

== by : $101$ ==
;Twenty Questions
== Solution ==
Successively ask:
:Does the number divided by $2$ leave a remainder?
:Does the quotient divided by $2$ leave a remainder?
:Does that quotient divided by $2$ leave a remainder?
Each of the $20$ questions will be the same.
Write down a string from right to left of:
:$1$ if the answer is yes
:$0$ if the answer is no.
The number sought in binary form is what you have written down.
== Sources ==
* : Solutions: $101$. -- Twenty Questions
* : Answers: $182$. Twenty Questions
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/102 - The Nine Barrels

== by : $102$ ==
;The Nine Barrels
:In how many different ways may these nine barrels be arranged in three tiers of three
::so that no barrel shall have a smaller number than its own below it or to the right of it?
:The first correct arrangement that will occur to you is $1 \ 2 \ 3$ at the top then $4 \ 5 \ 6$ in the second row, and $7 \ 8 \ 9$ at the bottom,
::and my sketch gives a second arrangement.
:How many are there altogether?
:
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Various Arithmetical and Algebraical Problems: $102$. -- The Nine Barrels
* : Arithmetical and Algebraical Problems: Digital Puzzles: $138$. The Nine Barrels
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/102 - The Nine Barrels/Solution

== by : $102$ ==
;The Nine Barrels
== Solution ==
There are $42$ different arrangements.
== Proof ==
Barrels $1$ and $9$ are fixed.
Let us place the $2$ immediately below the $1$.
If the $3$ is then below the $2$, there are $5$ arrangements, as the $4$ is now also fixed:
:$\begin{array}{ccc}
1 & 4 & 7 \\
2 & 5 & 8 \\
3 & 6 & 9
\end{array} \qquad
\begin{array}{ccc}
1 & 4 & 6 \\
2 & 6 & 7 \\
3 & 8 & 9
\end{array} \qquad
\begin{array}{ccc}
1 & 4 & 5 \\
2 & 6 & 8 \\
3 & 7 & 9
\end{array} \qquad
\begin{array}{ccc}
1 & 4 & 6 \\
2 & 5 & 7 \\
3 & 8 & 9
\end{array} \qquad
\begin{array}{ccc}
1 & 4 & 6 \\
2 & 5 & 7 \\
3 & 8 & 9
\end{array}$
If the $3$ is put to the right of the $1$, there are $5$ arrangements with the $4$ below the $2$:
:$\begin{array}{ccc}
1 & 3 & 5 \\
2 & 6 & 7 \\
4 & 8 & 9
\end{array} \qquad
\begin{array}{ccc}
1 & 3 & 5 \\
2 & 6 & 8 \\
4 & 7 & 9
\end{array} \qquad
\begin{array}{ccc}
1 & 3 & 6 \\
2 & 5 & 7 \\
4 & 8 & 9
\end{array} \qquad
\begin{array}{ccc}
1 & 3 & 6 \\
2 & 5 & 8 \\
4 & 7 & 9
\end{array} \qquad
\begin{array}{ccc}
1 & 3 & 7 \\
2 & 5 & 8 \\
4 & 6 & 9
\end{array}$
another $5$ arrangements with the $5$ below the $2$:
:$\begin{array}{ccc}
1 & 3 & 4 \\
2 & 6 & 7 \\
5 & 8 & 9
\end{array} \qquad
\begin{array}{ccc}
1 & 3 & 4 \\
2 & 6 & 8 \\
5 & 7 & 9
\end{array} \qquad
\begin{array}{ccc}
1 & 3 & 6 \\
2 & 4 & 7 \\
5 & 8 & 9
\end{array} \qquad
\begin{array}{ccc}
1 & 3 & 6 \\
2 & 4 & 6 \\
5 & 7 & 9
\end{array} \qquad
\begin{array}{ccc}
1 & 3 & 7 \\
2 & 4 & 8 \\
5 & 6 & 9
\end{array}$
another $4$ arrangements with the $6$ below the $2$:
:$\begin{array}{ccc}
1 & 3 & 4 \\
2 & 5 & 7 \\
6 & 8 & 9
\end{array} \qquad
\begin{array}{ccc}
1 & 3 & 4 \\
2 & 5 & 8 \\
6 & 7 & 9
\end{array} \qquad
\begin{array}{ccc}
1 & 3 & 5 \\
2 & 4 & 7 \\
6 & 8 & 9
\end{array} \qquad
\begin{array}{ccc}
1 & 3 & 5 \\
2 & 4 & 8 \\
6 & 7 & 9
\end{array}$
and another $2$ arrangements with the $7$ below the $2$:
:$\begin{array}{ccc}
1 & 3 & 4 \\
2 & 5 & 6 \\
7 & 8 & 9
\end{array} \qquad
\begin{array}{ccc}
1 & 3 & 5 \\
2 & 4 & 6 \\
7 & 8 & 9
\end{array}$
This is $21$ arrangements with the $2$ under the $1$.
By symmetry, there are another $21$ arrangements with the $2$ to the right of the $1$.
This accounts for all the arrangements.
== Sources ==
* : Solutions: $102$. -- The Nine Barrels
* : Answers: $138$. The Nine Barrels
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/103 - A New Cutting-out Puzzle

== by : $103$ ==
;A New Cutting-out Puzzle
:Cut the figure into four pieces that will fit together and form a square.
:
=== Click here for solution $1$ ===
=== Click here for solution $2$ ===
== Sources ==
* : Geometrical Problems: Dissection Puzzles: $103$. -- A New Cutting-out Puzzle
* : Geometrical Problems: Dissection Puzzles: $327$. A New Cutting-out Puzzle
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/103 - A New Cutting-out Puzzle/Solution 1

== by : $103$ ==
;A New Cutting-out Puzzle
== Solution ==
:
== Historical Note ==
== Sources ==
* : Solutions: $103$. -- A New Cutting-out Puzzle
* : Answers: $327$. A New Cutting-out Puzzle
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/103 - A New Cutting-out Puzzle/Solution 2

== by : $103$ ==
;A New Cutting-out Puzzle
== Solution ==
This solution is based on a tessellation:
:
== Sources ==
* : $9$: Dissections from Tessellations $\text I$: Fig. $9.2$
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/104 - The Square Table-Top

== by : $104$ ==
;The Square Table-Top
:A man had three pieces of beautiful wood, measuring $12$ units, $15$ units and $16$ units square respectively.
:He wanted to cut those into the fewest pieces possible that would fit together and form a small square table-tip $25$ units by $25$ units.
:How was he to do it?
 adds:
:I have found several easy solutions in six pieces, very pretty, but have failed to do it in five pieces.
:Perhaps the latter is not possible.
:I know it will interest my readers to examine the question.
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Dissection Puzzles: $104$. -- The Square Table-Top
* : Geometrical Problems: Dissection Puzzles: $328$. The Square Table-Top
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/104 - The Square Table-Top/Solution

== by : $104$ ==
;The Square Table-Top
== Solution ==
:
Piece $F$ is the $16$ by $16$ square.
Pieces $A$ and $B$ come from the $12$ by $12$ square.
Pieces $D$, $E$ and $F$ come from the $15$ by $15$ square.
== Sources ==
* : Solutions: $104$. -- The Square Table-Top
* : Ansers: $328$. The Square Table-Top
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/105 - The Squares of Veneer

== by : $105$ ==
;The Squares of Veneer
:A man has two squares of valuable veneer, each measuring $25$ units by $25$ units.
:One piece he cut, in the manner shown in our illustration, in four parts that will form two squares,
::one $20$ units by $20$ units, and the other $15$ units by $15$ units.
:Simply join $C$ to $A$ and $D$ to $B$.
:
:How is he to cut the other square into four pieces that will form again two other squares, with sides in exact units,
::but not $20$ and $15$ as before?
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Dissection Puzzles: $105$. -- The Squares of Veneer
* : Geometrical Problems: Dissection Puzzles: $329$. The Squares of Veneer
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/105 - The Squares of Veneer/Solution

== by : $105$ ==
;The Squares of Veneer
== Solution ==
The required squares are $24$ by $24$ and $7$ by $7$.
:
Pieces $A$, $B$ and $C$ go to form the $24$ by $24$ square, while piece $D$ is the $7$ by $7$ square cut out intact.
== Sources ==
* : Solutions: $105$. -- The Squares of Veneer
* : Answers: $329$. The Squares of Veneer
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/106 - Dissecting the Moon

== by : $106$ ==
;Dissecting the Moon
:In how large a number of pieces can this crescent moon be cut with five straight cuts of the knife?
:The pieces may not be piled or shifted after a cut.
:
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Dissection Puzzles: $106$. -- Dissecting the Moon
* : Geometrical Problems: Dividing-the-Plane Puzzles: $302$. Dissecting the Moon
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/106 - Dissecting the Moon/Solution

== by : $106$ ==
;Dissecting the Moon
== Solution ==
As many as $21$ pieces:
:
The maximum number of pieces a crescent can be cut into by $n$ cuts is $\dfrac {n^2 + 3 n} 2 + 1$.
For a circle, you can only make $\dfrac {n^2 + n} 2 + 1$ pieces.
== Sources ==
* : Solutions: $106$. -- Dissecting the Moon
* : Answers: $302$. Dissecting the Moon
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/107 - Dissecting the Letter E

== by : $107$ ==
;Dissecting the Letter $\text E$
:Can you cut this letter $\text E$ into only five pieces so that they will fit together to form a perfect square?
:All the measurements have been given so that there should be no doubt as to the correct proportions of the letter.
:
:In this case you are not allowed to turn over any piece.
=== Click here for solution ===
== Also see ==
* Puzzles and Curious Problems: $184$ - Dissecting the Letter $\text E$
== Sources ==
* : Geometrical Problems: Dissection Puzzles: $107$. -- Dissecting the Letter $\text E$
* : Geometrical Problems: Dissection Puzzles: $330$. Dissecting the Letter $\text E$
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/107 - Dissecting the Letter E/Solution

== by : $107$ ==
;Dissecting the Letter $\text E$
== Solution ==
:
== Sources ==
* : Solutions: $107$. -- Dissecting the Letter $\text E$
* : Answers: $330$. Dissecting the Letter $\text E$
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/108 - Hexagon to Square

== by : $108$ ==
;Hexagon to Square
:Can you cut a regular hexagon into $5$ pieces that will fit together to form a square?
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Dissection Puzzles: $108$. -- Hexagon to Square
* : Geometrical Problems: Dissection Puzzles: $331$. Hexagon to Square
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/108 - Hexagon to Square/Historical Note

== Historical Note on by : $108$ - Hexagon to Square ==
 claims that he was the first to publish this construction, citing the Weekly Dispatch for August $1901$.
However, reports that unearthed the information that in fact it had previously been discovered by .
This was published in Volume $\text {II}$ of in $1875$, on page $83$.
== Sources ==
* : Solutions: $108$. -- Hexagon to Square
* : Answers: $331$. Hexagon to Square
Category:Historical Notes