nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Puzzles and Curious Problems/206 - The Triangular Plantation/Solution

== by : $206$ ==
;The Triangular Plantation
== Solution ==
:$1216$
== Proof ==
The number of ways to select $3$ from $21$ is the binomial coefficient $\dbinom {21} 3$ which evaluates to $\dfrac {21 \times 20 \times 19} {3 \times 2 \times 1} = 1330$.
All of these form a triangle except where the trees are in a straight line.
Hence we need to eliminate all the latter.
:
Observe that the triangle of $21$ consists of lines of $1$, $2$, $3$, $4$, $5$ and $6$, which can be constructed in $3$ different ways, depending on which corner you place the $1$ at.
For each line of $6$, there are $\dbinom 6 3 = 20$ lines of $3$.
In each line of $5$, there are $\dbinom 5 3 = 10$ lines of $3$.
In each line of $4$, there are $\dbinom 4 3 = 4$ lines of $3$.
In each line of $3$, there is just the one line of $3$.
Then there are the $3$ vertical lines of $3$, each of which is one more straight line to be eliminated
This gives $20 + 10 + 4 + 1 + 3 = 38$ arrangements to be eliminated from the count.
Each of these $38$ straight lines are found in each of the $3$ orientations of the triangle of $21$ trees.
Hence the number of lines of $3$ is $3 \times 38 = 114$
So the total count of triangular arrangements is $1330 - 114 = 1216$.
== Sources ==
* : Solutions: $206$. -- The Triangular Plantation
* : Answers: $286$. The Triangular Plantation
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/207 - Six Straight Fences

== by : $207$ ==
;Six Straight Fences
:A man had a small plantation of $36$ trees, planted in the form of a square.
:Some of these died, and had to be cut down in the positions indicated by crosses in the diagram.
:
:How is it possible to put up $6$ straight fences across the field,
::so that every one of the remaining $20$ trees shall be in a separate enclosure?
:As a matter of fact, $22$ trees might be so enclosed by $6$ straight fences if their positions were a little more accommodating,
::but we have to deal with the trees as they stand in regular formation, which makes all the difference.
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $207$. -- Six Straight Fences
* : Geometrical Problems: Dividing-the-Plane Puzzles: $301$. Six Straight Fences
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/207 - Six Straight Fences/Solution

== by : $207$ ==
;Six Straight Fences
== Solution ==
:
== Sources ==
* : Solutions: $207$. -- Six Straight Fences
* : Answers: $301$. Six Straight Fences
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/208 - Dividing the Board

== by : $208$ ==
;Dividing the Board
:A man had a board measuring $10$ feet in length, $6$ inches wide at one end, and $12$ inches wide at the other,
::as shown in the diagram.
:
:How far from $B$ must the straight cut at $A$ be made in order to divide it into two equal pieces?
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $208$. -- Dividing the Board
* : Geometrical Problems: Triangle, Square & Other Polygon Puzzles: $272$. Dividing the Board
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/208 - Dividing the Board/Solution

== by : $208$ ==
;Dividing the Board
== Solution ==
Approximately $5.8114$ feet from $B$.
== Proof ==
Let $d$ be the length of the cut at $A$ from $B$.
The length of this cut is $6 + \dfrac {6 d} {10}$.
From Area of Trapezium we have (indirectly) that:
:$\paren {12 + 6 + \dfrac {6 d} {10} } \paren {10 - d} = \paren {6 + 6 + \dfrac {6 d} {10} } d$
After simplification, this gives:
:$d^2 + 20 d - 150 = 0$
Using the Quadratic Formula, this gives:
:$d = -10 \pm \sqrt {250}$
which has a positive root approximately equal to $5.8114$.
== Sources ==
* : Solutions: $208$. -- Dividing the Board
* : Answers: $272$. Dividing the Board
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/209 - A Running Puzzle

== by : $209$ ==
;A Running Puzzle
:$ABCD$ is a square field of $40$ acres.
:The line $BE$ is a straight path, and $E$ is $110$ yards from $D$.
:In a race Adams runs direct from $A$ to $D$,
::but Brown has to start from $B$, go from $B$ to $E$, and thence to $D$.
:Each keeps to a uniform speed throughout, and when Brown reaches $E$, Adams is $30$ yards ahead of him.
:
:Who wins the race, and by how much?
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $209$. -- A Running Puzzle
* : Geometrical Problems: Triangle, Square & Other Polygon Puzzles: $273$. A Running Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/209 - A Running Puzzle/Solution

== by : $209$ ==
;A Running Puzzle
== Solution ==
Brown wins by $8$ yards.
== Proof ==
First let us establish the geometry of the situation.
We have that the field is square, and of $40$ acres.
An acre is $4840$ square yards.
Hence the field is $40 \times 4840 = 440^2$ square yards, and so $440$ yards on a side.
So $AE = 440 - 110 = 330$ yards, and $\triangle ABE$ is a $3$-$4$-$5$ triangle.
Hence $BE = 550$ yards, and the total distance Brown is to run is $660$ yards.
Adams, however, is to run only $440$ yards, the length of the side of the field.
We have that Brown gets to $E$ after Adams has run $330 + 30 = 360$ yards.
Now let $v$ be the speed that Adams runs.
Then Brown runs $\dfrac {550 v} {360}$ times as fast as Adams.
So while Adams reaches $D$ after $\dfrac {440} v$, Brown reaches $D$ after $\dfrac {660 \times 360} {550 v} = \dfrac {432} v$, in appropriate units.
So Brown gets there first.
So Adams has only managed to get $432$ yards by the time Brown reaches the end.
That is, he has $8$ yards still to go.
== Sources ==
* : Solutions: $209$. -- A Running Puzzle
* : Answers: $273$. A Running Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/20 - Boys and Girls

== by : $20$ ==
;Boys and Girls
:Nine boys and three girls agreed to share equally their pocket-money.
:every boy gave an equal sum to every girl,
::and every girl gave another equal sum to every boy.
:Every child then possessed exactly the same amount.
:What was the smallest possible amount that each then possessed?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $20$. -- Boys and Girls
* : Arithmetical and Algebraical Problems: Money Puzzles: $26$. Boys and Girls
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/20 - Boys and Girls/Solution

== by : $20$ ==
;Boys and Girls
:Nine boys and three girls agreed to share equally their pocket-money.
:every boy gave an equal sum to every girl,
::and every girl gave another equal sum to every boy.
:Every child then possessed exactly the same amount.
:What was the smallest possible amount that each then possessed?
== Solution ==
Let us assume that the smallest unit of money is $1$.
On that basis, each boy possessed $12$ and he gave $1$ to each girl.
Every girl then had $36$, and gave $3$ to each boy.
Everybody then possessed $18$.
's repackaging of this puzzle in was answered on the basis of that unit being $1$ cent.
In England at the time this puzzle was set, the smallest unit of money was $\tfrac 1 4 \oldpence$
On that basis, each boy possessed $3 \oldpence$ and he gave $\tfrac 1 4 \oldpence$ to each girl.
Every girl then had $9 \oldpence$ and gave $\tfrac 3 4 \oldpence$ to each boy.
Everybody then possessed $4 \tfrac 1 2 \oldpence$
== Proof ==
== Sources ==
* : Solutions: $20$. -- Boys and Girls
* : Answers: $26$. Boys and Girls
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/210 - Pat and his Pig

== by : $210$ ==
;Pat and his Pig
:The diagram represents a field $100$ yards square.
:Pat is at $A$ and his pig is at $B$.
:The pig runs straight for the gateway at $C$.
:
:As Pat can run twice as fast as the pig, you would expect that he would first make straight for the gate and close it.
:But this is not Pat's way of doing things.
:He goes directly for the pig all the time, thus taking a curved course.
:Now, does the pig escape, or does Pat catch it?
:And if he catches it, exactly how far does the pig run?
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $210$. -- Pat and his Pig
* : Geometrical Problems: Plane Geometry Puzzles: $307$. Pat and his Pig
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/210 - Pat and his Pig/Solution

== by : $210$ ==
;Pat and his Pig
== Solution ==
Pat catches the pig after it has run $66 \tfrac 2 3$ yards towards the gate.
== Proof ==
This can be modelled using the technique of Differential Equation of Perpendicular Pursuit Curve.
Let us rotate the frame of reference so as to make:
:the initial position of the pig at the origin of a Cartesian plane
:the initial position of the farmer at the point $\tuple {c, 0}$ on this frame
:the initial position of the gate at the point $\tuple {0, c}$ on this frame.
Let $P$ be the position of the pig at time $t$.
Let $F$ be the position of the farmer Pat at time $t$.
From Differential Equation of Perpendicular Pursuit Curve, the differential equation describing the path taken by $F$ is:
:$\dfrac {\d y} {\d x} = \dfrac 1 2 \paren {\paren {\dfrac x c}^{1 / 2} - \paren {\dfrac c x}^{1 / 2} }$
So we need to solve this for a start.
Simplification is in order.
Thus we have:
This differential equation can be solved by Solution to Separable Differential Equation:
:$\ds \int \rd y = \dfrac 1 {2 \sqrt c} \int \dfrac {x - c} {\sqrt x} \rd x + C$
When $y = 0$ we have that $x = c$ and so:
leaving us with:
:$y = \dfrac {1 \sqrt x^3} {3 \sqrt c} + \sqrt {c x} + \dfrac {2 c} 3$
The farmer catches his pig when $x = 0$, that is:
:$y = 0 + 0 + \dfrac {2 c} 3$
That is, $\dfrac 2 3$ of the way to the gate, or $66 \tfrac 2 3$ yards from $B$.
== Sources ==
* : Solutions: $210$. -- Pat and his Pig
* : Answers: $307$. Pat and his Pig
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Category:Pursuit Curves
Henry Ernest Dudeney/Puzzles and Curious Problems/211 - The Twenty Matches

== by : $211$ ==
;The Twenty Matches
:The diagram shows how $20$ matches, divided into two groups of $14$ and $6$,
::may form two enclosures so that one space enclosed is exactly $3$ times as large as the other.
:
:Now divide the $20$ matches into two groups of $13$ and $7$,
::and with them again make two enclosures,
::one exactly three times as large as the other.
=== Click here for solution ===
== Also see ==
* $335$ - Three Times the Size, in the same collection
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $211$. -- The Twenty Matches
* : Match Puzzles: $496$. The Twenty Matches
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/211 - The Twenty Matches/Solution

== by : $211$ ==
;The Twenty Matches
== Solution ==
:
== Also see ==
* $335$ - Three Times the Size, in the same collection
As points out:
:There are other solutions.
== Sources ==
* : Solutions: $211$. -- The Twenty Matches
* : Answers: $496$. The Twenty Matches
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/212 - Transplanting the Trees

== by : $212$ ==
;Transplanting the Trees
:A man has a plantation of $22$ trees arranged in the manner shown.
:
:How is he to transplant only six of the trees so that they shall then form $20$ rows with $4$ trees in every row?.
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $212$. -- Transplanting the Trees
* : Combinatorial & Topological Problems: Point Alignment Puzzles: $437$. Transplanting the Trees
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/212 - Transplanting the Trees/Solution

== by : $212$ ==
;Transplanting the Trees
== Solution ==
The trees which were moved are in red, while the places where they were moved from are in white.
The $20$ lines of $4$ trees are shown as dotted lines.
:
== Sources ==
* : Solutions: $212$. -- Transplanting the Trees
* : Answers: $437$. Transplanting the Trees
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/213 - A Swastikaland Map

== by : $213$ ==
;A Swastikaland Map
:Here is a puzzle that the reader will probably think he has solved at almost the first glance.
:But will he be correct?
:Swastikaland is divided in the manner shown in our illustration.
:
:The Lord High Keeper of the Maps was ordered so to colour this map of the country
::that there should be a different colour on each side of every boundary line.
:What was the smallest number of colours that he required?
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $213$. -- A Swastikaland Map
* : Combinatorial & Topological Problems: Map Coloring Puzzles: $443$. A Swastikaland Map
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/213 - A Swastikaland Map/Solution

== by : $213$ ==
;A Swastikaland Map
== Solution ==
It would have been $2$, but for the glitch at the bottom left hand corner, where there happen to be three adjoining regions.
Thus $3$ colours are needed.
== Sources ==
* : Solutions: $213$. -- A Swastikaland Map
* : Answers: $443$. A Swastikaland Map
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/214 - Colouring the Map

== by : $214$ ==
;Colouring the Map
:Colonel Crackham asked his young son one morning to colour all the $26$ districts in this map
::in such a way that no two contiguous districts should be the same colour.
:
:The lad looked at it for a moment, and replied,
::"I haven't enough colours by one in my box."
:This was found to be correct.
:How many colours had he?
:He was not allowed to use black or white -- only colours.
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $214$. -- Colouring the Map
* : Combinatorial & Topological Problems: Map Coloring Puzzles: $444$. Coloring the Map
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/214 - Colouring the Map/Solution

== by : $214$ ==
;Colouring the Map
== Solution ==
The solver will be excused for giving the perfectly reasonable answer of $3$.
However, the tricksy answer given by is:
:Two!
== Proof ==
It is immediately apparent by inspection of the general central area that at least $4$ colours are needed.
By the Four Color Theorem it is known that no more than $4$ are needed.
Hence, being one short of $4$, there are only $3$ colours in the box.
's reasoning is that if he has two colours in his box, he can create a third colour by mixing the two, but cannot make a fourth.
Hence and so.
== Sources ==
* : Solutions: $214$. -- Colouring the Map
* : Answers: $444$. Coloring the Map
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/215 - The Damaged Rug

== by : $215$ ==
;The Damaged Rug
:A lady had a valuable Persian rug, $12$ feet by $9$ feet, which was badly damaged by fire.
:So she cut from the middle a strip $8$ feet by $1$ foot, as shown in the diagram,
::and then cut the remainder into two pieces that fitted together
::and made a perfectly square rug $10$ feet by $10$ feet.
:
:How did she do it?
:Of course, no allowance is to be made for turnings.
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $215$. -- The Damaged Rug
* : Geometrical Problems: Dissection Puzzles: $360$. The Damaged Rug
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/215 - The Damaged Rug/Solution

== by : $215$ ==
;The Damaged Rug
== Solution ==
:
== Sources ==
* : Solutions: $215$. -- The Damaged Rug
* : Answers: $360$. The Damaged Rug
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/216 - The Four Householders

== by : $216$ ==
;The Four Householders
:The diagram represents a square plot of land with four houses, four trees, a well (W) in the centre,
::and hedges planted across with four gateways.
:can you divide the ground so that each householder shall have an equal portion of land,
::one tree, one gateway, an equal length of hedge, and free access to the well without trespass?
:
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $216$. -- The Four Householders
* : Geometrical Problems: Dividing-the-Plane Puzzles: $297$. The Four Householders
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/216 - The Four Householders/Solution

== by : $216$ ==
;The Four Householders
== Solution ==
:The simplest, though not the only solution:
:
== Sources ==
* : Solutions: $216$. -- The Four Householders
* : Answers: $297$. The Four Householders
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/217 - The Three Fences

== by : $217$ ==
;The Three Fences
:A man had a circular field, and he wished to divide it into four equal parts by three fences, each of the same length.
:How might this be done?
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $217$. -- The Three Fences
* : Geometrical Problems: Circle Puzzles: $288$. The Three Fences
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/217 - The Three Fences/Solution

== by : $217$ ==
;The Three Fences
== Solution ==
:
== Also see ==
* Division of Circle into 4 Equal Parts by 3 Equal Length Lines
== Sources ==
* : Solutions: $217$. -- The Three Fences
* : Answers: $288$. The Three Fences
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Category:Division of Circle into 4 Equal Parts by 3 Equal Length Lines
Henry Ernest Dudeney/Puzzles and Curious Problems/218 - The Farmer's Sons

== by : $218$ ==
;The Farmer's Sons
:A farmer once had a square piece of ground on which stood $24$ trees, exactly as shown in the illustration.
:
:He left instructions in his will that each of his eight sons should receive the same amount of ground and the same number of trees.
:How was the land to be divided in the simplest possible manner?
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $218$. -- The Farmer's Sons
* : Geometrical Problems: Dividing-the-Plane Puzzles: $299$. The Farmer's Sons
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/219 - The Three Tablecloths

== by : $219$ ==
;The Three Tablecloths
:A person had $3$ tablecloths, each $4$ feet square.
:What is the length of the side of the largest square table top that they will cover together?
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $219$. -- The Three Tablecloths
* : Geometrical Problems: Triangle, Square & Other Polygon Puzzles: $274$. Three Tablecloths
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/219 - The Three Tablecloths/Solution

== by : $219$ ==
;The Three Tablecloths
== Solution ==
The three tablecloths will cover a table $5$ feet $1$ inch square, if laid like so:
:
$ABCD$ is the tabletop, and $1$, $2$, $3$ are the cloths.
Portions of $2$ and $3$ fall off the edge of the table.
== Sources ==
* : Solutions: $219$. -- The Three Tablecloths
* : Answers: $274$. Three Tablecloths
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/21 - The Cost of a Suit

== by : $21$ ==
;The Cost of a Suit
:Melville bought a suit.
:The jacket cost as much as the trousers and waistcoat.
:The jacket and two pairs of trousers would cost $\pounds 7, 17 \shillings 6 \oldpence$
:The trousers and two waistcoats would cost $\pounds 4, 10 \shillings$
:Can you tell me the cost of the suit?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $21$. -- The Cost of a Suit
* : Arithmetical and Algebraical Problems: Money Puzzles: $27$. The Cost of a Suit
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/21 - The Cost of a Suit/Solution

== by : $21$ ==
;The Cost of a Suit
== Solution ==
The jacket costs $\pounds 3, 7 \shillings 6 \oldpence$
The trousers cost $\pounds 2, 5 \shillings$
The waistcoat costs $\pounds 1, 2 \shillings 6 \oldpence$
In total: $\pounds 6, 15 \shillings$
== Proof ==
Let us do all this in shillings.
Hence:
Then we have:
where $J$, $T$ and $W$ are used in the obvious way.
The given solution shows these prices converted back into pounds, shillings and pence.
== Sources ==
* : Solutions: $21$. -- The Cost of a Suit
* : Answers: $27$. The Cost of a Suit
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/220 - The Five Fences

== by : $220$ ==
;The Five Fences
:A man owned a large, fenced-in field in which were $16$ oak trees, as depicted in the diagram.
:
:He wished to put up five straight fences so that every tree should be in a separate enclosure.
:How did he do it?
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $220$. -- The Five Fences
* : Geometrical Problems: Dividing-the-Plane Puzzles: $298$. The Five Fences
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/221 - The Fly's Journey

== by : $221$ ==
;The Fly's Journey
:A fly, starting from point $A$, can crawl around the four sides of the base of this cubical block in $4$ minutes.
:
:Can you say how long it will take to crawl from $A$ to the opposite corner $B$?
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $221$. -- The Fly's Journey
* : Geometrical Problems: Solid Geometry Puzzles: $319$. The Fly's Journey
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/221 - The Fly's Journey/Solution

== by : $221$ ==
;The Fly's Journey
:A fly, starting from point $A$, can crawl around the four sides of the base of this cubical block in $4$ minutes.
:
:Can you say how long it will take to crawl from $A$ to the opposite corner $B$?
== Solution ==
The shortest route would take the fly approximately $2.236$ minutes.
== Proof ==
The shortest route is that indicated by the dotted line:
:
Let $d$ be the length of one edge of the cube.
Then the length $l$ of the dotted line is given by Pythagoras's Theorem:
:$l = \sqrt {d^2 + \paren {2 d}^2} = d \sqrt 5$
As it takes $1$ minute to walk the length $d$, it takes $\sqrt 5$ minutes to walk the length $l$.
== Sources ==
* : Solutions: $221$. -- The Fly's Journey
* : Answers: $319$. The Fly's Journey
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions