# Math Help - Finding the equation of a circle given it pases through the 4 vertices of a pyramid?

1. ## Finding the equation of a circle given it pases through the 4 vertices of a pyramid?

In 3D Euclidian Vector geometry, if given the 4 vertices of a (Triangular) pyramid:

A (12,0,0) B(0,6,0) C(0,0,4) and O (origin: 0,0,0)

how can you find the sphere that passes through those vertices?

NB: this sphere is S: x² + y² + z² -12x -6y -4z = 0 but i want to know HOW you find this??

Help VERY appreciated thanks!

2. Originally Posted by Yehia
In 3D Euclidian Vector geometry, if given the 4 vertices of a (Triangular) pyramid:

A (12,0,0) B(0,6,0) C(0,0,4) and O (origin: 0,0,0)

how can you find the sphere that passes through those vertices?

NB: this sphere is S: x² + y² + z² -12x -6y -4z = 0 but i want to know HOW you find this??
let $(a,b,c)$ be the sphere's center of radius $r$. each vertex is equidistant from the center and yields the following four equations ...

$r^2 = (a-12)^2 + b^2 + c^2$

$r^2 = a^2 + (b-6)^2 + c^2$

$r^2 = a^2 + b^2 + (c-4)^2$

$r^2 = a^2 + b^2 + c^2$

using each of the first three equations and the last one, you get these equations ...

$144 - 24a = 0$ ... $a = 6$

$36 - 12b = 0$ ... $b = 3$

$16 - 8c = 0$ ... $c = 2$

sphere's equation is

$(x - 6)^2 + (y - 3)^2 + (z - 2)^2 = 49$

expand and simplify to get the equation you provided

3. Originally Posted by skeeter
let $(a,b,c)$ be the sphere's center of radius $r$. each vertex is equidistant from the center and yields the following four equations ...

$r^2 = (a-12)^2 + b^2 + c^2$

$r^2 = a^2 + (b-6)^2 + c^2$

$r^2 = a^2 + b^2 + (c-4)^2$

$r^2 = a^2 + b^2 + c^2$

using each of the first three equations and the last one, you get these equations ...

$144 - 24a = 0$ ... $a = 6$

$36 - 12b = 0$ ... $b = 3$

$16 - 8c = 0$ ... $c = 2$

sphere's equation is

$(x - 6)^2 + (y - 3)^2 + (z - 2)^2 = 49$

expand and simplify to get the equation you provided
Sorry, but how did you get those four equations? (Probably me just not seeing the wood for the trees).

Thanks for the help.

4. Originally Posted by Yehia
Sorry, but how did you get those four equations? (Probably me just not seeing the wood for the trees).

Thanks for the help.
distance formula between any two points in space ...

$r = \sqrt{(x-a)^2 + (y-b)^2 + (z-c)^2}$

5. Hello, Yehia!

I'll give it a try . . .

In 3-space, we are given the four vertices of a tetrahedron:
. . . . $A(12,0,0),\;B(0,6,0),\;C(0,0,4),\;O(0,0,0)$

Find the sphere that passes through those vertices

Answer: . $x^2 + y^2 + z^2 -12x -6y -4z \:=\:0$

The center $\,P$ is equidistant from all four points: . $PA \,=\, PB \,=\,PC\,=\,PD$

$\underbrace{\sqrt{(x-12)^2+y^2}}_{[1]} \:=\:\underbrace{\sqrt{x^2+(y-6)^2 + z^2}}_{[2]}$

. . . . . . . . . . . . . . $=\:\underbrace{\sqrt{x^2 + y^2 + (z-4)^2}}_{[3]} \:=\:\underbrace{\sqrt{x^2+y^2+z^2}}_{[4]}$

$\text{Set }[1] = [4]\!:\;x^2 - 24x + 144 + y^2 + z^2 \:=\:z^2+y^2+z^2$

. . . . . . . . . $-24x + 144 \:=\:0 \quad\Rightarrow\quad\boxed{ x \:=\:6}$

$\text{Set }[2] = [4]\!:\;x^2 + y^2 - 12y + 36 + z^2 \:=\:x^2 + y^2+z^2$

. . . . . . . . . $-12y + 36 \:=\:0 \quad\Rightarrow\quad\boxed{ y \:=\:3}$

$\text{Set }[3] = [4]\!:\;\;x^2+y^2+z^2-8z + 16 \:=\:x^2+y^2+z^2$

. . . . . . . . . $-8z+16 \:=\:0 \quad\Rightarrow\quad\boxed{ z \:=\:2}$

Hence, the center is: $P(6,3,2)$

The distance from $\,P$ to any vertex is the radius.

. . $r \:=\:PO \:=\:\sqrt{(6\!-\!0)^2 + (3\!-\!0)^2 + (2\!-\!0)^2} \:=\:\sqrt{36 + 9 + 4} \:=\:\sqrt{49} \;=\;7$

The equation of the sphere is: . $(x-6)^2 + (y-3)^2 + (z-2)^2 \:=\:7^2$

6. Skeeter and Soroban are using the fact that the distance from the center of the sphere to each of the given points is the same. Here's another way to argue that gives essentially the same equations: the general equation of a sphere is $(x- a)^2+ (y- b)^2+ (z- c)^2= r^2$.

To have (12,0,0) on the sphere, we must have
$(12- a)^2+ (0- b)^2+ (0- c)^2= r^2$.

To have (0,6,0) on the sphere, we must have
$(0- a)^2+ (6- b)^2+ c^2= r^2$.

To have (0,0,4) on the sphere, we must have
$(0- a)^2+ (0- b)^2+ (4- c)^2= r^2$.

To have (0,0,0) on the sphere, we must have
$(0- a)^2+ (0- b)^2+ (0- c)^2= r^2$.

That gives four equations to solve for the four values, a, b, c, and r.

When you multiply out the first three equations, each will have $a^2+ b^2+ c^2$ plus linear terms equal to $r^2$. If you replace $a^2+ b^2+ c^2$ by $r^2$ from the fourth equation, each equation immediately reduces to a simple equation.

7. Originally Posted by skeeter
distance formula between any two points in space ...

$r = \sqrt{(x-a)^2 + (y-b)^2 + (z-c)^2}$
Originally Posted by Soroban
Hello, Yehia!

I'll give it a try . . .

The center $\,P$ is equidistant from all four points: . $PA \,=\, PB \,=\,PC\,=\,PD$

$\underbrace{\sqrt{(x-12)^2+y^2}}_{[1]} \:=\:\underbrace{\sqrt{x^2+(y-6)^2 + z^2}}_{[2]}$

. . . . . . . . . . . . . . $=\:\underbrace{\sqrt{x^2 + y^2 + (z-4)^2}}_{[3]} \:=\:\underbrace{\sqrt{x^2+y^2+z^2}}_{[4]}$

$\text{Set }[1] = [4]\!:\;x^2 - 24x + 144 + y^2 + z^2 \:=\:z^2+y^2+z^2$

. . . . . . . . . $-24x + 144 \:=\:0 \quad\Rightarrow\quad\boxed{ x \:=\:6}$

$\text{Set }[2] = [4]\!:\;x^2 + y^2 - 12y + 36 + z^2 \:=\:x^2 + y^2+z^2$

. . . . . . . . . $-12y + 36 \:=\:0 \quad\Rightarrow\quad\boxed{ y \:=\:3}$

$\text{Set }[3] = [4]\!:\;\;x^2+y^2+z^2-8z + 16 \:=\:x^2+y^2+z^2$

. . . . . . . . . $-8z+16 \:=\:0 \quad\Rightarrow\quad\boxed{ z \:=\:2}$

Hence, the center is: $P(6,3,2)$

The distance from $\,P$ to any vertex is the radius.

. . $r \:=\:PO \:=\:\sqrt{(6\!-\!0)^2 + (3\!-\!0)^2 + (2\!-\!0)^2} \:=\:\sqrt{36 + 9 + 4} \:=\:\sqrt{49} \;=\;7$

The equation of the sphere is: . $(x-6)^2 + (y-3)^2 + (z-2)^2 \:=\:7^2$

Originally Posted by HallsofIvy
Skeeter and Soroban are using the fact that the distance from the center of the sphere to each of the given points is the same. Here's another way to argue that gives essentially the same equations: the general equation of a sphere is $(x- a)^2+ (y- b)^2+ (z- c)^2= r^2$.

To have (12,0,0) on the sphere, we must have
$(12- a)^2+ (0- b)^2+ (0- c)^2= r^2$.

To have (0,6,0) on the sphere, we must have
$(0- a)^2+ (6- b)^2+ c^2= r^2$.

To have (0,0,4) on the sphere, we must have
$(0- a)^2+ (0- b)^2+ (4- c)^2= r^2$.

To have (0,0,0) on the sphere, we must have
$(0- a)^2+ (0- b)^2+ (0- c)^2= r^2$.

That gives four equations to solve for the four values, a, b, c, and r.

When you multiply out the first three equations, each will have $a^2+ b^2+ c^2$ plus linear terms equal to $r^2$. If you replace $a^2+ b^2+ c^2$ by $r^2$ from the fourth equation, each equation immediately reduces to a simple equation.
Thank you guys very very much, yes I do understand it now, like I said, i couldn't see the wood for the trees. But if you don't mind I just have 3 other questions (still reffering to this question though).

How do you show that the centre of this sphere lies OUTSIDE the pyramid??
and also how do you prove that a point lies inside a sphere?
and one last thing: how would you find the point on a sphere that lies CLOSEST to a point P that is inside that sphere.

I hope those that sounds clear and again thanks very much!

8. Originally Posted by Yehia
In 3D Euclidean Vector geometry, if given the 4 vertices of a (Triangular) pyramid:

A (12,0,0) B(0,6,0) C(0,0,4) and O (origin: 0,0,0)

how can you find the sphere that passes through those vertices?

NB: this sphere is S: x² + y² + z² -12x -6y -4z = 0 but i want to know HOW you find this??

Help VERY appreciated thanks!
The intersection of the sphere with the $\displaystyle xy$ plane is a circle which passes through the three points: $\displaystyle A(12,\ 0,\ 0),\ B(0,\ 6,\ 0),\ O(0,\ 0,\ 0)\,.$ The line segment, $\displaystyle \overline{AO}$ is a chord of this circle, so the center of this circle lies along the perpendicular bisector of this chord, i.e. the center of this circle lies along the line $\displaystyle (6,\ y,\ 0)\,.$ Similarly, the perpendicular bisector of $\displaystyle \overline{BO}$ is the line $\displaystyle (x,\ 3,\ 0)\,.$ The these lines intersect at the point $\displaystyle (6,\ 3,\ 0)\,.$ Therefore the center of the sphere lies on the line $\displaystyle (6,\ 3,\ z)\,.$

Similarly, the intersection of the sphere with the $\displaystyle xz$ plane is also a circle. This circle passes through the points $\displaystyle A(12,\ 0,\ 0),\ C(0,\ 0,\ 4),\ O(0,\ 0,\ 0)\,.$ The center of this circle is located at $\displaystyle (6,\ 0,\ 2)\,,$ so the center of the sphere lies on the line $\displaystyle (6,\ y,\ 2)\,.$

The two lines intersect at $\displaystyle (6,\ 3,\ 2)\,,$ which is the location of the center of the sphere. The radius, $\displaystyle r$, of the sphere is the distance from point $\displaystyle O(0,\ 0,\ 0)\,.$ to the center of the sphere.

Thus, the sqare of the radius is: $\displaystyle r^2=6^2+2^2+3^2=49\, .$

The equation of the sphere is: $\displaystyle (x-6)^2+(y-2)^2+(z-3)^2= 49\, .$

A little bit of algebra provides the given answer.

9. Originally Posted by Yehia
...But if you don't mind I just have 3 other questions (still reffering to this question though).

How do you show that the centre of this sphere lies OUTSIDE the pyramid??
Here is a way but I'm sure that there must be a more elegant one:

1. Calculate the 3 heights of the pyramid which correspond to the 3 side planes. (Use the volume of the pyramid and the base area, calculated by the cross-product).
2. If r is greater than any of the 3 heights then the centre lies outside the pyramid.

and also how do you prove that a point lies inside a sphere?
Plug in the coordinates of the point into the equation of the sphere. If the left hand side of the equation is smaller than $r^2$ then the point lies inside the sphere.

and one last thing: how would you find the point on a sphere that lies CLOSEST to a point P that is inside that sphere.

...
Let M denote the midpoint of the sphere. Then the line MP intersects the sphere at the point which belongs to the sphere and is closest to P.


# Product of discontinuous functions

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Let $$\displaystyle f:\mathbb{R} \to \mathbb{R}$$ and $$\displaystyle g:\mathbb{R} \to \mathbb{R}$$ be discontinuous at a point $$\displaystyle c$$ . Give an example of a function $$\displaystyle h(x)=f(x)g(x)$$ such that $$\displaystyle h$$ is continuous at c.
$$\displaystyle f(x) = \begin{cases} 0 & \text{if } x \in \mathbb{Q} \\ 1 & \text{if } x \in \mathbb{R}-\mathbb{Q} \end{cases}$$

$$\displaystyle g(x) = \begin{cases} 1 & \text{if } x \in \mathbb{Q} \\ 0 & \text{if } x \in \mathbb{R}-\mathbb{Q} \end{cases}$$

$$\displaystyle f,g$$ are continous nowhere but $$\displaystyle h(x)=0 \,\,\, \, \forall \,\, x \in \mathbb{R}$$.

What other examples you might think of ?

#### Sudharaka

##### Well-known member
MHB Math Helper
Let $$\displaystyle f:\mathbb{R} \to \mathbb{R}$$ and $$\displaystyle g:\mathbb{R} \to \mathbb{R}$$ be discontinuous at a point $$\displaystyle c$$ . Give an example of a function $$\displaystyle h(x)=f(x)g(x)$$ such that $$\displaystyle h$$ is continuous at c.
$$\displaystyle f(x) = \begin{cases} 0 & \text{if } x \in \mathbb{Q} \\ 1 & \text{if } x \in \mathbb{R}-\mathbb{Q} \end{cases}$$

$$\displaystyle g(x) = \begin{cases} 1 & \text{if } x \in \mathbb{Q} \\ 0 & \text{if } x \in \mathbb{R}-\mathbb{Q} \end{cases}$$

$$\displaystyle f,g$$ are continous nowhere but $$\displaystyle h(x)=0 \,\,\, \, \forall \,\, x \in \mathbb{R}$$.

What other examples you might think of ?
Hi Zaid,

How about the set for functions, $$\{f,\,g\}$$ such that,

$f(x)=\begin{cases}a & \text{if } x \geq c \\b & \text{if } x < c\end{cases}$

$g(x)=\begin{cases}b & \text{if } x \geq c \\a & \text{if } x < c\end{cases}$

where $$a,\,b,\,c\in \Re$$ and $$a\neq b$$.

#### Deveno

##### Well-known member
MHB Math Scholar
The most general, and all-encompassing example I can think of, off the top of my head:

Let $a \neq b$ and define:

$f(x) = a,\ x \neq c$
$f(c) = b$

$g(x) = b,\ x \neq c$
$g(c) = a$

Clearly, neither $f$ nor $g$ is continuous at $c$, as can be proved straight from the definition (use an $0 < \epsilon < |b - a|$).

Just as clearly:

$fg(x) = ab,\ \forall x \in \Bbb R$, which is clearly continuous.

One can construct more "extravagant" examples, but the important part is that $a \neq b$, and that $f$ and $g$ "complement" each other. In fact, there is nothing special about the partition of $\Bbb R$ into the two disjoint sets $\{c\}$ and $\Bbb R - \{c\}$, you can use any partition (such as the Dedekind cut example Sudharaka gives, or the partition into the rationals and irrationals).

To me, this underscores the fact that continuity (of a function) is dependent on the DOMAIN OF DEFINITION of said function, not just the "rule itself" of said function.

In other words, a "continuous function" doesn't really MEAN anything, what IS meaningful is: a function continuous at all points of a set $A$. The underlying domain is important. Context is everything: a function that is perfectly continuous on the real numbers may suddenly spectacularly fail to be so on the complex numbers, for example (as is the case with:

$f(x) = \dfrac{1}{1 + x^2}$).

#### Amer

##### Active member
Characteristic function of A
$$\chi_A : \mathbb{R} \rightarrow \{0,1\}$$
$$\chi_A =\left\{ \begin{array}{lr} 1 &,x\in A \\ 0 &,x\in A^{c} \end{array} \right.$$
$$\chi_{A^{c}} = \left\{ \begin{array}{ir} 0 & , x\in A \\ 1 & , x\in A^{c} \end{array} \right.$$
Their product is zero function which is continuous


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# Why would a pendulum experiment give $g > 9.8\ \mathrm{m/s^2}$?

I am taking an introductory lab course in which we've done an experiment on the physical pendulum.

We've seen that for small oscillations, the period is

$$T=2\pi\sqrt{\dfrac{I_S}{Mgd_{cm}}}\tag{1}$$

where $S$ is the pivot point, $M$ is the total mass of the object, and $d_{cm}$ is the distance between $S$ and the pendulum's center of mass.

Now, varying $d_{cm}$, I've obtained seven different periods. I've calculated $g$ in terms of $T$ and $d_{cm}$ for those seven distinct values of $T$ and $d_{cm}$. So, $g$ can be expressed as

$$g=4\pi^2\dfrac{I_S}{T^2Md_{cm}}.\tag{2}$$

All the periods I've obtained were always greater than $1$, and the values of $g$ where between $10.15$ and $10.3$. I am trying to understand why is it that $g$ gave me always greater than $9.8$, the expected value, and not less than $9.8$. If I consider the air friction, I would expect the period to be greater; from what I've said and from equation (2), I would say that the values of $g$ should be less than $9.8$, contrary to the values I got.

Note that that $T$ is in seconds, $[g]=\dfrac{m}{s^2}$ and the angle from which the pendulum was released is of $25$ degrees approximately.

I would appreciate if someone could help me to understand the reason why I've obtained these values for $g$.

• When you say the period is greater than 1, what units are you using? Similarly when you report your value of $g$, I'm assuming you're using meters per second squared? You should include that. – David Z Mar 12 '14 at 0:17
• What's your release angle? That formula is only valid for small amplitudes. – Kvothe Mar 12 '14 at 0:36
• @Kvothe: Yes, but wouldn't applying the formula for larger angles (where the pendulum will take longer to swing back and forth than the simple formula implies) yields a smaller value for the gravitational acceleration -- assuming accurate measurements? – Keith Thompson Mar 12 '14 at 0:41
• @KeithThompson - You are correct. The first order correction for larger angles is $g = \frac{4\pi^2 L}{T^2} \left( 1 + \frac{1}{8} \theta_0^2 \right) + {\cal O} \ ( \theta_0^4 \ )$. Thus, the measured value using the first order formula is generically smaller. – Prahar Mar 12 '14 at 0:52
• Thanks for all your comments, guys; to @David Z: sorry for not clarifying that information, I have added the units and the angle. So, I still don't have the slightest idea why I got greater values than $9,8$, I can't think of any possible mistakes I could have made during the experiment. – user100106 Mar 12 '14 at 2:01

I) OP is using the period formula

$$\tag{1} T~=~2\pi\sqrt{\frac{I}{MgR}}$$

for a compound/physical pendulum (in the small amplitude limit) to estimate the gravitational acceleration constant

$$\tag{2} g~=~\left(\frac{2\pi}{T}\right)^2 \frac{I}{MR}.$$

Here $I$ is the moment of inertia around the pivot point; $R$ is the distance from CM to the pivot point; and $M$ is the total mass.

II) After doing the experiment OP finds values for $g$ that are 3-5% too big. (These results are close enough that OP likely did not make any elementary mistakes with units.) A finite amplitude of

$$\tag{3} \theta_0 ~\approx ~25^{\circ}~\approx~ .44~ {\rm rad}$$

makes the pendulum

$$\tag{4} \frac{\theta_0^2}{8}~\approx~ 2\%$$

slower, as compared to the ideal pendulum (1), cf. comment by Prahar. So correcting for a finite amplitude makes OP's estimates worse, 5-7% too big, as Keith Thompson points out in a comment above.

So the discrepancy is caused by something else. The culprit is likely that it is difficult to get a precise estimate for the moment of inertia $I$. All the other quantities $T$, $M$ and $R$ should be fairly easy to measure reliable. So OP's value for $I$ is likely too big. According to Steiner's theorem

$$\tag{5} I~=~MR^2+I_0,$$

where $I_0$ is the moment of inertia around the CM (and the actual quantity which is poorly known).

III) Below follows a suggestion. Plot OP's seven data points in an $(x,y)$ diagram with axes

$$\tag{6} x~:=~R^2 \quad\text{and}\quad y~:=~R\left(\frac{T}{2\pi}\right)^2.$$

Theoretically, the $(x,y)$ data points should then lie on a straight line

$$\tag{7} y~=~ax+b$$

with slope

$$\tag{8} a~=~\frac{1}{g}$$

and $y$-intercept

$$\tag{9} b=~~\frac{I_0}{gM}.$$

In other words, find the best fitting straight line. This method should hopefully produce a good estimate for $g$ without having to know $I_0$ a priori. (By the way, notice that we in principle also don't need to know the mass $M$, cf. the equivalence principle!)

IV) Finally, as always in experiments, estimate all pertinent uncertainties in the various measurements.

• Super clear and helpful answer, you're right about the moment of inertia, I had some difficulty calculating $I_0$ for the two "disks" and the rod. – user100106 Mar 16 '14 at 22:00


# Finding the minimal number of members

I've been working on the following problem

For every issue in the Blue's association, a commission with 10 members (belonging the Blue's) is formed in order to solve the problem. The only condition is

There can't be two commissions having more than one member in common

The Blue's association has formed this year 40 commissions.

What's the minimal amount of members in the Blue's association?

I've only found out the following

For any commission you can form $$\binom{10}{2}=45$$ different pairs and none of them can appear in another commission.

Since 40 different commissions are formed, the minimal number of pairs is $$45\times 40=1800$$.

Denote by $$n$$ the number of members. Thus $$\binom{n}{2}≥1800\Rightarrow n>60$$

The minimal amount of members has to be 100 or less.

You can observe a distribution for 100 members here

My question:

Is 100 the answer or is there an ever smaller possible amount of members? If so, how can I prove it?

• Different but related question, perhaps some of the ideas there are helpful: math.stackexchange.com/questions/2879313/… – Servaes Dec 2 '18 at 18:57
• If there are two members that are in $6$ commisions, then they share at most one commission, and to fill these $11$ commissions then requires another $10\times9+1\times8=98$ distinct members, yielding a total of $100$ members if we count the two members we started with. So any example smaller than the one you found has at most one member in more than $5$ commissions. – Servaes Dec 2 '18 at 20:02
• By a similar (but simpler) argument, no member can be in more than $11$ commisions; to fill $11$ commissions sharing one member requires another $11\times9=99$ distinct members, yielding a total of $100$ members. – Servaes Dec 2 '18 at 20:06
• @Servaes If $a$ and $b$ are two members that are in $6$ commissions, how do you know that those $10\times9+1\times6=98$ members are all distinct? Why can't a commission containing $a$ overlap with a commission containing $b$? – bof Dec 5 '18 at 9:50
• @Servaes Note that, if the points of a projective plane of order $9$ are our "members" and the lines are our "commissions", then we have $91$ members and $91$ commissions, and each member belongs to $10$ commissions. – bof Dec 5 '18 at 11:32

This post exhibits a solution with $$82$$ members. Combined with the excellent answer by @Song, this means $$82$$ is indeed optimal.

Motivation: The excellent answer by @Song and the followup comments by @Servaes make me wonder... perhaps if we look for 41 commissions (not 40) then there is a solution with a great deal of symmetry:

• (a) 82 members (the optimal answer we seek)
• (b) 41 commissions (exceeds OP requirement)
• (c) each member associated with exactly 5 commissions (not part of OP)
• (d) each commission associated with exactly 10 members (equals OP requirement)
• (e) each 2 members have exactly 1 common commission (not part of OP)
• (f) each 2 commissions have exactly 1 common member (exceeds OP requirement)

This would be like a finite projective plane, but with 82 points and 41 lines. However, in a finite projective (respectively: affine) plane, the no. of points and no. of lines are equal (respectively: almost equal), and this is probably why a solution based on FPP only reaches 84. So I decided to look at related structures called Block Designs, Steiner Systems, etc. where there are typically many more "lines" than "points". After quite a bit of digging, I think I found the right structure:

The solution: It is a Steiner $$S(t=2,k=5,n=41)$$ system. A Steiner system is defined by the following properties:

• there are $$n=41$$ objects (these are the commissions)

• there are $$b$$ blocks (these are the members), each block (member) being a subset of objects (i.e. the commissions he/she is associated with)

• each block has $$k=5$$ objects (each member is associated with 5 commissions)

• every $$t=2$$ objects is contained in exactly 1 block (every 2 commissions have exactly 1 common member)

So this already satisfies (b), (c) and (f). Next, quoting from https://en.wikipedia.org/wiki/Steiner_system#Properties we have:

• $$b = {n \choose t} / {k \choose t} = (41 \times 40) / (5 \times 4) = 41 \times 2 = 82$$, satisfying (a)

• $$r = {n-1 \choose t-1} / {k-1 \choose t-1} = 40 / 4 = 10$$, where $$r$$ denotes "the number of blocks containing any given object", i.e. the number of members associated with any given commission, satisfying (d).

Thinking more, I don't think (e) can be satisfied. However, (e) is not needed for the OP, so it doesn't matter.

So finally we just need to prove that such a Steiner $$S(t=2,k=5,n=41)$$ system exists. This existence is non-trivial, but luckily more digging reveals:

• https://math.ccrwest.org/cover/steiner.html has a list of Steiner systems that are known to exist. $$S(2,5,41)$$ (the webpage sometimes lists the 3 parameters in different order) is not part of any infinite families listed, but if you go further down the page it is listed as a standalone example; clicking on that link goes to...

• https://math.ccrwest.org/cover/show_cover.php?v=41&k=5&t=2 which exhibits the system, created via "Cyclic construction" whatever that means.

I did not check the numbers thoroughly, but if I understand the webpage correctly, there should be 82 rows (members / blocks), each containing 5 numbers (commissions), all the numbers being 1 through 41 inclusive (the 41 commissions), each number (commission) should appear in 10 rows, and every pair-of-numbers should appear in 1 row.

I am not an expert in any of these, so if I had a mistake or misunderstanding above, my apologies. Perhaps someone more expert can check my work?

• +1 Good ideas and good find! I will have a look at the second link. As a side note, indeed (e) is impossible; there are $\binom{82}{2}=3321$ pairs of members. But there are $\binom{10}{2}=45$ pairs of members per commission, so $41\times45=1845$ pairs of members that share a commission. Alternatively, every member is in $5$ commissions, and hence in a commission with $5\times9=45$ members, so $\frac{82\times45}{2}=1845$ pairs of members that share a commission. Either way, there are too many pairs of members to all share a commission. – Servaes Dec 14 '18 at 0:55
• *Another, simpler, way to see that (e) is impossible; this would imply that every member shares a commission with $81$ members. But every member is in only $5$ commissions, and hence shares a commission with only $5\times9=45$ members. – Servaes Dec 14 '18 at 0:57
• @Servaes - yes of course. in fact, more abstractly / vaguely speaking, perhaps insisting on both (e) and (f) is WHY a finite projective plane has the same no. of lines and points. so if we introduce imbalance in (a) vs (b), and hence (c) vs (d), we must necessarily imbalance (e) vs (f) as well. this is all very vague of course. :) – antkam Dec 14 '18 at 1:07
• With duality between lines & points (or equiv: rows & cols of an incidence matrix), there is really no need to translate, but it's up to you. anyway, here's a link I found explaining (I think) the "cyclic construction" of some Steiner systems, although it doesn't mention $S(2,5,41)$ specifically: pitt.edu/~kaveh/Chap7-MATH1050.pdf – antkam Dec 14 '18 at 1:53
• Dan Gordon migrated the La Jolla covering design repository to his new site, so it might be a good idea to update your links. – hardmath Mar 16 at 0:46

Let $$i$$ denote each member of Blue's association and assume that there are $$N$$ members in total, that is, $$i=1,2,\cdots, N.$$ And let $$j,k=1,2,\ldots, 40$$ denote each of 40 commission. We will show that $$N$$ is at least $$82$$.

Consider the set $$S=\{(i,j,k)\;|\;1\leq i\leq N, 1\leq j Let $$d_i$$ denote the number of commissions that $$i$$ joined. We will calculate $$|S|$$ using double counting method. First, note that $$|S|=\sum_{(i,j,k)\in S}1 = \sum_{1\leq j since for each $$j, there is at most one $$i$$ in common. On the other hand, $$|S| = \sum_{1\leq i\leq N} \sum_{(j,k):(i,j,k)\in S}1 = \sum_{1\leq i\leq N} \binom{d_i}{2},$$ since for each $$i$$, the number of pairs $$(j,k)$$ that $$i$$ joined is $$\binom{d_i}{2}$$. We also have $$\sum_{1\leq i\leq N}d_i = 400,$$by the assumption.
Finally, note that the function $$f(x)= \binom{x}{2} = \frac{x^2-x}{2}$$ is convex. Thus by Jensen's inequality we have that $$\binom{40}{2}\geq |S|=\sum_{1\leq i\leq N} \binom{d_i}{2}\geq Nf\left(\frac{\sum_i d_i}{N}\right)=N\binom{\frac{400}{N}}{2}.$$ This gives us the bound $$40\cdot 39 \geq 400\cdot(\frac{400}{N}-1),$$and hence $$N \geq \frac{4000}{49} = 81.63\cdots$$ This establishes $$N\geq 82$$. However, I'm not sure if this bound is tight. I hope this will help.

$$\textbf{Note:}$$ If $$N=82$$ is tight, then above argument implies that $$d_i$$'s distribution is almost concentrated at $$\overline{d} = 400/82 \sim 5$$.
EDIT: @antkam's answer seemingly shows that $$N=82$$ is in fact optimal.

• +1 Fantastic argument! – Servaes Dec 10 '18 at 10:28
• A corollary of this argument is that in the best solution every commission must share a member with some other commission. – Will Fisher Dec 10 '18 at 20:02
• @WillFisher You mean, if there is a solution with $N=82$? In this case, it also follows that there are precisely $72$ members that are in $5$ commissions, and $10$ that are in $4$ commissions. Also, that every commission has precisely $9$ members in $5$ commissions and $1$ member in $4$ commissions. – Servaes Dec 10 '18 at 20:38
• I don't really understand this step: "Finally, note that the function $x\to \binom{x}{2} = \frac{x^2-x}{2}$ is convex. Thus, we have that $$\binom{40}{2}\geq |S|=\sum_{1\leq i\leq N} \binom{d_i}{2}\geq N\binom{\frac{\sum_i d_i}{N}}{2}=N\binom{\frac{400}{N}}{2}.$$" – Dr. Mathva Dec 13 '18 at 23:35
• I should have mentioned that it is by Jensen's inequality. The statement is in en.m.wikipedia.org/wiki/Jensen%27s_inequality. @Dr. Mathva – Song Dec 13 '18 at 23:44

This is just a partial answer. I will show that $$85$$ members suffice; I don't know if $$85$$ is the minimum.

Recall that a projective plane of order $$n$$ exists if $$n$$ is a prime power: it has $$n^2+n+1$$ points and $$n^2+n+1$$ lines; each line has $$n+1$$ points, and there are $$n+1$$ lines through each point; any pair of lines meets in a unique point, and any pair of points determines a unique line.

Consider a projective plane of order $$9$$; it has $$9^2+9+1=91$$ points and $$91$$ lines; there are $$10$$ points on each line and $$10$$ lines through each point. A set of points is in general position if no three of the points are collinear. Note that, if we have a set of $$t$$ points in general position, then the lines determined by those points (taken two at a time) cover a total of at most $$t+8\binom t2$$ points; as long as $$t\le5$$ then the number of covered points is at most $$5+8\binom52=85\lt91$$, so we can add another point to the set and still have them in general position. Thus we can find a set $$S$$ of $$6$$ points in general position.

Let the members of the Blue's association be the $$91-6=85$$ points which are not in $$S$$. The commissions are the lines which do not meet $$S$$; they have $$10$$ members each, and any two have exactly one member in common. Finally, by the in-and-out formula, the number of commissions is $$91-\binom61\cdot10+\binom62\cdot1=46.$$

P.S. Let $$m$$ be the minimum possible number of members. I showed above that $$m\le85$$. On the other hand, I have a small improvement on your lower bound $$m\ge61$$.

Suppose the $$i^\text{th}$$ member belongs to $$d_i$$ commissions; then $$\sum_{i=1}^md_i=400$$ since there are $$40$$ commissions with $$10$$ members each. Moreover $$d_i\le9$$ since $$m\le85\lt91$$. Let $$k=|\{i:d_i\ge5\}|$$. Then $$400=\sum_{i=1}^md_i\le4(m-k)+9k=4m+5k\le340+5k,$$ whence $$k\ge12$$; i.e., there are at least $$12$$ members who are on at least $$5$$ commissions. Choose two members $$i$$ and $$j$$ who are on at least $$5$$ commissions.

Case 1. There is a commission containing both $$i$$ and $$j$$.

First, there are $$10$$ members on the commission which $$i$$ and $$j$$ both belong to. Next $$i$$ belongs to $$4$$ more commissions, with $$36$$ additional members. Finally, $$j$$ belongs to $$4$$ more commissions, each of which contains at most one member of each of the $$5$$ commissions containg $$i$$, and at least $$5$$ members who haven't been counted yet, for a total of $$20$$ new members. This shows that $$m\ge10+36+20=66$$.

Case 2. There is no commission containing both $$i$$ and $$j$$.

In this case a similar argument shows that $$m\ge67$$.

This proves that $$m\ge66$$. Combining this with the upper bound shown earlier, we have $$66\le m\le85.$$

• +1 Great progress and nice ideas! Note that for every prime power $q$ there are arcs of $q+1$ points in $\Bbb{P}(\Bbb{F}_q)$. In particular you can take your set $S$ to contain $7$ points, yielding $42$ commissions and $m\leq84$. With $S$ containing $8$ points we get $39$ commissions and $m\leq83$, perhaps another commission can be fitted in somewhere? – Servaes Dec 5 '18 at 16:53
• Your bottom argument can be extended to show $m\ge 73$. – Will Fisher Dec 5 '18 at 17:04
• @Servaes I don't know much about projective planes. That fact about "arcs of $q+1$ points", is it easy or hard to prove? Is it true for all projective planes of prime power order, or just the field planes? – bof Dec 6 '18 at 10:20
• @WillFisher Nice. How do you prove it? – bof Dec 6 '18 at 10:21
• @bof In any projective plane over a finite field, any nondegenerate conic is an example of an arc of $q+1$ points. I've also checked that the same construction with a set $S$ of $8$ or more points cannot fit any other commission directly (i.e. without adjusting some of the existing $39$ or fewer commissions). – Servaes Dec 6 '18 at 12:46

Allow me to summarize and slightly refine the results in the current answers (if only to straighten out my own thoughts); they show that the minimum number of members $$m$$ satisfies $$82\leq m\leq84$$. They also imply strict conditions on any solution with $$m=82$$.

I also include my result that if $$m=83$$, then no member is in more than $$7$$ commissions. Much more can be said, but I do not have a definitive proof for the cases $$m=82$$ or $$m=83$$.

The upper bound $$m\leq84$$ comes from bof's construction in the projective plane of order $$9$$; the projective plane $$\Bbb{P}^2(\Bbb{F}_9)$$ consists of $$91$$ points on $$91$$ lines, with $$10$$ points on each line and $$10$$ lines through each point. Importantly, each pair of lines meets in precisely one point, and each pair of points is on precisely one line.

For $$7$$ distinct points in general position (no $$3$$ on a line, e.g. points on a smooth conic) there are precisely $$7\times10-\binom{7}{2}\times1=49$$ lines containing these points. Removing these $$7$$ points and the $$49$$ lines containing them leaves $$84$$ points and $$91-49=42$$ lines each containing $$10$$ points, and any pair of lines meets in at most one point. That is, we have $$84$$ members in $$42$$ commissions, with no $$2$$ commissions sharing more than one member, so $$m\leq84$$.

The lower bound $$m\geq82$$ comes from Song's answer; the number of pairs of commissions that share a member is at most $$\binom{40}{2}$$, as there are $$40$$ commissions. As every commission shares at most one member, this can also be counted as the number of pairs of commissions that each member is in. If the $$i$$-th member is in $$d_i$$ commissions, then it is in $$\binom{d_i}{2}$$ pairs of commissions and hence $$\sum_{i=1}^m\binom{d_i}{2}\leq\binom{40}{2}.\tag{1}$$ As there are $$40$$ commissions with $$10$$ members each, we also have $$\sum_{i=1}^md_i=400$$. In the inequality above we can bound the left hand side from below using the fact that for all positive integers $$x$$ we have $$\binom{x-1}{2}+\binom{x+1}{2}=2\binom{x}{2}+1.$$ This allows us to even out the $$d_i$$'s to find that $$\sum_{i=1}^m\binom{d_i}{2}\geq(m-n)\binom{x}{2}+n\binom{x+1}{2},\tag{2}$$ for some $$x$$ and $$n$$ with $$0\leq n, where $$(m-n)x+n(x+1)=\sum_{i=1}^md_i=400.$$ The latter simplifies to $$mx+n=400$$, which shows that $$x=\lfloor\frac{400}{m}\rfloor$$ and $$n=400-mx$$. Plugging this back in shows that $$\begin{eqnarray*} \binom{40}{2}&\geq&\sum_{i=1}^m\binom{d_i}{2} \geq(m-n)\binom{x}{2}+n\binom{x+1}{2}\\ &=&(m-(400-m\lfloor\tfrac{400}{m}\rfloor))\binom{\lfloor\frac{400}{m}\rfloor}{2}+(400-m\lfloor\tfrac{400}{m}\rfloor)\binom{\lfloor\frac{400}{m}\rfloor+1}{2}\\ &=&-\frac{m}{2}\lfloor\tfrac{400}{m}\rfloor^2-\frac{m}{2}\lfloor\tfrac{400}{m}\rfloor+400\lfloor\tfrac{400}{m}\rfloor. \end{eqnarray*}$$ The latter is strictly decreasing for $$m$$ in the interval $$[1,84]$$. The inequality is satisfied if and only if $$m\geq82$$, which proves the lower bound.

Let $$S$$ denote the number of times we need to apply the identity $$\binom{x-1}{2}+\binom{x+1}{2}=2\binom{x}{2}+1$$ to reduce the left hand side of $$(2)$$ to the right hand side. We can then write $$(2)$$ more precisely as $$\sum_{i=1}^m\binom{d_i}{2}=(m-n)\binom{x}{2}+n\binom{x+1}{2}+S.$$ Knowing that $$82\leq m\leq84$$ simplifies the above significantly, as then $$x=\lfloor\tfrac{400}{m}\rfloor=4$$ and $$n=400-4m$$. We find that $$780\geq\sum_{i=1}^m\binom{d_i}{2}=1600-10m+S.$$ In particular, for $$m=82$$ we find that $$S=0$$ and hence that there are precisely $$n=400-82\times4=72$$ members that are in $$4$$ commissions and $$10$$ members that are in $$5$$ commissions. We also see that we have equality in $$(1)$$, meaning that every pair of commissions shares a member. This implies $$\sum_{i\in C}(d_i-1)=39$$ for every commission $$C$$, from which it follows that every commission has precisely $$1$$ member that is in $$4$$ commissions, and $$9$$ members that are in $$5$$ commissions.

If $$m=83$$ then $$S\leq10$$, and there are at most $$10$$ pairs of commissions that do not share a member.

Here are a few unincorporated observations that may or may not be helpful. These concern restrictions on minimal examples with $$m<84$$, i.e. $$m=82$$ or $$m=83$$. They are all subsumed by the observations above for $$m=82$$, so I prove them only for $$m=83$$.

Observation 1: For all $$i$$ we have $$d_i\leq9$$.

To fill the commission of member $$i$$ requires $$9d_i+1$$ distinct members, including member $$i$$. We have $$9d_i+1\leq m=83$$ and hence $$d_i\leq9$$.

Observation 2: For all $$i$$ we have $$d_i\leq8$$.

To fill the commission of a member $$i$$ with $$d_i=9$$ requires $$9d_i+1=82$$ distinct members, leaving $$1$$ member remaining as $$m=83$$. Each of the remaining $$40-d_i=31$$ commissions has at most one member from teach of the $$d_i$$ commissions of $$i$$, and hence contains the remaining member. But this member is in at most $$9$$ commissions by observation $$1$$, a contradiction.

Observation 3: For any pair $$i$$, $$j$$ of members in a commission we have $$d_i+d_j\leq14$$.

If the inequality does not hold then without loss of generality $$d_i=8$$ and $$d_j\geq7$$. To fill the shared commission requires another $$8$$ members, and to fill the remaining $$7$$ commissions of member $$i$$ requires another $$9\times7=63$$ members. Each of the $$d_j-1$$ remaining commissions of $$j$$ contains at most $$7$$ members from the $$7$$ commissions of $$i$$, and hence at least $$2$$ new members. Hence we have a total of $$2+8+9\times(d_i-1)+2\times(d_j-1)\geq2+8+63+2\times6=85,$$ members, contradicting $$m=83$$.

Observation 4: For all $$i$$ we have $$d_i\leq7$$.

Suppose toward a contradiction that $$d_i=8$$ for some member $$i$$. To fill these $$d_i=8$$ commissions requires $$9d_i+1=73$$ distinct members, including member $$i$$, leaving $$10$$ members. Each of the remaining $$32$$ commissions has at most $$8$$ members from the $$d_i=8$$ commissions, hence at least $$2$$ members from the remaining $$10$$. Numbering these $$1$$ throught $$10$$ we find that $$\sum_{k=1}^{10}d_k\geq2\times32=64.$$ We distinguish two cases:

Case 1: If $$d_j=8$$ for some $$1\leq j\leq10$$ then $$j$$ shares a commission with at least $$8$$ other of these $$10$$ members, hence they all have $$d_k\leq6$$ by observation $$3$$. To satisfy the inequality there must be one more member $$j'$$ with $$d_{j'}=8$$, and the other $$8$$ have $$d_k=6$$.

We have $$11$$ members, including $$i$$, that together take up $$8+64=72$$ spots in the $$40$$ commissions. The remaining $$83-11=72$$ members then take up $$400-72=328$$ spots. As noted before, the sum $$\sum\binom{d_i}{2}$$ ranging over the remaining $$72$$ members is minimal when the values $$d_i$$ differ by at most $$1$$. This happens precisely when $$d_i=5$$ for $$40$$ members and $$d_i=4$$ for $$432$$ members. Then $$\sum_{k=1}^{83}\binom{d_i}{2}\geq3\binom{8}{2}+8\binom{6}{2}+40\binom{5}{2}+32\binom{4}{2}=796,$$ which exceeds the bound of $$\binom{40}{2}=780$$ we found before, a contradiction.

Case 2: If $$d_j\neq8$$ for all $$10$$ remaining members, then to satisfy $$\sum_{k=1}^{10}d_k\geq64$$ there must be a t least $$4$$ members with $$d_k=7$$. We also have $$\sum_{k=1}^{10}\leq70$$, and we proceed as before.

We have $$5$$ members, including $$i$$, that together take up $$8+28=36$$ spots in the $$40$$ commissions. Hence the remaining $$83-5=78$$ members take up $$400-36=364$$ spots. The sum $$\sum\binom{d_i}{2}$$ over the remaining $$78$$ members is minimized when the $$d_i$$ differ by at most $$1$$. This happens precisely if $$d_i=5$$ for $$52$$ members and $$d_i=4$$ for $$26$$ members, and we $$\sum_{k=1}^{83}\binom{d_k}{2}\geq\binom{8}{2}+4\binom{7}{2}+52\binom{5}{2}+26\binom{4}{2}=788,$$ again contradicting the upper bound of $$\binom{40}{2}=780$$.

Much more can be said, but my computer is already freezing up at this big an answer.

Here's a partial answer that increases the lower bound for any (not necessarily optimal) solution to $$m\ge 74$$.

Suppose there is a solution with $$m$$ members and we know that there are two members each in $$l+1$$ commissions, then $$m\ge 9(l+1)+(8-l)(l+1)+2.$$ This is because if member one is in $$\ge l+1$$ commissions, each commission needs to be filled with $$9$$ new members since these $$l+1$$ commissions already have maximum overlap. For the commissions that member two is in, each needs $$9$$ more members to be accounted for. We can't have any overlap between these commissions because they already have maximum overlap (that being member two). We can at best choose one member from each group with member one in it, giving us $$l+1$$ members, but the other $$9-(l+1)=8-l$$ are new. This gives $$9(l+1)+(8-l)(l+1)$$ members other than the two we started with. (Note that this is the best bound in $$l$$ possible).

Now, suppose $$m$$ members has a solution to the problem. Let $$d_i$$ be the how many commissions the $$i$$-th member is in. First note that $$m\ge 9d_i+1$$ for every $$i$$, so $$d_i\le \lfloor (m-1)/9\rfloor$$. Let $$k_l=|\{i\; :\; d_i>l\}|$$. Then $$400=\sum_{i=1}^md_i\le l(m-k_l)+\lfloor (m-1)/9\rfloor k_l.$$ Hence $$k_l\ge \frac{400-lm}{\lfloor (m-1)/9\rfloor -l}.$$ Since $$k_l$$ is an integer, if $$\frac{400-lm}{\lfloor (m-1)/9\rfloor -l}>1$$ then $$k_l\ge 2$$, meaning that there is at least $$2$$ members in at least $$l+1$$ commissions so by the above $$m\ge 9(l+1)+(8-l)(l+1)+2$$. Note that $$\frac{400-lm}{\lfloor (m-1)/9\rfloor -l}>1$$ exactly when $$\frac{400-\lfloor (m-1)/9\rfloor}{m-1}>l$$. Hence we have that for all $$\frac{400-\lfloor (m-1)/9\rfloor}{m-1}>l$$ we have $$m\ge 9(l+1)+(8-l)(l+1)+2.$$ For this to be satisfied gives $$m\ge 74.$$

• Very nice! I had some trouble understanding the part from "Hence we have that for all...". Perhaps you could make more explicit that if this inequality holds for $l$, then the next inequality must also hold for that $l$. – Servaes Dec 7 '18 at 14:29
• @Servaes I edited the answer. It is basically just solving for the $l$'s such that $k_l\ge 2$, letting the first paragraphs argument follow. – Will Fisher Dec 7 '18 at 16:06


# Am I more likely to roll at least one six if I have more dice?

• May 28th 2011, 05:31 PM
theyThinkImWrong
Am I more likely to roll at least one six if I have more dice?
Hi,

I've had a debate with some friends, about whether I'm more likely to roll at least a single 6 when I have 1000 dice, than I am to roll one with only 1 dice.

I say, the more dice I have, the more likely it is to roll at least a single 6. The others try to convince me, that the possibility to roll at least one 6 is always 1/6.

Now my thoughts for this are as follows:
At first, it doesn't matter, whether I roll them one after another, or one at a time.
The chances to roll a six on the first dice that I roll a 6, is 1/6.

Now, I don't care about the others if I did roll a 6, but I do care if I didn't (which is in 5/6 of all cases)

So for the second dice roll a six, the probability is still 1/6
However, I'm wondering about the probability that the first dice does NOT show a six, but the second does, so the probability for that to happen is 5/6*1/6.

So the probability to roll at least one 6 with two dice is:
1/6 + 1/6*5/6

If I was to expand this for 1000 dice, I would end up with:

$\sum_{k=0}^{999}\left (\frac{5}{6} \right )^k \times \frac{1}{6}$
The probability will get ever more closely to 100%, the more dice I have, correct?
• May 28th 2011, 06:52 PM
Wilmer
Why get 1000 dice? Same thing if you have 1 dice: just keep rolling it.
• May 28th 2011, 06:55 PM
theyThinkImWrong
Well, originally I asked this: If you were to roll 1000 dice at once, how likely is it, that there is at least one 6?

But it doesn't really matter :)
• May 28th 2011, 07:01 PM
bryangoodrich
The possibility of rolling a 6 with one die is 1/6, and for each die you have you increase your changes of rolling 1/6. By your friend's reasoning, you're no more liking to have "at least one head (H) in two coin flips" than in just one coin flip. Let us use this much more basic example. The probability of flipping a head of an unbiased coin is 1/2 because we split the event space {T, H} evenly into two. When we flip two coins we have a new event space, namely {TT, TH, HT, HH}. Here the events are equally likely of 1/4, but notice "at least one H" can turn out in three different events. Thus, the probability of "at least one H" is 3/4.

The same works for dice. With one die we have an event space {1, 2, 3, 4, 5, 6}. With two dice the size of the event space is 36:

Spoiler:
11 12 13 14 15 16
21 22 23 24 25 26
31 32 33 34 35 36
41 42 43 44 45 46
51 52 53 53 55 56
61 62 63 64 65 66

Now, in how many ways can we have "at least one six?" In this small, but sufficient, example, we can just look at see. The bottom row and the right-most column all include a 6. This is 11/36, which is almost 0.31, compared to 1/6 =0.167. So yes, increasing the die increases your chances of rolling a six.
• May 28th 2011, 07:24 PM
theyThinkImWrong
Thanks, that would confirm my point of view.

They: But how is that possible? The probability never changes? So it should always be 1/6?

Also, we'd value second and third opinions :) Thank you very much in advance!
• May 28th 2011, 07:58 PM
bryangoodrich
The probability of a 6 emerging on a single die does not ever change, but that is not the event we're looking at when we have more than one die. When we are using, say, two dice we're looking at a different event space with each event having 1/36 chance of occurring. Of those thirty-six, eleven of them will have a 6 appear. This is undeniable. This has no impact on the probability of a six appearing on a single die. Your friends seem to be ignoring the fact that the possible ways of a 6 occurring change when you have two dice. With one die there is one, and only one, way of a 6 occurring. Thus, the probability is 1/6. With two dice we have 11 different ways, each with 1/36 = (1/6)(1/6) chance of happening. The probability of the individual dice does support this result, as we can see, but you cannot ignore the fact our event space is different when you include another die.

I can go even further and define it this way by the inclusion-exclusion rule of probability theory. Let X be the outcome of the first die and Y the outcome of the second. What we want then is:

$P(X = 6\ or\ Y=6)=P(X=6)+P(Y=6)-P(X=6\ \&\ Y= 6)=\frac{1}{6}+\frac{1}{6}-\frac{1}{36}=\frac{11}{36}$

Do you see why these values work? The probability of the single die being six does not change, but we have to sum the two probabilities together. We then have to subtract the event that is common to both of them, otherwise we're double counting that event. This happens only once.
• May 28th 2011, 11:48 PM
mr fantastic
Quote:

Originally Posted by theyThinkImWrong
Hi,

I've had a debate with some friends, about whether I'm more likely to roll at least a single 6 when I have 1000 dice, than I am to roll one with only 1 dice.

I say, the more dice I have, the more likely it is to roll at least a single 6. The others try to convince me, that the possibility to roll at least one 6 is always 1/6.

Now my thoughts for this are as follows:
At first, it doesn't matter, whether I roll them one after another, or one at a time.
The chances to roll a six on the first dice that I roll a 6, is 1/6.

Now, I don't care about the others if I did roll a 6, but I do care if I didn't (which is in 5/6 of all cases)

So for the second dice roll a six, the probability is still 1/6
However, I'm wondering about the probability that the first dice does NOT show a six, but the second does, so the probability for that to happen is 5/6*1/6.

So the probability to roll at least one 6 with two dice is:
1/6 + 1/6*5/6

If I was to expand this for 1000 dice, I would end up with:

$\sum_{k=0}^{999}\left (\frac{5}{6} \right )^k \times \frac{1}{6}$
The probability will get ever more closely to 100%, the more dice I have, correct?

Let X be the random variable 'Number of 6's rolled'.

Then X ~ Binomial(n, p = 1/6).

Calculate Pr(X > 0) = 1 - Pr(X = 0). Then it is obvious that as n increases so does the probability.

I suggest you review the binomial distribution if you don't know it ( I do not plan to give a tutorial).


# Any closed subset in a separable metric space is the union of a perfect set and a set that is at most countable. Why does this matter?

I am writing regarding problem 2.28 in Rudin's "Principles of Mathematical Analysis". The statement the reader is asked to prove is as follows:

Every closed subset of a separable metric space is the union of a perfect set and a set that is at most countable.

This statement sounds like it should be important.

What came to mind to me was the definition of the functional limits and the ensuing definitions of continuity we encounter later in an undergraduate analysis course. The definition of continuity works "intuitively" with limit points in the domain, but less "intuitively" at isolated points. Since a perfect set is made up entirely of limit points (and is closed), it seems like this result should let us say something about "weird things" happening at only countably many places for functions defined on closed subsets of a separable metric space. I'm not really sure if this matters at all or is really comprehensible.

Thank you.

• It could be what you say. But also if the space is complete, then its perfect subsets have the same cardinal as $\mathbb{R}$, therefore this theorem proves that the continuum hypothesis is valid for closed subsets of $\mathbb{R}$. One can even use this to show it holds for borelian subsets Jan 12, 2018 at 7:30
• The various text-snippets you can read from a google-books search for Cantor-Bendixson should give you an idea of the importance of this result in mathematics. Jan 12, 2018 at 14:43

By the time you read the statement

$\tag 1 \text{Every closed subset of a separable metric space}$ $\qquad \qquad \text{is the union of a perfect set and a set that is at most countable.}$

you are far along the path of mathematically 'sculpturing' topological concepts. it is instructive to 'reverse engineer' this statement. I think you can safely say that if you do this, you will be forced to discover the foundations of point-set topology.

We all agree that the integers $\mathbb Z$ is a closed set of isolated points on the real number line $\mathbb R$. But this is just too easy to do, since you have the integers extending out to both $-\infty$ and $+\infty$. So can you find a countable (infinite) closed set of isolated points constrained to be inside the interval $[-1,+1]$?

Answering this question and the natural questions that arise as you proceed on your journey will be an awe inspiring mathematical challenge.

To answer the OP's question, reading and understanding (1) in a math book makes me think of mountaineers, and images like

The work, training and challenges of the accomplishment is safely behind them now.

The reader might be interested in exploring the perfect set property where they will find the following:

The Cantor–Bendixson theorem states that closed sets of a Polish space $X$ have the perfect set property in a particularly strong form; any closed set $C$ may be written uniquely as the disjoint union of a perfect set $P$ and a countable set $S$. Thus it follows that every closed subset of a Polish space has the perfect set property. In particular, every uncountable Polish space has the perfect set property, and can be written as the disjoint union of a perfect set and a countable open set.

Example: Consider $\text{2-dim Euclidean space, } \mathbb R^2$, which is a polish space. Define

$\quad C = \text{The Unit Circle } \cup \{(\pm \frac{1}{n},\, 0) \;|\; n \ge 1\} \cup \{(0,\, \pm \frac{1}{n}) \;|\; n \ge 1\} \cup \{(0,\,0)\}$

The set $C$ is closed in $\mathbb R^2$, and it can be written (uniquely) as the union of the unit circle, a perfect set, and the remaining countable points in $C$ with norm less than $1$. All these points not on the unit circle are isolated except for $(0,\,0)$, the origin.


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Prove convergence using $\varepsilon$-$N$ definition

This is an example presented by professor in class. I understand the idea behind this kind of definition, but I'm having trouble following my professor's thought process.

We must prove that the $\displaystyle \lim_{n\to\infty} \frac{2n-1}{n^2} = 0$ using the $\varepsilon$-$N$ definition of a limit of a sequence.

So, the foregoing limit is true iff $\forall \varepsilon>0,\ \exists N(\varepsilon)>0 : \forall n \ ( n \geq N \Rightarrow |f(n) - 0| < \varepsilon)$.

Now, the way I understand it is that if we prove the conditional statement by choosing an appropriate $N$, we've proven the limit.

So first we fix $\varepsilon>0$ since it's given, and play around with $$\displaystyle \left| \frac{2n-1}{n^2} \right| < \varepsilon$$ $$|2n-1| < \varepsilon \ n^2 .$$

We note that $|2n-1|<|2n|$ which sits right with me.

However, my professor then reasons that $|2n-1|<|2n|<\varepsilon \ n^2$ and so $2n < \varepsilon \ n^2$.

How do we know that $2n < \varepsilon \ n^2$? If epsilon is really small, wouldn't there be a point when this inequality is no longer true? This is what is bothering me.

From that we conclude that $\displaystyle n>\frac{\varepsilon}{2}$ and choose $\displaystyle N=\frac{\varepsilon}{2}$ so that

$$n \geq \frac{\varepsilon}{2} \Rightarrow \left| \frac{2n-1}{n^2} \right| < \varepsilon$$

is always true. And that is true iff the limit is true, so the limit is true.

I'd like to know if my understanding of this definition is correct, and if anyone can explain to me the reasoning behind the inequalities above.

Thank you very much!

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The point is that because $2n<\varepsilon n^2\iff 2<\varepsilon n$, no matter how small $\varepsilon$ is, you can find an $n$ such that the inequality will be true, (and more importantly it will be true for $n, n+1, n+2, \ldots$. Does this help? – Git Gud Oct 19 '13 at 18:15
Oh yes! Just like our original inequality is contrived, we still can find an $n$ large enough for it to be true, and then let that large $n$ be $N$. So we simply introduce $|2n|$, assume it to be less than $\varepsilon n^2$, and instead use that to find a large $N$ since $|2n-1|<|2n|$. Correct? – Andrey Kaipov Oct 19 '13 at 18:33
Spot on.${{{}}}$ – Git Gud Oct 19 '13 at 18:39

I'll give you another way to think about it. Instead of putting $\epsilon$ there from beginning, my approach is always to simply find an upper bound for $|a_n - L|$, and then look at it to figure it out which conditions put over $n$ assure that this bound is less than $\epsilon$.

Your sequence is $(a_n)$ where $a_n = (2n-1)/n^2$. Now, we want to bound $|a_n - 0|$ and look at that bound to see what $n$ must be bigger than in order to make that less than $\epsilon$.

Since $n \geq 1$ we have that $2n - 1 > 0$ and so $|a_n| = a_n$, in other words, we can look just to $(2n-1)/n^2$. Now, this is the same as

$$\dfrac{2n-1}{n^2}=2\dfrac{1}{n}-\dfrac{1}{n^2},$$

Now, look that $-1/n^2 < 0$ always, so that

$$\dfrac{2n-1}{n^2} <\dfrac{2}{n}$$

Now, what must $n$ be greather than in order to make that less than $\epsilon$. Of course, if $n > 1/\epsilon$, then $1/n < \epsilon$, however there's a $2$ there, so that to take care of it we make $n > 2/\epsilon$, now when we divide by $n$ we get $1/n < \epsilon/2$, and hence the condition follows. So, given $\epsilon > 0$, if $n > 2/\epsilon$ we have that

$$\left|\dfrac{2n-1}{n^2}-0\right|=\dfrac{2n-1}{n^2}=\dfrac{2}{n}-\dfrac{1}{n^2}<\dfrac{2}{n} <\epsilon$$

and hence $a_n \to 0$ when $n\to \infty$.

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Right so we simply set the simpler expression, in this case the upper bound $\frac{2}{n}$, to be less than epsilon, because that will make $\frac{2n-1}{n^2} < \epsilon$. Then once we solve for $n$ in terms of $\epsilon$, we know that is what $n$ should be given some $\epsilon$ to make the inequality true. In this case your $N$ would be $\frac{2}{\epsilon}$, which is interesting to see, since it's different than the $N$ in the original post. My understanding of this has definitely improved. Thank you. – Andrey Kaipov Oct 19 '13 at 19:11

Your professor is not concluding that $|2n|<\varepsilon n^2$, but is instead noting that it is sufficient to have $|2n|<\varepsilon n^2.$

It's certainly not true for all $n,$ but when it is true, you will have $\left|\frac{2n-1}{n^2}\right|<\varepsilon.$ In particular, it is true whenever we have $n>\frac2\varepsilon.$

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# how to find rational numbers satisfying the binary quadratic equation $x^2+3xy+5y^2=4$

I am looking for a generalisation of the solution of $x,y$ wich are rational numbers,they could be infinite,how can i find such solutions,integer solutions are obvious

I have found that

$\Delta=-11,u=1,d=-11$

also

$(x+\frac{3+\sqrt{-11}y}{2})(x+\frac{3-\sqrt{-11}y}{2})=4$

• By obvious you mean Pell equations? – chubakueno Apr 5 '14 at 16:45
• yes ,i am trying to say in this case i am not interested in integer solutions,but rational – Jonas Kgomo Apr 5 '14 at 16:52

We have the simple solution $x=2$, $y=0$. Now take the line though $(2,0)$ with slope $m$, and find where it meets the ellipse. That will give a rational parametrization of the ellipse.

Details: The line has equation $y=m(x-2)$. Substitute. We get $x^2+3xm(x-2)+5m^2(x-2)^2=16$. This simplifies to $(1+3m+5m^2)x^2-(6m+20m^2)x+20m^2-4=0$. The product of the roots is $\frac{20m^2-4}{1+3m+5m^2}$. But one of the roots is $2$, so the other is $\frac{10m^2-2}{1+3m+5m^2}$. Now we can compute the corresponding $y$. Any rational $m$ will give us a rational solution of the original equation.

Remark: The procedure was general. In particular, let $ax^2+bxy+cy^2=d$ be an ellipse with $a,b,c,d$ rational. If we know a rational point on the ellipse, then we can find a parametric expression for all rational points. The most important special case is the circle.

And the basic idea generalizes, importantly, to elliptic curves.

• Your remark is what makes your answer better than mine. But I would like to know: Is there an efficient method to get a rational point? Can we know when does it exist, or some bound on the denominator and numerator? – chubakueno Apr 5 '14 at 17:52
• Yes, there is a huge amount of theory of ternary quadratic forms (it is useful to homogenize, and consider $ax^2+bxy+cy^2=4z^2$.) The details are complicated, they go back to Legendre and Gauss, and are a large chapter in the history of number theory. – André Nicolas Apr 5 '14 at 17:59
• i am actually interested in that,more about binary and not tenary ,can i say that solution is for both rational and integers – Jonas Kgomo Apr 5 '14 at 18:07
• The integer solutions of the ternary (homogenized) form are in a simple way connected to the rational solutions of the original equation. – André Nicolas Apr 5 '14 at 18:10
• i would appreciate if both of you can help me with this as well math.stackexchange.com/questions/739752/… – Jonas Kgomo Apr 5 '14 at 18:14

This is a variation of Hecke's answer $$(2x+3y)^2+11y^2=4^2$$ $$\left(\frac{2x}y+3\right)^2+11=\left(\frac4y\right)^2$$ $$11=\left(\frac4y\right)^2-\left(\frac{2x}y+3\right)^2$$ $$11=\left(\frac{4-2x}y-3\right)\left(\frac{4+2x}y+3\right)$$ $$\frac{4-2x}y-3=t, \frac{4+2x}y+3=\frac{11}t$$ $$\implies x = -\frac{2(t^2+6t-11)}{t^2+11}, y = \frac{8t}{t^2+11}, t\neq0$$

My comment contained an imprecision. This works for every equation that can be transformed into the form: $$(ax+by)^2+cy^2=\left(\frac pq\right)^2$$ Where $a,b,c,p,q\in\mathbb Z, q\neq 0$

• this is excellent,it seems like it will absolutely work for any binary equation – Jonas Kgomo Apr 5 '14 at 18:11
• is $t$ an integer – Jonas Kgomo Apr 5 '14 at 18:16
• @Jonas12: $t$ is any rational number. – TonyK Apr 5 '14 at 18:18
• @Jonas12 Actually we were quite lucky :) .When we completed the square in the LHS multiplying by $4$, $4^2$ resulted to be a square. If you change that $4$ by $5$(or any non-square number) in your equation we would have a problem. In particular, this works with the general equation $ax^2+bxy+cy^2=d$, when $d$ is a perfect square. André Nicolas solutions helps us here if it is not, if we can find a rational solution. – chubakueno Apr 5 '14 at 18:23
• the manner of the solution proposed by the proffesor,is in the manner of the solution on page 47-48 for the problem $x^2+3xy-5y^2=65$ i am wondering if this approach and @Andre Nicolas solutions will be acceptable – Jonas Kgomo Apr 5 '14 at 18:42


# System of equations

#### anemone

##### MHB POTW Director
Staff member
Solve in real numbers the system of equations

$(3a+b)(a+3b)\sqrt{ab}=14$

$(a+b)(a^2+14ab+b^2)=36$

#### HallsofIvy

##### Well-known member
MHB Math Helper
The first can be "reduced" to a fifth degree polynomial equation and the second to a third degree equation.

#### Opalg

##### MHB Oldtimer
Staff member
Solve in real numbers the system of equations

$(3a+b)(a+3b)\sqrt{ab}=14$

$(a+b)(a^2+14ab+b^2)=36$
Let $x = a+b$, $y = \sqrt{ab}$. Then the equations become $$y(3x^2 + 4y^2) = 14,$$ $$x(x^2 + 12y^2) = 36.$$ Multiply the first equation by 2, so they become $$x^3 + 12xy^2 = 36,$$ $$6x^2y + 8y^3 = 28.$$ Add them to get $(x+2y)^3 = 36+28=64$, and subtract them to get $(x-2y)^3 = 36-28 = 8.$ Take the cube roots: $x+2y=4$, $x-2y=2.$ Thus $x=3$, $y=1/2$. So $a+b=3$, $ab = 1/4.$ Hence $a,b$ are the roots of the equation $\lambda^2 - 3\lambda + 1/4 = 0$, namely $\frac12(3\pm\sqrt8).$ (Clearly $a$ and $b$ are interchangeable, so those values can be taken either way round.)

#### anemone

##### MHB POTW Director
Staff member
The first can be "reduced" to a fifth degree polynomial equation and the second to a third degree equation.
Hi HallsofIvy,

According to the guidelines thread for posting in this subforum:

This forum is for the posting of problems and puzzles which our members find challenging, instructional or interesting and who wish to share them with others. As such, the OP should already have the correct solution ready to post in the event that no correct solution is given within at least 1 week's time...Responses to these topics should be limited to attempts at a full solution, or requests for clarification addressed to the OP if the problem statement is vague or ambiguous.
When people post problems here in this subforum, they are not seeking help, but rather posting problems for the enjoyment of and challenge to the members of MHB. I do however appreciate the fact that you are trying to help. So, go ahead and have fun with these problems if you like and post full solutions.

Let $x = a+b$, $y = \sqrt{ab}$. Then the equations become $$y(3x^2 + 4y^2) = 14,$$ $$x(x^2 + 12y^2) = 36.$$ Multiply the first equation by 2, so they become $$x^3 + 12xy^2 = 36,$$ $$6x^2y + 8y^3 = 28.$$ Add them to get $(x+2y)^3 = 36+28=64$, and subtract them to get $(x-2y)^3 = 36-28 = 8.$ Take the cube roots: $x+2y=4$, $x-2y=2.$ Thus $x=3$, $y=1/2$. So $a+b=3$, $ab = 1/4.$ Hence $a,b$ are the roots of the equation $\lambda^2 - 3\lambda + 1/4 = 0$, namely $\frac12(3\pm\sqrt8).$ (Clearly $a$ and $b$ are interchangeable, so those values can be taken either way round.)
What a nice skill and great solution, Opalg! Thanks for participating!


# Integrating exponent with polynomial

http://i.stack.imgur.com/4tXNr.jpg

$e^{x^2/2}\int e^{-x^2/2}(-x^3+x)\ dx$ turns out to be equal to $e^{x^2/2}[e^{-x^2/2}(x^2+1)]$

Is there a easier method of integrating such functions? I can't grasp how text book was able to integrate it so easily. They dont show the steps, but rather go straight to integrated function. How would you tackle such integration? Integration by parts of these fuctions proven to be tedious and time consuming as they require another integration by parts to be performed.

Sorry for the link, it said I need 10 rep points to post the photo

• I don't understand what you're asking. If you have a product of functions of a single variable, you pretty much need to integrate by parts unless one of the functions is the derivative of the other. – Edward Evans Jun 8 '16 at 1:32
• for such product of functions, only method possible is integration by parts? – Sysnaptic Jun 8 '16 at 1:36
• Look up "Reduction Formulae". With products of trigonometric functions and exponentials, you'll end up in an endless loop. – Edward Evans Jun 8 '16 at 1:41
• You'll need to post in Latex, otherwise your post is likely to get down voted and/or closed. See here for formatting help. – mattos Jun 8 '16 at 1:44

Hopefully you recognize $e^{-x^2}$ as a function which can not be integrated using regular functions. As such using 'by parts' as your first step with your parts being the exponential and the polynomial will be unsuccessful. Note that the derivative of $x^2$ contains $x$ and the second part of your integral has this as a factor so:

$$\int e^{-x^2/2}(-x^3+x)\ dx=\int e^{-x^2/2}(-x^2+1)x\ dx$$

Let $u=\frac{x^2}{2}$ so $du=x\ dx$.

$$=\int e^{-u}(-2u+1)du$$

$$=\int e^{-u}\ du-2\int u e^{-u}\ du$$

Then use 'by parts' on the second integral

$$=-e^{-u} - 2\left(-u e^u-\int -e^u\ du\right)$$

$$=-e^{-u} +2u e^{-u}+2e^{-u} +c$$

$$=2u e^{-u}+e^{-u} + c$$

$$=x^2e^{x^2/2}+e^{-x^2/2}+c$$

$$=e^{x^2/2}(x^2+1)+c$$

• Wow! Thank you so much for your answer. This answers everything I was curious about – Sysnaptic Jun 8 '16 at 2:55

If, as in this case, the terms of the polynomial are all odd, the substitution $u = x^2$ works.

$$J = \int e^{-x^2/2} (-x^3 + x)\; dx = \int e^{-u/2} \dfrac{-u+1}{2}\; du$$

Next, using the method of undetermined coefficients, a form for the antiderivative should be

$$J = e^{-u/2} (a u + b) + C$$

Taking the derivative, we need $$\dfrac{dJ}{du} = e^{-u/2} \left(-\frac{a u + b}{2} + a\right) = e^{-u/2} \dfrac{-u+1}{2}$$ Thus $a = 1$ and $a-b/2 = 1/2$, so $b = 1$. The result is

$$J = e^{-u/2} \left(u + 1 \right) + C= e^{-x^2/2} \left( x^2 + 1\right) + C$$