Mafiascum.net

In post 99, Mitillos wrote:Lend the dead man a camel.

In post 101, Mitillos wrote:Since you insist: You can't. By the time the will takes effect, the trader is dead. It is impossible to change the number of camels, assuming you have provided all pertinent information, thus at least one camel will have to be chopped up.
If we allow silly solutions like lending a dead man a camel, he now has 18 camels, which can be divided as you have requested. The sum I gave shows that one camel will be left over, which can then be returned to the lender.

In post 102, vizIIsto wrote:You don't need to 'lend a camel' to solve it, but adding an extra camel is how you will be able to divide the 17 camels fair and square.

In post 5, Ircher wrote:What exactly is a function in terms of a graph? As in, what inputs is it mapping? I may give this a try, but I would.need to know the previous.

In post 120, Wickedestjr wrote:I think I got it...

Spoiler:

Let's call the vertices of the triangle A, B, and C. And we'll call four of the extended points D, E, F, and G. I've also labeled the measures of the angles of the triangle: M_a, M_b, and M_c.

Notice that CD and EC have the same total length. This means that triangle CED is an isosceles triangle and angles CDE and DEC are equal to each other (as marked in the diagram). But we also know that the sum of the angles of triangle CED is 180 degrees. So:
180 = <DCE + <DEC + <CDE
180 = M_c + 2*<DEC
<DEC = <CDE = 90 - 0.5*M_c

Similarly, we can find that:
<BFG = <BGF = 90 - 0.5*M_b

Next, notice that triangle ADG is also isosceles, so <ADG = <AGD. Also we can see that <DAG and <BAC are equivalent angles. So <DAG = M_a and that means:
<ADG = <AGD = 90 - 0.5*M_a

Finally, calculate the sum of <FGD + <BED:
[<FGD] + [<BED]
= [(<BGF) + (<AGD)] + [<CED]
= [(90 - 0.5*M_b) + (90 - 0.5*M_a)] + [90 - 0.5*M_c]
= 270 - 0.5*M_a - 0.5*M_b - 0.5*Mc
= 270 - 0.5*(M_a + M_b + M_c)
= 270 - 0.5*180 = 270 - 90 = 180

So <FGD + <BED = 180 and it can be similarly shown that <EDG + <CFG = 180 as well. This means that quadrilateral DEFG is cyclic, so it's vertices must all fall on the same circle.

And this same method can be used to show that the other two extended points fall on the same circle as well.

Spoiler:

Let's label up.


First, if the hypothesis is true, where would the centre of the circle have to be? Since UW, VY, and XZ have the same length, their closest points to the centre are equidistant. Therefore, you can draw a smaller concentric circle tangent to all three chords, and the centre must be the triangle's incentre.

With that, we go back to the triangle ABC. We find the incentre and we draw a circle so that the sides extended to chords have length equal to the perimeter of the triangle. Labeled up, we get this:

Let BC = a, CA = b, and AB = c

Since the (in)centre lies on the angle bisectors of <CAB, <ABC, and <BCA, then by a symmetry argument:

AY = AZ = a'
BW = BX = b'
CU = CV = c'

Now all we need to prove is that a = a', b = b', and c = c'. This can easily be done using the equations below:

a + b' + c' = 2s(WU)
a' + b + c' = 2s(VY)
a' + b' + c = 2s(XZ)
a + b + c = 2s(definition of s)