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From wikipedia description, the Austin procedure goes as follows

  1. Alice places one knife on the left of the cake and a second parallel to it on the right where she judges it splits the cake in two.
  2. Alice moves both knives to the right in a way that the part between the two knives always contains half of the cake's value in her eyes (while the physical distance between the knives may change).
  3. George says "stop!" when he thinks that half the cake is between the knives. How can we be sure that George can say "stop" at some point? Because if Alice reaches the end, she must have her left knife positioned where the right knife started. The Intermediate value theorem establishes that George must be satisfied the cake is halved at some point.
  4. A coin is tossed to select between two options: either George receives the piece between the knives and Alice receives the two pieces at the flanks, or vice versa. If partners are truthful, then they agree that the piece between the knives has a value of exactly 1/2, and so the division is exact.

In the third step, it is said that this procedure ensures that George will always say stop at a some moment, this is proved using the Intermediate value theorem (IVT), I have been trying to link between IVT and this procedure to understand why it is ensured that a position at some time will satisfy George but i failed to establish a relation.

How is IVT used to proof this procedure always satisfy George?

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Assume George thinks there is less than half the cake between the knives at the start. At the end the other piece is between the knives, so George thinks there is more than half between the knives. As George's estimate of the value of the cake between the knives is a continuous function of the position of the knives, it must cross $\frac 12$ at some point. The IVF is used to say there is some point George's estimate is exactly $\frac 12$

The case where George thinks there is more than half between the knives at the start works the same.

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    $\begingroup$ "At the end the other piece is between the knives, so George thinks there is more than half between the knive" this is very intuitive but didn't cross my mind, i was focusing on the first part, i understand more now, thanks for pointing it out. $\endgroup$ – markeb Jan 1 at 13:24
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Before Alice moves the knives, the area of the cake in between the knives is $\int_{-r}^a2\sqrt{r^2-x^2}\text{ }dx=A$, where r is the radius of the cake and $x=0$ is the middle of the cake, and after Alice moves the knives, the area of the cake in between the knives is $\int_{a}^r2\sqrt{r^2-x^2}\text{ }dx=\pi r^2-A$ because if Alice perceived $a$ as the midpoint of the cake, she would stick to that as being her lower bound as the other part of the cake must've also been half in her perspective.

If she perceived $a \ne 0$, then the area between the knives at the beginning ($A$) was either larger than half ($a>0$) or smaller than half ($a<0$) so at the end, the area between the knives ($\pi r^2-A$) would be either smaller than half ($a>0$) or larger than half ($a<0$) in each of those scenarios, respectively. Thus, the area between the knives must have equaled half at some point as the area is a continuous function and the interval of areas over the movement of the knives has endpoints that are less than and greater than half.

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    $\begingroup$ Thanks for the explanation, you answer clarifies well but i dont have enough reputation to upvote it, $\endgroup$ – markeb Jan 1 at 13:26
  • $\begingroup$ That is fine I am just glad that I was able to help you :) $\endgroup$ – Soham Konar Jan 1 at 15:12

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