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Different 'number couples' like to dine at Domino's Pizza on the weekends.
Different 'number couples' visit at different times (HH:MM format).
In case of each 'number couple', HH is the value of one of the numbers and MM is the value of the another number.
As a tradition, each 'number couple' carves out the first slice of their pizza with an arc length equal to the function of their values.
All the prepared pizzas have radii of 18 cm.


A certain 'number couple' which is a pair of consecutive even numbers carves out the first slice of their pizza with an arc length equal to 54.97 cm.


A certain 'number couple' which is a pair of consecutive odd numbers carves out the first slice of their pizza with an arc length equal to 16.49 cm.


A certain 'number couple' which is a pair of consecutive perfect squares carves out the first slice of their pizza with an arc length equal to 5.49 cm.


What will be the arc length when the 'number couple' which is a pair of first two prime numbers carves out the first slice of their pizza?
All the numbers lie in the range [1,25]

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    $\begingroup$ @AlainReve I have modified it. Thanks for the feedback. Hope it is clear now. $\endgroup$ Aug 26, 2022 at 11:56
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    $\begingroup$ Are the given numbers rounded, or is the exact value? $\endgroup$
    – bobble
    Aug 26, 2022 at 13:53
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    $\begingroup$ @bobble Rounded to two decimal places. $\endgroup$ Aug 26, 2022 at 14:54
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    $\begingroup$ Is the function the same for each couple, with only the two inputs differing? $\endgroup$
    – Ed Murphy
    Aug 26, 2022 at 14:58
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    $\begingroup$ @EdMurphy Yes. The method of deriving the final arc length remains the same for each couple according to their values. $\endgroup$ Aug 26, 2022 at 15:16

2 Answers 2

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The arc length ...

... of the pizza for the first two prime numbers 2 and 3 will be 13.67 cm.

The rule is:

Each of the number pais represent a time. One of the numbers is the hours part, the other number is the minute part, but the order isn't specified. The angle of the pizza slice is the angle between the hours and minute hand when that time is shown on a 12-hour clock.

The even numbers are 08:10 with an angle of 175°.
The odd numbers are 03:05 with an angle of 62.5°.
The square numbers are 16:25 with an angle of 17.5°.
The first two primes are 02:03 with an angle of 43.5°.

Clock faces or pizza slices? You decide.

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First of all, I'm not sure this is exactly the correct answer based on two issues, which I will highlight as they appear.

The first observation is...

that a domino is divided into two parts by exactly one line.

This means

The pizza is not sliced into standard "triangular" sectors, but rather it is sliced by a chord.
As shown here:
enter image description here

Now we can find

the angle of each section using the formula $\theta = \frac{s}{r}$ where $s$ is the arclength and $r$ is the radius.

First couple: $\theta_1 = \frac{54.97\ \rm{cm}}{18\ \rm{cm}} = 3.0539\ \rm{rad}$
Second couple: $\theta_2 = \frac{16.49\ \rm{cm}}{18\ \rm{cm}} = 0.9161\ \rm{rad}$
Third couple: $\theta_3 = \frac{5.49\ \rm{cm}}{18\ \rm{cm}} = 0.3050\ \rm{rad}$

Now we calculate

the area of these slices, using the formula $A = \frac{r^2}{2}(\theta-\sin\theta)$.

First couple: $A_1 = \frac{(18\ \rm{cm})^2}{2}(3.0539-\sin3.0539) = 486.09\ \rm{cm}^2$
Second couple: $A_2 = \frac{(18\ \rm{cm})^2}{2}(0.9161-\sin0.9161) = 145.82\ \rm{cm}^2$
Third couple: $A_3 = \frac{(18\ \rm{cm})^2}{2}(0.3050-\sin0.3050) = 48.55\ \rm{cm}^2$

Note that while $\theta$ is found in radians, the $\sin$ function is evaluated as though its argument is in degrees. This is the first major issue; I only initially considered this by a mistake in my calculator settings. Regardless, I think this path notable because scaling our pizzas to the unit circle, we get

First couple: $A_1 = \frac{1}{2}(3.0539-\sin3.0539) = 1.5003$
Second couple: $A_2 = \frac{1}{2}(0.9161-\sin0.9161) = 0.45006$
Third couple: $A_3 = \frac{1}{2}(0.3050-\sin0.3050) = 0.14983$

Which are all extremely close to a ratio of small integers. Unfortunately, I had no luck trying to construct them using numbers given the couples' constraints (eg. consecutive odd).

However,

Looking back at the sector area in $\rm{cm}^2$, we find a rather simple means of constructing the area from two numbers:

Shift the smaller of the numbers one place to the left, then add.
This works for each of the given couples:
First couple: $\{44,46\}\Rightarrow 440+46 = 486\approx486.09\ \rm{cm}^2$
Second couple: $\{13,15\}\Rightarrow 130+15 = 145\approx145.82\ \rm{cm}^2$
Third couple: $\{4,9\}\Rightarrow 40+9 = 49\approx48.55\ \rm{cm}^2$

Note the second major issue--the first couple uses numbers which are outside the given allowable range of $[1,25]$.

With a function for determining the arclength identified, we can now find the arclength for the first two primes.

$\{2,3\}\Rightarrow 20+3 = 23\ \rm{cm}^2$

Solving $23\ \rm{cm}^2=\frac{(18\ \rm{cm})^2}{2}(\theta-\sin\theta)$:
$\theta = 0.1445\ \rm{rad}$

$s = r\theta = (18\ \rm{cm})0.1445 = 2.60\ \rm{cm}$

Therefore, the first two prime numbers share a slice with an arclength of

2.60 centimeters

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  • $\begingroup$ Thanks for taking your time in attempting. By carving out a slice, they obtain a normal triangular sector itself. It is nowhere mentioned otherwise. Convert the sector angles you have obtained in radians into degrees. You might be able to get a different perspective then :) $\endgroup$ Aug 27, 2022 at 15:05

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