# Challenge Riddles and Puzzles: Extend the following to a valid equation

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#### lpetrich

Problem 71:
Let us consider one of the circles, and let us give it radius 1. The squares next to it in size are the inscribed square, the smaller one, and the circumscribed square, the larger one.

The inscribed square has its vertices on the circle. By symmetry, its center coincides with the center of the circle, O. Two neighboring vertices V1 and V2 have distances OV1 = OV2 = 1, and their angle is V1OV2 = 90d. By Pythagoras's theorem, distance V1V2 = sqrt(2).

The circumscribed square has the circle tangent to its edges. Consider two neighboring tangent points, T1 and T2. The tangent line for OT1 is perpendicular to it, and thus parallel to OT2. Likewise, the tangent line for OT2 is parallel to OT1. The intersection of those tangent lines I will call X. Angles T1OT2, XT1O, and OT2X are all 90d, making T2XT1 also 90d. This makes quadrangle T1OT2X a rectangle, and since OT1 and OT2 both have length 1, T1OT2X is a square with all sides having length 1. Since the tangent lines extend in both directions, the distance from one intersection to a neighboring one is thus 1+1 = 2.

Thus, the circumscribed square has side length 2, sqrt(2) times the inscribed square's side length. Taking the square gives the area, meaning that the circumscribed square has an area 2 times that of the inscribed square.

Thus, the ratio of the areas of two consecutive squares in this construction is 2.

#### fresh_42

Mentor
2018 Award
Here is an easy way to see / solve it:

#### mfb

Mentor
What is complicated about the adoption?

Is the year the same?

A mistake in the sheet the teacher got?

Mentor
2018 Award

#### mfb

Mentor
Well, they could be half-brothers (shared father). Not technically an adoption I guess but similar.
They could have the same father and their biological mothers could be twins. That would make them genetically as similar as (non-identical) twins and it would be the result of questionable family planning choices.

"They lied" is a very simple possible answer, but I guess that is not the intended answer.

#### fresh_42

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2018 Award
"They lied" is a very simple possible answer, but I guess that is not the intended answer.
Right. The real reason is completely natural and the easiest possible.

#### fresh_42

Mentor
2018 Award
Let me phrase it with a mathematical anecdote my mentor once told me.
He listened to a dialogue between two students in the university's elevator.
One student was rather upset about the exam she just has had.
"This idiot asked me how often the constant function can be differentiated and I said once. This stupid question ruined a better grade."

So is the constant function exactly twice differentiable?

#### DavidSnider

Gold Member
72. Summer is over and there are two new students in the class who are hard to tell apart. "You are certainly twins?", asks the teacher.

"No," answer the two boys.

The teacher is surprised and looks again in her documents: both have the same birthday and the same parents. How can that be?

D88
Surrogacy?

#### fresh_42

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2018 Award
No. Or let's say: It could have happened in ancient Greece.

Edit: No Disney fans here?

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#### fresh_42

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2018 Award
73. $45$ must be divided in four parts, such that we get the same number if we add $2$ to the first part, subtract $2$ from the second, multiply the third by $2$ and divide the fourth by $2$.

D89

#73
8, 12, 5, 20

#### fresh_42

Mentor
2018 Award
74. Enter the numbers $1,2,\ldots,10$ into the circles, such that the sums of all numbers along the three inner triangles are equal.

D89

#### mfb

Mentor
So is the constant function exactly twice differentiable?
Triplets (or more) then.

#### lpetrich

Proof of #73
In symbolic form, $45 = n_1 + n_2 + n_3 + n_4$ where $n_1 + 2 = n$, $n_2 - 2 = n$, $2n_3 = n$, and $n_4/2 = n$ for some number n.

One can easily find n1 to n4 from these: $n_1 = n - 2$, $n_2 = n + 2$, $n_3 = n/2$, $n_4 = 2n$. Substituting into the equation for 45, I find
$$45 = (n-2) + (n+2) + (n/2) + 2n = 9n/2$$
Its solution is $n = 10$ and the four parts are $8, 12, 5, 20$.

#### fresh_42

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2018 Award
75. Two trains passed each other in opposite directions, one at a speed of 36 km / h, the other at 45 km / h. A passenger on the second train counted and found, that the first train to pass him needed six seconds.
How long was the train?

D90

#### mfb

Mentor
74. Enter the numbers $1,2,\ldots,10$ into the circles, such that the sums of all numbers along the three inner triangles are equal.
The center is part of the sum for all triangles, we can fill it last with whatever number is remaining and ignore it from now on. For every pair of triangles there are two numbers that are shared between these triangles, and one number that is exclusive to the other triangle. We can find a solution if we can find three mutually exclusive triples that satisfy a+b=c where a+b are put in the shared numbers and c is put on the opposite side. That way the sum in every triangle will be the sum of all numbers apart from the three outer numbers.
Doesn't need much searching to find 1+9=10, 2+6=8, 3+4=7.
The leftover number for the center is 5 and the sum in every triangle is 30, which means the average per field is 5 as well. Nice symmetry around 5 despite the asymmetric start (average number we fill in is 5.5).

#### fresh_42

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2018 Award
Very interesting. I have a solution with 38 instead of 30:

I didn't think that more than one solution is possible. Maybe I should have asked how many sums are possible.

#### lpetrich

For #74, I wrote a short program that does a brute-force search. The number of permutations searched was 10! = 3628800, and I found 6528 solutions. Because none of the solutions have any symmetries, the total number of inequivalent solutions is the total number divided by the size of the triangle symmetry group (6): 1088.

I wrote it in C++, and I used STL <algorithm> function next_permutation() to iterate through the possibilities. The Standard Template Library is *great*.

#### fresh_42

Mentor
2018 Award
For #74, I wrote a short program that does a brute-force search. The number of permutations searched was 10! = 3628800, and I found 6528 solutions. Because none of the solutions have any symmetries, the total number of inequivalent solutions is the total number divided by the size of the triangle symmetry group (6): 1088.

I wrote it in C++, and I used STL <algorithm> function next_permutation() to iterate through the possibilities. The Standard Template Library is *great*.
Have you calculated the possible sums in every solution?

#### lpetrich

Have you calculated the possible sums in every solution?
I have just done so. The number of solutions for each sum value is;
28: 96 16
29: 192 32
30: 480 80
31: 864 144
32: 1248 208
33: 768 128
34: 1248 208
35: 864 144
36: 480 80
37: 192 32
38: 96 16
with the total number and the number to within symmetries.

#### mfb

Mentor
There is more symmetry. There is the factor 6 for the arrangement of the three triangles, but there is also a factor 8 from swapping numbers on the diagonals - different pattern but effectively the same solution. This is the reason all your numbers are divisible by 8. Removing this symmetry we end up with 136 options.

There is a more subtle symmetry: Replacing 1,2,...,10 by 10,9,...,1 leads to a solution as well (with triangle sum 66-x instead of x). That leads to the symmetric pattern of options you found. Removing this symmetry as well we end up with 68 options.

Is there a solution that also has the same sum for the big outer triangle?

#### lpetrich

There is more symmetry. There is the factor 6 for the arrangement of the three triangles, but there is also a factor 8 from swapping numbers on the diagonals - different pattern but effectively the same solution. This is the reason all your numbers are divisible by 8. Removing this symmetry we end up with 136 options.
How does that symmetry work? I find it difficult to picture.

#### mfb

Mentor
Just swap e.g. 3 and 4 in the solution I posted. You can also swap 1 and 9 or 2 and 6.

#### lpetrich

So it's swap outer and middle along each diagonal.
Is there a solution that also has the same sum for the big outer triangle?
Indeed there is.

Inner-triangle length-equal solutions
28: 96 16 2
29: 192 32 4
30: 480 80 10
31: 864 144 18
32: 1248 208 26
33: 768 128 16
34: 1248 208 26
35: 864 144 18
36: 480 80 10
37: 192 32 4
38: 96 16 2
Total: 6528 1088 136
(raw, triangle symmetry, diagonal-interchange symmetry)

Equal to the outer-triangle length
31: 48 8
32: 120 20
34: 120 20
35: 48 8
Total: 336 56
(raw, triangle symmetry)

mfb

#### fresh_42

Mentor
2018 Award
76. $n^3+ n^2 u +n^2 v+n^2w+nuv+nuw+nvw+uvw = 27,673,509,091$ with $u<v<w$.
What is $u\cdot n^2\,?$ (All numbers are non negative integers and $n$ maximal among all solutions.)

D91

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