No Possible Triangles

I'm not so sure how to make it rigorous but my intuition is the following: let $l_1, l_2, \ldots, l_{10}$ be the 10 distances from least to greatest. then $l_1+l_2\leq l_3$ and $l_n\geq l_{n-1}+l_{n-2}$ . Then if $l_n\geq 3$ , we can show by induction that $l_n \geq f_{n-1 }l_1+f_n l_2$ , where $f_n$ is the nth fibonacci number. Thus $l_{10}\geq 21 l_1 +34 l_2$ . This sum is minimized for $l_1=1$ and $l_2=2$ in this case we get $l_{10} \geq 89$ .

We know that this is possible, by using the Fibonacci sequence to create the lengths of the rods, $l_n = f_{n+1}$ .

Sort the rods from shortest to longest. Then the triangle condition is equivalent to: no rod may be shorter than the sum of any previous two. Since those are sorted we may rephrase this bound as the sum of the two immediately preceding rods.

We see easily by induction that each rod must be at least as long as the corresponding term of the Fibonacci-Lucas sequence generated by the first two rods.

Since the first two rods have distinct lengths, they must be at least 1 and 2, respectively, so the minimum rod length follows a (shifted) Fibonacci sequence, giving us a 10th term of 89.

without any loss of generality arrange the rods in ascending order

a < b < c < d < e < f < g < h < i < j where the letters stand for the lengths of the rods

as Will is not able to form a triangle

therefore by triangle inequality any side should be greater than or equal to the sum of the other 2 sides(only then a triangle cannot be formed)

therefore j is greater than or equal to i+h. so the shortest length of j is i+h.similarly the least value of i is g+h.

continuing the logic we get the shortest length of c is a+b but the least value of

a+b is 1+2(as they are different integer sided rods)

c=1+2=3

d=c+b=3+2=5

e=d+c=8

f=e+d=13

g=e+f=21

h=g+f=34

i=g+h=55

j=i+h=89

thus proved

**To, make a non degenerate triangle the sum of any 2 side of the triangle must be bigger than the other one.

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We have to find the smallest value of the largest rod.

So, * we have to make a collection of 10 integer numbers in which the largest number must be Equal(as we have to find the least value) than the sum of the nearest 2 numbers of the largest number. *

So, this is * 1,2,3,5,8,13,21,34,55,89 *

For example 89 is equal to 34+55=89

34 and 55 are the previous two big numbers of 89.

So, we * cannot make a triangle. *

So * 89 * is answer.

The solution can be derived from iteratively adding a new rod to a set S of rods that cannot already be used to form a triangle. At each step, to form a triangle, you either use the new rod or you don't. If you don't, you have to use 3 rods from S, which by assumption cannot be used to form a triangle. So we only need to find a way to add a new rod so that it cannot be used with any two rods in S to form a triangle.

To do so, we apply the triangle inequality that says that the longest side of a triangle must be more than the sum of the two other sides. The sum of lengths of any 2 rods in S is at most the sum of lengths of the 2 longest rods, so the shortest rod that can be added to S is the sum of the 2 longest rods in S.

To begin, we let S contain rods of length 1 and 2, because these are the two smallest length rods with different lengths. Then, we add 8 more rods by the above method. This follows the fibonacci sequence. So, the new rods are of lengths 3, 5, 8, ..., 55, 89. So the longest rod is of length 89.

Let $s = \{ a_1 , a_2 , \ldots , a_{10} \}$ be the lengths of the 10 rods, given in an increasing order. It is known that a triangle can be made by sides of lengths $a$ , $b$ and $c$ such that:

• $a + b > c$
• $a + c > b$
• $b + c > a$

It is obvious that the length of the longest side of a triangle has to be smaller than the sum of the lengths of the 2 others. In every triple that we chose from the set $s$ , there is a largest integer. If in one chosen triple, $a_n (3 \leq n \leq 10)$ is the largest integer, then the other two integers are $a_i$ and $a_j$ such that $1 \leq i, j < n$ (because $s$ is given in an increasing order, if $i > n$ (or $j > n$ ), then $a_i > a_n$ (or $a_j > a_n$ ), and this contradicts the statement that $a_n$ is largest integer in the chosen triple) and $i \neq j$ . This triple does not represent a triangle if and only if $a_n \geq a_i + a_j$ . The largest values we can get for $a_i$ and $a_j$ are $a_{n-1}$ and $a_{n-2}$ , so this leads us to the conclusion: $a_n \geq a_{n-1} + a_{n-2} ( 3 \leq n \leq 10 )$ . We need the smallest length for the longest rod, and that's why we choose $a_n = a_{n-1} + a_{n-2}$ . Because $s$ must contain different integers, we let $a_1=1$ and $a_2=2$ (the smallest possible different integer lengths for the rods). Finally, we get: $s = \{ 1, 2, 3, 5, 8, 13, 21, 34, 55, 89 \}$ . The smallest possible length of the largest rod is 89.

Note: The set $s$ resembles the Fibbonaci Series .

We try to find the length of each rod in the configuration where all the rods are as short as possible (i.e. It is impossible to substitute any element with a smaller one and get a valid sequence again). We note that if we have such a sequence with $n$ elements, in order to find such a sequence with $n+1$ elements we just need to add to the previous sequence another number that is as small as possible.

We will call $l_1$ the length of the shortest rod in this sequence, $l_2$ the second shortest and so on...

$l_1$ must be $1$ , as $1$ is the smallest positive integer. $l_2$ must be $2$ as it is the smallest positive integer different from $1$ .

To choose $l_3$ we will use the triangular inequality, that states:

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It is possible to construct a triangle of sides $a$ , $b$ and $c$ with $a \geq b\geq c$ iff: $a< b+c$

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therefore in order to be unable to construct a triangle of sides $l_1$ , $l_2$ and $l_3$ , we have to choose $l_3$ bigger or equal to $l_1+l_2$

Therefore the smallest $l_3$ that we can choose is: $3$ .

In the same way, if we take a sequence of $n$ different elements such that we are unable to construct any triangle with any three of them as side-length (and such that they are as small as possible, i.e. It is impossible to substitute any element with a smaller one and get a valid sequence again), and say that $a$ and $b$ are the biggest elements of the sequence, the smallest element that we can add to the sequence in order to be unable to construct any triangle with any three of them as side-length is $a+b$ .

Therefore we can define the sequence as:

$\begin{cases} l_1=1 \\ l_2=2 \\ l_n=l_{n-1}+l_{n-2} \end{cases}$ This is the Fibonacci sequence shifted by 1. Therefore the answer will be its $11$ -th element: $\fbox{89}$

The longest side of a nondegenerate triangle must be smaller than the sum of the other two sides.

The smallest possible integer length stick is 1 unit long; the second smallest possible integer length stick is 2 units long.

The next stick would have to be at least 2+1 = 3 units long to avoid forming a nondegenerate triangle. The next after that would have to be at least 2+3 = 5 units long.

Preceding this way (adding the two longest sticks in the set to obtain the next stick length) I made a chart, filling out ten rows. I find that the tenth stick must be 89.

The condition for a triangle to be formed is sum of any two sides should be greater than the third side, and the absolute difference between any two sides should be lesser than the third side. So, we need the integral values that disqualify the above mentioned conditions. That can be done if we follow fibonacci sequence which starts with 1, and Nth term = sum of N-1 & N-2th term. So our sequence would be 1, 1, 2, 3, 5.......and so on, but we also know that all rods have different integral values. In that case, we have this series: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89. So our answer would be 89! :)

The longest side of a triangle should be no more than the sum of the 2 other sides. It is still possible to make the longest side to be just a very little less of the sum, but if it is equal, then it is impossible to make triangle, for the longest side will now coincides with the other two.

Knowing that, the question wants us to find the shortest length possible. So we should start counting by assuming the 2 shortest rods from all, they should also be the smallest integer there are existed. They should be 1 and 2.

The third shortest rod should be smaller than the sum of 1 and 2 (=3) to make a triangle, but since we know from the question that we can't make any triangle, then it should be the opposite : larger of equal than the sum. But then since we want the shortest length possible, we should pick the possibility of "equal to the sum of the other 2 sides", which in this rod, is 1+2=3.

Based on the same logic above, the 4th shortest should be the sum of the previous two rods : 2+3=5. 5th shortest = 3+5=8. 6th shortest = 5+8=13. 7th shortest = 8+13=21. 8th shortest = 13+21=34. 9th shortest = 21+34=55. 10th shortest/longest = 34+55=89. (Extra note : After knowing the logic above, we can also conclude that it is following the Fibonacci pattern)

So, the shortest length possible for the longest rod should be 89.

An easy solution will be if we find a set of numbers(the lengths of the rods) such that for any 3 numbers from the set (a, b, c) this holds: c >= a+b and c > b > a (because if c>=a+b than the numbers can't be lengths of sides of a triangle). Using this condition we can simulate the set. In order to get the smallest numbers (thus minimising the biggest number) we start with the smallest positive integers (the shortest possible integer lengths): 1 & 2, then we add a number to the set which is equal to the 2 largest numbers from the existing set. simulation: set = {1, 2} add 3 = (1+2) set = {1, 2, 3 } add 5 = (2+3) set = {1, 2, 3, 5 } add 8 = (3+5) set = {1, 2, 3, 5, 8 } add 13 = (5+8), and so on until we get: set = {1, 2, 3, 5, 8, 13, 21, 34, 55, 89 } So there you have it, the answer to the question is 89 .

=>For the longest rod to have the shortest possible length, the shortest two rods must also have the shortest possible lengths which is 1 and 2 =>The length of each new rod = sum of the longest two rods we have, =>The length of the 3rd rod = 3 (2+1) =>The length of the 4th rod = 5 (3+2) =>The length of the 5th rod = 8 (5+3) =>The length of the 6th rod = 13 (8+5) =>The length of the 7th rod = 21 (13+8) =>The length of the 8th rod = 34 (21+13) =>The length of the 9th rod = 55 (34+21) =>The length of the 10th rod = 89 (55+34)

Let the lengths of the rods be Li ; i=1,2,3……,10 ; Li is an integer such that L1 < L2 < L3< ……… < L10

So, the minimum possible value of L1 = 1 and L2=2

Now for L1,L2&L3 to form a non-degenerate triangle, L1+L2>L3

But we know that no triangles can be formed with any 3 rods.

Therefore, L3<=3.

Since no 2 rods are of same length, min(L3)=3

Where min(x)=minimum value of x.

Similarly, L4 can be 5 or 4.

But, If L4=4, L4,L3&L2 will form a triangle.

Therefore min(L4)=5

Also, since L3,L2>L1 the possibility of L1,L2&L4 would already be ruled out.

Similarly, working out for L5, min(L5)=8

So we can summarize as,

For any Li, Li=[L(i-1)+L(i-2)]

So, we can see the lengths of the rods form a Fibonacci series(A series where a sum of 2 consecutive terms =3rd term).

Therefore , min(L10)=89.

NOTE: Li is NOT L * i but simply a variable name. If you still don’t get what a Fibonacci series is you can work out the answer directly by the logic.

Suppose there are 3 rods, The 3 rods would have to be of length 1,2,3. Let us add another rod, The fourth rod must be >= to 2+3, to minimize we choose the one =2+3 or 5 Similarly, we pick the fifth one to be 3+5 = 8 It will just yield us a fibonacci sequence missing the first 1, thus the 10th rod must be the 11th term of the fibonacci sequence which is 89

In order to the triangle to not exist, the sum of the smaller rods must be smaller than the third rod, then, in order to obtain the smaller possible length to the biggest rod:

1) length = 1 2) length = 2 ... A n) length = A (n-1) + A_(n-2) ... 10) length = 89

First notice that for any triangle with side lengths a,b and c, we have a+b>c, b+c>a, c+a>b by the Triangle Inequality Theorem.

To tackle this problem, obviously assume the first 2 rods are of length 1 and 2. Then the 3rd rod must have length less than or equal to 1+2=3. Since each rod has a different length, the 3rd rod has length 3.

To calculate the length of the 4th rod and above, we must apply the Triangle Inequality to the two largest lengths of the pool of rods whose length we have already determined to find an upper bound. If any other set of two lengths are used, a non-degenerate triangle can be formed by using the two largest lengths and the new length equal to the sum of the two lengths used. For example, for the 4th rod, we must use 2+3=5 to generate the 4th rod's length. If we use another pair of lengths, say 1+3=4, then 2,3,4 forms a triangle since 2+3>1+3=4.

Since the two largest lengths in the pool of rods are the last two that are calculated, we simply have the (n+2)th rod length = (n+1)th rod length + nth rod length.

This turns out to be the Fibonacci sequence, where the nth rod has length F(n+1) so the length of rod 10 is F11 = 89.

Choose 1 and 2 for the lengths of the two shortest rods. For each successive rod, you want its length to be equal to the sum of the lengths of the next two biggest rods, so that the triangle inequality a+b > c (which has to be satisfied by the side lengths a, b, and c of a triangle) isn't satisfied, and the rod's length is minimized. Turns out the lengths of the rods give you terms of the Fibonnacci sequence, defined recursively by F(n+2) = F(n+1) + F(n).

We take the first rod and give it the minimum possible value, 1. Then the next rod can be 2. Then the 3rd rod can be 1+2=3 to make a triangle of area 0. Then we continue as such, the lengths are 1, 2, 3, 5, 8, 13, 21, 34, 55, 89.

Given two lengths,a and b, if we don't want the third, length c to form a triangle, c has a lower bound of a+b. Since we want to minimize the longest rod, c=a+b. It is known that in the sequence of Fibonacci numbers starting from 1,2,3,5..., no sum of two numbers can exceed another number. So we start with 2 shortest possible values, 1 and 2. Continue to get the next rod of the smallest possible length by adding a and b. Then we get a series of Fibonacci numbers, 1,2,3,5,8...The 10th number is 89.

Lowest two positive integers are 1 and 2, To make a triangle with positive area the third rod requires a length greater than 1+2=3. If the length is 3 it will contribute a triangle with 0 area ( it is the minimum case). Thus the length of the rods with which the triangle of positive area cannot be made is just Fibonacci series.Thus the length of the rods are 1,2,3,5,8,13,21,34,55,89. Thus 10th number or the greatest no of these is 89

Since the rods have to have different integer lengths, the first two must be 1 and 2. Since the third side must be greater than or equal to their sum to not satisfy the triangle inequality, the third is 3. Following this procedure, the next is 2 + 3 = 5. At this point it is apparent we are dealing with a form of the Fibonacci sequence. Once realizing this, we see the first rod in our sequence is the second in the Fibonacci sequence, the second rod is the third, and so the nth rod is the (n+1)th Fibonacci number. So, the tenth rod is the eleventh Fibonacci number, being 89.