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 #6
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Jan 6, 2019
 #1
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1. Let the first term be a, and let the common difference be d. Then the four positive integers are a, a+d, a+2d, and a+3d. The sum of these four positive integers is 4a+6d=46, so 2a+3d=23. Solving for d, we find that =(23-2a)/3. The third term is a+2d=a+2(23-2a)/3=(46-a)/3. Thus, to maximize this expression, we minimize a, so a must be 1. Substituting 1 in, we find that the largest possible third term is 15.

 

2. 94 =F + 26D,  
27 =F + 93D, solve for F, D
F =120 - First term
D = -1 - Common difference

 

3.  As we begin, we note that no side of the triangle can exceed 29 units in length. Let’s start at the 20–20–20 triangle. We first count the number of triangles that have 2 sides of length less than 20. Let the third side be of length 20+k where 2≤k≤9. This we do by taking away k units in length from the first and the second sides, by subtracting at least one unit from each. For each k, We can do it in ⌊k/2⌋ ways. Thus, counting for each possible k, we have 1+1+2+2+3+3+4+4=201+1+2+2+3+3+4+4=20 cases.

Let’s now go back to the 20–20–20 triangle and count the number of cases where only one side has a length <20<20 units. We can take away 1 to 18 units from this one side and redistribute it to the other two such that no side exceeds 29 units in length. Let us take away k units from this one side, such that 1≤k≤9. In order for the triangles to be distinct, the redistribution can be done in ⌊k/2⌋+1⌊k/2⌋+1 ways. Then, for all possible values of k in this range, we have 1+2+2+3+3+4+4+5+5=29 cases. Now, let us consider the cases where 9
Finally, we add up all our cases:

1+20+29+25=751+20+29+25=75 distinct triangles. Also, subtract one because of the equilateral one.

 

4. I really can't think of another way other than listing.

Jan 5, 2019