New York City Mathematics League Part 2

[Forte]

I thought you guys would be interested in giving these a shot. I'll post the senior division questions if I can get them.

1.) Matthew is choosing from the dimensions of tiles he will use. He can have three tiles. One with length 12, another with 10, and another with 15. The widths of the tiles will be 9,8,7. He can pair any length with any width. Compute the maximum total area of the tiles he chooses.
2.) Each student in a group of 12 math students is to be assigned into one of three teams. One team will have 3 students, another will have 4, and one will have 5. Find the number of such assignments.

3.) Find all ordered pairs (x,y) of positive integers such that a number, x, squared, minus a number, y, squared is equal to 55.
4.) Compute the square root of 20 plus the square root of 20, plus the square root of 20... and so on.

5.) The sum of 5 consecutive integers, all of which are greater than 1000, is divisible by 15. Compute the least possible sum of these 5 integers.
6.) In trapezoid ABCD with bases AB and CD, E is on AD, and F is on BC so that EF and AB are parallel. The area of ABFE and EFCD are equal. IF AB= 6, BC=9, CD=10, and AD=7, compute EF.

[Ravi B]

Which division are these from (Soph-Frosh, Junior, Senior B, Senior A)? Are they from this school year? If so, which round? Thanks.

[darkdieuguerre]

Problem 1

Problem 3

implies . There are two cases, with and with . The two solutions are and

Problem 4

Let be the given sum. Note that . Equivalently, , or . Since , .

Problem 5

Let the smallest integer be . We want to be divisible by . Therefore, modulo . Since , . The smallest possible sum:

[Forte]

Yes they are from this school year. These are the jr. division problems.

[Forte]

Click to reveal hidden content

Problem 2
From 12 students, for the first team we can make 12C3 arrangements. For the second team we can make 9C4 arrangements. And for the last team we can make 4C4 arrangements. 12C3= 220. 9C4= 126. And 4C4= 1. 220*126*1= 27,720 ways.

[Forte]

If anyone could provide a non-trig solution for number 6, that would be greatly appreciated.

[Ravi B]

Hint for Problem 6

Extend the two legs of the trapezoid to complete a triangle. Then compare the area of similar triangles.

[Forte]

Ravi can you post your solution? I get to the part with the area ratio. But, I cant figure out how we can use that to find EF.

[Ravi B]

Start of Solution for Problem 6

Let be the point at which and would meet when extended. Let . The areas of , , and are in the ratio . So the areas of and are in the ratio to .

[Forte]

I read somewhere a while ago that EF can be represented by the adding the squares of the bases, dividing by two, and taking the square root.

So for this question the bases are 6 and 10. 6²+10²= 136. 136/2= 68. Square root of 68 is 2√17. So EF is 2√17.

If someone can give me the name of the theorem that would be a great help.