## Online Course Discussion Forum

### AMC 10 8 wk prep course Week 4 Homework

I need help with problems 9 and 10.

The only thing I've done is to draw out each of the problems.

For #9, from what I have drawn, the diagonals are perpendicular (maybe?) so I assume we would have to find the lengths of the diagonals, but again, I don't know where to start.

For #10, I have the smallest side as a, and the other sides as a+1 and a+2

We also know that for each n after the previous, the perimeter will be half the previous.

Also, for the triangle to not exist, for triangle inequality, a<1 (from a+a+1<a+2)

That's all I know, I need some help to move on from there.

For 9, we are given the diagonals are perpendicular. (Its diagonals AC¯¯¯¯¯¯¯¯AC¯ and BD¯¯¯¯¯¯¯¯BD¯ are perpendicular and intersect at point EE.)

Given AE, you can use Pythagorean Theorem to find the rest of the _E lengths (BE, CE, and DE).

We can then find the areas of each individual right triangle, and summing them up.

I also wanted to ask about #10.

Hey Charles, Happy New Year

Your hint on #9 helped me figure the problem out in 10 secs, thanks for that.

I figured out #10 after drawing out different triangles after one another; if you do this, you will notice that each triangle has 1/2 the perimeter of the previous. Also, using triangle inequality, and if you label the smallest side s, the only for the triangle to not exist would be if s+s+1<s+2. So, s<1: which means your smallest side is smaller than 1. Using this drawing out each triangle, assuming you start at n+1 (the triangle after the one with dimensions 2011,2012, and 2013) then you see that the perimeter is 3018/2^k, where k is the # of triangles after. After writing all the triangles out until s<1 (which took a while, but it is doable) then you should get your solution. Let me know what you got for an answer. Hope this helped!

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