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 and are perpendicular and intersect at point .)
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!
Hey Duy! Happy New Year :D
I followed what you said, and proved the intermediate steps (like why all T's are in the form of s, s+1, s+2, why T_n+1 has half the perimeter of T_n, yatta yatta yatta)
I got D in the end, finding that 1509/128 > 6 and 1509/256 < 6.