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!
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.
社交网络