Ding! AMC MATH Q

Discus: SAT/ACT Tests and Test Preparation: June 2004 Archive: Ding! AMC MATH Q
 By Loop123 (Loop123) on Thursday, June 17, 2004 - 06:44 pm: Edit

Given the nine-sided regular polygon A1A2A3A4A5A6A7A8A9, how many distinct equilateral triangles in the plane of the polygon have at least two vertices in the set {A1,A2,...,A9}?

First to answer correctly gets ten points!

 By Caddyshack (Caddyshack) on Thursday, June 17, 2004 - 07:01 pm: Edit

3+9+9= 21?

 By Daromanian (Daromanian) on Thursday, June 17, 2004 - 08:39 pm: Edit

72?

 By Jbaby262 (Jbaby262) on Thursday, June 17, 2004 - 08:51 pm: Edit

um...i think 66

 By Optimizerdad (Optimizerdad) on Thursday, June 17, 2004 - 09:06 pm: Edit

Let x denote a vertex that doesn't belong to {A1,A2,..,A9}

2*9 triangles of the type A1-A2-x, A2-A3-x, ..., A9-A1-x; the '2*' comes from the fact that x can lie within the polygon or (mirror image) outside it, and

2*9 triangles of the type A1-A3-x, A2-A4-x, ..., A9-A2-x, and

6 triangles A1-A4-A7, A1-A4-x, A2-A5-A8, A2-A5-x, A3-A6-A9, A3-A6-x, and

2*9 triangles of the type A1-A5-x, A2-A6-x, ..., A9-A4-x, and

2*9 triangles of the type A1-A6-x, A2-A7-x, ..., A9-A5-x

Total: 2*9 + 2*9 + 6 + 2*9 + 2*9 = 78

 By Optimizerdad (Optimizerdad) on Thursday, June 17, 2004 - 09:11 pm: Edit

Nah, correct that; the 6 triangles above should really be 4+4+4 (e.g. A1-A4-A7, A1-A4-x, A4-A7-x,A7-A1-x plus 2 other quadruples).

Answer is, I think, 18+18 +12 + 18+18 = 84

 By Loop123 (Loop123) on Thursday, June 17, 2004 - 09:40 pm: Edit

BING! Jbaby got it, congrats. That was a tough problem. I was fooled by the answer 72. My logic was that any one of the 9 points could connect to 8 others, and then any of the remaining could connect to 7 others, etc. That would make 8+7+6+5+4+3+2+1 = 36 triangles. Double that to account for the mirror images of the triangles and it = 72. Apparently your method actually works however (lol). Share your method for an additional 10 points!!!

 By Etsrep78328 (Etsrep78328) on Friday, June 18, 2004 - 07:54 am: Edit

Find the positive integer B for which logB(2/3) + logB(3/4) + logB(4/5) + ... + logB(31249/31250) = -6.

Note: logB(x) means log base B of x -- since I cannot format the text on this post for subscripts.

 By Optimizerdad (Optimizerdad) on Friday, June 18, 2004 - 08:05 am: Edit

Doesn't that simplify to logB(2) - logB(31250)=-6
or logB(2/31250) = -6
B^(-6) = 2/31250 i.e. B^6 = 31250/2 = 15625
so B=5

True?

 By Optimizerdad (Optimizerdad) on Friday, June 18, 2004 - 08:09 am: Edit

Ah, I messed up in my calculation of 84 triangles. I think one of those groups, the last 2*9, is a double-count. Answer then works out to
18+18 + 12 +18 = 66.

Loop123, your explanation is cleaner and it works as well. Just remember that 6 of the 72 are double counts, remove 'em, and you are home free.

 By Loop123 (Loop123) on Friday, June 18, 2004 - 12:16 pm: Edit

Thanks Optimizerdad, but how exactly would I know during the test to remove the six double-counts? Looking at a diagram of a regular nonagon, it seemed to me that each of the triangles created was unique. Each triangle would have a different base, and their reflections would be off of that same unique base, thus making them different. I'm sure you're right, but could you please explain to me just for future reference?

 By Optimizerdad (Optimizerdad) on Friday, June 18, 2004 - 12:53 pm: Edit

Bear with me if the explanation is a trifle long-winded; I tend to think out aloud, and my memory ain't what it used to be :-)

1. We can construct a regular 9_sided polygon from three equilateral triangles A1-A4-A7, A2-A5-A8 and A3-A6-A9. Imagine these triangles stacked one on top of the other, then rotate the A2 triangle clockwise by (360/9) = 40 degrees, and the A3 triangle clockwise by 80 degrees. The points A1,A2,...A9 should then describe the vertices of our 9_sided polygon.

2. You can generate 8 triangles A1-A2-x, A1-A3-x,A1-A4-A7 (important!), A1-A5-x, A1-A6-x, A1-A7-A4 (discard, clone of A1-A4-A7), A1-A8-x, A1-A9-x.
So we get 7 distinct triangles using A1; multiply by 2 for mirror images, to get 14 distinct triangles.
Similarly, as you generate triangles A2-A3-x, A2-A4-x and so on, you'll run into other clone triangles you will need to discard. We discarded 1*2 from the A1... triangles; I believe there will be two other instances like this. You'll finish up with 2*(8+7+6+5+4+3+2+1) - (2+2+2) = 66.

 By Loop123 (Loop123) on Friday, June 18, 2004 - 01:31 pm: Edit

Thanks Optimizerdad, I stupidly failed to realize that some of the triangles would have the third vertex on the nonagon. I guess I should've paid more attention to the part in the quesiton where it says "how many distinct equilateral triangles in the plane of the polygon have AT LEAST two vertices". Thanks again Optimizerdad.

 By Loop123 (Loop123) on Friday, June 18, 2004 - 01:33 pm: Edit

Thanks Optimizerdad, I stupidly failed to realize that some of the triangles would have the third vertex on the nonagon. I guess I should've paid more attention to the part in the quesiton where it says "how many distinct equilateral triangles in the plane of the polygon have AT LEAST two vertices". Thanks again Optimizerdad.

 By Forsakn4 (Forsakn4) on Friday, June 18, 2004 - 01:36 pm: Edit

what is the amc
is it good to take it to report for colleges?