How it Works › Forums › Advanced Problems (AMC 8 Problems 20-25 and beyond) › Counting Cleverly Advanced Problems
This topic contains 5 replies, has 2 voices, and was last updated by Ayush 1 year, 4 months ago.
October 20, 2016 at 2:05 pm #382
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Hey Math Circle Students!
Here are the advanced problems for the next $3$ days! These are more difficult than AMC 8 Problems 15-20 but they fit the range of $20-25$ and some are even AMC 10 level. Each of these problems are worth $10$ points for your leaderboard score!
If you don’t understand one of the questions, feel free to post in the questions forum here
1. How many rectangles are present in a 5×5 chessboard. This includes rectangles of any size: 1×1, 1×2, …etc.
2. We color the vertices of an octagon with 4 different colors where there are 2 vertices of each color. What is the maximum number of diagonals of the octagon connecting two vertices of differing colors? What if this was a dodecagon (12 sides) instead?
3. In a $10$-by-$10$ rectangular grid, how many triangles are present in the grid? (Remember: The whole point is to count this cleverly)
Remember that you can post your solutions below to these problems and I will keep a running leaderboards for the top 10 students who participate the most! I especially appreciate the same problem being solved in different ways!
October 20, 2016 at 10:01 pm #393
- This topic was modified 1 year, 4 months ago by admin.
A 5 by 5 chess board would have 1 by 1, 1 by 2, …, 2 by 1, 2 by 2, … , 5 by 5.
1 * 1: 25
1 * 2: 20
1 * 3: 15
It’s going down 5 each time. So the numbers would be 25, 20, 15, 10, and 5 respectively.
It’s a different pattern for 2:
2 * 1: 20
2 * 2: 16
This pattern is going down 4 each time.
Then, the next goes down 3 each time, starting at 15.
The next goes down 2 each time, starting at 10.
The next goes down 1 each time, starting at 5.
If you add all these numbers up, you would get a final answer of
This response is written by Yash AgarwalOctober 21, 2016 at 2:24 pm #394
Great work Yash! You have been awarded 10 points for a terrific solution.
However, there is a much more efficient way to do this problem instead of just searching of patterns.
This is also a more valid proof, think about what determines a rectangle and count by that.October 23, 2016 at 7:23 am #399
I am not sure about the second one but I think it is 100 choose 3 because we can choose any three random points and join themOctober 23, 2016 at 7:48 am #400
1. recall about how many translation a side with a certain dimension can go through before it touches the border. for example a rectangle with side 1*2 can go 4 side or 5 up before it touches the border.to find the total we just multiply the two to get 4*5=20(20 1*2 rectangles)
case I – they are rectangles
since a 4*5 rectangle I the same as 5*4 rectangle in a square grid we just consider the two as one and then multiply two to the result
the quick way is by making a list of how much a dimension can go
since we just have to consider 10 dimensions and multiply the result by two we get 2(85)=170
case 2-they are squares
since there are 5(6)(11)/6 we get 55 possible squares
addin the two cases we get 55+170 =225 possible rectanglesOctober 25, 2016 at 6:35 am #406
Great work Pratyush! Excellent approach, but there may be a better way to say it mathematically. I will reveal the solution in 2 days so stay tuned.