Count the number of triangles in the picture above.
Hint: The answer is not 4. There's a faster way than simply counting them all!
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The technique of using combinations does often work in general, but specific problems often have special cases that need to be considered. You might consider the general method to be more of a application of the Rule of Product that accounts for duplicates with division (which is itself simply a generalization of combinations).
Suppose we want to count the number of rectangles on an 8 by 8 grid. We'll do this by picking two points that represent the opposite corners of a rectangle.
Since there are 9 × 9 = 8 1 corner points, we can naively suppose ( 2 8 1 ) = 6 4 8 0 different rectangles.
One issue, though, is we may have picked our corners on the same row and column! So really our first choice has 81 points, but our second needs to exclude not only the point picked but everything on the same row and column; this represents 8 points on the row, 8 points on the column, so 8 1 − 1 − 8 − 8 = 6 4 choices. This means we have 8 1 × 6 4 = 5 1 8 4 ways of picking the two corners.
However, also note we have duplicated each rectangle 4 times. Accounting for that brings our count to 5 1 8 4 / 4 = 1 2 9 6 rectangles, which is the correct answer.
Can you explain how the 9 x 9 corner points when you are counting the number of rectangles on an 8 by 8 grid?
Does this work for every problem like this?
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I think so. But in this problem there is only one layer of triangles but if you add more layers the formulation can become quite complicated. Post more problems like this then I can try again.
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What about this ?
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@Munem Shahriar – The answer is:
Every side of the big square gives 5 triangles, therefore, 4 sides give 5 × 4 = 2 0 .
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@Chew-Seong Cheong – Is this the same concept?
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@Munem Shahriar – Same concept. I can always say ( 1 4 ) × 5 = 2 0
hi, loved the method but could you please explain what the (5/2) means as well as how you got 5 * 4 and 2 * 1 Thanks
Neat solution!
Cool solution! You linked my first thought with combinatorics: ( 2 n ) is the n'th triangle number!
The best solution here is to count the triangles manually, and that is the easiest way. Using other methods (formulas) are complicated.
There is an easier way to count the rectangles with no fear of repetition: each rectangle is determined by two horizontal lines (lines are formed by edges of the small squares) to be chosen from the 9 you see, as well as two verticals. So the number of possibilities is 9C2 times 9C2 = 36 x 36 = 1296.
Here the total points are 6 out of which 5 are collinear and to form a triangle we need 3 points so total no of triangles formed is 6C3 (combination=C) but it also include the triangles formed by collinear points which cannot form a triangle but a straight line so we have to subtract 5C3 from 6C3.
Small triangles = 4
Triangles made by combining 2 triangles = 3
Triangles made by combining 3 triangles = 2
The big triangle = 1
Sum = 4 + 3 + 2 + 1 = 10
Answer: 10 triangles
I thought the small triangles could be 2 triangles since the angle you see them, that is why counted more.
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Can you explain what you meant better? Especially "since the angle you see them"? It is very unclear to me what you mean
there are 4 small triangles......................[from the left side--1, 2, 3, 4]
there is a 1 big triangle......................... [combining all(1+2+3+4)]
3.there are 3 triangles................................ [combining two------from the left side -(1+2), (2+3), (3+4)]
4.there are 2 triangles................................. [combining three---from the left side -(1+2+3), (2+3+4)]
so, there are--4+1+3+2=10 triangles
I can count 11 each time , am I counting the middle pair twice please?
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look carefully, then count the small triangles and then the biggest triangle.you have got already 5 triangles.
now, take 2 small triangles and count from left or right sides[as there are 4 small triangles you can count in this way---(1+2) ,(2+ 3) ,(3+4) but you can not count (1+3) or,(1+4) because that does not make any triangles]
now, take 3 small triangles and count in this same way.
at last take 4 small triangles together .that is the biggest triangles which you have taken before. so, there is no more than 10 triangles.
if it helps you then tell me ,please
fun...missed by 2....
I kind of agree with you but wouldn't it just be easier to add up the one side and multiply it by two. That's what I did and I got the right answer.
I'm dull, wrong app
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why?bro ,why?
Everyone's a beginner at some point in their life. Keep working on it! ;)
I think the answer is not 10 but seven since: there is a big triangle that's =1
:then 4 triangles inside the bigger =4
combining the two triangles from right =2
A total of seven triangles. Winnie Jelagat;Mpesa Foundation
Academy.
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you counted in the right way but forgot to count the 3 triangles combining two.
i mean, the big triangle........(you counted)
4 small triangles..............(you counted)
2 triangles from right or left combining three.......(you counted)
3 triangles from right or left combining two........(you didn't count)
if need any other query,let me help you.
Thank you for that; agree with you: Winnie Jelagat
But what if there are too many to count?
The amount of small numbers is 4 and you can work out how many it is quickly using 4!
1+2+3+4=10
didn't understand.
remember that 4! is the product of 4,3,2,1 and not the sum 4 ! = 4 × 3 × 2 × 1 = 2 4
I didn't got it.. 4! Isn't 10 man.. its 4×3×2×1=24 😲
Can you explain what you mean? What are these small numbers that you are talking about?
Chew-Seong's answer is better, but I think it can be solved recursively. (n is the number of points between the two vertices on the hypotenuse, n ≥ 0 )
f ( 0 ) = 1
f ( n ) = f ( n − 1 ) + n + 1
cool, and non-recursively with: f ( n ) = 2 n 2 + 3 n + 2 EDIT: I just learned from brilliant to use "explicit" instead of "non-recursive" :)
Fascinating!
Glad to see a generalisation. Maybe you could explain how you formed the recurrence?
In the following diagram, a triangle can be formed from any 3 of the 6 vertices.So it is the number of ways of choosing 3 vertices from 6 vertices.
But since 5 points lie on the same line(a triangle can't be formed by 3 points which lie on he same line), we should subtract the number of ways of choosing 3 vertices from 5 of those vertices.
So the answer is 6C3 - 5C3 = 20 - 10 = 10
It turns out that ( 3 6 ) − ( 3 5 ) = ( 2 5 ) . Is it in general true that, ( k n ) − ( k n − 1 ) = ( k − 1 n − 1 ) ?
@Agnishom Chattopadhyay It is true. https://en.wikipedia.org/wiki/Pascal%27s_rule
I don't have much of a solution, more of a remark: It's pretty neat that the number of triangles in that bigger triangulated triangle added up to 10, the fourth triangular number!
Neat observation indeed. Have you tried explaining this connection?
There are 6 relevant points in the figure, and 5 colinear points. Thus the number is 6C3 - 5C3 = 20 - 10 = 10
Since we know that a triangle requires 3 points to form but it should contain one left corner bottom point so we need to only select 2 points out of 5.Thus total ways of selecting 2 points (as one point is already selected) = 5C2 = 10
Five in red and 5 in orange
I think you wanted to attach a diagram.
There are 6 points out of which 5 points lie on the same line. So, the answer is 6C3 - 5C3.
I don't agree with the hint in this problem. The best way is to count the triangles.
Really? But what if there were 10 small triangles instead of 4 small triangles?
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Relevant wiki: Binomial Coefficient
We note that a triangle is formed with two points on the slanting side and the point at the left-hand bottom vertex.
Therefore, the problem is equivalent to finding the number of ways to choose two points out of five on the slanting side which is N = ( 2 5 ) = 2 × 1 5 × 4 = 1 0
Notation: ( k n ) = k ! ( n − k ) ! n ! denotes the binomial coefficient .