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8 solutions
Hey Zhou ZeHao I like ur question and way u solve its quite simple can put some of the more problems like this plzz. So that i can get a firm grip over these types of questions plz
great... thanks for the knowledge men...
I am up to 19
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i also counted 19 but i figured out my mistake, the bottom 3 is also the blue triangle.
answer should be 19 because the larger one should also to be counted
You counted the blue triangle twice.
I agree, but they don't want to count the big, all-encompassing one.
To everyone who is answering 19, The overall triangle is apart of this rule. Since none of the angles of the different levels of triangles change the proportions of the triangles must remain the same meaning that the areas of the triangles are increasing by a factor of three. Count each of the increments that come together to create a triangle. Start by only counting the triangles that are made up of one horizontal row of the smallest increments, there should be three triangles of increasing magnitude in each of the three rows, then when you have finished, count the triangles that are made up with two horizontal rows of the smallest increments, there should be three triangles of increasing magnitude in each of the two overlapping rows, then count the number of triangles when you consider the triangles made of three horizontal rows of the smallest increments. The overall triangle is not left out as it is made of three sections of three rows of triangles made of the smallest increments.
i didn't get it..please help me out any 1
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In total there are 4 vertical & 3 horizontal lines. Triangle needs three (mutually) intersecting lines. If we look in the given figure, the only way to make a triangle is by selecting 2 vertical lines and one horizontal line. Hence choosing 2 out of 4 vertical lines & then selecting 1 out of 3 horizontal lines = ( 2 4 ) × ( 1 3 )
The overall triangle is apart of this rule. Since none of the angles of the different levels of triangles change the proportions of the triangles must remain the same meaning that the areas of the triangles are increasing by a factor of three. Count each of the increments that come together to create a triangle. Start by only counting the triangles that are made up of one horizontal row of the smallest increments, there should be three triangles of increasing magnitude in each of the three rows, then when you have finished, count the triangles that are made up with two horizontal rows of the smallest increments, there should be three triangles of increasing magnitude in each of the two overlapping rows, then count the number of triangles when you consider the triangles made of three horizontal rows of the smallest increments. The overall triangle is not left out as it is made of three sections of three rows of triangles made of the smallest increments.
There are three sections of triangles, the three small ones, two overlapping medium sized ones and one large one. Each of the six triangles in each of the three sections are increasing in area by a factor of three. For each of the three triangles in the first section equaling 1/3 of the overall triangle, there are three increments of magnitude, this means that there are six triangles in the first section of the smallest triangle.
In the second section of triangles containing two overlapping medium sized triangles with an area equaling 2/3 of the overall triangle, there are three increments of magnitude meaning that there are six triangles in the second section of triangles.
In the third section of triangles containing one large triangle with an area of 3/3, there are three sections of magnitude. The overall triangle is the third section of magnitude.
How would you generalize this question? What if there were n horizontal lines and m vertically diagonal lines?
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I tried and derived one formula and it works. Let m be the vertical lines (including left and right sides of the triangle). & n be the horizontal lines (including the base).
Then, the formula to calculate number of triangles in figures of given type is
m * ( m -1 ) * n / 2
Putting the values m = 4 & n = 3 as in above question we can get answer 18
I would suggest mC2 n , given that n is the number of horizontal line and m is the number of vertically diagonal line. In this case, 4C2 3. Which yield 18.
first count triangles with no division then with two divisions ans so on till you get triangle with maximum divisions this makes it very simple
answer is 18; to further illustrate the solution, if below are the points of the triangle; A B C D E F G H I J K L M
count them as always referring to the tip which is triangle A; to enumerate, triangles are below; ABC, ACD, ADE, ABD, ACE, ABE = 6 triangles do this with the other 2 bases and will give an answer of 6 x 3 = 18.
m n (m-1)*(n-1)/4
There is a pattern if you look at it as triangles just getting bigger, then at every level, there is the following formula 1+3+2....and then its a matter of how many levels you have total 3 bases 3 levels
I have counted them and still only get 15.
Why can I only see four triangles?
I count 19..
3h(v+1) V= is the number of vertical lines & H=for the horizontal lines
Vertically we have three triangles and they are triplets. So if you just count how many triangles we have in the top smallest triangle, then you can see that has 6. 6 times 3 equal 18.
Better count it.
1. Select the uper portion of the triangle (as the triangle is divided into 3 part horizontaly).
2. Count the number of triangle in that portion, it will be equal to 6.
1. whole protion will give total 1 triangle
2. this is agian divided vertically into 3 part, so we got total of 3 triangle.
3. will cant get two differnt triangle by combining left and middle
portion and by middle and right portion.
3. Add the (1 + 3 + 2) = 6
The divide a triangle vertically into 3(A,B,C) Triangles 1.A 2.B 3.C 4.AB 5.BC 6.ABC
Vertically divided into 3(X,Y,Z) So 6X3 = 18
What about the big one?
a triangle need 3 sides, 1.)fixed one line(blue) on either side (one side fixed), for other two sides (green ): -> each parallel horizontal line and one of the converging vertical line yield one triangle. 2. )now fix two green vertical line as two fixed sides, now each (green) horizontal parallel line gives a triangle. Finally add the bigger blue triangle
You can say that a triangle cut with x lines going from side A to vertex a creates (x+1)+(x+0)+(x-1)+(x-2)...+1=y triangles as a general statement, in this case simply 3+2+1=6 triangles, and then for each "horizontally" (assuming the others to be primarily vertical) cutting line (a line that becomes another possible base for a triangle where two of its edges are the outermost edges of the "original" triangle) you do y*b=z where b is the number of bases and z is your total number of triangles! While this all may sound more complicated than simply counting, it does provide a general solution if there were many more triangles. And it is quite quick, so we quickly find that here we have (3+2+1)(3)=18 triangles.
Choose two of four vertical diagonal lines. This can be done in ( 2 4 ) ways. Now, for each choice of these two lines, there are exactly 3 choices for the horizontal line for its base giving ( 2 4 ) ⋅ 3 = 1 8 triangles.