Triangle Friangle

How many triangles are there?????????


The answer is 64.

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5 solutions

Let us name the 2 base points P and Q

There are 5 lines generating from point P : { P 1 , P 2 , P 3 , P 4 , P 5 P_{1}, P_{2}, P_{3}, P_{4}, P_{5} }

Similarly, there are 5 lines generating from point Q : { Q 1 , Q 2 , Q 3 , Q 4 , Q 5 Q_{1}, Q_{2}, Q_{3}, Q_{4}, Q_{5} }

Now, out of these two sets of lines one line is common to both the sets. Let us say, they are { P 5 P_{5} or, Q 5 Q_{5} }

If we select any 2 lines from the set of { P k P_{k} } and 1 line from set of { Q k Q_{k} }, we will form a triangle

This can be done in 5 C 2 × 4 C 1 = 40 w a y s 5C2 \times 4C1 = 40 ways

Also, If we select any 2 lines from the set of { Q k Q_{k} } and 1 line from set of { P k P_{k} }, we will form a triangle

This can also be done in 5 C 2 × 4 C 1 = 40 w a y s 5C2 \times 4C1 = 40 ways

But, there are few triangles which are counted double.

As, P 5 P_{5} and Q 5 Q_{5} are essentially the same line, so when we select

  • P 5 P_{5} or Q 5 Q_{5}
  • 1 line from the other 4 lines from the set of { P k P_{k} }
  • 1 line from the other 4 lines from the set of { Q k Q_{k} }

These triangles have been counted in either of the 40 cases

Such triangles can be formed in 4 C 1 × 4 C 1 = 16 w a y s 4C1 \times 4C1 = 16 ways

Total ways = 40 + 40 16 = 64 40 + 40 - 16 = \boxed{64}

I can give another explanation to this:

There are 9 lines here ... Selecting one of them in 9C3 ways

But, concurrent lines will not form triangles

two sets of 5 concurrent lines:

Subtracting those cases:

9 C 3 ( 2 × 5 C 3 ) = 64 9C3 - (2 \times 5C3) = 64

Soumya Chakraborty - 7 years, 3 months ago

if you take 5c2 x 4c1= 40. then why 4c1 x 4c1. I think 5c2 x 4c1= 40 plus p and 5c2 x 4c1= 40 for q and sub 4c1. 40 + 40 - 4 =76.

A Former Brilliant Member - 7 years, 3 months ago

a little bit complicated

Jeet Dubey - 7 years, 3 months ago

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I mean explaining it was complicated ... But the process is pretty straight forward.

Soumya Chakraborty - 7 years, 3 months ago

I counted it manually. I have started from the left corner, I counted 40 triangles so I assumed that from the right corner it would also be 40... then I notice that there are triangles that is similar on both points.... So 40+40-similarities... The problem is I just saw four similarities so my answer was 76.

Adrian Hansen - 7 years, 3 months ago

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hai i also counted manually and got 76

A Former Brilliant Member - 7 years, 3 months ago

no of similarities = no of vertices 4x4 which is 16

Soumya Sharma - 7 years, 3 months ago

I have a doubt. please dont mistake me.why 5c2 4C1. It should be 5c2 5c1.

A Former Brilliant Member - 7 years, 3 months ago

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Can't be ... you can not take 3 lines from the set of P to get a triangle ... When you say 5 C 2 × 5 C 1 5C2 \times 5C1 ... You are counting the cases where you select the base line in both the scenarios

Soumya Chakraborty - 7 years, 3 months ago

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please can you use the same method for calculating number of triangles using 6 lines. without the middle 3 lines emerging from point p.

A Former Brilliant Member - 7 years, 3 months ago

5c2 x 5c1 insted of 5c2 x 4c1

A Former Brilliant Member - 7 years, 3 months ago

dot all the point of intersections,,,, now count no. of all lines originating from them , add them, , subtract it from 5 (common sense , in a triangle three point of intersection also called vertex, and total six lines originating from them i.e two from each vertex therefore we subtract 5 from it since we know there is one triangle , therefore doing the same for this one and got it right) THINK ANALOGOUS , SMALL THINGS BUILD BIG ONE.

Shahnawaz Iqubali - 7 years, 3 months ago

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Makes no sense ... Lot of assumptions ... and absolutely illogical ...

Using you method, how many triangles are there in a tetrahedron ??

Soumya Chakraborty - 7 years, 3 months ago

64 triangles

Rishikesh Bahety - 7 years, 3 months ago

can you kindly explain what you meant by the equation 5C2x4C1. what does C stand for?

nawal hussain - 7 years, 3 months ago
Vaibhav Agarwal
Feb 25, 2014

Let the triangle be ABC with the vertex being A

First take BC and see How many triangles are formed. You will get the no. to be 16

Then form the points on AB and vertex at C, 6 triangles will be formed. Go on moving inside and you will get 24 triangles in total. Similarly 24 for the other side (AC)

Therefore total no. of triangles 16+24+24=64.

2X4X(5C2)-4X4=64

Subhrajyoti Sinha - 7 years, 3 months ago

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can u explain howzz its done ?

g j - 7 years, 3 months ago

Damn! I'm in the 90% by a small margin! :D Somehow I counted up to 62 but couldn't get 2 more! Anyways, nice question. ;)

Kou$htav Chakrabarty - 7 years, 3 months ago

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Thank You Kou$htav Chakrabarty

Shubh Agarwal - 7 years, 3 months ago
Gaurav Sharma
Mar 4, 2014

there are two types of lines in the diagram . one originating from left-down vertex B and other originating from right-down vertex C. Base line BC is common to the both types. We can take it to be in any type. Now , a triangle is formed by taking either two lines from left side and one from right side or one from left side and two from right side. But we cannot make any triangle by taking single BC line from left side. So number of triangles with single line from left side and any combination of two lines out of 4 lines from right side is equal to 4 * (4c2) .= 24 Number of triangles with any combination of two lines out of 5 lines from left and one line from right is (5c2)*4= 40 So total triangles is 64

can some one count the number of triangles from the picture. i get total of 76. sorry if it is wrong.

A Former Brilliant Member - 7 years, 3 months ago

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that's my answer too. Manually counting is more accurate, I believe.

Adrian Hansen - 7 years, 3 months ago

I belive the answer given is wrong.

A Former Brilliant Member - 7 years, 3 months ago

count the whole triangle manually don't just assume

Shubh Agarwal - 7 years, 3 months ago

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No need for counting . Any three lines on a plane which are not concurrent , make a triangle. If you have counted 76 triangles, name them and reply.

Gaurav Sharma - 7 years, 3 months ago

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i have counted 76 triangles. I just can't why there are 16 similarities where I only counted 4. Counting manually is , i think, more accurate.

Adrian Hansen - 7 years, 3 months ago

Ohh... apologies. I saw the other 12 triangles... you're right...

Adrian Hansen - 7 years, 3 months ago

If you believe that the similarities is 16, then can you please point where they are?

Adrian Hansen - 7 years, 3 months ago

I saw it already... you're right

Adrian Hansen - 7 years, 3 months ago
Jimit Vyas
Mar 10, 2014

64 is the ans

Aravind Raj
Mar 4, 2014

Total number of triangles in this picture=9C3-5c3-5c3 =84-10-10=64

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