'98% Fail' - How Many Triangles Are There? Viral Bollywood Puzzle

This problem caught the attention of Bollywood celebs, including Sonam Kapoor, Aditi Rao Hydari, and Diana Penty, after fashion magazine editor Jitesh Pillai tweeted it. How many triangles are there? This is not just a simple test of observational ability. This is a common test question in India, and there is a clever mathematical method to count the triangles quickly and accurately. In the video I explain and prove the formula.

My blog post for this video

How many triangles? When Bollywood celebs tried to solve a math puzzle. April 9, 2018 by Shreya Das.

Twitter moment how many triangles

How to count triangles in this shape trick (this is one of many similar videos that explains the pattern)

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[iverburl]

All triangles share the vertex on top. Look only at the lowest horizontal segment. It is the base of three small triangles (base = 1), two medium triangles (base = 2), and one large triangle (base = 3), for six total. Each additional horizontal segment is a new base, repeating this pattern to add 6 more. Three bases here, 18 triangles.

[ragnar]

4C2 × 3C1 = 18. A triangle can be made by choosing any two slanting lines and 1 horizontal line. So all we have to do is to choose 2 lines from the 4 slanting lines and 1 from the 3 horizontal lines.

[vissevolker]

Logic rules: Separate your counts into groups. For triangle shapes in the drawing there are 3 widths and 3 lengths. 1 Full length, 3 single length and 2 double. = 6 triangles per length X 3 different lengths = 18.

[calesr]

This triangle is just for the purpose of illustrating how to solve this problem mathematically, an actual test question could contain a triangle with more than 4 slanted lines and 3 horizontal lines (e.g. 25 horizontal and 26 slanted lined triangle for instance would take longer than 30 seconds to count).

[masked_mizuki]

"Its impossible for them to count these in a test situation."

... what

Dude its 18, counting that takes 30 seconds

[akhilpathak]

It's much easier to do it with combinatrics it's simply choosing any 2 of the 4 slant line giving us 4C2 and choosing any of the 3 horizontal lines giving us 3C1 hence our answer : *4C2×3C1=18*

[Nightcoffee]

Counting isn’t “insurmountable” it was 6 in the top section and the horizontal divisions don’t create more triangle ‘apexes’ so it’s just 6*3. That took me about 12 seconds.

[varadchiplunkar]

I remember solving such problems in school for a state scholarship exam. Many people would get the answer wrong or at the least need more time if you just made the bottom lines not parallel, some variations had the bottom lines intersecting to create more vertices. This was way befor you get taught the concept of a general solution in Indian classrooms. What's funny was we were asked to skip such time consuming questions to score more on the test. Feels great to revisit them with a greater understanding of math, alot more time and some nostalgia to boot. Thanks Presh 😊

[NetAndyCz]

This is super easy to count because there is a point which is shared by all the triangles.

[AnonimityAssured]

To those who insist that counting is faster and better, would you try counting if there were, say, 12 lines from the apex and 7 horizontal lines? It would be a lengthy and potentially error-prone process. In contrast, a mathematical approach would allow you to determine the total even if the figure were subdivided hundreds or thousands of times. Presh's specific solution here is perhaps not the most intuitive. If he had based his reasoning on either the number of lines emanating from the apex or the number of non-overlapping triangles (i.e., the smallest ones), he could have derived a slightly simpler formula. Nevertheless, his approach is, as always, systematic and effective, and he reminds us, for the umpteenth time, of a general piece of good advice: examine the simplest cases and find the pattern.

[codebeard]

First, observe every triangle must use the top vertex, one horizontal line and two lines from the vertex. There are three horizontal lines from which to choose, so (3 choose 1) = 3. There are four radial lines from which to choose, (4 choose 2) = 6. (3 choose 1) × (4 choose 2) = 3 × 6 = 18. Doesn't need a 5 minute video to explain.

[msalih]

You can calculate this with the method that take 2 vertical and 1 horizontal lines.
Which is
(4, 2) = 6
(3, 1) = 3

6x3 = 18

[pritamyawale]

in such problems, there's a pattern. Just count in how many parts these slanted lines inside a triangle are divided into and multiply it by 3 . So here two slanted lines inside a triangle are divided into 6 parts. So 6* 3 =18 triangles.

[WahranRai]

We could say that for 1 triangle we need 2 obliques and 1 horizontal.
The total is : (4 choose 2 = 6) pairs of oblique multiply by 3 horizontals = 18

[Reto]

In this version with two lines inside the big triangle, only, counting is easy. However counting would have been challenging, if there were 4, 5 or even more lines, which split the big triangle. In such cases one would appreciate the algebraic solution shown in this video. Thank you, Mr. Talwalker 👍

[StevenTorrey]

For those of us (me) in the slow group, is this how you set up the math problem: (6 + 1) (6 + 2) / 2? 6 being the number of slanted lines. But that give 28 trianagles. So how do you set up the math problem?

[LughSummerson]

0:49 – 0:59 Counts all the triangles in ten seconds.
1:24 – 4:41 Takes over three minutes to work it out mathematically.

Even if it takes just five seconds to remember and apply the general formula, you would have to encounter the problem 40 times in exams before it would be worth having done it that way.

[ymskillzz]

"It's impossible for people to count the amount of triangles in an exam..."


No it's not. It's really easy to count to 18.

[mudassiralvi]

I discovered much simpler formula :
No. Of triangles = m*n*(n+1)/2
Where
m = no. Of horizontal lines
n = no. Of vertical sections

In this shown problem
m=3 and n=3
That gives
No of triangles = 3*3*4/2 = 18 !!!!

[GermansEagle]

Using ideas from probability and permutations. Being A the left guy, B, the middle and C the right. You can't have nul, nor AC, so:
A, B, C = 2^3-2=6
AA, BB, CC=2^3-2=6
AAA, BBB, CCC=2^3-2=6

18