How Many Triangles?
How many triangles are there in the figure above?
The answer is 47.
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6 solutions
Moderator note:
There's a lot of temptation on this kind of problem to list all the solutions and be done with it, but since the sizes of triangles don't follow an easy-to-order pattern (like in a find-the-square problem where each square must have a set side length), a complete solution really does need to establish all the solutions have been found.
How do we make sure that we have found all types of triangles and haven't missed any?
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This is a very important question.
I started with the simplest triangles, those not intersected by any lines. Then I went to triangles with one line through them (made out of two of the simple triangles). The way I did that was for each type of simple triangle I would take a representative and see what it can make with any one of its neighbors. Some of those shapes turned out not to be triangles, but some were. The same way you can go to two neighbors (and possibly neighbors of neighbors), three, etc. It gets more complicated for larger numbers and harder to keep track of.
You can go at it from the other side as well, looking at lines and trying to see what triangles they are part of. This approach is a bit easier to formalize, so I added that to the solution.
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Oh, these are nice ways to count them all systematically.
While using the first method, we would have to consider trapeziums as well (although there are none in this problem) because trapezium + triangle can also be a triangle. I found the second method easier to implement, especially when there are triangles made of 3 or more triangles. I also liked Dan's method of keeping track of triangles using vertices.
Consider each point, and all of the possibilities of selecting 2 other points from that given point. Group together those which are geometrically the same, like Marta has done. Takes a while:)
The way I established that I do have all the triangles was to list all the segments and find all triangles which have that segment as a side. Additional check came from the fact that those triangles made of sides of different types each showed up three times in this listing, and those with two side types twice. Only two triangles, those made of a single type of sides, showed up only once.
It is too late to post a solution.
It is confusing to look at the diagram to count but simple to make a sure count.
Label all the vertices 1..10. List all the triples (x,y,z) that are triangles and x < y < z (a straight forward directed search procedure) Count them. Done.
This set of triples covers all triangles so no triangle is omitted. Since they are ordered, none can be duplicated. (A duplicate would have the same three numbers, but one would be out of order.)
A graphic for this on my Facebook page. https://www.facebook.com/photo.php?fbid=1350873644958196&set=a.1133769940001902.1073741832.100001067207093&type=3&theater
I am a bit surprised at the 5% pass rate, the lowest I have ever seen here, considering there are 3 tries.
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Yes, this is a sure way of counting all triangles. Did you mean "consider all un ordered triples" instead of "ordered triples"? Because there are only 120 unordered triples compared to 720 ordered triples, and considering all ordered triples would take way too much time.
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Have I cleaned it up enough? (The original version was meant settle the question comprehensively in theory, in practice one searches for the triangles starting with 1,2,? 1,3,? etc (in nested loops in a program version), and never really looks at all the triples.) This puzzle amounts to a special case of the problem of finding all distinct 3-circuits in a connected graph. The procedure above may be tedious but it doesn't get cross-eyed looking at a picture. There is probably a more optimized example somewhere. As described above it can be applied to any triangle counting puzzle one is likely to see on the internet. In the same Facebook album there is another example that does yield in a general way to a rather awkward but generalized summation scheme. If there is a cleaner formula on the example below that I failed to discover, let me know.
https://www.facebook.com/photo.php?fbid=990978260947738&set=a.990978234281074.1073741830.100001067207093&type=3&theater
I think, 12 should be another valid answer, as the question leaves open whether it refers to only the basic pieces or to all compound triangles, too. Of course the former is a pretty boring question, nonetheless a possible interpretation of the given question. Suggestion for a maybe better wording: "How many, possibly overlapping, triangles are there in the figure above?" This would make it unambiguous.
Very visually pleasing set up:)
Is there is any formula to get the answer
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Not that I know of. But there are some regularities. For example, those triangles made out of three lines all of which are of the same type appear only once. Those made of segments two of which are of the same type appear three times. Those made out of three types of segments appear six times.
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This example seems a bit of a "one-off". One triangle, all the midpoints and corners connected, and the bisecting lines from the corners meet in the center to provide the final corner for some more triangles. What do you propose as the general case here?
@ Brilliant - This was put in the wrong category! Based on the sheer number of types of triangles and of the total triangles to keep straight, this problem should have been placed in the "Intermediate" category.
Consider the points marked in red. Each pair of red points are connected by a line. With some exceptions, any three red points chosen form a triangle. This gives ( 3 7 ) = 3 5 triangles. However, they will not form a triangle when colinear, so we subtract 6 to get 2 9 triangles.
Now, consider the points in green. If a green point is a vertex of a triangle, then the triangle must be right. This makes it easy to determine that for each green point, there are 6 triangles. This gives 3 ∗ 6 = 1 8 triangles.
2 9 + 1 8 = 4 7
Great solution, Bryan. I'll add some explanation on how each green points gives 6 triangles in case it's not clear to someone. Consider the green point in the middle, vertical line. We can choose any of the three red points on the vertical line to be another endpoint of the triangle. Finally, we have to choose one of the two red points that connect to it via the horizontal line. The first part gives you 3 choices and each has 2 more cases. Therefore, each green point gives 6 triangles.
A bit tedious, but principled. There are ten points that can potentially form triangles, ( 3 1 0 ) = 1 2 0 point triplets in all. Not all of them can form triangles, of course. There are 18 triplets of points that lie on the same line (3 for the big triangle, 3 for the small triangle inside, and 3 × ( 3 4 ) for three bisectors, which leaves us with 1 2 0 − 1 8 = 1 0 2 triplets. Now we can think of the missing segments. We see that the points on the sides of the small triangle are not connected to each other and to two vertices of the big triangle each. That leaves us with 9 missing segments (2 for each vertex of the big triangle, and additional 3 for the middle points). Every missing segment produces 8 superfluous triplet points, therefore we should subtract 9 × 8 . However, if we do this, we will overcount since some of superfluous triangles contain two missing segments at once, and some consist of them entirely. Using the inclusion-exclusion principle we substract 72 then add 21 (triangles having two missing segments) and subtract 4 (triangles consisting entirely of missing segments). ( 3 1 0 ) − 1 8 − 7 2 + 2 1 − 4 = 4 7 .
What I did was to enumerate the types of node and the possible combinations of three of them, and then determine how many possible triangles there are of each.
The four types of node are: V (vertex), M (midpoint of side), X (the points at which two lines intersect at right angles) and C (centre). There are 20 possible raw combinations of the four letters. We can eliminate 4 of these since there's only one centre, and a further 4 since no two X points are connected, and furthermore VVX and VXC are impossible. For each of the remaining combinations, we can readily see how many triangles are possible.
They are VVV × 1, VVM × 6, VVC × 3, VMM × 9, VMX × 6, VMC × 6, MMM × 1, MMX × 6, MMC × 3, MXC × 6, giving 4 7 in total.
The image above shows the answer. What's left is to show that there is no other answer possible. One way to do that is to draw all the possible triangles for each of the segments. There are three types of lines. Medians have three segments each giving us 6 combinations to explore, sides have two identical segments with two total segments to check out, and the remaining line has two identical segments as well. Segments being investigated are drawn in black below, with each possible triangle drawn in color. If there are two triangles possible of the same type, only one is shown. There are obviously overlap between the drawings (the same triangle, or one of the same type (shown in the same color), will be produced from up to three different sides), but the purpose is to demonstrate that all the solutions have been found.