The problem is as follows:

In the following sequence of figures. Which is the total number of right trapezoids which do make up $\textrm{figure 10}$?

The figure is below:

Sketch of the problem

The alternatives given are:

  • 66
  • 60
  • 56
  • 64
  • 72

What I tried to do is to account for the number of trapezoids:

$\textrm{In figure 1:}$

  • There is just $1$

$\textrm{In figure 2:}$

This is some tricky:

I thought that there are $4$

  • $2$ make up a small individual unit

  • $1$ accounts for the other two

  • $1$ accounts for the all of them

$\textrm{In figure 3:}$

  • $3$ are consisting of only one figure.

  • $2$ are two figures.

  • $1$ has three figures.

  • $1$ accounts for the three trapezoids in horizontal arrangement and the first oblique bar next to them.

  • $1$ accounts for the previous set plus the second oblique bar next to them.

Therefore in the figure are $8$

$\textrm{In figure 4:}$

  • $4$ are sets of 1 figure.

  • $3$ are sets of 2 figures.

  • $2$ are sets of 3 figures.

  • $1$ is a set of 4 figures.

  • $1$ is a set of the previous 4 figures in horizontal direction plus the first oblique bar appearing next to the trapezoids.

  • $1$ is the previous set and the second oblique bar.

  • $1$ is the previous set and the third oblique bar.

This accounts for $13$ figures.

So in total the series is comprised of.

$\textrm{1, 4, 8, 13,...}$

From this series I found that the recursion formula is stated as:


By replacing with the fourth term it seems to check.

$$T_{4}=\frac{1}{2}\left( 4 \right )^{2}+\frac{3}{2}\left( 4 \right )-1=8+6-1= 13$$

So if what it is being asked is figure number $10$ then by replacing in the previous equation:

$$T_{10}=\frac{1}{2}\left( 10 \right )^{2}+\frac{3}{2}\left( 10 \right )-1=50+15-1= 64$$

Therefore I would choose $64$ as the number of figures in the $\textrm{10th figure}$. I'm not very certain if is it correct?

The answer appears within the alternatives but the process to find the figures was very tedious and it took me a long time to find it, needless to say that to find the recursive formula was another problem as well but eased due the fact which I had some idea of how to find it, as a result.

Is there any other method to speed up the calculations or to find an answer in this situation or to avoid miscounting or double counting the figures?


This looks correct to me. As for the more general question, I would approach the problem by asking 'how can I make up a right trapezoid from these figures?' This breaks naturally into two pieces: trapezoids that use the 'slanting' shapes on the right of the figure, and ones that don't.

If I'm going to use any of the shapes on the right then it's clear I have to use all the ones on the left, and I have to use the ones on the right from left to right; thus, there are as many ways of doing this as there are trapezoids on the right of the figure, i.e. $n-1$.

On the other hand, if I'm using on the sub-trapezoids on the left, then it has to be a contiguous block of them (e.g., the second through the fourth would be valid, but the first and third would not), but any such contiguous block will do; you should be able to see that the number of contiguous blocks is the same as the number of ways of choosing a number $i$ (the 'bottom block') between 1 and $n$ and a number $j$ (the 'top block') between $1$ and $i$; there are exactly $n(n+1)/2$ of these.

So summing up, I get a total of $n(n+1)/2+n-1 = \frac12n^2+\frac32n-1$ possible trapezoids, matching your answer.

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  • $\begingroup$ First of when I solved this problem I totally ignored the fact that the figures on the right side were also trapezoids but I was "saved" by the fact the problem asked about counting right trapezoids. Since I'm not good with geometry, do those are also trapezoids?. They look like rectangles which had been tilted to one of their sides, anyway. I'm still lost at where does this $n-1$ comes from. Why not $n-2$?. Most of what I did was by let's say brute force or intuition not really paying attention on details, hence why It took me hours to decode this thing. $\endgroup$ – Chris Steinbeck Bell Apr 6 '18 at 17:48
  • $\begingroup$ The paragraph where you explain about the subtrapezoid thing I'm stuck at where you mention about $\frac{n(n+1)}{2}$ How do I get to this?. All I know it is that it looks the sum from integers from $1$ to $n$ does this has something to do with the problem?. Maybe if you could rework the explanation a bit better I could understand. $\endgroup$ – Chris Steinbeck Bell Apr 6 '18 at 17:48

I think you got it.

Counting the stack of $n$ trapezoids in the left part of each diagram is the $n$th triangular number -- normally denoted as $T_n$ but I'll denote it $t_n$ so as not to clash with your notation. The tenth triangular number $t_{10} = 55.$

As additional explanation, let's look at the fourth diagram in your question. Let's label the stack of short, fat trapezoids (looks like a geometrical stack of pancakes) on the left part of that diagram $A,B,C,D$ from top to bottom.

You have $4$ trapezoids with the topmost trapezoid at the top ($A$ by itself, $A+B$, $A+B+C$, and $A+B+C+D$). You'd have $3$ trapezoids with the second-highest trapezoid on top ($B$ by itself, $B+C$, and $B+C+D$). And so on. So the total number of trapezoids, of any height in this stack, is $4+3+2+1 = 10$, which is the fourth triangular number.

Similarly, in the tenth diagram, it would be $10 + 9 + 8 + ... + 2 + 1 = t_{10} = 55$.

A different but completely equivalent way to look at it is that there is $1$ trapezoid that is stacked $10$ high, $2$ trapezoids that are stacked $9$ high, $3$ that are stacked $8$ high ... and $10$ that are stacked $1$ high. It still adds up to $55$.

Since the side pieces extend only the trapezoid that represents the entire stack, you add $n-1$ of those. (From the diagrams you have, the third diagram has two additional trapezoids from the side pieces, hence the $n-1$ formula.)

Adding these two disjoint sets of trapezoids gives your answer $55+(10-1)=64$.

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  • $\begingroup$ Okay but how do I know it was a triangular number?. Why is it a triangular number?. Reading solely to the article you mentioned it has gave me an idea but it is not very clear. i.e the second figure has only two right trapezoids in the left, so by calling that triangular number thing there are three figures only? as the second term is three?. Why should be added $n-1$?. Where does it come from?. Can you explain this with more words please? $\endgroup$ – Chris Steinbeck Bell Apr 6 '18 at 17:33
  • $\begingroup$ Added some more explanation. $\endgroup$ – John Apr 6 '18 at 17:39
  • $\begingroup$ Maybe I'm too dumb but I can't picture what you just mentioned. $10$ trapezoid with the topmost thin on the top. What does it mean?. I tried to relate their numbers with the triangular numbers again. But it does not check with the fourth term. It says $10$ but in the 4th figure there are $13$. If I ignore some of them and regard the ones in the left plus those in the right would be $4$+$4$ so there are $8$ or Am I understanding it wrong?. I revisited your answer different times but I got tangled up with how should I picture that in my mind, sorry. $\endgroup$ – Chris Steinbeck Bell Apr 6 '18 at 18:39
  • $\begingroup$ No, you're not too dumb! I tried again. See if that's clearer. $\endgroup$ – John Apr 6 '18 at 18:58
  • $\begingroup$ Thanks for adding that explanation now I can "see" it better. But, $55$ is not the answer, so I understand that it is required to add something, Why this something to be added has to be $n-1$? you mentioned because it is one unit less of what it is on the stack, so in this portion we do not need to perform the triangular number thing because due the position of those figures with the others makes it possible to know the number by counting them normally (hence getting their total directly) as they will not add up as if they were stacked horizontally?. $\endgroup$ – Chris Steinbeck Bell Apr 6 '18 at 20:44

ETA: I've added a somewhat shoddy diagram to illustrate what I mean. I've also changed the notation somewhat to be (a bit) more logical.

I'm not sure this is much simpler than what you did, but I numbered the vertices down the left-hand side $B_{0, 0}, B_{1, 0}, \ldots, B_{n, 0}$, then rightward from $B_{0, 0}$ were $B_{0, 1}, B_{0, 2}, \ldots, B_{0, n}$, and rightward from $B_{n, 0}$ were $B_{n, 1}, B_{n, 2}, \ldots, B_{n, n}$. Finally, down the leftmost diagonal, between $B_{0, 1}$ and $B_{n, 1}$, were $B_{1, 1}, B_{2, 1}, \ldots, B_{n-1, 1}$.

enter image description here

Hopefully, it'll be clear that the right trapezoid must contain two vertices of the form $B_{i, 0}$ and $B_{j, 0}$ (without loss of generality, assume $i < j$), since the right angles only live there. The other two vertices must then be $B_{i, k}$ and $B_{j, k}$, for some value of $k$.

Then, there are only two basic classes of right trapezoids:

  • If $k = 1$, then $i$ and $j$ can be any values such that $0 \leq i < j \leq n$.

  • If $2 \leq k \leq n$, then $i$ and $j$ must be $0$ and $n$, respectively.

There are $\binom{n+1}{2}$ of the former, and $n-1$ of the latter, which gives us a count of

$$ \binom{n+1}{2}+n-1 = \frac{n^2+3n-2}{2} $$

which gives us $64$ when $n = 10$, as expected.

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  • $\begingroup$ I do not get clearly the idea of counting from the corners. I mean okay, count from the corners but I don't understand the notation you have used. To me this kind of answers would be greatly improved if it had some visual aid as I'm not good with imagination. This binomial $\binom{n+1}{2}$ where does it comes from and why $n-1$?. $\endgroup$ – Chris Steinbeck Bell Apr 6 '18 at 17:39
  • $\begingroup$ @ChrisSteinbeckBell: I'll add a diagram when I get a chance. The $\binom{n+1}{2}$ examples of the first class come from the fact that $i$ and $j$ can be any two of the integers $0$ through $n$, which is $n+1$ of them. The second class is $n-1$ and not $n$ because the first class already includes the trapezoid $A_0, B_{0, 1}, B_{n, 1}, A_n$, so that $k$ can only range from $2$ through $n$, which is $n-1$ possible values. $\endgroup$ – Brian Tung Apr 6 '18 at 17:41
  • $\begingroup$ Thanks for the diagram it really eased my understanding but. Where is $j$?. I assume you mean by $i$ the top bottom corner on the left side and $j$ the $1$ seen in the corners starting from $B_{n,1}$?. I can get more or less that $n-1$ is because if in the right side there are $n$ then in the left must be $n-1$ as it is seen in the diagrams. Or is it because of other reason? I'm still confused at how do I know that the total is that specific binomial?. Because is the sum from 1 to n? But if that's the case $n-1$ solely is not the sum of those trapezoids. Can you help me with that? $\endgroup$ – Chris Steinbeck Bell Apr 6 '18 at 18:32
  • $\begingroup$ @ChrisSteinbeckBell: $i$ and $j$ indicate the upper and lower horizontals of the trapezoid, respectively. Does that help enough? $\endgroup$ – Brian Tung Apr 6 '18 at 22:33

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