A Sunday Morning Geometry Problem

It’s 6.15 am on a Sunday morning, 18th October to be exact. The newspapers have not been delivered and the kids are still sleeping. Some peace and quiet for me to do some reading while sipping my morning coffee … hmmm … here’s a nice problem*.

In the diagram, the three sides of a triangle have been split into thirds, then the lines have been added. What is the area of the smaller shaded triangle as a fraction of the area of the large triangle?


This looks familiar. Now, where have I seen it before? I don’t recall an immediate solution or approach to this. Let’s try to understand the problem. I need to draw. A cursory search on the table and floor revealed a piece of paper and a purple coloured magic marker. From the way the question is asked and previous experience with problems of this sort, the proportion is an invariant. Can I draw a special triangle that helps to solve the problem immediately? An equilateral triangle? A right-angled triangle? I cannot seem to control the shape of the inner triangle. It looks like another approach is needed.

Clearly triangle BCD** has 1/3 of the area, as do the other pieces ACE and BFA. Ah … inclusion and exclusion. If I take away these three pieces I would get the shaded portion but I’m double counting. This would mean that the shaded area = sum of the areas of the 3 small triangles at the vertices. Is that right? Let me check that again. … Yes, that’s right. Now let’s label the areas.

Hmm… how to proceed? … Can’t seem to get anywhere? … I wonder what if I draw another three lines joining the vertices with the other dots G, H and I. Would I get something nice here? I should. It looks like I would get a hexagon. Ok… must focus on solving the simpler triangle problem first.

Uh-oh, I hear the door opening, the boy must be awake. I remember reading that Euler could write mathematics with kids on his lap and/or running circles around him. But he is Euler. He could even write mathematics half-blind. Then again, Euler didn’t have my kids.

Good Morning, Dad!

Ok, time to put this problem on hold.
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It’s half an hour to noon. The kids are happy painting away in the living room and I steal some time to return a couple of emails. That ratio has been on my mind the whole morning even as I was playing and watching cartoons with the monkeys. What is that number? Since the computer is switched on, time to call in the heavy artillery — geogebra. I plotted a general triangle. Dividing the sides into three equal parts turns out to be not trivial. An easy way out is to use the distance function and move the points around until the ratio is 1:3. Geogebra automatically calculates the areas of the two triangles and I can compute the ratio directly in the input box to get the number 0.14. Hmm … looks weird. In rational terms that would be 7/50. Can this be right? How to split the triangles into 50 parts? Anyway I distorted the triangle and as expected the ratio remains the same 0.14. Hmm … . Oh time for lunch. I’ll leave the computer on and maybe return to it later.
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It’s 3.30 in the afternoon. The girl is napping but the boy, as usual, cannot sleep and is playing with lego. Time to work on my geometry problem again. The purple marker and the sketch from the morning is still on the table. Let me make a neater sketch which may help me spot something that I missed. … Yes! I can draw an auxiliary line and the area will be 2X. Doing this for Y and Z, I now have 4 (X+Y+Z) with 3 more regions to compute. The 7/50 puzzles me, I cannot see how the fractions will sum to 7/50.

But it looks like I have enough the solve the problem. The unknown regions at the lower left and right have area A/3 -3Z-Y and A/3 – 3X-Z respectively. Now by focusing on the correct inner triangle (i.e. AKB in Diagram 1) we have
[tex] (X+Y+Z)+ (\frac{A}{3}-3Z-Y) +2Z = 2 ( \frac{A}{3} -3X-Z +Z) [/tex]
[tex] \Rightarrow 7X = \frac{A}{3}. [/tex]

By symmetry (in this case algebraic symmetry in the sense of the same computations), we must have X=Y=Z. So in fact
[tex] \frac{(X+Y+Z)}{A}=\frac{3X}{A} = \frac{1}{7} [/tex].
So I was deceived by geogebra. The ratio was not 0.14 but 0.142857 … which was rounded to two decimal place. There is a lesson here about using computers/software and not paying attention to the assumptions and possible pitfalls.

Ok problem solved. Very nice. There is the natural generalization of whether the problem has an invariant solution for other ratios besides 1:2 but that is for another day. The girl has woken up from her nap and it’s playtime again.
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Monday, 19th Oct, 10.30 am. Lunch with colleagues. Between the few of us, we rummaged up a pen and some useless piece of paper and I explained the problem. D quickly recognized it as a problem he has previously solved, took over the paper and furiously calculated away. E quickly worked out the relation depicted in diagram 2. While D computed on, we mused about generalisations and whether adaptations to arbitrary quadrilaterals can be solved and by and by the conversation drifted to another topic.
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12 noon. D came into my office and presented the following generalization. Given a triangle of size 1, with one side divided into the ratio 1:t and another into the ratio 1:s, where we only require t, s to be positive. By drawing the auxiliary line from F, the area A (and hence B) can always be computed.

We have the following pair of equations:
[tex] (t+1)A+B=\frac{1}{1+s}[/tex]
[tex] tA + (s+1)B = \frac{t}{1+t} [/tex]
The system can be solved if and only if
[tex] \frac{t+1}{t} \ne \frac{1}{1+s} [/tex].
The above is clear since the left side is strictly greater than 1 while the right is strictly less than 1. Let me check by substituting t=s=2, … Yes, A= 1/21.
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20th Oct, Tuesday morning at the office. Perhaps it would be helpful to record down the thoughts and processes that went through my head over the past two days while thinking about the problem on my own and with colleagues. Writing is an integral part of discovery because it allows one to rethink the problem, consolidate the understanding, check the solutions and possibly further generalize — what Polya calls Look Back. … In both of the solutions, the drawing of the auxiliary line seems to be critical. As an experiment, let me try to label the three unknown regions in Diagram 2 as L, M, N. Assuming the total area is 1, I now have six variables and I can form 7 linear equations, for example of the form
[tex] X +L + Z =\frac{1}{3} [/tex]
[tex] M + Y+ (X+Y+Z)+ N =\frac{2}{3} [/tex]
[tex] X+Y+Z+L+M + N + (X+Y+Z) =1 [/tex].
Let’s see if this system can be solved. Oops, that last one there is just the sum of the previous two. Well I still have six equations with six variables. Instead of doing it by hand, I use matlab. … Ok, it does not work. The matrix has rank 3. … Back to writing things down. … Hey in D’s general solution, although the system is consistent would it be possible to end up with non-positive solutions? Ok, might as well try to get a closed form for A and B. … Uh-oh, time to go for a meeting. Have to put this aside again.
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Tuesday 6.10 pm. At the classroom with the part-time Masters students. I just passed out quiz 4 and the students are working on it. Good. I have some time to continue with my calculations … The algebra is slightly messy but the end result is nice.
[tex] A = \frac{1}{(1+t)(1+s+st)}, B= \frac{st}{(1+s)(1+s+st)} [/tex]
Quite neat. Better check again for s=t=2. Yes, it agrees. How about s=t=1? A=B=1/6. That’s right, since 3 medians of a triangle are concurrent and results in six triangles of equal area. … 20 minutes have passed, time to put this aside again and get ready to collect the quiz and continue with the lesson.
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Wednesday 21 Oct, 7.40 am at the office. No classes today, maybe I can finish up my writing before embarking on the other boring admin duties. I’m pretty sure the closed form I obtained was right, but let us substitute it into the equation to be sure. … Yes, the solutions check out. … Oh, I still have my hexagon generalization unresolved. Let’s try. … Hey, the general formula still works!

Using the ratios t=1, s=1/2, area of the three red triangles are all equal to A=2/15.
Similarly, using the ratios t=1/2, s=2, the area of the three green triangles are all equal to sB=1/6. Hence the hexagon has area = 1-1/2-2/5 = 1/10. Just to be sure, let me use geogebra to check on a general triangle.

I’m almost ready to move on to other things, but maybe it is a good time to do a search on google. The search terms “area triangle 1/7″ look promising. … As expected the problem is well known. The problem is sometimes called Feynman’s Triangle after the physicist but it certainly dates back further. In fact, there is a general theorem called Routh’s theorem. In diagram 4, assuming the third side is partitioned into the ratio 1:r, then the inclusion-exclusion argument would give the area of the middle triangle as
[tex] 1- \frac{1}{1+t} -\frac{1}{1+s} – \frac{1}{1+r} + \frac{1}{(1+t)(1+s+st)} + \frac{1}{(1+s)(1+r+rs)}+ \frac{1}{(1+r)(1+t+tr)} [/tex]
[tex] = \frac{(rst-1)^2}{(1+s+st)(1+r+rs)(1+t+tr)} [/tex].
In the above, I used Maple to simplify the algebra. Wikipedia has a very good write-up on the theorem which appears in Routh’s text from 1891, together with links to the corresponding pages of the text. The Feynman connection is also mentioned and many other references including several ingenious proofs of the 1/7 result can be found by following the various links.

At this point, 21st Oct 12:19pm, the narrative caught up with the writing in real time and I think I should stop here.

*I saw the problem in Issue 1 of chalkdustmagazine.
** There were no labellings in the original diagram and they were included for clarity.

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