[previously: favorite problems 1 – pirates]

Two summers ago at Twitter Math Camp 14, I saw Dylan Kane (@math8_teacher) present a “My Favorite” about a problem he had enjoyed pondering during his summer adventures. You can read his post about it on his blog here. I believe the original problem came from the Five Triangles blog (which is now password protected. 😦 )*Given three “lined up” congruent equilateral triangles, each with area 20 cm^2, what is the area of the shaded region?*

Dylan was using the problem to tell a story, but I couldn’t help myself and immediately started sketching out ideas for solutions. If you’ve been to TMC, you know the sheer volume of good ideas, interactions with rockstar teachers, and mathy things to ponder is overwhelming, so after my initial attempt at a solution I got distracted and set the problem aside for the rest of the conference.

Once I finished driving home from Jenks, the first thing I did that night was sit down and start thinking about the problem again. I told my sister Lindsey about it and she got in on the fun too. Here we are, lounging in my room past midnight doodling triangles and getting stumped.

The week got busy as preparations for the 2014-2015 school year got underway, and I think I forgot about the problem for a bit. Lucky for me, I mentioned the problem to Dawson over brunch one Sunday, and by mid-afternoon he was texting me about his solution.

[Note: Don’t read ahead if you don’t want to get spoiled on the problem!]

Dawson:Solved the triangle problem (I think/hope)! It’s great!! Although I thought it was leading me toward a different answer than I ultimately got

Rachel:i think i did too

did you do it in terms of 60 units sq?

can we compare answers?

and i agree…. i was almost annoyed at what it turned out to be

(if i did it right)

Dawson:Heading out to run errands

Happy to discuss later but…

Rachel:okay no worries!

Dawson:With each eq triangle 20 (total of 60)

I got 50/3

Rachel:ME TOO!

Dawson:Thought I was headed to 20 and that would have been awesome

Rachel:haha, yes.

okay let’s chat later about methods. i’m relieved we got the same thing though!

Dawson:Me too! Talk soon

Extensions to the triangle problem are amazing

I knew part of what was happening after 4 triangles…but 5 triangles really clarified it for me

Rachel:oh goodness, dawson. i guess i have to do that now…

Dawson:No no would love to talk about whatever you have later on

[Note: Do you see what he did there?!?! Fast-forward to “later on.”]

Rachel:what did you get for 4, 5? did you do it in terms of 20 sq units per triangle? or just proportion of one whole triangle?

both are illuminating, but proportion of one whole triangle has a nicer pattern… i think… if i’m doing it right

(2n-1) / 6 * area of 1 triangle??

Dawson:Proportion of one triangle

Rachel:n = number of triangles

Dawson:Yes

Rachel:sweet

Dawson:Did you look at each individual small triangle?

Rachel:just one, then scaled down area

like my work for the first one was 2/3 + (1/2)^2 * (2/3)

and the second 3/4 + (2/3)^2 * (3/4) + (1/3)^2* (3/4)

Dawson:Ok…

Rachel:once i had thoroughly convinced myself the first shaded region was 2/3, i was good and set

Dawson:If you start with 5 eq triangles of area 20…

What’s the area of each individual part?

That’s what got me hooked

Rachel:when you look at the individual pieces is there something pretty?

Dawson:YES

Rachel:i just continued on… 4/5 + (3/4)^2 * (4/5) + (2/4)^2 * (4/5) + (1/4)^2 * (4/5)

Dawson:I think we did this WAY differently

At least in approach

Take your (3/4)^2 * 4/5 and simplify

Rachel:9/20…. okay okay give me a sec

Dawson:You bet…I needed a long time!!

Rachel:WHAT THE HELL WHY ARE THEY PERF SQUARES

give me ANOTHER second

Dawson:Yes!!!!!!!

It’s awesome

Rachel:holy cow what is going on??? so 4 triangles is 9/12+4/12+1/12???

WHERE ARE THESE NUMBERS COMING FROM

Dawson:Yes

It’s beautiful

Rachel:holy cow

and those are triangular numbers on the denom??

does that continue?

Our text conversation turned into an email thread that included Dylan, Justin Lanier (@j_lanier), Nathan Kraft (@nathankraft1), Glenn Waddell (@gwaddellnvhs), Mimi Yang, Mark Greenaway, Jed Butler (@MathButler), and Tina Cardone (@crstn85), all of whom sent out their own version of the solution. It’s completely fascinating to me how many different approaches people took. People employed so many different strategies from using similar triangles and scale factor of sides/area, using ratios of sides in special right triangles and the formula for area of a triangle, writing equations of lines and finding points of intersection, to using trig and law of sines.

Several folks tackled the extension Dawson prompted above, and some worked on generalizing the problem for congruent isosceles triangles, or even congruent scalene triangles.

I have a folder where I’ve saved all of the submitted solutions so far, and I’m still collecting more, if you want to share! Mine and Dawson’s are both posted below. Not quite as high-tech as Justin’s typed-up solution, or Jed’s impressive iMovie solution, but they get the job done. I especially enjoy Dawson’s for his use of colored pens, and the running self-deprecating commentary throughout. Enjoy. 🙂

One last thing: I used this problem as a “warm up” problem in a math department meeting a few months ago. One of my colleagues emailed me with his solution after the meeting, along with what might be my favorite quote in recent memory.

“It took me about 15 minutes beyond the mtg,

and yes, after one sees it, one feels slightly ashamed, as with all good puzzles.“

Indeed. 😀

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What to do post AP/IB exam? – A Recursive Processon 05.01.2016at 8:46 pm

I’m gonna have to try this. I’ve avoided reading too much of the solutions so far…

You might like this one which might be related… http://i.imgur.com/ESCloC9.png “Does alpha + beta = gamma?”

Related as far as “3 lined up shapes” and diagonals at least…

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Scotton 05.05.2016at 5:13 am

Nice! I think I have seen that angles problem before, but had forgotten it. I’ll definitely give it a try.

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rdkpickleon 05.05.2016at 1:00 pm

You got me hooked on it for the past week. I’ll show all my scratchwork later but I got into creating a transformation based solution here:

Now finally looking at the solutions you showed above its very nice to see a few different methods! neat! Now I want to dig into that extension where you go towards other lengths.

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scottfarraron 05.12.2016at 4:28 pm

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