In February, I was throwing together a geometry test for my 12-year-olds. I wanted a standard angle-chasing problem, but – and here’s the trick – I’m lazy. So I grabbed a Google image result, checked that I could do it in my head, and pasted it into the document.

But when I started writing up an answer key, I ran into a wall. Wait… *how* did I solve this last time? I trotted out all the standard techniques. They weren’t enough. A rung of the logical ladder seemed to have vanished overnight, and now I was stuck, grasping at air.

Eventually, some colleagues and I solved it by with industrial-strength tools: the Law of Sines, the Law of Cosines, and some fairly sophisticated algebra. Yet the problem looked so elementary, I felt sure that our approach was overkill—that we were bashing down the door, whereas a defter hand could simply pick the lock.

Apparently, I’d actually chosen a famous gem of recreational mathematics, born in 1922 from the mind of Robert Langley, and known since as “Langley’s Adventitious Angles.”

And, as I suspected, the niftiest solution requires no trigonometry or algebra, just a single ingenious move: construct a line here, at a 20^{o} angle to the base.

It’s a masterstroke. It pops the lid right off of the problem. And, at least to me, it’s utterly un-guessable. If you had a thousand monkeys with a thousand typewriters and a thousand protractors, you’d get full verses of Shakespeare long before any of our furry friends stumbled upon this solution.

Sometimes the general techniques fail, and you need a sneaky trick. That’s life.

To me, a “trick” is a disposable insight, a funny little key that opens one particular closet door. And a “technique” is something more useful: a skeleton key, fitting the lock for a whole hallway of rooms.

But is the difference always so clear?

There’s a certain algebraic move that I first encountered in high school, which you might call “multiplying by the conjugate.”

You use it here, to simplify an expression with square roots:

And here, with complex numbers:

And here, again with square roots, to solve a limit (without recourse to the heavy machinery of L’Hopital’s Rule):

And here, to boil down a trigonometric expression:

At first glance, it feels like a trick: a little too slick, a little too cute. It’s hard to imagine it blossoming into a general technique. But it eventually proves useful in a shocking variety of situations, cutting across the whole secondary mathematics curriculum. This funny little key, which barely looks like it should open a single door, turns out to open hundreds.

Ultimately, I can’t find any categorical difference between a trick and a technique. They’re both problem-solving innovations, lying along a continuum that runs from “almost never useful” to “useful all the darn time.” A technique is simply a trick that went viral, and a trick is simply a technique that fizzled out after a single use.

It makes me wonder how to define the job of the mathematician (and of her little spiritual cousin, the mathematics student). Is it to master existing techniques? To cook up new tricks from scratch? Or to recognize how the tricks of today can be nurtured and expanded into the techniques of tomorrow?

Yes to all of that. But of all the aspects of the mathematician’s job, I think the ability to birth truly new ideas is overrated. Leaf through the pages of history, from Pythagoras to the present, and you’ll find a stunning supply of tricks, techniques, and everything in between. In the five millennia we’ve been working on this, humanity has accrued a remarkable wealth of mathematical ideas. Yes, we want our fresh young minds to make new deposits into the bank of our knowledge; but we also want them to know how to access the existing funds.

I would never, in a million years, come up with the trick to solve Langley’s Adventitious Angles. But I don’t need to. Someone else already has.

By the way, as promised, here’s the rest of the solution to Langley’s Adventitious Angles:

It’s always just the one piece of insight that you’re missing, then it all fits together once you get it, isn’t it? It becomes even more impressive when the insight is a “trick” like that.

I’m missing that insight for a similar but much harder problem that I’ve been going at for years. Maybe you’d like to take a crack at it. Replace the 60° and 50° angles with 70° and 60° respectively and adjust the other angles accordingly (i.e. 20° and 30° become 10° and 20° respectively).

http://thinkzone.wlonk.com/MathFun/Triangle.htm

You should just work with sums. You can easily get all the angles besides the following ones: CDE, CED, x=EDA, DEB. I refer to the figure on this site. You don’t know the angles but you know the sums. You work it out from that.

Exactly, that’s how I just solved it. You have four equations – one from the upper triangle, one from the middle triangle, one from the left side, and one from the right side. All of these must add to 180 degrees.

Yea see this isvwhat makes math so infuriating and godawful..no one would have VEVER EVER thoight of this..not Ramanujan not me not Einstein not Newton not Euler..someone just stumbled on it or something..how can anyone possibly do math all the tone or spend years working on this craps?

“A method is a device you use twice.” —G. Polya,

How To Solve ItPolya! Always writing my blog posts in 8 words, decades before I get around to the much less pithy version.

The solution, if I see that correctly, works only in this particular case. The general case seems to be impossible with SIMPLE means. Interesting problem!

Here’s a more accurate picture of the configuration of lines: https://dl.dropboxusercontent.com/u/8592391/trick.png

Thanks! Mine is admittedly pretty terrible.

So I spent two hours half asleep from the crack of dawn on this, with the hint line. I did see the original triangle and two more triangles as isosceles, and then an equilateral triangle appeared. i returned after a nap, two or three more isosceles triangles popped up, one of which had the looked for angle – so damn simple, in the end! The “trick” line is not a trick, it is a result of seeing that the problem is about isosceles triangles. I agree with Rene Grothman.

What appears to onlookers as a trick is often the end result of lots of trial and error, headbanging, and coffee, the results of which end in the bin. Nobody really wants anyone to know that they didn’t “see” the magic line (in this case) instantly.

Here is another special case, wich can be solved by lookint at it closely and some geometry:

http://i.imgur.com/kLbSMFW.png

Is the answer 44?

Yup, because the quadrilateral is cyclic 🙂

How do you know it is cyclic?

Is that exact?

By replacing the 31 with 20 and the 44 with 45, I get ? = 45 degrees.

It’s easiest to see this by constructing the figure by placing the bottom-left to top-right diagonal on the X axis.

It produces two pairs of congruent triangles: an isosceles pair with apex 30 degrees, and a scalene pair with angles 30 degrees, 45 degrees and 105 degrees.

Geogebra is one tool you can use. The main pitfall I encounter with Geogebra is my unintentionally click-creating, and then relying on, extraneous points that look as though they are on a line, but are in fact slightly off.

Sorry, typo. It should read: ‘replacing the 31 with 30’.

I saw the given hint but got no further . I had noticed the two isoceles triangles ADapex and ABE which give us two radii with centres at A and D.

I related the ratio of one radius to the other using the Sine Rule in triangles ABD and AED and then equated these to eliminate the radii.

Let R=AD = dapex and r=EA=AB

IN ABD

R/sinABD =r/sinADB yielding

R/r=sin80/sin40

In AED

r/sinx=R/sin(160-x) yielding

R/r=sin(20+x)/sinx

Equating these :

sin(20+x)/sinx=sin80/sin40 so

sin20.cosx+cos20.sinx=sinx.sin80/sin40

sin20.cosx=sinx(sin80/sin40-cos20)

sinx/cosx=tanx=sin20/(sin80/sin40-cos20) yielding

tanx= 0.577…

Thus x=30deg

The idea is simple enough but the workings are not so ‘clean’ . However , this leads me to think there might well be some vaguely elegant expressions for the sines and cosines of the multiples of 10 for this particular answer to drop out . Can anybody help ?

I didn’t spend a lot of time on this, but with some guess and check, multiple (even infinite) answers seem to lead to agreement on all fronts. But that doesn’t mesh with the fact that the full triangle is fixed, line AD is fixed, and line BE is fixed (therefore line ED must be fixed and the angles in the triangle must be fixed). I tried a system of equations using the 4 unknown variables, but it also led to the conclusion that infinite answers would lead to full agreement with all other numbers in the figure. What am I missing here? If you plug in an incorrect answer, what do you measure it against to see that it is incorrect? This problem seems paradoxical in that sense, because there can only be one correct answer, but multiple (infinite) answers agree with all other numbers. For example, make x equal 10, then fill in the other missing angle measures accordingly, and all numbers still work. Try 35. Still works. ????????

“I would never, in a million years, come up with the trick to solve Langley’s Adventitious Angles. But I don’t need to. Someone else already has.” So I’m familiar with this problem but don’t really remember the soln but it came up in a book I’m working through i.e. the author thought it was doable. And I think given enough time and playing around that its much more approachable than you give yourself credit for.

A small riff on all of this: https://mymathclub.blogspot.com/2019/05/langleys-adventitious-angles.html