Playing with Right Triangles

An interesting set-up of right triangles allows us to prove radical identities.

I was asked by a colleague last week to prove an identity involving radicals. The two expressions arise when considering cosine of the angle pi/12. Normally, one would apply a sum or difference formula

doubleangle

and this would simplify to

(1)

sqrt1

However, when one of his students used a calculator, the calculator returned back an unusual expression:

(2)

sqrt2

He and the student were able to verify that these expressions evaluated to a similar decimal expansion, so must be equivalent. But then his student asked him how to prove the equivalence of expressions like this. He tried for a bit, unsuccessful – then he tormented me with this problem all Easter weekend. Eventually, I was able to show the equivalence using an old right triangle trick I saw a few years back.

Attach two right triangles together in such a way so that the right leg of the second, and the bottom leg of the first meet at a right angle. On the hypotenuses of the smaller triangles write root 6 and root 2, respectively. This is done so that the hypotenuse of the larger right triangle is root 6 + root 2 – matching up with the numerator of expression (1).

triangle1

Our goal is to apply the Pythagorean Theorem on the large right triangle, so we need to determine the legs of the larger triangle. To do this, we will determine the legs of the smaller right triangles. For the root 2 triangle, we have the obvious choice of making the legs (1, 1). For the root 6 triangle, we could make the legs (root 2, root 4), (root 3, root 3) or (root 1, root 5). Notice that in expression (2), we have a root 3. This suggests we might want to try the (root 3, root 3) combination for the root 6 triangle. This shows us that the legs of the larger right triangle are both root 3 + 1.

triangle2.gif

Now we can apply the Pythagorean Theorem on the large right triangle.

pyth1

pyth2

pyth3

Taking the square root of both sides gives

pyth4

pyth5

And finally, dividing both sides of the equation by 4 yields the desired result.

pyth6.gif

I suppose that the moral of the story here, besides seeing some really interesting mathematics, is that I never would have solved this problem unless I had seen the previous problem involving something similar. In general, I believe it is safe to state that in order to be successful solving problems, one should be exposed to many different types of problems (ever wonder how those Math Olympiad contestants get so “smart”?). From a cognitive science perspective this makes sense – it allows us to create problem archetypes (schemata) that we can draw upon to help solve future problems. And the more well-connected these schemata become, the easier it becomes to solve problems.

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