# External bisectors of the angles of ABC triangle form a triangle $A_1B_1C_1$ and so on

If the external bisectors of the angles of the triangle ABC form a triangle $A_1B_1C_1$,if the external bisectors of the angles of the triangle $A_1B_1C_1$ form a triangle $A_2B_2C_2$,and so on,show that the angle $A_n$ of the $n$th derived triangle is $\frac{\pi}{3}+(-\frac{1}{2})^n(A-\frac{\pi}{3})$,and that the triangles tend to become equilateral.

My attempt:The angle $A_1$ of triangle $A_1B_1C_1$ is $\frac{\pi}{2}-\frac{A}{2}$,angle $A_2$ of triangle $A_2B_2C_2$ is $\frac{A}{4}-\frac{3\pi}{4}$ but my answer is nowhere look like resembling final answer.Is my approach correct.If not,what is the correct way to solve this question?Can someone guide me?

You are on the right track, you just need to continue. You have a correct formula for $A_1$, so format it differently.

\begin{align} A_1 &= \frac{\pi}2-\frac A2 \quad\text{(your formula)}\\ &= \frac{\pi}3+\frac{\pi}6+\left(-\frac 12\right)A \\ &= \frac{\pi}3+\left(-\frac 12\right)\left(A-\frac{\pi}3\right) \\ &= \frac{\pi}3+\left(-\frac 12\right)^1\left(A-\frac{\pi}3\right) \\ \end{align}

Your formula for $A_1$ from $A$ also gives $A_{n+1}$ from $A_n$, so combining your formula with my derivation above and using induction we get

\begin{align} A_{n+1} &= \frac{\pi}3+\left(-\frac 12\right)\left(A_n-\frac{\pi}3\right) \\ &= \frac{\pi}3+\left(-\frac 12\right)\left[\frac{\pi}3+ \left(-\frac 12\right)^n\left(A-\frac{\pi}3\right)-\frac{\pi}3\right] \\ &= \frac{\pi}3+\left(-\frac 12\right)\left[ \left(-\frac 12\right)^n\left(A-\frac{\pi}3\right)\right] \\ &= \frac{\pi}3+\left(-\frac 12\right)^{n+1}\left(A-\frac{\pi}3\right) \\ \end{align}

which finishes the proof by induction.

So the given formula is correct. As we let $n\to\infty$ we see that $\left(-\frac 12\right)^n\to 0$ and thus $A_n\to\frac{\pi}3$. We can do the same to $B$ and $C$ to get $B_n\to\frac{\pi}3$ and $C_n\to\frac{\pi}3$, so all three angles approach $60°$, and the (increasingly large) triangles $\triangle A_nB_nC_n$ approach equilateral.

Here is a quick proof of your formula $A_1 = \frac{\pi}2-\frac A2$, though I used Greek letters for angles in this diagram.

This should be self-explanatory, and the final value for $A_1= \frac{\pi}2-\frac A2$ comes directly from $A_1=\frac{\beta+\gamma}2$ and $\alpha+\beta+\gamma=\pi$.

Your approach was a good way to start working on this problem. By working the first two or three steps, you can see what calculations are involved. Your idea of comparing your results to the desired results was also a good one. It gives you a chance to check yourself in case you go down a wrong path.

But while it's true that the values you calculated, namely, $A_1 = \frac{\pi}{2} - \frac A2$ and $A_2 = \frac A4 - \frac{3\pi}{4}$, do not look much like $\frac{\pi}{3}+\left(-\frac12\right)^n \left(A - \frac{\pi}{3}\right)$, that alone should not disturb you; general formulas very often look quite different from the results you get by direct calculations. That's part of what made it so interesting and useful when people discovered such formulas. It often takes some art to "see" a pattern in a sequence of values from some procedure such as $A_1$, $A_2$, $A_3$, and so forth.

For example, the sum of the first $n$ positive integers is $\frac12(n^2 + n)$. But the sum of the first $4$ positive integers is $1+2+3+4 = 10$, and $10$ does not look like $\frac12(n^2 + n)$. It doesn't even look like $\frac12(4^2 + 4)$, but it turns out that if you do the multiplications and additions of that formula, you can see that $\frac12(4^2 + 4) = \frac12(16 + 4) = \frac12(20) = 10$.

That is, in order to compare a particular result of yours (the values you found for $A_1$ and $A_2$) against the general formula, you can plug the appropriate value of $n$ into the formula, do whatever operations the formula says to do, and see what comes out. For example, your own calculation found that $A_1 = \frac{\pi}{2} - \frac A2$. The general formula for $A_n$ should produce $A_1$ when $n = 1$. So let's set $n = 1$ and see what happens:

\begin{align} A_n & = \frac\pi3 + \left(-\frac12\right)^n \left(A - \frac\pi3\right) \\ A_1 & = \frac\pi3 + \left(-\frac12\right) \left(A - \frac\pi3\right) & \text{(because we set $n = 1$)} \\ & = \frac\pi3 - \frac12 A + \frac\pi6 \\ & = \frac{\pi}{2} - \frac A2 \end{align}

So your calculation of $A_1$ looks OK. For $A_2$, we set $n = 2$:

\begin{align} A_n & = \frac\pi3 + \left(-\frac12\right)^n \left(A - \frac\pi3\right) \\ A_2 & = \frac\pi3 + \left(-\frac12\right)^2 \left(A - \frac\pi3\right) & \text{(because we set $n = 2$)} \\ & = \frac\pi3 + \frac14 \left(A - \frac\pi3\right) \\ & = \frac\pi3 + \frac14 A - \frac{\pi}{12} \\ & = \frac\pi4 + \frac A4 \end{align}

This is a different from the result $\frac A4 - \frac{3\pi}{4}$ that you calculated for $A_2$. So maybe there was a problem with your method, or maybe you just made an arithmetic error.

But once you figure out what is going on with $A_2$, you still need to prove the general formula for all $n$, and for that you may need something like the other answer (which used induction). But if you can see how to get $A_1$ from $A$ and how to get $A_2$ from $A_1$, that can help when you have to get $A_{n+1}$ from $A_n$, as the inductive proof requires.

• This is a good response, and the method you outline here is close to the one I used to get my answer. +1! – Rory Daulton Aug 3 '15 at 14:43