Find the angles of the triangle if $∠BAC = 60°$ and $AB + BA' = AB' + BB'$.

Question

Let $AA′$ and $BB′$ be the bisectors of the angles $∠BAC$ and $∠ABC$ of the triangle $ABC$. Find the angles of the triangle if $∠BAC = 60°$ and $AB + BA' = AB' + BB'$.

Hi, can anybody solve this problem for me?

Thanks

Answer 1

One possible solution is $\angle ABC=80°$ (see diagram below). In that case $BB'C$ is an isosceles triangle and $BB'=B'C$. In addition, if $C'$ is the point on $AC$ such that $\angle CA'C'=40°$, then triangles $ABC$ and $AA'C'$ are congruent, while $A'C'C$ is isosceles. We have then: $$ AB'+BB'=AB'+B'C=AC $$ and $$ AB+BA'=AC'+A'C'=AC'+C'C=AC. $$ It follows that this triangle satisfies the given property and is indeed a solution.

I proved that no other solutions exist, but through a complicated trig equation, which cannot be solved in a simple way. I'm sure that some more simple and elegant proof can be found.

Answer 2

To reduce notational clutter, define $\alpha := A/2$, $\beta := B/2$, $c := |\overline{AB}|$, $e := |\overline{BB^\prime}|$, $p := |\overline{A^\prime B}|$, $q := |\overline{AB^\prime}|$. We won't yet impose the condition $A = 60^\circ$.

By the Law of Sines, we have $$ \frac{e}{\sin 2\alpha} = \frac{c}{\sin(2\alpha+\beta)} \qquad \frac{p}{\sin\alpha} = \frac{c}{\sin(\alpha+2\beta)} \qquad \frac{q}{\sin\beta} = \frac{c}{\sin(2\alpha+\beta)} \tag{1}$$

The condition equating sums of lengths says $$c\;\frac{\sin\alpha + \sin(\alpha+2\beta)}{\sin(\alpha+2\beta)} = c+p = e+q = c\;\frac{\sin 2\alpha + \sin \beta}{\sin(2\alpha+\beta)} \tag{2}$$

Thus,

$$\left(\;\sin\alpha + \sin(\alpha+2\beta)\;\right) \sin(2\alpha+\beta) = \left(\;\sin 2\alpha + \sin\beta\;\right)\sin(\alpha+2\beta) \tag{3}$$

In the absence of particularly-keen insight, there really isn't much to do but expand the trig expressions and hope for simplification. These days, this is the kind of thing I leave to a computer algebra system like Mathematica. As it turns out, we get a very nice simplification:

$$4\;\sin\alpha\;\sin\left(\alpha+\frac{1}{2}\beta\right)\;\left( \cos\beta - \frac{1}{2} \right)\;\cos\left(\alpha+\frac{3}{2}\beta\right) = 0 \tag{4}$$

Our options are these:

$$\begin{align} \sin\alpha= 0: &\quad\to\quad \alpha = 0^\circ \quad\text{or}\quad \alpha = 180^\circ \quad\to\quad\text{degenerate triangle} \tag{5a}\\ \sin\left(\alpha+\frac{1}{2}\beta\right) = 0: &\quad\to\quad \alpha+\frac{1}{2}\beta = 180^\circ \quad\to\quad\text{impossible: $\alpha+\beta < 90^\circ$} \tag{5b} \\ \cos\beta = \frac{1}{2}: &\quad\to\quad \beta = 60^\circ \quad\to\quad B = 120^\circ \tag{5c} \\ \cos\left(\alpha+\frac{3}{2}\beta\right) = 0: &\quad\to\quad \alpha+\frac{3}{2}\beta = 90^\circ \quad\to\quad B = 120^\circ - \frac{2}{3}A\tag{5d} \end{align}$$

Of the two viable solutions, we dismiss $B = 120^\circ$ only because the previously-ignored condition $A = 60^\circ$ would imply a degenerate triangle. This leaves only (5d), which, with the ignored condition, gives

$$B = 80^\circ \qquad C = 40^\circ \tag{6}$$

Done. $\square$

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That the brute-force expansion of (3) is so very messy, yet simplifies so very well to (4), suggests that there may be something very interesting going on here. Of course, invoking the condition $\alpha = 30^\circ$ at that stage would have reduced the symbolic complexity in the expansion, but I don't think the problem gets appreciably easier from an algebraic standpoint; indeed, as it probably would have been more difficult to glean options (5b) and (5d) from a mess of $1/2$s and $\sqrt{3}/2$s, I'm led to believe that the condition must be exploitable geometrically. Moreover, with $B$ being related to two-thirds of $A$, perhaps some sneaky trisection is at work in the general case. For the moment, whatever clever approach these facts may be hinting-at has escaped me.