Let the triangle ABP be where A(0,0) and B(4,0). If its interior angles α and β adjacent to the base AB are such that α+β= π/4

Question

I have the following problem: Let the triangle ABP be where A(0,0) and B(4,0). If its interior angles α and β adjacent to the base AB are such that α+β= π/4, find the equation of the locus of the points P that are in the first quadrant. My idea is this: α+β+mAPB=180° then mAPB=135° AP => y=m_ABx BP => y=m_BP(X-4) But I will have the following: m_ABx=m_BP(X-4) tan(α)x=tan(β)(x-4)

Answer 1

Set $\alpha=\beta=\frac{\pi}{8}$. Then $\Delta ABP$ is an isosceles triangle. By the law of sines, one of the two identical unknown sides is calculated as $$4\frac{sin(\pi/8)}{sin(\pi/8)cos(\pi/8)+sin(\pi/8)cos(\pi/8)} = 4\frac{sin(\pi/8)}{2sin(\pi/8)cos(\pi/8)} = \frac{2}{cos(\pi/8)}$$ $$=4/(\sqrt{2+\sqrt{2}})$$

By the Pythagorean theorem, the height of the triangle is equal to:$$\sqrt{16/(2+\sqrt{2})-4}=\sqrt{(16-8-4\sqrt{2})/(2+\sqrt{2})}=\sqrt{(8-4\sqrt{2})/(2+\sqrt{2})} = \sqrt{4(2-\sqrt{2})/(2+\sqrt{2})}=2\sqrt{(2-\sqrt{2})/(2+\sqrt{2})}$$.

Since $2-\sqrt{2}$ and $2+\sqrt{2}$ are conjugates, this simplifies nicely to $\frac{4-2\sqrt{2}}{\sqrt{2}}$, and rationalizing the denominator leads to $4\frac{\sqrt{2}-1}{2} =2(\sqrt{2}-1)$. Very neat, in my opinion!

The next step is to use the fact that every triangle has a circumcircle and to leverage the intersecting chords theorem to find its radius. Let $C$ be the circumcircle of $\Delta ABP$ and let $r_C$ be its radius. Let $\overline{AB}$ be a chord of $C$. It is perfectly clear that an altitude dropped from $P$ bisects $\overline{AB}$ (since $\Delta ABP$ is isosceles). Call the point at which the altitude intersects $\overline{AB}$, $M$. Then $\overline{AM}= \overline{MB} = 2$. Extend $\overline{PM}$ so that it intersects $C$ and name this point $I$. Then, by the intersecting chords theorem, the following identity holds: $$\overline{AM} \cdot \overline{MB} = \overline{PM} \cdot \overline{MI}$$

We have $\overline{AM}, \overline{MB}$ and $\overline{PM}$ (which is just the height of $\Delta ABP$, remember, equal to $2(\sqrt{2}-1)$). All that's left is $\overline{MI}$, which is the "gap" between $\overline{AB}$ and the minor arc of $C$; equivalently, it is the "missing" part of the diameter of $C$, where $\overline{PM}$ is the known part. It follows immediately that $r_C = \frac{\overline{PM} + \overline{MI}}{2}$.

Fortunately, $\overline{MI}$ is easy to find. The product of $\overline{AM}$ and $\overline{MB}$ is $4$, given that they are both equal to $2$, and we know that $\overline{PM} = 2(\sqrt{2}-1)$. Therefore: $$\overline{MI} = \frac{4}{2(\sqrt{2}-1)} = \frac{2}{\sqrt{2}-1} = 2(\sqrt{2}+1).$$

Now we find $r_C:$

$r_C = \frac{\overline{PM} + \overline{MI}}{2} = \frac{2(\sqrt{2}-1)+2(\sqrt{2}+1)}{2} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}.$

Man! Neat expressions just keep falling out of this problem, don't they? Having found $r_C$, the task now is to find the centre of $C$. This is pretty easy in the case of the $x$-component, which is $2$ (since the centre and point $M$ lie on the same line, and we know $M$ lies $2$ units to the right of the origin). The $y$-component requires a bit more computation, but it's not too hard: using the fact that $\overline{AB}$ is coincident with the line $y=0$ and that the centre of of $C$ and the point $M$ are contained in the same line, we can use the Pythagorean theorem again to calculate the height above the $x$-axis at which the centre of $C$ is located, which turns out to be $\sqrt{8-4} = 2$.

We now have everything we need for the equation of $C.$ Without further ado, it is $(x-2)^2+(y-2)^2=8.$

INTERJECTION: I've taken the time to graph all of this on Geogebra, and it is absolutely correct, only it's all backwards lol. That's very easy to do in geometry, especially when working in a semi-synthetic manner. If you graph this all out yourself, with all the points mentioned so far (and I encourage you to), then the "correct" triangle we are considering — "correct" as in "satisfies the constraints" — will be labelled $\Delta ABI$ as per the nomenclature I've introduced. So keep that in mind (or just swap the labels $P$ and $I$ in the diagram).

RESUME PROBLEM: we're nearly there. All we need now is to use the inscribed chord theorem, which states that the inscribed angle subtending a chord is constant on the same side of said chord. In other words, you can slide $I$ anywhere along the minor arc of $C$ without changing the value of $\alpha + \beta$. It therefore suffices to specify the range of $x$ and $y$ values along that arc. I'll leave the details to you, but the answer you seek is given as the below set:

$\{(x,y)|(x-2)^2+(y-2)=8, 0 \lt x \lt 4, (2-2\sqrt{2}) \le y \lt 0\}$ (we disallow $x=0, x=2$ and $y=0$ as these result in a degenerate triangle).

Anyway, that's it! It would be possible to write the set of points in terms of angles — with trig functions basically — but I think this is simpler. The equation of the locus of points is simply the above expression, only without the curly brackets and the "$(x,y)|$" at the beginning. I hope it helps, and I do plan to come back and clean up my calculations. For now though, I'm exhausted, and as long as you follow my answer closely you shouldn't go wrong!