euclidean-geometrygeometry
$ABCD$ is a line segment, trisected by the points $B$ and $C$. $P$ is any point on the circle where $BC$ is its diameter. If the angles $\angle APB$ and $\angle CPD$ are respectively $\alpha$ and $\beta$, prove that $$4\tan\alpha ⋅ \tanβ = 1$$
I tried dropping perpendiculars from $B$ and $C$ to $AP$ and $DP$ to get $tan\alpha$ and $tan\beta$ in terms of sides but I don't know how to utilise that. I also can't think of any other way to get $\tan\alpha$ and $\tan\beta$
I wanted to give you some hints, but it soon got too complicated, so I'm posting the full solution. There are other approaches, but I like the one using reflections the most: I really find it simpler than the one at Wikipedia page, and also it is my own invention (reinvention probably, though).
I'm assuming that $X$, $Y$, and $Z$ are the orthogonal projections of $D$ onto $AB$, $BC$ and $CA$. If so, then you are trying to prove the existence of Simson line. Consider the following picture:
$\hspace{20pt}$
where $AA'$ is the diameter and $D_{AB}$, $D_{BC}$, $D_{CA}$ are reflections of $D$ across $AB$, $BC$ and $CA$ respectively. From properties of reflection (composition of two reflections is a rotation around the intersection of their axes) we know that $D_{AB}$ is an image of $D_{BC}$ in rotation around $B$ by angle $2\angle CBA$ (i.e. the triangle $\triangle D_{AB}BD_{BC}$ is isosceles), hence $$\angle D_{AB}D_{BC}B = 90^\circ-\angle ABC = \angle A'BC.$$ For similar reasons, $\angle D_{CA}D_{BC}C = \angle BCA$. Moreover, since $D_{BC}$ is a reflection of $D$ via $BC$, we have $\angle BD_{BC}C = \angle BDC$. Obviously $D_{AB}$, $D_{BC}$ and $D_{CA}$ are collinear if and only if green and blue angles sum up to $180^\circ$, in other words $\angle BDC = \angle BA'C$, but that happens if and only if $P$ belongs to the circumcircle of $ABC$.
Finally, points $X$, $Y$ and $Z$ are images of $D_{AB}$, $D_{BC}$ and $D_{CA}$ respectively in homothety centered at $D$ and ratio $\frac{1}{2}$ (that is, for example, $X$ is the midpoint of $DD_{AB}$). It follows that $X$, $Y$ and $Z$ are collinear if and only if $D_{AB}$, $D_{BC}$ and $D_{CA}$ are collinear, that is, if and only if $D$ belongs to circumcircle of $ABC$.
It is worth noting, that if $D_{AB}$, $D_{BC}$ and $D_{CA}$ are collinear, then they are also collinear with the orthocenter of $ABC$, e.g. see here. Check out also the Wikipedia and MathWorld.
I hope this helps ;-)
Set the center of the circle to be $\left(0,0\right)$. Then the set of points on the circle are those such that $x^2+y^2=r^2$ and the set of points on the line segment are solutions to $y=mx+b$ from some $m$ and $b$ (these are easily found given the two endpoints of the line segment). Substituting, we have \begin{equation} x^2+(mx+b)^2=r^2\Longleftrightarrow (m^2+1)x^2+2mbx+(b^2-r^2)=0 \end{equation} Then this is just a quadratic you can solve with the quadratic formula giving you $x$, and you can get $y$ by $y=mx+b$. However, this gives you two solutions; you must pick the right one (find which is between the two endpoint). Fortunately finding the angle after this is pretty easy: it's just simple trigonometry. If $(x,y)$ is the solution, $\tan^{-1}\left(\frac{y}{x}\right)$ gives you the angles measure counterclockwise from the $x$-axis. You asked for the angle from the $y$-axis, which is then $\frac{\pi}{2}-\tan^{-1}\left(\frac{y}{x}\right).$
A worked example
I'll work out the example you have given. We have radius $r=\frac{3}{2}$, center $\left(3,4\right)$ and a line segment from $(3,3)$ to $(6,7)$. First we'll move the center to $(0,0)$. We do this by subtracting $(3,4)$ from each point. This gives us the center (as desired) of $(0,0)$, and a line segment from $(0,-1)$ to $(3,3)$. Then the line between the two points is $y=\frac{4}{3}x-1$. The equation for the circle is $x^2+y^2=\frac{9}{4}$. Substituting the equation for the line in, we have \begin{equation} x^2+\left(\frac{4}{3}x-1\right)^2=\frac{25}{9}x^2-\frac{8}{3}x+1=\frac{9}{4} \end{equation} which gives the quadratic \begin{equation} \frac{25}{9}x^2-\frac{8}{3}x-\frac{5}{4}=100x^2-96x-45=0 \end{equation} Then the quadratic equation gives $x=\frac{96\pm \sqrt{96^2+4\cdot 100\cdot 45}}{200}=-0.345,1.305$. We're going to want the solution with positive $x$, so $x=1.305$. Then $y=\frac{4}{3}(1.305)-1=0.740$. To find the desired angles, we merely take $90-\tan^{-1}\left(\frac{0.740}{1.305}\right)=90-29.56=60.44$. I'm not sure what the discrepancy between my answer and the one you provided is; it could be rounding errors, but it is more likely I just made a calculation error somewhere (please do point it out if you see it).
I know it seems like a lot, but the technique is relatively straight forward, the calculation is just some hard work. I hope this answer has shed some light on the process for you.
Best Answer
Since triangles $ABP$, $BCP$ and $CDP$ have common altitude from vertex $P$ and $AB = BC = CD$, these three triangles have same area: $$ \frac{1}{2}|AP|\cdot|BP|\sin\alpha = \frac{1}{2}|BP|\cdot|CP| = \frac{1}{2}|CP|\cdot|DP|\sin\beta $$ which implies $$ |BP| = |DP|\sin\beta, \quad |CP| = |AP|\sin\alpha. $$
On the other hand, we have $S_{APD} = S_{ABP} + S_{BCP} + S_{CDP} = 3S_{BCP}$: $$ \frac{1}{2}|AP|\cdot|DP|\sin\left(\frac{\pi}{2} + \alpha + \beta\right) = \frac{3}{2}|BP|\cdot|CP| = \frac{3}{2}|DP|\sin\beta\cdot|AP|\sin\alpha $$ which gives us $$ \sin\left(\frac{\pi}{2} + \alpha + \beta\right) = 3\sin\alpha\sin\beta $$ $$ \cos(\alpha + \beta) = 3\sin\alpha\sin\beta $$ $$ \cos\alpha\cos\beta = 4\sin\alpha\sin\beta $$ $$ 4\tan\alpha\tan\beta = 1 $$
Q.E.D.