This should be a simple problem to solve, but I cannot seem to get my head around it. I would like to find the third vertex of the triangle shown below $(x_i, y_i)$ given the following information:

  • I know the position of two vertices of a triangle: $(x_a, y_a)$ and $(x_b, y_b)$.
  • I can calculate the angle $\theta=\arctan\left(\frac{y_b-y_a}{x_b-x_a}\right) $
  • I can also calculate the length of one side of the triangle: $\sqrt{(x_b-x_a)^2+(y_b-y_a)^2}$.
  • The other two sides are of length $s_b\Delta t$ and $s_a\Delta t$ where $s_a$ and $s_b$ are both known, but $\Delta t$ is not.
  • I also know the angle $\phi$, as shown.

What isn't clear to me is if I have enough information for a solution. Can this be solved and, if so, can you point me in the right direction of how to do so? (My old math teacher's voice is in my head muttering "Law of cosines!", which I could apply to get angles within my triangle but I'm drawing a blank on how to proceed from there.) Any advice would be greatly appreciated!

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  • $\begingroup$ Lyngbakr, $\phi$ is probably the angle opposite to side it's vertices given. If so, you have to find the equation of perpendicular bisector of this side .The measure of the arc connecting these vertices is two times the measure of $\phi$, this show the position of the center of the circle passing the vertices. Then third vertex can be any ponit on the large section of circle. $\endgroup$
    – sirous
    Feb 19, 2021 at 1:55

2 Answers 2



We would like to emphasize that the information given by the OP is enough to solve this problem.

Let us describe what is known as the circle of Apollonius. Given one side of a triangle and the ratio of the lengths of the other two sides, the locus of the third vertex is the Apollonius circle. Sounds familiar? Well, this is exactly what you need to construct to find the sought third vertex mentioned in your problem statement. Probably, you may not need the construction illustrated in $\mathrm{Fig.\space 1}$, if you decide to determine the third vertex analytically. However, we recommend you to study it, because it sheds light on the problem you are trying to solve.

Draw the line segment $AB$ to make an angle $\theta$ with the horizontal as requested in the problem. We assume for the time being $s_A > s_B$. Find the point $E$ on $AB$, such that $AE\space \mathbf{:}\space EB = s_A\space \mathbf{:}\space s_B$. Draw a line segment $BD$, so that $BD > BA$. After joining $AD$, construct a parallel line to it through $E$ to intersect $BD$ at $E$. Draw a circle $\Gamma_1$ with radius $BF$ and center at $B$. Draw the circle $\Gamma_2$ with radius $FD$ and center at $A$. These two circles intersect each other at two points, but we need only one of them, say $G$. Join $EG$ and construct its perpendicular bisector to meet the extended $AB$ beyond $B$ at $O$. This, in fact, is the center of the mentioned Apollonius circle. Complete the construction by drawing a circle having the radius $OE$.

Finally, draw a line through $B$ to make the given angle $\phi$ with the horizontal to meet the Apollonius circle at $C=\left(x_C,\space y_C\right)$, which is the sought third vertex of the triangle.

Please note that, if $s_A < s_B$, the center of the Apollonius circle is located on the extended $BA$ beyond $A$.

Here is the recipe for analytical determination of the third vertex. As shown in $\mathrm{Fig.\space 2}$, we let $\measuredangle CAB = \omega$ and $\measuredangle BCA = \beta$, where $\beta = 180^o + \phi - \theta - \omega$. You have already derived the following couple of equations and, therefore, they need no explanations.

$$AB =\sqrt{\left(x_A – x_B\right)^2 + \left(y_A – y_B\right)^2} \tag{1}$$ $$\theta = \tan^{-1} \left(\frac{y_A – y_B}{x_A – x_B}\right) \tag{2}$$

We apply the law of sines to the triangle $ABC$. $$\frac{CA}{\sin\left(\theta - \phi\right)} = \frac{AB}{\sin\left(\beta\right)} = \frac{BC}{\sin\left(\omega\right)} \tag{3}$$

$$\therefore\quad \frac{BC}{CA} = \frac{s_B\Delta t}{s_A\Delta t} = \frac{\sin \left(\omega\right)}{\sin\left(\theta - \phi\right)} \quad\rightarrow\quad \omega = \sin^{-1}\left(\frac{s_B}{s_A}\sin\left(\theta - \phi\right)\right). \tag{4}$$

Similarly, from (3) we have, $$BC=AB\frac{\sin \left(\omega\right)}{\sin \left(\theta + \omega - \phi\right)}. \tag{5}$$

Since $AB$ and $\omega$ are known to us from(1) and (4) respectively, we can determine $BC$ from (5). Once we know $BC$, we can find the $x$- and $y$-coordinates of $C$ from $$x_C = x_B – BC\cos\left(\phi\right)\quad\space\mathrm{and}\space\quad y_C = y_B – BC\sin\left(\phi\right).$$

  • $\begingroup$ Nice work. The construction of an extra suitable ratio point on the Apollonian circle is interesting, but I think I would use $r(\Gamma_1) = 2|BE|, r(\Gamma_2) = 2|AE|$ by doubling distances along the $AB$ line. $\endgroup$
    – Joffan
    Feb 23, 2021 at 9:53
  • $\begingroup$ This is a great help and is much appreciated! I'm curious – what package did you use to produce those beautiful diagrams? $\endgroup$
    – Dan
    Feb 23, 2021 at 14:40
  • 1
    $\begingroup$ @Lyngbakr The name of the package is called GeoGebra. It is free. $\endgroup$
    – YNK
    Feb 23, 2021 at 15:41
  • $\begingroup$ @Joffan Thanks for the suggestion. It has 2 construction steps less than mine. $\endgroup$
    – YNK
    Feb 23, 2021 at 15:44
  • $\begingroup$ @YNK No problem. In fact both our methods could fail if $E$ is sufficiently close to $B$. After a little thought, a much better method is to construct a circle $\Gamma_1$ on $E$ through $B$, make the perpendicular to $AB$ on $B$ and the tangent from $A$ to $\Gamma_1$ meeting that perpendicular at $G$. Then $AG$ is also tangent to the Apollonian Circle so a perpendicular from $G$ will find the centre $O$. $\endgroup$
    – Joffan
    Feb 23, 2021 at 18:24


Working with complex numbers. Denote by $a$ the point $(x_a,y_a)$, similarly for $b$ and $m = (x_i,y_i)$ the point to be find. Also denote $\lambda$ the positive real $\Delta t$.

You have $m = b - s_b\lambda e^{i \phi}$ and $\vert m-a \vert = s_a \lambda$. Which leads to the equation

$$\vert b - s_b \lambda e^{i \phi} -a \vert^2 =s_a^2 \lambda^2.$$

This is one quadratic equation in $\lambda$ that you can solve.

  • $\begingroup$ Thanks for the hint! It's much appreciated. Stupid question: when combining your expressions why did you square them? $\endgroup$
    – Dan
    Feb 18, 2021 at 16:37
  • 1
    $\begingroup$ I did that because squared complex modulus avoids using square roots. $\endgroup$ Feb 18, 2021 at 17:34

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