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I have a triangle, vertices $ABC$, and a fixed point, $P$. Point $P$ is connected to each triangle vertex by lines $AP$, $BP$ and $CP$ respectively. The lengths of the sides of the triangle ($AB$, $BC$ and $CA$) are known, as are the angles $\angle APB$, $\angle BPC$ and $\angle CPA$.

Given this information I would like to find the lengths of lines $AP$, $BP$ and $CP$.

I intuitively believe that, given the constraints, the geometry of this scenario should not be variable - that is the position of point $P$ must be fixed relative to the triangle vertices. I cannot however seem to figure out how to calculate the position of point $P$.

enter image description here

What I have tried so far:

As I have one side and the corresponding angle of triangles $APB$, $APC$ and $BPC$ so I have attempted to use the sine rule. I know that, for example:

$$ \frac{\sin\angle APB}{AB}=\frac{\sin\angle ABP}{AP}=\frac{\sin\angle BAP}{BP} $$

And that these angles are related by:

$\angle APB + \angle ABP +\angle BAP = 180$

From here I have tried substituting various permutations of these equations into one another to isolate one unknown variable but have not had any success.

Any pointers in the right direction gratefully appreciated!

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    $\begingroup$ Please use law of cosine. You will get $3$ equations in $3$ variables. For example, $PA^2 + PB^2 - 2 PA \cdot PB \cos (\angle APB) = AB^2$ $\endgroup$
    – Math Lover
    Commented Jul 6, 2021 at 14:36
  • $\begingroup$ Hi Math Lover, could you elaborate on this please? I understand the cosine law, its permutations and how to use it but I can't see how it can solve my problem? I can see how you could use it to calculate the main angles of the triangle, e.g. $\angle ABC$ but not how this could be used to solve the problem. $\endgroup$ Commented Jul 6, 2021 at 15:28
  • $\begingroup$ You know $\angle APB$ and $AB$, yes? So what I have written gives you an equation in $PA, PB$, correct? Now similarly make equations in $PB, PC$ and $PC, PA$. $\endgroup$
    – Math Lover
    Commented Jul 6, 2021 at 15:32
  • $\begingroup$ I'm still failing to see how the cosine rule will work here. I don't think the equation you have given can be solved as there are still two unknown variables ($PA$ and $PB$). Am I missing something? $\endgroup$ Commented Jul 6, 2021 at 20:16
  • $\begingroup$ Are you reading my comments fully? That is one equation. You need to make two more equations. Please read my last comment. $\endgroup$
    – Math Lover
    Commented Jul 6, 2021 at 20:19

3 Answers 3

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Given a line segment AB, the loci of points P such that the angle APB has a constant value are circular arcs that pass through the points A and B. As a special case of this we have the well-known fact that if APB is a right angle, then the locus of points is the circle with the diameter AB.

https://www.mathpages.com/home/kmath173/kmath173.htm

(I reached this page by doing a search on "locus of points having given angle to given line segment".)

If you can find three points that satisfy the constraint of $\angle APB$ being a fixed value, you should be able to construct a circle from those points[1]. Once you've done this for all three angles, you'll have three circles, which will have either one or two points of common intersection (or zero, if the given angles do not correspond to any valid solution).

[1] I believe that $A$ and $B$ lie on this circle, which should simplify the construction a bit. One option for finding the third point is to solve for the point on the perpendicular bisector of $AB$.

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$P$ lies on the circle through $B$, $C$ whose center $O_A$ sees a half of $BC$ under an angle $180°-\alpha$ (where I set $\alpha=\angle BPC$). If we also set $a=BC$ then the radius of this circle is $r_A=a/(2\sin\alpha)$.

But $P$ also lies on the circle through $A$, $C$ whose center $O_B$ sees a half of $AC$ under an angle $180°-\beta$ (where I set $\beta=\angle APC$). If we also set $b=AC$ then the radius of this circle is $r_B=b/(2\sin\beta)$.

From the sides of triangle $ABC$ one can also compute $\gamma=\angle ACB$, and we have $\angle O_ACO_B=\alpha+\beta+\gamma-180°$. But $PC$ is twice the altitude of triangle $O_ACO_B$ with respect to base $O_AO_B$, that is:

$$ PC=2{-r_Ar_B\sin(\alpha+\beta+\gamma)\over\sqrt{r_A^2+r_B^2+2r_Ar_B\cos(\alpha+\beta+\gamma)}}. $$ Make the obvious changes to find $PA$ and $PB$.

enter image description here

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  • $\begingroup$ Wow, this is suuuuper! I had started barking up the tree of a similar inscribed circle theorem route but didn't get anywhere with it. This is a beautiful solution that I doubt I would have ever got to on my own. Thank you very much! $\endgroup$ Commented Jul 6, 2021 at 23:24
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Upon further reading, it seems my question is actually a formulation of the Weber problem - https://en.wikipedia.org/wiki/Weber_problem

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