Solving for point P within the edge AC, given points A, B and C, and length from B to P?

Question

(EDIT: one of the comments noted that AC is too short in the image, this is completely my fault for not checking before rounding, therefor replace the magnitude of AC to, say 10)

So given points A B and C we can calculate the magnitudes of AB, AC, and CB, and we know there is a point in the line of AC that has the distance of, say 60 like in the image, to the point B. What are the necessary steps to solve for point P?

Answer 1

This is the Stewart’s theorem in its puriest form:

\begin{align} |BC|^2\cdot|AP|+ |AB|^2\cdot(|AC|-|AP|) =|AC|\cdot(|BP|^2+|AP|\cdot(|AC|-|AP|)) \tag{1}\label{1} , \end{align}

\begin{align} |AP|&= \frac{|AB|^2+|AC|^2- |BC|^2\pm\sqrt{\Big(|AB|^2-|BC|^2\Big)^2-|AC|^2\cdot\Big(2|BC|^2+2|AB|^2-|AC|^2-4|BP|^2\Big)}} {2|AC|} \tag{2}\label{2} , \end{align}

for given $|AB|=61$, $|BC|=58$, $|BP|=60$, $|AC|=10$
\eqref{2} gives

\begin{align} |AP|&=\tfrac1{20}(457\pm\sqrt{160449}) \end{align}

with the only valid root

\begin{align} |AP|&=\tfrac1{20}(457-\sqrt{160449}) \approx 2.82 . \end{align}

Edit

Note that for some set of side lengths there are indeed two valid solutions, for example if we have $|AB|=61$, $|BC|=58$, $|BP|=55$, $|AC|=60$, the expressions

\begin{align} |AP_+|&=\tfrac1{40}\,(1319+\sqrt{626161}) \approx 52.7576 \\ \text{and }\quad |AP_-|&=\tfrac1{40}\,(1319-\sqrt{626161}) \approx 13.1924 \end{align} are valid and give two points as the answer:

\begin{align} P_+&=A+\frac{|AP_+|}{|AC|}\cdot(C-A) ,\\ P_-&=A+\frac{|AP_-|}{|AC|}\cdot(C-A) , \end{align}

see the picture:

Answer 2

The 'basic triangle calculus' you are referring to are the Law of Sines and Cosines. In this particular example, we apply the Law of Cosines twice. We have:

$$AB^2 = AC^2+CB^2-2AC\cdot CB \cos\angle ACB$$

which gives

$$\cos \angle ACB = \frac {10^2+58^2-61^2}{2\cdot10\cdot58}=-\frac {257}{1160}$$

Now we also have from Law of Cosines:

$$BP^2 = PC^2+CB^2-2PC\cdot CB\cos\angle ACB$$

with $BP=60, BC =58$, the equation above is just a quadratic equation in $PC$, which we can solve by the quadratic formula. Here is WolframAlpha taking a crack at it.

Answer 3

Note \begin{align} AP &= AB \cos A -BP \cos \angle BPC \\ &= AB \cos A -BP \sqrt{ 1- \sin^2 \angle BPC}\\ &= AB \cos A -BP \sqrt{ 1- \sin^2 \angle BPA} \\ &= AB \cos A -BP \sqrt{ 1- \frac{AB^2}{BP^2}\sin^2A}\>\>\>\>\>\>\>(sine \>rule)\\ &= AB \cos A -\sqrt{ BP^2 - {AB^2}(1-\cos^2A)}\\ \end{align} where $AB$ and $BP$ are given and from the cosine rule $$\cos A =\frac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC} $$