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I have been struggling finding an analytic solution to the following geometry problem:

Consider an acute triangle $ABC$ (with all sides and angles known), with some unknown point $X$ in its interior. Denoting the (unknown) distances from the vertices to this interior point by $AX$, $BX$, and $CX$ (for simplicity assume $AX < BX < CX$), we are given the values of $s = BX-AX$, and $t = CX-BX$. We want to find $AX$. (I believe this should entirely pin down everything, see discussion later).Triangle Picture 1

A way to illustrate this is to draw three circles with center $X$ and radii $AX$, $BX (= AX + s)$ and $CX (= AX + s + t)$ respectively.Triangle Picture 2

Alternatively we can draw these circles with their centerpoints at $A$, $B$ and $C$ instead, which gives us $X$ as their intersection point: Triangle Picture 3

Now if we imagine varying the radius of the smallest circle $(AX)$, and always keeping the radii of the other two circles $AX + s$ and $AX + s + t$ respectively, then there should be(?) a unique value for $AX$ where all three circles intersect, and thus the point $X$ is uniquely determined. So although I'm pretty convinced there should be a unique solution, I'm struggling to actually derive an analytic formula for the answer.

I did try using pythagoras' theorem a bunch on a picture like this: Triangle Picture 4

However I didn't seem to get very far, I just got kinda lost in the formulas. Is it possible to find a formula for the distance $AX$?

As a sidenote, during digging through the formulas I discovered that in this setup (interior point in acute triangle with normals splitting all three sides), we get that $a^2 + b^2 + c^2 = (a')^2 + (b')^2 + (c')^2$ is satisfied. Does this result have a name? Can it be generalized? (What happens with non-interior points, obtuse triangles, is this condition satisfied if and only if the sides are split by a point in such a way?)

Hodge
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2 Answers2

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The formula to be derived are so messy that I can only write down the general ideas and illustrate the work with a special example.

enter image description here

Refer to the figure, let $p=s+t$.

From $\Delta ABX$,

$$ XB^2=AX^2+AB^2-2(AX)(AB)\cos \alpha$$ $$(r+s)^2=r^2+c^2-2rc\cdot \cos \alpha$$ $$r^2+2rs+s^2=r^2+c^2-2rc\cdot \cos \alpha$$ $$2r(s+c \cdot \cos \alpha)=c^2-s^2 \tag{1}$$

Similarly from $\Delta ACX$, $$2r(p+ b \cdot \cos(A-\alpha))=b^2-p^2 \tag{2}$$

From $(1), (2)$, we have $$(c^2-s^2)(p+b \cdot \cos(A-\alpha))=(b^2-p^2)(s+c\cdot \cos \alpha)$$

$$(c^2-s^2)(p+b \cdot (\cos A \cos \alpha+ \sin A \sin \alpha))=(b^2-p^2)(s+c\cdot \cos \alpha)$$

The equation above can be reduced to the form:

$$m \cdot \cos \alpha + n \cdot \sin \alpha = k \tag{3} $$

From $(3)$, we can find $\alpha$ by standard method Solving $a \cos x + b \sin x = c$ and hence $r$ from either $(1)$ or $(2)$.

Example: Suppose we want to solve for $r$ with the information given in the figure.

enter image description here

From $\Delta ABX$, $$(r+\sqrt 3-1)^2=r^2+2^2-4r\cos \alpha$$

$$2r((\sqrt 3 -1)+2 \cdot \cos \alpha) = 4-(\sqrt 3 - 1)^2 $$

$$r((\sqrt 3 -1)+2\cos \alpha) = \sqrt 3 \tag{1}$$

Similarly from $\Delta ACX$, $$(r+\sqrt {13}-1)^2=r^2+16-8r\cos(120^{\text o}-\alpha)$$

$$2r((\sqrt {13}-1)+4 \cdot \cos (120^{\text o}-\alpha))= 16-(\sqrt {13}-1)^2$$

$$r((\sqrt {13}-1)+4\cos (120^{\text o}-\alpha))= 1+ \sqrt {13} \tag{2}$$

$(1), (2) \implies$

$$\sqrt 3\left( \sqrt {13}-1 +4 \cos(120^{\text o}-\alpha) \right)=(\sqrt {13}+1)\left(\sqrt 3 -1 + 2 \cos \alpha \right)$$

$$\sqrt {13}-2\sqrt 3+1=(2\sqrt{13}+2)\cos \alpha -4\sqrt 3(\cos 120^{\text o} \cos \alpha + \sin 120^{\text o}\sin \alpha )$$

$$(2\sqrt {13}+2\sqrt 3+2)\cos \alpha - 6 \sin \alpha = \sqrt {13}-2\sqrt 3 +1 \tag{3}$$

$(3)$ is of the form $$m \cdot \cos \alpha + n \cdot \sin \alpha = k $$ which can be solved by standard method.

Indeed the solution of $(3)$ is $$\alpha = 60^{\text o}$$

Once we have $\alpha = 60^{\text o}$, we can find the value of $r$ from $(1)$ $$r((\sqrt 3 -1)+2\cos \alpha) = \sqrt 3$$ which gives $$r=1.$$

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It is possible to give an analytic solution, using the $p,q$ method. It turns out to be a rather complicated function of your variables $s,t$ and the side lengths--too long to post here but I can describe the rather simple method.

We denote the side lengths by $a,b,c$ as usual. Then there is a unique pair $p,q$ so that your triangle is directly congruent to one with vertices $(0,0)$, $(c,0)$ and $(p,q)$. You can determine $p,q$ as functions of the side lengths by solving (by hand or using, say, Mathematica) $$p^2+q^2=b^2, (p-c)^2+q^2=a^2.$$ Now suppose that $X$ is $(x ,y)$ . You can solve your equations for $x$ and $y$ in terms of $p,q,c$ (I used Mathematica for this task--it coped with ease). You can now use the first solution to express this one in terms of the side lengths to get the required result. As explained it is too long for me to post here but is the work of a few minutes with the aid of Mathematica.