Two Triangles $ABC$ and $DEF$ are superimposed such that the points $D$ and $E$ fall on $AC$ and $BC$ respectively. $DEF$ is an isosceles triangle where $DF$ is equal to $EF$
Given the base and height of $DEF$ and the coordinates of points $(A_x,A_y)$, $(B_x,B_y)$, $(F_x,F_y)$:
find $(C_x,C_y)$ such that $DEC$ is also an isosceles triangle.
This seems like a relatively straight forward geometry problem, but I'm stumped. Any and all advice is welcome.
On
This is an attempt to work through following Arentino's approach. It is incomplete at this point, and any and all help is appreciated. The basic geometry of the situation looks like this.
The equations of the two circles are given by $$(M_x−F_x)^2+(M_y−F_y)^2=h^2\tag{1}$$ and $$(D_x−F_x)^2+(D_y−F_y)^2=(L_x−F_x)^2+(L_y−Fy)^2=l^2\tag{2}$$ we also know that the points $D$ and $E$ fall on a line tangent to the first circle at point $M$. The slope from $FM$ is $$m_1 = \frac{M_y-F_y}{M_x-F_x}$$ therefore the slope of the tangent line is $m_2 = \frac{-1}{m_1}=-\frac{M_x-F_x}{M_y-F_y}=\frac{F_x-M_x}{M_y-F_y}$ and the equation of the line is given by $$y-M_y = m_2*(x-M_x)=\frac{F_x-M_x}{M_y-F_y}(x-M_x)\tag{3}$$
We know that $(D_x, D_y)$ and $(E_x, E_y)$ are both points which satisfies the two equations: $$y-M_y = (x-M_x)\frac{F_x-M_x}{M_y-F_y}\tag{3}$$ $$(x−F_x)^2+(y−Fy)^2=l^2\tag{4}$$
Solving $(3)$ for $y$ we get $$y= (x-M_x)\frac{F_x-M_x}{M_y-F_y}+M_y \tag{3}$$
To keep the calculations cleaner, let's say $m_3 = \frac{F_x-M_x}{M_y-F_y}$ leaving us with:
$$y= m_3(x-M_x)+M_y \tag{3}$$
Substituting $(3)$ into $(4)$ we get: $$(x−F_x)^2+(m_3(x-M_x)+M_y−Fy)^2=l^2\tag{8}$$
Which when solved for $x$ gives us two solutions which represent the $x$ coordinates $D_x$ and $E_x$ as a function of $M_x$ and $M_y$:
$$E_x = \frac{(-\sqrt{(l^2-M_x^2+2*F_x*M_x-F_x^2)*m^2+((2*F_x-2*M_x)*F_y+(2*M_x-2*F_x)*M_y)*m+l^2-F_y^2+2*M_y*F_y-M_y^2}-M_x*m^2+(M_y-F_y)*m-F_x}{m^2+1}\tag{8}$$ $$D_x = \frac{(\sqrt{(l^2-M_x^2+2*F_x*M_x-F_x^2)*m^2+((2*F_x-2*M_x)*F_y+(2*M_x-2*F_x)*M_y)*m+l^2-F_y^2+2*M_y*F_y-M_y^2}+M_x*m^2+(F_y-M_y)*m+F_x}{m^2+1}$$
Substituting $\pm(8)$ into $y= m_3x-m_3M_x+M_y \tag{3}$ to get equations for $E_y$ and $D_y$ we find:
$$E_y= m_3(\frac{(-\sqrt{(l^2-M_x^2+2*F_x*M_x-F_x^2)*m^2+((2*F_x-2*M_x)*F_y+(2*M_x-2*F_x)*M_y)*m+l^2-F_y^2+2*M_y*F_y-M_y^2}-M_x*m^2+(M_y-F_y)*m-F_x}{m^2+1})-m_3M_x+M_y \tag{9}$$
$$D_y= m_3(\frac{(\sqrt{(l^2-M_x^2+2*F_x*M_x-F_x^2)*m^2+((2*F_x-2*M_x)*F_y+(2*M_x-2*F_x)*M_y)*m+l^2-F_y^2+2*M_y*F_y-M_y^2}+M_x*m^2+(F_y-M_y)*m+F_x}{m^2+1})-m_3M_x+M_y$$
So the theory is that we now know $A$ and $B$ because they are given. We also now have equations for the $(E_x, E_y)$ and $(D_x, D_y)$ as functions of $(M_x, M_y)$. We can define equations for the lines $AD$ and $BE$ as:
$$y =\frac{A_y-D_y}{A_x-D_x}(x-A_x)+A_y\tag{5}$$ $$y =\frac{B_y-E_y}{B_x-E_x}(x-B_x)+B_y\tag{6}$$
Plugging $D_y$ and $D_x$ into $(5)$ gives us
$$y =\frac{A_y-(m_3(\frac{(\sqrt{(l^2-M_x^2+2*F_x*M_x-F_x^2)*m^2+((2*F_x-2*M_x)*F_y+(2*M_x-2*F_x)*M_y)*m+l^2-F_y^2+2*M_y*F_y-M_y^2}+M_x*m^2+(F_y-M_y)*m+F_x}{m^2+1})-m_3M_x+M_y)}{A_x-(\frac{(\sqrt{(l^2-M_x^2+2*F_x*M_x-F_x^2)*m^2+((2*F_x-2*M_x)*F_y+(2*M_x-2*F_x)*M_y)*m+l^2-F_y^2+2*M_y*F_y-M_y^2}+M_x*m^2+(F_y-M_y)*m+F_x}{m^2+1}}(x-A_x)+A_y\tag{10}$$
Plugging $E_x$ and $E_y$ into $(6)$ gives us:
$$y =\frac{B_y-(m_3(\frac{(-\sqrt{(l^2-M_x^2+2*F_x*M_x-F_x^2)*m^2+((2*F_x-2*M_x)*F_y+(2*M_x-2*F_x)*M_y)*m+l^2-F_y^2+2*M_y*F_y-M_y^2}-M_x*m^2+(M_y-F_y)*m-F_x}{m^2+1})-m_3M_x+M_y)}{B_x-(\frac{(-\sqrt{(l^2-M_x^2+2*F_x*M_x-F_x^2)*m^2+((2*F_x-2*M_x)*F_y+(2*M_x-2*F_x)*M_y)*m+l^2-F_y^2+2*M_y*F_y-M_y^2}-M_x*m^2+(M_y-F_y)*m-F_x}{m^2+1})}(x-B_x)+B_y\tag{11}$$ Next steps:
$(10)$ and $(11)$ represent the solutions to the lines $AD$ and $BE$ which intersect at point $C$ so we can set $(10)=(11)$ and solve for $x$.
$$\frac{A_y-(m_3(\frac{(\sqrt{(l^2-M_x^2+2*F_x*M_x-F_x^2)*m^2+((2*F_x-2*M_x)*F_y+(2*M_x-2*F_x)*M_y)*m+l^2-F_y^2+2*M_y*F_y-M_y^2}+M_x*m^2+(F_y-M_y)*m+F_x}{m^2+1})-m_3M_x+M_y)}{A_x-(\frac{(\sqrt{(l^2-M_x^2+2*F_x*M_x-F_x^2)*m^2+((2*F_x-2*M_x)*F_y+(2*M_x-2*F_x)*M_y)*m+l^2-F_y^2+2*M_y*F_y-M_y^2}+M_x*m^2+(F_y-M_y)*m+F_x}{m^2+1}}(x-A_x)+A_y=\frac{B_y-(m_3(\frac{(-\sqrt{(l^2-M_x^2+2*F_x*M_x-F_x^2)*m^2+((2*F_x-2*M_x)*F_y+(2*M_x-2*F_x)*M_y)*m+l^2-F_y^2+2*M_y*F_y-M_y^2}-M_x*m^2+(M_y-F_y)*m-F_x}{m^2+1})-m_3M_x+M_y)}{B_x-(\frac{(-\sqrt{(l^2-M_x^2+2*F_x*M_x-F_x^2)*m^2+((2*F_x-2*M_x)*F_y+(2*M_x-2*F_x)*M_y)*m+l^2-F_y^2+2*M_y*F_y-M_y^2}-M_x*m^2+(M_y-F_y)*m-F_x}{m^2+1})}(x-B_x)+B_y$$
This equation then needs to be solved for $M_x$ which is likely a nightmare. The result is then plugged into $y = \frac{M_y-F_y}{M_x-F_x}(x-F_x)-F_y\tag{7}$ which can then be solved for $M_y$
In theory this is going to work, but the odds that I haven't made a mistake in there somewhere are zero. Because the results aren't meaningful until the last step I can't check my work. I'm going to try again starting from the other side where I know $C$ and try to find $F$
On
Here's another approach.
We'll take $\triangle ABF$ to be the primary triangle; in what follows, $\angle A$ and $\angle B$, and $a$ and $b$, will indicate elements of that triangle. Let $A^\prime$, $B^\prime$, $D^\prime$, $E^\prime$ be feet of perpendiculars to the line through $F$ that is parallel to $\overline{AB}$. We'll take $\triangle DEF$ to be determined, not by its base and height, but equivalently, by its leg-length $r$ and vertex angle $2\phi$. Let the angle bisector of $\angle DFE$ meet $\overline{AB}$ at $G$, and introduce parameter $\theta = \angle AGF = \angle B^\prime F G$.
Our goal is to determine $\theta$, which is established by the fact that $\overleftrightarrow{AD}$, $\overleftrightarrow{BE}$, and $\overleftrightarrow{FG}$ meet at a common point, $C$. By the trigonometric form of Ceva's Theorem, this occurs if and only if
$$\frac{\sin\angle BAD}{\sin\angle DAF} \cdot \frac{\sin\angle FBE}{\sin\angle EBA} \cdot \frac{\sin\angle AFG}{\sin\angle GFB} \;=\; 1 \tag{$\star$}$$
We have ...
$$\begin{align} \angle AFG = \pi - \theta - \angle A &\quad\to\quad \sin\angle AFG = \sin(\theta + A) \\ \angle GFB = \phantom{\pi - \,}\theta - \angle B &\quad\to\quad \sin\angle GFB = \sin(\theta - B) \end{align}$$ ... and, by the Law of Sines, ...
$$\begin{align} \frac{\sin \angle DAF}{r} = \frac{\sin \angle DFA}{d} &\quad\to\quad \sin\angle DAF = \frac{r}{d}\,\sin\angle DFA = \frac{r}{d}\,\sin(\theta+\phi+A) \\[6pt] \frac{\sin\angle FBE}{r} = \frac{\sin\angle EFB}{e} &\quad\to\quad \sin\angle FBE = \frac{r}{e}\,\sin\angle EFB = \frac{r}{e}\,\sin(\theta-\phi-B) \end{align}$$
... and also (finally utilizing all those perpendiculars) ...
$$\begin{align} \sin \angle BAD = \frac{|\overline{DD^{\prime\prime}}|}{d} = \frac{|\overline{AA^{\prime}}|-|\overline{DD^\prime}|}{d} = \frac{b \sin A - r \sin(\theta+\phi)}{d} \\[6pt] \sin \angle EBA = \frac{|\overline{EE^{\prime\prime}}|}{e} = \frac{|\overline{BB^{\prime}}|-|\overline{EE^\prime}|}{e} = \frac{a \sin B - r \sin(\theta-\phi)}{e} \end{align}$$
With these, and a little cancellation, $(\star)$ becomes ...
$$\begin{align} &\sin(\theta + A)\;\sin(\theta-\phi-B)\;\left(b \sin A - r \sin(\theta+\phi)\right) \\[4pt] =\; &\sin(\theta-B)\;\sin(\theta+\phi+A)\;\left(a \sin B - r \sin(\theta-\phi)\right) \end{align} \tag{$\star\star$}$$
We an refine this slightly by introducing $p$ and $q$ as the circumdiameters of $\triangle ABF$ and $\triangle DEF$, respectively. The Law of Sines allows us to write ... $$a = p \sin A \qquad b = p \sin B \qquad r = q \sin\angle FDE = q\cos\phi$$
Therefore, we have ...
$$\begin{align} &\sin(\theta + A)\;\sin(\theta-\phi-B)\;\left(p \sin A \sin B - q \cos\phi \sin(\theta+\phi)\right) \\[4pt] =\; &\sin(\theta-B)\;\sin(\theta+\phi+A)\;\left(p \sin A \sin B - q \cos\phi \sin(\theta-\phi)\right) \end{align} \tag{$\star\star\star$}$$
Solving this equation for $\theta$ is a considerable chore that I must postpone for now.
On
This is an attempt to try to find a solution by working backwards. That means starting with values for $A,B$, the lengths of $AC$ and $BC$ and $b$ and $h$ for the small triangle and trying to find $F$. The reason for doing it that way is that I can test my answers at each step along the way to catch mistakes.
So we know $(A_x, A_y)$, $(B_x, B_y)$, $L_a$, and $L_b$ which are the lengths of $AC$ and $BC$. We also know that $A_y = B_y$ which may simplify things a little. If we declare $A$ as the origin then we can say $AB$ is of length $B_x$
By the law of cosines, we can say
$$\theta_{cab} = \arccos\left(\frac{L_a^2+B_x^2-L_b^2}{2L_aB_x}\right)\tag{1}$$
We can then say that
$$C_x = L_a\cos(\theta_{cab}) = L_a\left(\frac{L_a^2+B_x^2-L_b^2}{2L_aB_x}\right)\tag{2}$$
and
$$C_y = L_a\sin(\theta_{cab}) = L_a\sin\left(\arccos\left(\frac{L_a^2+B_x^2-L_b^2}{2L_aB_x}\right)\right)$$
By the identity $\sin(arccos(x))= \sqrt{1-x^2}$ we simplify that to:
$$C_y = L_a\sqrt{1-\left(\frac{L_a^2+B_x^2-L_b^2}{2L_aB_x}\right)^2}\tag{3}$$
We also know that $$\theta_{acb} = \arccos\left(\frac{L_a^2+L_b^2-B_x^2}{2L_aL_b}\right)\tag{4}$$
and that the lengths of $CD$ and $CE$ are given by
$$l_{CD}=l_{CE}=\pm\frac{b}{\sqrt2*\sqrt{1-\cos(\theta_{acb})}}=\pm\frac{b}{\sqrt2*\sqrt{1-\left(\frac{L_a^2+L_b^2-B_x^2}{2L_aL_b}\right)}}\tag{5}$$
We know this from simplifying the law of cosines for a equilateral triangle.
Where $$m_{AC} = \frac{A_y-C_y}{A_x-C_x} = \frac{A_y-L_a\sqrt{1-\left(\frac{L_a^2+B_x^2-L_b^2}{2L_aB_x}\right)}}{A_x-L_a\left(\frac{L_a^2+B_x^2-L_b^2}{2L_aB_x}\right)}$$
Because $A_x = A_y = 0$ we get some cancellations and end up with
$$m_{AC} = \frac{L_a\sqrt{1-\left(\frac{L_a^2+B_x^2-L_b^2}{2L_aB_x}\right)}}{\frac{L_a^2+B_x^2-L_b^2}{2L_aB_x}}\tag8$$
I was going to try to keep everything as one equation and solve for $l_a$ and $l_b$ at the end but that's not looking good I'm going to stop writing every equation as a function of known variables now.
We know that $$D_x = C_x - \frac{l_{CD}}{\sqrt{m_{AC}^2 + 1}}\tag6$$ $$D_y = m_{AC}*C_x\tag7$$
Why is this true? These equations are from old and messy notes. Where did I derive them from? $(7)$ comes from solving $m_{AC} = \frac{Ay-Dy}{Ax-Dx}=\frac{Dy}{Dx} \to D_y=m_{AC}D_x$
Similarly: $$M_{BC} = frac{B_y-C_y}{B_x-C_x}=\frac{C_y}{C_x-B_x}\tag9$$ $$E_x = C_x - \frac{l_{CD}}{\sqrt{m_{BC}^2 + 1}}\tag{10}$$ $$E_y = m_{BC}*(E_x-B_x)\tag{11}$$
$M$ is the midpoint of $DE$ so:
$$M_x = \frac{D_x + E_x}{2}\tag{12}$$ $$M_y = \frac{D_y + E_y}{2}\tag{13}$$
The slope of the line $MF$ is given by $$m_{MF}=\frac{-1}{m_{DE}}=\frac{-1}{\frac{D_y-E_y}{D_x-E_x}} = \frac{E_x-D_x}{D_y-E_y}\tag{14}$$
We can now find $F_x$ and $F_y$ because we know the height of the triangle $DEF$, $M$ and the slope $m_MF$.
$$F_x = M_x - \frac{h}{\sqrt{m_{MF}^2 + 1}}\tag{15}$$ $$F_y = M_y - (M_x-F_x)M_{MF}\tag{16}$$
Now that we have good values to test against we can check our work along the way as we work in the other direction.
On
Another partial solution that I'm still working on.
As per the previous partial-solutions, we construct $M$ as the midpoint of $\overline{DE}$, so $C$ lies on $\overleftrightarrow{MF}$. We're given the "base and height" of $\triangle DEF$ and we assume the base is side $\overline{DE}$. Hence, we also know the height, $h$, which is the length of $\overline{FM}$.
The lines $\overleftrightarrow{AC}$ and $\overleftrightarrow{BC}$ respectively have equations:
$$y=\frac{C_y-A_y}{C_x-A_x}(x-A_x)+A_y\quad(1)$$ $$y=\frac{C_y-B_y}{C_x-B_x}(x-B_x)+B_y\quad(2)$$
Let $c_1$ be the circle with centre $F$ and radius $q=FD=FE$. Hence, the equation for $c_1$ is:
$$(x-F_x)^2+(y-F_y)^2=q^2\quad(3)$$
Let $c_2$ be the circle with centre $C$ and radius $r=CD=CE$. Hence, the equation for $c_2$ is:
$$(x-C_x)^2+(y-C_y)^2=r^2\quad(4)$$
Given that we know the length of $\overline{DE}$, we can say that $s=\overline{DE}$ and hence,
$$\sqrt{(D_x-E_x)^2+(D_y-E_y)^2}=s\quad(5)$$
As we are given $h=FM$, we may say that:
$$\sqrt{(M_x-F_x)^2+(M_y-F_y)^2}=h\quad(6)$$
Given that $c_1$ and $c_2$ intersect $\overleftrightarrow{AC}$ and $\overleftrightarrow{BC}$ at $D$ and $E$ respectively, we can say that $(D_x,D_y)$ and $(E_x,E_y)$ are the points that satisfy $(3)$, $(4)$ and $(5)$, with the additional criteria that $(D_x,D_y)$ satisfies $(1)$ and $(E_x,E_y)$ satisfies $(2)$.
Substituting $(1)$ into $(4)$ and using a computer a computer-algebra-system to solve for $x=D_x$ gives us that:
$$D_x=\frac{\pm r(A_x-C_x)\sqrt{(A_y-C_y)^2+(A_x-C_x)^2}+C_x(A_y-C_y)^2+C_x(A_x-C_x)^2}{(A_y-C_y)^2+(A_x-C_x)^2}$$
We can label the distance of $\overline{AC}$ as $AC$ and simplify this to:
$$D_x=\frac{\pm r(A_x-C_x)}{AC}+C_x$$
WLOG, we can say:
$$E_x=\frac{\pm r(B_x-C_x)}{BC}+C_x$$
We may also easily find each $D_y$, $E_y$ by substitution into $(1)$ and $(2)$.
At the moment, I'm getting a computer-algebra-system to combine the equations and will get back to you with any progress.
Let $M=(M_x,M_y)$ be the midpoint of $DE$: point $M$ belongs to the circle of center $F$ and given radius $FH=h$, so that $(M_x-F_x)^2+(M_y-F_y)^2=h^2$. Once $M$ is chosen on that circle, then the coordinates of points $D$ and $E$ can be found, because they are the intersections between the line through $M$ perpendicular to $FM$ and the circle of center $F$ and given radius $FD=FE=l$.
You can then write the equations of lines $AD$ and $BE$, as a function of $M_x$ and $M_y$: their common point $C$ must lie on line $FM$, and that condition allows you to find $M$ and thus $C$.