Prove that the equation of the cone is $yz(\frac{b}{c}+\frac{c}{b})+zx(\frac{c}{a}+\frac{a}{c})+xy(\frac{a}{b}+\frac{b}{a})=0$

Question

The plane $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ cuts the coordinate axes in $A,B,C$. Prove that the lines passing through the origin and intersecting the circle $ABC$ generate the cone with equation

$$yz(\frac{b}{c}+\frac{c}{b})+zx(\frac{c}{a}+\frac{a}{c})+xy(\frac{a}{b}+\frac{b}{a})=0$$

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My Attempt:
The equation of the generator line passing through the origin and direction cosines $l,m,n$ is $\frac{x}{l}=\frac{y}{m}=\frac{z}{n}$

Cone is a surface generated by the lines called generators passes through fixed point called vertx on the fixed curve called guiding curves.

I am not able to find the equation of the guiding curve the circle $ABC$.Hence i could not find out the equation of the cone.

Please help.

Answer 1

Let $(X,Y,Z)$ be the center of the circle $ABC$. This is on the plane $$\frac xa+\frac yb+\frac zc=1\tag1$$ so, we can write $$Z=C\left(1-\frac Xa-\frac Yb\right)$$ Solving $$(X-a)^2+Y^2+\left(C\left(1-\frac Xa-\frac Yb\right)\right)^2=X^2+(Y-b)^2+\left(C\left(1-\frac Xa-\frac Yb\right)\right)^2=X^2+Y^2+\left(C\left(1-\frac Xa-\frac Yb\right)-C\right)^2$$ gives $$X=\frac{a^3(b^2+c^2)}{2(a^2b^2+b^2c^2+c^2a^2)},\quad Y=\frac{b^3(c^2+a^2)}{2(a^2b^2+b^2c^2+c^2a^2)},\quad Z=\frac{c^3(a^2+b^2)}{2(a^2b^2+b^2c^2+c^2a^2)}$$

Since the radius is given by $$\sqrt{\left(\frac{a^3(b^2+c^2)}{2(a^2b^2+b^2c^2+c^2a^2)}-a\right)^2+\left(\frac{b^3(c^2+a^2)}{2(a^2b^2+b^2c^2+c^2a^2)}\right)^2+\left(\frac{c^3(a^2+b^2)}{2(a^2b^2+b^2c^2+c^2a^2)}\right)^2}=\frac 12\sqrt{\frac{(a^2+b^2)(b^2+c^2)(c^2+a^2)}{a^2b^2+b^2c^2+c^2a^2}}$$ the circle $ABC$ is represented as $$\left(x-\frac{a^3(b^2+c^2)}{2(a^2b^2+b^2c^2+c^2a^2)}\right)^2+\left(y-\frac{b^3(c^2+a^2)}{2(a^2b^2+b^2c^2+c^2a^2)}\right)^2+\left(z-\frac{c^3(a^2+b^2)}{2(a^2b^2+b^2c^2+c^2a^2)}\right)^2=\frac{(a^2+b^2)(b^2+c^2)(c^2+a^2)}{4(a^2b^2+b^2c^2+c^2a^2)}\tag2$$ and $(1)$.

Any point on the line $$\frac xl=\frac ym=\frac zn\tag3$$ is represented as $(lr,mr,nr)$.

So, from $(1)$, we can determine $r$ : $$\frac{lr}a+\frac{mr}b+\frac{nr}c=1\implies r=\frac{1}{\frac la+\frac mb+\frac nc}$$ From $(2)$, using the $r$ and letting $d=2(a^2b^2+b^2c^2+c^2a^2)$, $$\left(\frac{l}{\frac la+\frac mb+\frac nc}-\frac{a^3(b^2+c^2)}{d}\right)^2+\left(\frac{m}{\frac la+\frac mb+\frac nc}-\frac{b^3(c^2+a^2)}{d}\right)^2+\left(\frac{n}{\frac la+\frac mb+\frac nc}-\frac{c^3(a^2+b^2)}{d}\right)^2=\frac{(a^2+b^2)(b^2+c^2)(c^2+a^2)}{2d},$$ i.e. $$\left(\frac{1}{\frac 1a+\frac{m}{lb}+\frac{n}{lc}}-\frac{a^3(b^2+c^2)}{d}\right)^2+\left(\frac{1}{\frac{l}{ma}+\frac{1}{b}+\frac{n}{mc}}-\frac{b^3(c^2+a^2)}{d}\right)^2+\left(\frac{1}{\frac{l}{na}+\frac{m}{nb}+\frac{1}{c}}-\frac{c^3(a^2+b^2)}{d}\right)^2=\frac{(a^2+b^2)(b^2+c^2)(c^2+a^2)}{2d}$$

Eliminating $l,m,n$ by using $(3)$, $$\left(\frac{1}{\frac 1a+\frac{y}{xb}+\frac{z}{xc}}-\frac{a^3(b^2+c^2)}{d}\right)^2+\left(\frac{1}{\frac{x}{ya}+\frac{1}{b}+\frac{z}{yc}}-\frac{b^3(c^2+a^2)}{d}\right)^2+\left(\frac{1}{\frac{x}{za}+\frac{y}{zb}+\frac{1}{c}}-\frac{c^3(a^2+b^2)}{d}\right)^2=\frac{(a^2+b^2)(b^2+c^2)(c^2+a^2)}{2d},$$ i.e. $$\left(\frac{xabc}{xbc+yac+zab}-\frac{a^3(b^2+c^2)}{d}\right)^2+\left(\frac{yabc}{xbc+yac+zab}-\frac{b^3(c^2+a^2)}{d}\right)^2+\left(\frac{zabc}{xbc+yac+zab}-\frac{c^3(a^2+b^2)}{d}\right)^2=\frac{(a^2+b^2)(b^2+c^2)(c^2+a^2)}{2d},$$ i.e. $$\frac{a^2b^2c^2}{(xbc+yac+zab)^2}(x^2+y^2+z^2)-\frac{2abc}{(xbc+yac+zab)d}\left(xa^3(b^2+c^2)+yb^3(c^2+a^2)+zc^3(a^2+b^2)\right)-\frac{2a^2b^2c^2}{d}=0$$ Multiplying the both sides by $\frac{(xbc+yac+zab)^2d}{2abc}$, we get $$(a^2b^2+b^2c^2+c^2a^2)abc(x^2+y^2+z^2)-(xbc+yac+zab)\left(xa^3(b^2+c^2)+yb^3(c^2+a^2)+zc^3(a^2+b^2)\right)-abc(xbc+yac+zab)^2=0$$ Expanding the LHS, we get $$c(a^2+b^2)(a^2 b^2+b^2c^2+c^2a^2)xy+a(b^2+c^2)(a^2 b^2+b^2c^2+c^2a^2)yz+b(a^2+c^2)(a^2 b^2+b^2c^2+c^2a^2)zx=0$$ Finally, dividing the both sides by $abc(a^2b^2+b^2c^2+c^2a^2)\not=0$ gives $$\left(\frac ab+\frac ba\right)xy+\left(\frac bc+\frac cb\right)yz+\left(\frac ca+\frac ac\right)zx=0$$ as desired. (as some commented, the coefficient of $yz$ should be $\frac bc+\frac cb$.)

Answer 2

I do not know if it is really helpful, but the center $P$ of the circle has (horrible) coordinates $\dfrac{1}{2(a^2b^2+b^2c^2+c^2a^2)}\begin{pmatrix} a^3(b^2+c^2) \\ b^3(c^2+a^2) \\ c^3(a^2+b^2) \end{pmatrix}$.

Indeed, in the plane, we want to find the circle passing through the vertices of the triangle $ABC$. The center $P$ of this circle is the intersection of the perpendicular bisectors of $[AB]$ and $[AC]$. The middle $M$ of $[AB]$ has coordinates $\begin{pmatrix} \frac{a}{2} \\ \frac{b}{2} \\ 0 \end{pmatrix}$. We want the vectors $\overrightarrow{PM}$ and $\overrightarrow{AB}$ to be perpendicular, so we get the equation $ax-by = \dfrac{a^2-b^2}{2}$. Similarly for $[AC]$ we get $ax-cz = \dfrac{a^2-c^2}{2}$. Do not forget the third equation $\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$. Thus we have a linear system to solve, and the unique solution is the aforementioned one.

ADDENDUM after edit:

Let $M=\begin{pmatrix} x\\ y\\ z\end{pmatrix}$ and $t=\dfrac{x}{a}+\dfrac{y}{b} + \dfrac{z}{c}$. The line $(OM)$ cuts the plane if and only if $t \neq 0$. If so, the point of intersection is $M'=\dfrac{1}{t} M_0$. Then $M'$ is on the circle if and only if $\|P-M'\|^2 = \|P-A\|^2$. Now we have (with the obvious abuse of notation about norms of vectors and scalar product) \begin{align*} &\|P-M'\|^2 = \|P-A\|^2 \iff P^2-2P\cdot\dfrac{M}{t} + \dfrac{M^2}{t^2} = P^2 - 2aP_1+a^2 \\ &\iff M^2-2tP\cdot M = (a^2-2aP_1)t^2 \\ &\iff x^2+y^2+z^2-\dfrac{\dfrac{x}{a}+\dfrac{y}{b} + \dfrac{z}{c}}{a^2b^2+b^2c^2+c^2a^2}\left(xa^3(b^2+c^2)+yb^3(c^2+a^2)+zc^3(a^2+b^2)\right) = \dfrac{a^2b^2c^2}{a^2b^2+b^2c^2+c^2a^2}\left(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2} + \dfrac{2xy}{ab}+\dfrac{2yz}{bc}+\dfrac{2zx}{ca}\right)\end{align*}

You can check that the terms in $x^2$, $y^2$ and $z^2$ vanish, and that the remaining terms give you the desired equation. For example, when putting all the terms in the RHS, the coefficient of $xy$ is $\dfrac{1}{a^2b^2+b^2c^2+c^2a^2}\left(2abc^2+\dfrac{b^3(c^2+a^2)}{a} + \dfrac{a^3(b^2+c^2)}{b}\right) = \dfrac{1}{a^2b^2+b^2c^2+c^2a^2} \dfrac{2a^2b^2c^2+b^4(c^2+a^2)+a^4(b^2+c^2)}{ab} = \dfrac{a^2+b^2}{ab} = \dfrac{a}{b}+\dfrac{b}{a}$.

I have to admit this is a bit cumbersome...