Question about substitution

[ins-]

Do there exists a substitution that makes the following equation reciprocal / Quasi-palindromic equation:

$a^2 b c^2 + (-2 a^2 b c - a^2 c^2 - a b c^2) x + (a^2 b - a b c - a c^2) x^2 + (2 a b + b c) x^3 + b x^4=0$

or some other way to solve it avoiding Ferrari's formula?

[DeBill]

ins-
$b$ входит линейно. Так что единственная надежда - разложить на множители.
И вот если бы перед $abc^2$ был знак "плюс" - то $x=-a$ - подходит....

[ins-]

DeBill

Thank you for the idea. If one of the members in the coefficient before $x$ is missing it will be a reciprocal equation. I'll recheck my calculations. It is a part of a bigger problem:

If in a triangle $ABC$ is chosen a point $P$ and the lines $AP$, $BP$, $CP$ intersect the opposite to vertices sides at the points $A'$, $B'$, $C'$ and the areas of $APC'$, $BPA'$, $CPB'$ are $a$, $b$, $c$ - find the area of $ABC$.

I solved it using 3 different methods but the lowest possible degree of the equations I obtained is 4-th. The equation I posted is one of the equations I have as an intermediate result. $x$ is the area to one of the triangles formed by ABC's cevians.

[DeBill]

ins-
Недавно, на Израильской олимпиаде, была задача: доказать, что если $a=b=c$, то и три другие площади равны.
Задачу эту можно было свести к симметричной системе из трех уравнений (приводящую к уравнению высокой степени), которую можно было решить из соображений монотонности (если $x<y$, то $y<z$, и тогда $z<x$, противоречие). У вас система будет асимметричной. Но, если угадать решение, то тем же методом можно бы доказать и отсутствие других решений.... Но....

(Оффтоп)

Легко получить $abc = xyz, ab+bc+ac =xy+yz+zx$. Но что дальше?

[ins-]

About the problem from Israel - it can be proven from $a=b=c$ that $ABC$ is equilateral.
I also have proofs for the facts $abc=xyz$ and $ab+bc+ca=xy+yz+zx$ but don't know how to proceed further.
For a=84, b=35, c=40 I'm capable to solve the problem in 3 ways, but for the common case using software I found 2 kilometer long roots exist. It is the reason to wonder is there something simpler as an expression.
If we denote the ratios in which the cevians divide the triangle's sides with $m$, $n$, $k$ and the areas are $a$, $b$, $c$ the following system is correct:

$mnk=1$

$\frac{1+n}{m(k+1)}=\frac{a}{b}$

$\frac{1+m}{k(n+1)}=\frac{b}{c}$

$\frac{1+k}{n(m+1)}=\frac{c}{a}$

but again ugly 4-th degree equations can be found.

I wonder if we express the ratios $x/y$, $y/z$, $z/x$ where $x$, $y$, $z$ are the unknown areas in terms of $k$, $m$, $n$ and try to eliminate $m$, $n$, $k$ what will happens.

[DeBill]

ins- в сообщении #1184040 писал(а):

it can be proven from $a=b=c$ that $ABC$ is equilateral.

Нет, только равенство $m=n=k=1$

[ins-]

DeBill

Thank you for spending time to this problem and notifying me about my mistake. I solved it in 5 different ways, but wasn't capable to find an equation with a degree less than 4. I composed another problem with areas that leads in the common case to an equation with degree 3 with (I suppose) not a nice root $x^3+(a+b+c)x^2-4abc=0$. There are similar problems.