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Heron's formula

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Source: Serbia and Montenegro 2004

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2238 posts

#1 h Aug 11, 2005, 9:20 AM • 1 Y

Y by Adventure10

The side lengths of a triangle are the roots of a cubic polynomial with rational coefficients. Prove that the altitudes of this triangle are roots of a polynomial of sixth degree with rational coefficients.

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2238 posts

#2 h Aug 14, 2005, 10:49 AM • 1 Y

Y by Adventure10

hayyyyyyy,,no one

it's very beatiful

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2105 posts

#3 h Aug 14, 2005, 11:19 AM • 2 Y

Y by Adventure10, Mango247

ehsan2004

please!
could you send the solution??
This was also in greek IMO Tst 2005
(problem 1)

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enescu

#4 h Aug 14, 2005, 12:13 PM • 2 Y

Y by Adventure10, Mango247

Suppose that $a,b,c$ are the roots of the polynomial
$f(x)=x^3-mx^2+nx-p,$
with $m,n,p\in \mathbb{Q}$ and let $S$ denote the triangle's area.
Heron's formula and standard manipulations with Viete's formulas yield
$S^2=-m^4+4m^2n-8mp,$
hence $S=\sqrt{A}$ , for some $A\in \mathbb{Q}.$
We have $h_a=\frac{2S}{a}$ , hence $a=\frac{2\sqrt{A}}{h_a}$ . Write that $f(a)=0$ , group the terms containing $\sqrt{A}$ and square again. After clearing the denominators, we find that $h_a$ is the root of a 6th degree rational polynomial.

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#5 h Aug 14, 2005, 3:30 PM • 4 Y

Y by Adventure10, Chokechoke, Mango247, NicoN9

I have a good solution.
Because the coefficients are rational we write the equation as
$x^3-Ax^2+Bx-C=0$
Now let a,b,c the lengths of the triangle.
a+b+c=2t=A
ab+bc+ca=B
abc=C
We get all these from Vieta.
And from Iron we have that $E^2=t(t-a)(t-b)(t-c)=t(t^3+At^2+Bt-C)$ which is rational.
Also we have that
$\upsilon_a=\frac{2E}{a}$
$\upsilon_b=\frac{2E}{b}$
$\upsilon_c=\frac{2E}{c}$
are roots of $(x^2-\frac{4E^2}{a^2})(x^2-\frac{4E^2}{b^2})(x^2-\frac{4E^2}{c^2})=0$
This equation is 6th degree with rational coefficients.
There are rational because $\frac{4E^2}{b^2}+\frac{4E^2}{c^2}+\frac{4E^2}{a^2}=\frac{4E^2}{C^2}(B^2-2AC)$ is rational
Also ${ \frac{16E^4}{b^2a^2}+\frac{16E^4}{b^2c^2}+\frac{16E^4}{c^2a^2}= \frac{16(E^2)^2}{C^2}(A^2-2B)}$
The last is obviously rational so all are rational and we are done.

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NewAlbionAcademy

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NewAlbionAcademy

#6 h Dec 26, 2013, 11:47 PM • 2 Y

Y by Adventure10, Mango247

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#7 h Sep 26, 2022, 11:51 PM • 1 Y

Y by Mango247

This problem is not as daunting as it may seem.

Suppose $r_1, r_2, r_3$ are roots of some cubic $ax^3 + bx^2 + cx + d$ . Then we have that by Vieta, $r_1 + r_2 + r_3 = \frac{-b}{a} \in \mathbb{Q}$ , and $r_1r_2 + r_2r_3 + r_3r_1 = \frac{c}{a} \in \mathbb{Q}$ , and finally $r_1r_2r_3 = -d/a \in \mathbb{Q}$ . Hence each $r_i, 1 \leq i \leq 3$ is rational also.

Note that the altitudes are given by $2[\text{area}]/r_i$ . Now the area is simply $\sqrt{s(s - r_1)(s - r_2)(s - r_3)}$ , where $s$ is the semiperimeter (this is due to Heron's formula). Now for simplicity, let $X = \sqrt{s(s - r_1)(s - r_2)(s - r_3)}$ . Consider the polynomial $\left(x^2 - \frac{4X^2}{r_1^2}\right) \left(x^2 - \frac{4X^2}{r_2^2} \right) \left(x^2 - \frac{4X^2}{r_3^2} \right).$ Here each $\frac{4X^2}{r_i^2}$ is clearly rational. Hence this new polynomial will have rational coefficients, and also have degree $6$ . $\blacksquare$

This post has been edited 1 time. Last edited by th1nq3r, Sep 27, 2022, 2:02 PM

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NicoN9

#8 h Apr 29, 2025, 12:01 PM

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Given triangle $ABC$ , let $a=BC$ (and similarly for $b, c$ ), $s=(1/2)(a+b+c)$ , and $AA_1$ , $BB_1$ , $CC_1$ be altitudes. Note that $s$ is rational. we claim that $K(x)=(x^2-AA_1^2)(x^2-BB_1^2)(x-CC_1^2)$ works. Indeed, obviously $\deg K=6$ , $K(AA_1)=K(BB_1)=K(CC_1)=0$ , so it is suffice to prove that $K$ is in rational coefficients. By Heron's, we must have $AA_1^2=\frac{4}{a^2}\cdot (s(s-a)(s-b)(s-c))$ and we can easily confirm that $K$ is rational.

This post has been edited 2 times. Last edited by NicoN9, Apr 30, 2025, 6:36 AM

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