Find values of $a b+a c+b c$

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Find values of $a b+a c+b c$

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Al-Khwarizmi IJMO

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Source: 2nd Al-Khwarizmi International Junior Mathematical Olympiad 2024, Day2, Problem6

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NJAX

#1 h May 31, 2024, 12:08 PM • 1 Y

Y by seifbaba

Let $a, b, c$ be distinct real numbers such that $a+b+c=0$ and $a^{2}-b=b^{2}-c=c^{2}-a.$ Evaluate all the possible values of $a b+a c+b c$ .

Proposed by Nguyen Anh Vu, Vietnam

This post has been edited 1 time. Last edited by NJAX, May 31, 2024, 12:34 PM

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#2 h May 31, 2024, 3:06 PM

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Case 1) $a,b,c \not = 0$ . Put $b=-(a+c)$ . Then $a^2+a+c=(a+c)^2-c=c^2-a$ , so $2ac+a^2-c=-a \Rightarrow c(2a-1)=-(a^2+a)$ . Note that if $2a-1=0$ , then $a^2+a=0$ , which is impossible. By this reason we have $c=\frac{a^2+a}{1-2a}$ and after putting this equality in $a^2+a+c=c^2-a$ we get $3a(a^3-3a+1)=0 \Rightarrow a^3-3a+1=0$ . Since the condition is cyclic onto $a,b,c$ , we similarly get $b^3-3b+1=0$ and $c^3-3c+1=0$ . Note that if $P(x)=x^3-3x+1$ , then $gcd(P,P')=1$ , so it is possible situation that $a,b,c$ are distinct (i.e. there is no equals roots of $P(x)$ ). By Viet's theorem we get $ab+bc+ca=-3$ .
Case 2) $c=0$ . In this case we have $b+c=0$ and $-b=b^2-c=c^2$ . Put $b=-c$ , then $c=c^2-c=c^2 \Rightarrow c=0$ , which is impossible ( $a \not = c$ by condition).

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#3 h May 31, 2024, 9:29 PM • 3 Y

Y by Mathological03, orangesyrup, NJAX

There is a very elegant solution to this problem.

Let $x:=a^2-b=b^2-c=c^2-a$ . First note that

$a^2-b^2=b-c$ $b^2-c^2 = c-a$ $c^2-a^2 = a-b$ Multiplying the above equations we get
$(a+b)(b+c)(a+c)=1$ hence $abc=-1$ . The next step is to note that
$0=ax+bx+cx=a^3+b^3+c^3-ab-bc-ac$ hence
$ab+bc+ac=a^3+b^3+c^3=3abc=-3$

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17 posts

#4 h Jun 1, 2024, 6:08 AM

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Ciel_vert wrote:

There is a very elegant solution to this problem.

Let $x:=a^2-b=b^2-c=c^2-a$ . First note that

$a^2-b^2=b-c$ $b^2-c^2 = c-a$ $c^2-a^2 = a-b$ Multiplying the above equations we get
$(a+b)(b+c)(a+c)=1$ hence $abc=-1$ . The next step is to note that
$0=ax+bx+cx=a^3+b^3+c^3-ab-bc-ac$ hence
$ab+bc+ac=a^3+b^3+c^3=3abc=-3$

how can abc=-1?

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41765 posts

sqing

#5 h Jun 2, 2024, 1:58 PM

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NJAX wrote:

Let $a, b, c$ be distinct real numbers such that $a+b+c=0$ and $a^{2}-b=b^{2}-c=c^{2}-a.$ Evaluate all the possible values of $a b+a c+b c$ .

Proposed by Nguyen Anh Vu, Vietnam

Let $a, b, c$ be distinct real numbers such that $a^{2}-b=b^{2}-c=c^{2}-a$ and $a+b+c=k-1$ $(k\in N^+)$ . Prove that $ab+bc+ca=-k^2-2$

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129 posts

#6 h Jun 2, 2024, 3:35 PM

Y by

nai0610 wrote:

Ciel_vert wrote:

There is a very elegant solution to this problem.

Let $x:=a^2-b=b^2-c=c^2-a$ . First note that

$a^2-b^2=b-c$ $b^2-c^2 = c-a$ $c^2-a^2 = a-b$ Multiplying the above equations we get
$(a+b)(b+c)(a+c)=1$ hence $abc=-1$ . The next step is to note that
$0=ax+bx+cx=a^3+b^3+c^3-ab-bc-ac$ hence
$ab+bc+ac=a^3+b^3+c^3=3abc=-3$

how can abc=-1?

note that (x+y+z)(xy+yz+xy)=xyz+(x+y)(x+z)(y+z)

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57 posts

#7 h Jun 2, 2024, 7:18 PM

Y by

Ciel_vert wrote:

There is a very elegant solution to this problem.

Let $x:=a^2-b=b^2-c=c^2-a$ . First note that

$a^2-b^2=b-c$ $b^2-c^2 = c-a$ $c^2-a^2 = a-b$ Multiplying the above equations we get
$(a+b)(b+c)(a+c)=1$ hence $abc=-1$ . The next step is to note that
$0=ax+bx+cx=a^3+b^3+c^3-ab-bc-ac$ hence
$ab+bc+ac=a^3+b^3+c^3=3abc=-3$

my solution in the exam was exactly like this

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17 posts

#8 h Jun 5, 2024, 6:27 PM

Y by

orangesyrup wrote:

nai0610 wrote:

Ciel_vert wrote:

There is a very elegant solution to this problem.

Let $x:=a^2-b=b^2-c=c^2-a$ . First note that

$a^2-b^2=b-c$ $b^2-c^2 = c-a$ $c^2-a^2 = a-b$ Multiplying the above equations we get
$(a+b)(b+c)(a+c)=1$ hence $abc=-1$ . The next step is to note that
$0=ax+bx+cx=a^3+b^3+c^3-ab-bc-ac$ hence
$ab+bc+ac=a^3+b^3+c^3=3abc=-3$

how can abc=-1?

note that (x+y+z)(xy+yz+xy)=xyz+(x+y)(x+z)(y+z)

you didnt calculate ab+bc+ca though?

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1075 posts

#9 h Jun 5, 2024, 6:29 PM

Y by

nai0610 wrote:

Ciel_vert wrote:

There is a very elegant solution to this problem.

Let $x:=a^2-b=b^2-c=c^2-a$ . First note that

$a^2-b^2=b-c$ $b^2-c^2 = c-a$ $c^2-a^2 = a-b$ Multiplying the above equations we get
$(a+b)(b+c)(a+c)=1$ hence $abc=-1$ . The next step is to note that
$0=ax+bx+cx=a^3+b^3+c^3-ab-bc-ac$ hence
$ab+bc+ac=a^3+b^3+c^3=3abc=-3$

how can abc=-1?

We get $(a+b)(b+c)(c+a)=1$ . Note that $a+b=-c, b+c=-a, c+a=-b$ , so $-abc=1 \Rightarrow abc=-1$ .

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2 posts

#10 h Sep 15, 2024, 3:12 PM

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a^2-b=b^2-c=c^2-a=x
ax+bx+cx=0=a^3+b^3+c^3-ab-ac-bc
ab+bc+ac=a^3+b^3+c^3=3abc
a^2-b^2=b-c
b^2-c^2=c-a
c^2-a^2=a-b
Multiple all =>(a+b)(b+c)(a+c)=1=-abc
ab+bc+ac=3abc=-3
Answer:-3

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36 posts

#11 h Jan 27, 2025, 4:30 PM

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I solved this problem in the Olympiad

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#12 h Apr 1, 2025, 4:21 PM

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$1) a^2-b=b^2-c=c^2-a=y$ $ay+by+cy=0=a^3+b^3+c^3-ab-ac-bc$ $ab+bc+ac=a^3+b^3+c^3=3abc$ $a^2-b^2=b-c; b^2-c^2=c-a; c^2-a^2=a-b$ $(a+b)(b+c)(a+c)=1=-abc$ $ab+bc+ac=3abc=-3$

This post has been edited 1 time. Last edited by Baimukh, Apr 1, 2025, 4:38 PM
Reason: I had a mistake

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