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Difference between revisions of "1992 IMO Problems/Problem 1"

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The solutions are: <math>(a,b,c)=(2,4,8)</math> and <math>(a,b,c)=(3,5,15)</math>
 
The solutions are: <math>(a,b,c)=(2,4,8)</math> and <math>(a,b,c)=(3,5,15)</math>
  
−
~ Tomas Diaz
+
~ Tomas Diaz. orders@tomasdiaz.com
  
 +
==Solution 2==
 +
Let <math>x = a-1, y = b-1, z = c-1</math>. So <math>1 \leq x < y < z</math>. We're asked to solve
 +
<cmath>
 +
(k+1)xyz = (x+1)(y+z)(z+1)-1
 +
</cmath>
 +
where <math>k</math> is a non-negative integer.. Dividing by <math>xyz</math>
 +
<cmath>
 +
\begin{align*}
 +
k+1 &= \Big(1+\frac{1}{x}\Big)\Big(1+\frac{1}{y}\Big)\Big(1+\frac{1}{z}\Big) - \frac{1}{xyz} \\
 +
    &< (1+1)\Big(1+\frac{1}{2}\Big)\Big(1+\frac{1}{3}\Big) = 4
 +
\end{align*}
 +
</cmath>
 +
So <math>k=0,1</math> or <math>2</math>. Simplifying the first equation
 +
<cmath>
 +
kxyz = xy + yz + zy + x + y + z.
 +
</cmath>
 +
<math>k=0</math> is impossible. Case <math>k=2</math>: if <math>x \geq 2</math>, dividing the previous equation by <math>xy</math>
 +
<cmath>
 +
\begin{align}
 +
2z &= 1 + \frac{1}{x} + \frac{1}{y} + z\Big(\frac{1}{x} + \frac{1}{y} + \frac{1}{xy}\Big) \\
 +
  &\leq 1 + \frac{1}{2}+\frac{1}{3} + z\Big(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\Big) \\
 +
  &< 2 + z \nonumber
 +
\end{align}
 +
</cmath>
 +
So <math>z < 2</math> which is impossible. If <math>x = 1</math> its easy to show <math>y=2</math> leads to a contradiction. Solving <math>(1)</math> for <math>z</math>
 +
<cmath>
 +
z = 2 + \frac{5}{y-2}
 +
</cmath>
 +
which can only work if <math>y = 3, z = 7</math>. So <math>x=1,y=3,z=7</math> is a solution. Case <math>k=1</math>: considering the version of <math>(1)</math> with <math>z</math> on the LHS instead of <math>2z</math>, if <math>x = 1</math> then
 +
<cmath>
 +
\begin{align*}
 +
z &= 1 + \frac{1}{x} + \frac{1}{y} + z \Big(\frac{1}{x} + \frac{1}{y} + \frac{1}{xy}\Big) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3) \\
 +
  &= 2 + \frac{1}{y} + z\Big(1+\frac{1}{y}+\frac{1}{xy}\Big)
 +
\end{align*}
 +
</cmath>
 +
which is impossible. Similarly to <math>(2)</math>, if <math>x \geq 3</math> then
 +
<cmath>
 +
z \leq 1 + \frac{1}{3} + \frac{1}{4} + z \frac{8}{12}
 +
</cmath>
 +
i.e. <math>z < 5</math> which is impossible since <math>3 \leq x < y < z</math>. So <math>x= 2</math>. Its easy to show <math>y=3</math> leads to a contradiction. Solving <math>(3)</math> for <math>z</math> with <math>x=2</math> gives
 +
<cmath>
 +
z = 3 + \frac{11}{y-3}
 +
</cmath>
 +
which can only work if <math>y=4, z=14</math>. So <math>x=2,y=4,z=14</math> is a solution. Our two solutions give <math>a = 2,b=4,c=8</math> and <math>a=3,b=5,c=15</math>.
  
 +
~not_detriti
  
 +
{{alternate solutions}}
  
 +
==See Also==
  
−
{{alternate solutions}}
+
{{IMO box|year=1992|before=First Question|num-a=2}}
 +
[[Category:Olympiad Geometry Problems]]
 +
[[Category:3D Geometry Problems]]

Latest revision as of 09:45, 11 July 2025

Problem

Find all integers $a$, $b$, $c$ satisfying $1 < a < b < c$ such that $(a - 1)(b -1)(c - 1)$ is a divisor of $abc - 1$.

Solution

$1<\frac{abc-1}{(a-1)(b-1)(c-1)}<\frac{abc}{(a-1)(b-1)(c-1)}$

$1<\frac{abc-1}{(a-1)(b-1)(c-1)}<\left(\frac{a}{a-1}\right) \left(\frac{b}{b-1}\right) \left(\frac{c}{c-1}\right)$

With $1<a<b<c$ it implies that $a \ge 2$, $b \ge 3$, $c \ge 4$

Therefore, $\frac{a}{a-1}=1+\frac{1}{a-1}$

which for $a$ gives: $\frac{a}{a-1} \le 1+\frac{1}{2-1}$, which gives :$\frac{a}{a-1} \le 2$

for $b$ gives: $\frac{b}{b-1} \le 1+\frac{1}{3-1}$, which gives :$\frac{b}{b-1} \le \frac{3}{2}$

for $c$ gives: $\frac{c}{c-1} \le 1+\frac{1}{4-1}$, which gives :$\frac{c}{c-1} \le \frac{4}{3}$

Substituting those inequalities into the original inequality gives:

$1<\frac{abc-1}{(a-1)(b-1)(c-1)}<\left( 2 \right) \left(\frac{3}{2}\right) \left(\frac{4}{3}\right)$

$1<\frac{abc-1}{(a-1)(b-1)(c-1)}<4$

Since $\frac{abc-1}{(a-1)(b-1)(c-1)}$ needs to be integer,

then $\frac{abc-1}{(a-1)(b-1)(c-1)}=2$ or $\frac{abc-1}{(a-1)(b-1)(c-1)}=3$

Case 1: $\frac{abc-1}{(a-1)(b-1)(c-1)}=2$

$abc-1=2(a-1)(b-1)(c-1)=2abc-2(ab+bc+ac)+2(a+b+c)-2$

Case 1, subcase $a=2$:

$2bc-1=2bc-2(b+c)+2$ gives: $2(b+c)=3$ which has no solution because $2(b+c)$ is even.

Case 1, subcase $a=3$:

$3bc-1=4bc-4(b+c)+4$

$bc-4b-4c+5=0$

$(b-4)(c-4)=11$

$b-4=1$ and $c-4=11$ provides solution $(a,b,c)=(3,5,15)$

Case 2: $\frac{abc-1}{(a-1)(b-1)(c-1)}=3$

$abc-1=3(a-1)(b-1)(c-1)=3abc-3(ab+bc+ac)+3(a+b+c)-3$

Case 2, subcase $a=2$:

$2bc-1=3bc-3(b+c)+3$

$bc-3b-3c+4=0$

$(b-3)(c-3)=5$

$b-3=1$ and $c-3=5$ provides solution $(a,b,c)=(2,4,8)$

Case 2, subcase $a=3$:

$3bc-1=6bc-6(b+c)+6$

Since $(3bc-1$) mod $3 = -1$ and $(6bc-6(b+c)+6)$ mod $3 = 0$, then there is no solution for this subcase.

Now we verify our two solutions:

when $(a,b,c)=(2,4,8)$

$abc-1=(2)(4)(8)-1=63$ and $(a-1)(b-1)(c-1)=(1)(3)(7)=21$

Since $21$ is a factor of $63$, this solutions is correct.

when $(a,b,c)=(3,5,15)$

$abc-1=(3)(5)(15)-1=224$ and $(a-1)(b-1)(c-1)=(2)(4)(14)=112$

Since $112$ is a factor of $224$, this solutions is also correct.

The solutions are: $(a,b,c)=(2,4,8)$ and $(a,b,c)=(3,5,15)$

~ Tomas Diaz. orders@tomasdiaz.com

Solution 2

Let $x = a-1, y = b-1, z = c-1$. So $1 \leq x < y < z$. We're asked to solve \[(k+1)xyz = (x+1)(y+z)(z+1)-1\] where $k$ is a non-negative integer.. Dividing by $xyz$ \begin{align*} k+1 &= \Big(1+\frac{1}{x}\Big)\Big(1+\frac{1}{y}\Big)\Big(1+\frac{1}{z}\Big) - \frac{1}{xyz} \\ &< (1+1)\Big(1+\frac{1}{2}\Big)\Big(1+\frac{1}{3}\Big) = 4 \end{align*} So $k=0,1$ or $2$. Simplifying the first equation \[kxyz = xy + yz + zy + x + y + z.\] $k=0$ is impossible. Case $k=2$: if $x \geq 2$, dividing the previous equation by $xy$ \begin{align} 2z &= 1 + \frac{1}{x} + \frac{1}{y} + z\Big(\frac{1}{x} + \frac{1}{y} + \frac{1}{xy}\Big) \\ &\leq 1 + \frac{1}{2}+\frac{1}{3} + z\Big(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\Big) \\ &< 2 + z \nonumber \end{align} So $z < 2$ which is impossible. If $x = 1$ its easy to show $y=2$ leads to a contradiction. Solving $(1)$ for $z$ \[z = 2 + \frac{5}{y-2}\] which can only work if $y = 3, z = 7$. So $x=1,y=3,z=7$ is a solution. Case $k=1$: considering the version of $(1)$ with $z$ on the LHS instead of $2z$, if $x = 1$ then \begin{align*} z &= 1 + \frac{1}{x} + \frac{1}{y} + z \Big(\frac{1}{x} + \frac{1}{y} + \frac{1}{xy}\Big) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3) \\ &= 2 + \frac{1}{y} + z\Big(1+\frac{1}{y}+\frac{1}{xy}\Big) \end{align*} which is impossible. Similarly to $(2)$, if $x \geq 3$ then \[z \leq 1 + \frac{1}{3} + \frac{1}{4} + z \frac{8}{12}\] i.e. $z < 5$ which is impossible since $3 \leq x < y < z$. So $x= 2$. Its easy to show $y=3$ leads to a contradiction. Solving $(3)$ for $z$ with $x=2$ gives \[z = 3 + \frac{11}{y-3}\] which can only work if $y=4, z=14$. So $x=2,y=4,z=14$ is a solution. Our two solutions give $a = 2,b=4,c=8$ and $a=3,b=5,c=15$.

~not_detriti

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

1992 IMO (Problems) • Resources
Preceded by
First Question 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
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