Can three points form a square?

[Ackbach]

Here is this week's POTW:

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Let $A, B, C$ denote distinct points with integer coordinates in $\mathbb R^2$. Prove that if \[(|AB|+|BC|)^2<8\cdot [ABC]+1\]
then $A, B, C$ are three vertices of a square. Here $|XY|$ is the length of segment $XY$ and $[ABC]$ is the area of triangle $ABC$.

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[Ackbach]

Re: Problem Of The Week # 243 - Dec 05, 2016

This was Problem A-6 in the 1998 William Lowell Putnam Mathematical Competition.

No one answered this week's POTW. The solution, attributed to Kiran Kedlaya and his associates, follows:

Spoiler

Recall the inequalities $|AB|^2 + |BC|^2 \geq 2|AB||BC|$ (AM-GM) and $|AB||BC| \geq 2[ABC]$ (Law of Sines). Also recall that the area of a triangle with integer coordinates is half an integer (if its vertices lie at $(0,0), (p,q), (r,s)$, the area is $|ps-qr|/2$), and that if $A$ and $B$ have integer coordinates, then $|AB|^2$ is an integer (Pythagoras). Now observe that
\begin{align*}
8[ABC] &\leq |AB|^2+|BC|^2 + 4[ABC] \\
&\leq |AB|^2 + |BC|^2 + 2|AB| |BC| \\
&< 8[ABC]+1,
\end{align*}
and that the first and second expressions are both integers. We conclude that $8[ABC] = |AB|^2+ |BC|^2+4[ABC]$, and so $|AB|^2+|BC|^2 =2|AB| |BC|= 4[ABC]$; that is, $B$ is a right angle and $AB=BC$, as desired.