1968 IMO Problems/Problem 1
Problem
Prove that there is one and only one triangle whose side lengths are consecutive integers, and one of whose angles is twice as large as another.
Solution 1
In triangle , let
,
,
,
, and
. Using the Law of Sines gives that
Therefore . Using the Law of Cosines gives that
This can be simplified to . Since
,
, and
are positive integers,
. Note that if
is between
and
, then
is relatively prime to
and
, and
cannot possibly divide
. Therefore
is either the least of the three consecutive integers or the greatest.
Assume that is the least of the three consecutive integers. Then either
or
, depending on if
or
. If
, then
is 1 or 2.
couldn't be 1, for if it was then the triangle would be degenerate. If
is 2, then
, but
and
must be 3 and 4 in some order, which means that this triangle doesn't exist. therefore
cannot divide
, and so
must divide
. If
then
, so
is 1, 2, or 4. Clearly
cannot be 1 or 2, so
must be 4. Therefore
. This shows that
and
, and the triangle has sides that measure 4, 5, and 6.
Now assume that is the greatest of the three consecutive integers. Then either
or
, depending on if
or
.
is absurd, so
, and
. Therefore
is 1, 2, or 4. However, all of these cases are either degenerate or have been previously ruled out, so
cannot be the greatest of the three consecutive integers.
This shows that there is exactly one triangle with this property - and it has side lengths of 4, 5, and 6.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
Solution 2
[Incomplete, please edit]
In a given triangle , let
,
$\sin^2 A = \sin^2 2B = 2 \sin B \cos B \Sin 2B = \sin B(\sin B + \sin 3B) = \sin B(\sin B + \sin C)$ (Error compiling LaTeX. Unknown error_msg)
Hence,
(with assumptions. This needs clearing up)
If
is the shortest side,
No other permutation of
,
and
in terms of size gives integral values to
(show). So there is only one such triangle. $\qed$ (Error compiling LaTeX. Unknown error_msg)
Solution 3
NO TRIGONOMETRY!!!
Let be the side lengths of a triangle in which
Extend to
such that
Then
, so
and
are similar by AA Similarity. Hence,
. Then proceed as in Solution 2, as only algebraic manipulations are left.
See Also
1968 IMO (Problems) • Resources | ||
Preceded by First Problem |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |