Difference between revisions of "1968 IMO Problems/Problem 1"
Line 56: | Line 56: | ||
numbers are ordered. The six possibilities are: | numbers are ordered. The six possibilities are: | ||
− | 1: <math>a = b - 2, c = b - 1, b</math> | + | 1: <math>\ \ a = b - 2, c = b - 1, b</math> |
− | 2: <math>c = b - 2, a = b - 1, b</math> | + | 2: <math>\ \ c = b - 2, a = b - 1, b</math> |
− | 3: <math>a = b - 1, b, c = b + 1</math> | + | 3: <math>\ \ a = b - 1, b, c = b + 1</math> |
− | 4: <math>c = b - 1, b, a = b + 1</math> | + | 4: <math>\ \ c = b - 1, b, a = b + 1</math> |
− | 5: <math>b, a = b + 1, c = b + 2</math> | + | 5: <math>\ \ b, a = b + 1, c = b + 2</math> |
− | 6: <math>b, c = b + 1, a = b + 2</math> | + | 6: <math>\ \ b, c = b + 1, a = b + 2</math> |
For each case, we could substitute <math>a, c</math> in <math>(*)</math>, get an | For each case, we could substitute <math>a, c</math> in <math>(*)</math>, get an | ||
Line 127: | Line 127: | ||
That means we will have to consider three possibilities: | That means we will have to consider three possibilities: | ||
− | 1: <math>\frac{x}{\sin \alpha} = \frac{x + 1}{\sin 2\alpha} = \frac{x + 2}{\sin (\pi - 3\alpha)}</math> | + | 1: <math>\ \ \frac{x}{\sin \alpha} = \frac{x + 1}{\sin 2\alpha} = \frac{x + 2}{\sin (\pi - 3\alpha)}</math> |
− | 2: <math>\frac{x}{\sin \alpha} = \frac{x + 1}{\sin (\pi - 3\alpha)} = \frac{x + 2}{\sin 2\alpha}</math> | + | 2: <math>\ \ \frac{x}{\sin \alpha} = \frac{x + 1}{\sin (\pi - 3\alpha)} = \frac{x + 2}{\sin 2\alpha}</math> |
− | 3: <math>\frac{x}{\sin (\pi - 3\alpha)} = \frac{x + 1}{\sin \alpha} = \frac{x + 2}{\sin 2\alpha}</math> | + | 3: <math>\ \ \frac{x}{\sin (\pi - 3\alpha)} = \frac{x + 1}{\sin \alpha} = \frac{x + 2}{\sin 2\alpha}</math> |
Using the identities <math>\sin (\pi - A) = \sin A</math>, <math>\sin 2A = 2\sin A \cos A</math> | Using the identities <math>\sin (\pi - A) = \sin A</math>, <math>\sin 2A = 2\sin A \cos A</math> | ||
Line 137: | Line 137: | ||
by <math>\sin \alpha</math> the three cases become | by <math>\sin \alpha</math> the three cases become | ||
− | 1: <math>x = \frac{x + 1}{2\cos \alpha} = \frac{x + 2}{4\cos^2 \alpha - 1}</math> | + | 1: <math>\ \ x = \frac{x + 1}{2\cos \alpha} = \frac{x + 2}{4\cos^2 \alpha - 1}</math> |
− | 2: <math>x = \frac{x + 1}{4\cos^2 \alpha - 1} = \frac{x + 2}{2\cos \alpha}</math> | + | 2: <math>\ \ x = \frac{x + 1}{4\cos^2 \alpha - 1} = \frac{x + 2}{2\cos \alpha}</math> |
− | 3: <math>\frac{x}{4\cos^2 \alpha - 1} = x + 1 = \frac{x + 2}{2\cos \alpha}</math> | + | 3: <math>\ \ \frac{x}{4\cos^2 \alpha - 1} = x + 1 = \frac{x + 2}{2\cos \alpha}</math> |
+ | |||
+ | Each case is a system of two equations in two unknowns, <math>x, \cos \alpha</math>. | ||
+ | We will solve each system, obtain all possible solutions, and chose those | ||
+ | values for which <math>x, x + 1, x + 2</math> and <math>\alpha, 2\alpha, \pi - 3\alpha</math> | ||
+ | can be the sides and angles of a triangle. | ||
Revision as of 13:47, 22 August 2024
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.
Solution 2
(Note: this proof is an expansion by pf02 of an outline of a solution posted here before.)
In a given triangle , let
,
, and
.
Then
Hence,
Indeed, we know from the Law of Sines that
.
Denote this ratio by ; we have
.
Substitute in
and simplify by
.
We get
.
At this point, notice that is equivalent to the equality
from Solution 1. Indeed, the latter
can be rewritten as
, and we know
that
. So we could simply quote the fact (proven in
Solution 1) that if
are consecutive integers and
, then
is the only solution
which could be the sides of a triangle.
For the sake of completeness, and for fun, I give a slightly different proof here.
We have six possibilities, depending on how the three consecutive numbers are ordered. The six possibilities are:
1:
2:
3:
4:
5:
6:
For each case, we could substitute in
, get an
equation in
, solve it, and get all the possible solutions.
As a shortcut, notice that (*) implies that
. If
are consecutive integers, then they are relatively prime, so
can not be true unless
. In this case the
triangle would have sides
, which is impossible.
This eliminates cases 2, 3, 4 and 5.
In case 1, becomes
, or
.
This has solutions . The value
is impossible.
The value
yields
, which is impossible.
In case 6, becomes
, or
.
The solutions are . The value
is impossible.
Thus, we get the unique triangle
.
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.
Solution 4
Note: Adding this 4th solution is justified by the fact
that it is extremely straightforward, and by the fact
that in fact, it does not use the hypothesis that the
sides of the triangle are integers. We only need the
fact that the sides differ by (i.e. they are
for some
), and the condition
on the angles (that one is twice the other).
So, let us start by assuming that two angles are
and the sides are
.
We will want to apply the Law of Sines:
The angles should be so that
,
but we don't know how to map
to
. One thing we know, is that
. Indeed, if
the
inequality is true because
is increasing on
.
Now note that
since otherwise
could not be the angle of a triangle. So, if
then
and
.
That means we will have to consider three possibilities:
1:
2:
3:
Using the identities ,
and
and simplifying
by
the three cases become
1:
2:
3:
Each case is a system of two equations in two unknowns, .
We will solve each system, obtain all possible solutions, and chose those
values for which
and
can be the sides and angles of a triangle.
TO BE CONTINUED. I SAVE MIDWAY SO I DON'T LOSE WORK DONE SO FAR.
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 |