1960 IMO Problems/Problem 3
Problem
In a given right triangle , the hypotenuse
, of length
, is divided into
equal parts (
an odd integer). Let
be the acute angle subtending, from
, that segment which contains the midpoint of the hypotenuse. Let
be the length of the altitude to the hypotenuse of the triangle. Prove that:

Solution
Using coordinates, let ,
, and
. Also, let
be the segment that contains the midpoint of the hypotenuse with
closer to
.
Then, , and
.
So, , and
.
Thus,
.
Since ,
and
as desired.
Solution 2
Let be points on side
such that segment
contains midpoint
, with
closer to
and (without loss of generality)
. Then if
is an altitude, then
is between
and
. Combined with the obvious fact that
is the midpoint of
(for
is odd), we have
Solution 3
Let
Next, we shall denote line
We know that
As
Simplifying:
Using the identity
Thus:
Now, we know that:
Substituting this into the equation:
Factoring out
Thus:
By performing similar steps with
Courtesy of Gordon Freeman
Solution 4
Let and
be points on
such that
,
is closer to
than to
, and
is closer to
than
. Furthermore, let
and
, and let
and
. Applying Law of Sines on
yields
. Next, we use the sum of tangents formula:
, we need to find
. This is equal to
, which is the reciprocal of
, or
. Since
due to triangle areas and
due to the Pythagorean Theorem, we conclude that the expression is equal to
, which simplifies to
.
~eevee9406
See Also
1960 IMO (Problems) | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 • 7 | Followed by Problem 4 |