2003 USAMO Problems/Problem 4
Problem
Let be a triangle. A circle passing through
and
intersects segments
and
at
and
, respectively. Lines
and
intersect at
, while lines
and
intersect at
. Prove that
if and only if
.
Solutions
Solution 1
Extend segment through
to
such that
.
![2003usamo4-1.png](https://wiki-images.artofproblemsolving.com//8/85/2003usamo4-1.png)
Then if and only if quadrilateral
is a parallelogram, or,
. Hence
if and only if
, that is,
.
Because quadrilateral is cyclic,
. It follows that
if and only if
that is, quadrilateral
is cyclic, which is equivalent to
Because
,
if and only if triangles
and
are similar, that is
or
.
Solution 2
We first assume that . Because
and
, triangles
and
are similar. Consequently,
.
![2003usamo4-2.png](https://wiki-images.artofproblemsolving.com//2/25/2003usamo4-2.png)
Because quadrilateral is cyclic,
. Hence
implying that
, so
. Because quadrilateral
is cyclic,
. Hence
Because
and
, triangles
and
are similar. Consequently,
, or
. Therefore
implies
.
Now we assume that . Applying Ceva's Theorem to triangle
and cevians
gives
implying that
, so
.
Consequently, . Because quadrilateral
is cyclic,
. Hence
Because
and
, triangles
and
are similar. Consequently,
, or
.
Combining the above, we conclude that if and only if
.
Solution 3
We will construct a series of equivalent statements. First:
where the second equivalence follows from SAS similarity (because
).
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See also
2003 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.