2021 CIME I Problems/Problem 14
Problem
Let be an acute triangle with orthocenter
and circumcenter
. The tangent to the circumcircle of
at
intersects lines
and
at
and
, and
. Let line
intersect
at
. Suppose that
, and
for positive integers
where
is not divisible by the square of any prime. Find
.
Solution 1 by TheUltimate123
Let be the orthocenter of
, and let
,
be the feet of the altitudes from
. Also let
be the antipode of
on the circumcircle and let
, as shown below:
Disregarding the condition
, we contend:
In general,
is cyclic.
Recall that
, so the claim follows from Reims' theorem on
With , it follows that
is an isosceles trapezoid. In particular,
and
. Since
, we have
But note that
, so
i.e.\
. We are given
, and by the law of sines,
, so
, and
, so
.
Solution 2
Let and
be the feet of the altitudes from
to
and from
to
, respectively. It is clear that
, as
and
. We conjecture that quadrilateral
is cyclic because it looks cyclic. Indeed, note that for this to be true we need
, but
reduces this condition to
, which is obvious as
is cyclic with
.
Thus, note that cyclic along with
are enough to imply that it is an isosceles trapezoid and
. Then, going back to
, it is evident that
and
are directly similar and the ratio in which they are similar is given by
Consider the geometric transformation consisting of a homothety centered at
with ratio
followed by a reflection about the bisector of
, under which
maps to
. Then, point
(which lies on both
and
) maps to the point
on
such that
and
are isogonal wrt.
. But we know that the circumcenter and orthocenter of
are isogonal conjugates, and this is enough to imply
lies on
. Now
where the final step comes from undoing the homothety and reflection, thereby dividing all lengths by the ratio
. We are given
and
, thus
. It follows that, as
is an acute triangle,
. Finally,
and the answer is
.
~StressedPineapple
See also
2021 CIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All CIME Problems and Solutions |