2013 USAMO Problems/Problem 1
Contents
[hide]Problem
In triangle , points
lie on sides
respectively. Let
,
,
denote the circumcircles of triangles
,
,
, respectively. Given the fact that segment
intersects
,
,
again at
respectively, prove that
.
Solution 1
In this solution, all lengths and angles are directed.
Firstly, it is easy to see by that concur at a point
(the Miquel point). Let
meet
again at
and
, respectively. Then by Power of a Point, we have
Thusly
But we claim that
. Indeed,
and
Therefore,
. Analogously we find that
and we are done.
courtesy v_enhance, minor clarification by integralarefun
Solution 2
Diagram Refer to the Diagram link.
By Miquel's Theorem, there exists a point at which intersect. We denote this point by
Now, we angle chase:
In addition, we have
Now, by the Ratio Lemma, we have
(by the Law of Sines in
)
(by the Law of Sines in
)
by the Ratio Lemma.
The proof is complete.
Solution 3
Use directed angles modulo .
Lemma.
Proof.
Now, it follows that (now not using directed angles)
using the facts that
and
,
and
are similar triangles, and that
equals twice the circumradius of the circumcircle of
.
Solution 4
We will use some construction arguments to solve the problem. Let
and let
We construct lines through the points
and
that intersect with
at the points
and
respectively, and that intersect each other at
We will construct these lines such that
Now we let the intersections of with
and
be
and
respectively. This construction is as follows.
We know that Hence, we have,
Since the opposite angles of quadrilateral add up to
it must be cyclic. Similarly, we can also show that quadrilaterals
and
are also cyclic.
Since points and
lie on
we know that,
and that
Hence, the points and
coincide with the given points
and
respectively.
Since quadrilateral is also cyclic, we have,
Similarly, since quadrilaterals and
are also cyclic, we have,
and,
Since these three angles are of and they are equal to corresponding angles of
by AA similarity, we know that
We now consider the point We know that the points
and
are concyclic. Hence, the points
and
must also be concyclic.
Hence, quadrilateral is cyclic.
Since the angles and
are inscribed in the same arc
we have,
Consider by this result, we can deduce that the homothety that maps to
will map
to
Hence, we have that,
Since and
hence,
as required.
Solution 5 (Simple Rotational Homotethy)
We begin again by noting that the three circumcircles intersect at point by Miquel's theorem. In addition, we state that the angle
, hence
, as well as
, from which follows that
, so
, and
.
We shall prove that the triangles
and
and
are similar, which will imply a rotational homotethy with angle
about the point
, that takes
to
, thus proving the problem. (In essence, just imagine we rotate
around M and lengthen things out and get
- the ratios will remain identical.)
We do this by angle chasing. Denote angle . From the angles labeled before, we now know
. In addition,
. So the angles in triangle
are
,
and thus
, with
at the point
. In addition,
, so
, so
. Since
, the triangle
has angles
also, with
at
. Finally, In triangle
, we already know the angle
to be
; we also can find that
, so
, so
, so
. Thus, the three triangles are similar have a common point
, which proves that there is a rotational homotethy around
that maps
to
as desired.
Solution by SimilarTriangle.
See Also
2013 USAMO (Problems • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
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