Difference between revisions of "2016 USAJMO Problems/Problem 1"
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<math>D</math> is the center of the circle <math>A I_B B</math> (the <math>P-</math> excenter of <math>PAB</math> is also on the same circle). Therefore <math>D I_B = DB</math>. Similarly <math>E I_C = EC</math>. Since <math>AB=AC</math>, <math>DB=EC</math>. Therefore <math>D I_B = E I_C</math>. Obviously <math>ME = MD</math> and <math>\angle MEI_C = \angle MEP = \angle MDP = \angle MDI_B</math>. Thus by SAS, <math>\triangle MEI_C \cong \triangle MDI_B</math>. | <math>D</math> is the center of the circle <math>A I_B B</math> (the <math>P-</math> excenter of <math>PAB</math> is also on the same circle). Therefore <math>D I_B = DB</math>. Similarly <math>E I_C = EC</math>. Since <math>AB=AC</math>, <math>DB=EC</math>. Therefore <math>D I_B = E I_C</math>. Obviously <math>ME = MD</math> and <math>\angle MEI_C = \angle MEP = \angle MDP = \angle MDI_B</math>. Thus by SAS, <math>\triangle MEI_C \cong \triangle MDI_B</math>. | ||
− | Hence <math>\angle I_B | + | Hence <math>\angle I_B M I_C = \angle DME = \angle DPE = \angle I_B P I_C</math>, so <math>MPI_B I_C</math> is cyclic and we are done. |
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 15:11, 28 June 2020
Problem
The isosceles triangle , with
, is inscribed in the circle
. Let
be a variable point on the arc
that does not contain
, and let
and
denote the incenters of triangles
and
, respectively.
Prove that as varies, the circumcircle of triangle
passes through a fixed point.
Solution 1
We claim that (midpoint of arc
) is the fixed point.
We would like to show that
,
,
,
are cyclic.
We extend to intersect
again at R.
We extend
to intersect
again at S.
We invert around a circle centered at with radius
(for convenience).
(I will denote X' as the reflection of X for all the points)
The problem then becomes: Prove
,
, and
are collinear.
Now we look at triangle . We apply Menelaus (the version where all three points lie outside the triangle).
It suffices to show that
By inversion, we know for any point
and
for any points
and
.
Plugging this into our Menelaus equation we obtain that it suffices to show
We cancel out the like terms and rewrite. It suffices to show
We know that
is the diameter of
because
is isosceles and
is the angle bisector. We also know
so
and
are symmetric with respect to
so
.
Thus, it suffices to show .
This is obvious because
.
Therefore we are done.
Solution 2
We will use complex numbers as mentioned here. Set the circumcircle of to be the unit circle. Let
such that
We claim that the circumcircle of
passes through
This is true if
is real. Now observe that
so
is real and we are done.
Solution 3
Let be the midpoint of arc
. Let
be the midpoint of arc
. Let
be the midpoint of arc
.
Then
collinear. And
collinear.
We'll prove is cyclic. (Intuition: we'll show that
is the Miquel's point of quadrilateral
.
is the center of the circle
(the
excenter of
is also on the same circle). Therefore
. Similarly
. Since
,
. Therefore
. Obviously
and
. Thus by SAS,
.
Hence , so
is cyclic and we are done.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See also
2016 USAJMO (Problems • Resources) | ||
First Problem | Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |