Difference between revisions of "2012 IMO Problems/Problem 1"
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For simplicity, let <math>A, B, C</math> written alone denote the angles of triangle <math>ABC</math>, and <math>a</math>, <math>b</math>, <math>c</math> denote its sides. | For simplicity, let <math>A, B, C</math> written alone denote the angles of triangle <math>ABC</math>, and <math>a</math>, <math>b</math>, <math>c</math> denote its sides. | ||
− | Let <math>R</math> be the radius of the A-excircle. Because <math>CM = CL</math>, we have <math>CML</math> isosceles and so <math>\angle{CML} = \dfrac{\angle{C}}{2}</math> by the Exterior Angle Theorem. Then because <math>\angle{FBS} = 90^\circ - \dfrac{B}{2}</math>, we have <math>\angle{BFM} = \dfrac{angle{A}}{2}</math>, again by the Exterior Angle Theorem. | + | Let <math>R</math> be the radius of the A-excircle. Because <math>CM = CL</math>, we have <math>CML</math> isosceles and so <math>\angle{CML} = \dfrac{\angle{C}}{2}</math> by the Exterior Angle Theorem. Then because <math>\angle{FBS} = 90^\circ - \dfrac{B}{2}</math>, we have <math>\angle{BFM} = \dfrac{\angle{A}}{2}</math>, again by the Exterior Angle Theorem. |
Notice that <math>\angle{BJM} = \dfrac{\angle{B}}{2}</math> and <math>\angle{CJM} = \dfrac{\angle{C}}{2}</math>, and so | Notice that <math>\angle{BJM} = \dfrac{\angle{B}}{2}</math> and <math>\angle{CJM} = \dfrac{\angle{C}}{2}</math>, and so |
Revision as of 18:54, 8 February 2015
Problem
Given triangle the point
is the centre of the excircle opposite the vertex
This excircle is tangent to the side
at
, and to the lines
and
at
and
, respectively. The lines
and
meet at
, and the lines
and
meet at
Let
be the point of intersection of the lines
and
, and let
be the point of intersection of the lines
and
Prove that
is the midpoint of
.
Solution
First, because
and
are both tangents from
to the excircle
. Then
. Call the
the intersection between
and
. Similarly, let the intersection between the perpendicular line segments
and
be
. We have
and
. We then have,
. So
. We also have
. Then
. Notice that
. Then,
.
. Similarly,
. Draw the line segments
and
.
and
are congruent and
and
are congruent. Quadrilateral
is cyclic because
. Quadrilateral
is also cyclic because
. The circumcircle of
also contains the points
and
because there is a circle around the quadrilaterals
and
. Therefore, pentagon
is also cyclic. Finally, quadrilateral
is cyclic because
. Again,
is common in both the cyclic pentagon
and cyclic quadrilateral
, so the circumcircle of
also contains the points
,
, and
. Therefore, hexagon
is cyclic. Since
and
are both right angles,
is the diameter of the circle around cyclic hexagon
. Therefore,
and
are both right angles.
and
are congruent by ASA congruency, and so are
and
. We have
,
,
, and
. Since
and
are tangents from
to the circle
,
. Then, we have
, which becomes
, which is
, or
. This means that
is the midpoint of
.
QED
--Aopsqwerty 21:19, 19 July 2012 (EDT)
Solution 2
For simplicity, let written alone denote the angles of triangle
, and
,
,
denote its sides.
Let be the radius of the A-excircle. Because
, we have
isosceles and so
by the Exterior Angle Theorem. Then because
, we have
, again by the Exterior Angle Theorem.
Notice that and
, and so
after converting tangents to sine and cosine. Thus,
It follows that
. By the Law of Sines on triangle
and
and the double-angle formula for sine, we have
Therefore, triangle
is congruent to a right triangle with hypotenuse length
and one angle of measure
by SAS Congruence, and so
. It then follows that triangles
and
are congruent by
, and so
. Thus,
lies on the perpendicular bisector of
. Similarly,
lies on the perpendicular bisector of
, and so
is the circumcenter of
. In particular,
lies on the perpendicular bisector of
, and so, because
is perpendicular to
,
must be the midpoint of
, as desired.
--Suli 17:53, 8 February 2015 (EST)