Difference between revisions of "2012 IMO Problems/Problem 1"
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--[[User:Aopsqwerty|Aopsqwerty]] 21:19, 19 July 2012 (EDT) | --[[User:Aopsqwerty|Aopsqwerty]] 21:19, 19 July 2012 (EDT) | ||
+ | |||
+ | ==Solution 2== | ||
+ | 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. | ||
+ | |||
+ | Notice that <math>\angle{BJM} = \dfrac{\angle{B}}{2}</math> and <math>\angle{CJM} = \dfrac{\angle{C}}{2}</math>, and so | ||
+ | <cmath>a = R \tan \frac{B}{2} + R \tan \frac{C}{2} = R \frac{\sin \frac{B+C}{2}}{\cos \frac{B}{2} \cos \frac{C}{2}}</cmath> | ||
+ | after converting tangents to sine and cosine. Thus, | ||
+ | <cmath>R = a \cos \frac{B}{2} \cos \frac{C}{2} \sec \frac{A}{2}.</cmath> | ||
+ | It follows that <math>BM = a \sin \dfrac{B}{2} \cos \dfrac{C}{2} \sec \frac{A}{2}</math>. By the Law of Sines on triangle <math>BFM</math> and <math>ABC</math> and the double-angle formula for sine, we have | ||
+ | <cmath>BF = BM \cdot \frac{\sin \frac{C}{2}}{\sin \frac{A}{2}} = a \sin \frac{B}{2} \cdot \frac{\sin C}{\sin A} = c \sin \frac{B}{2}.</cmath> | ||
+ | Therefore, triangle <math>BFA</math> is congruent to a right triangle with hypotenuse length <math>c</math> and one angle of measure <math>90^\circ - \dfrac{B}{2}</math> by SAS Congruence, and so <math>\angle{BFA} = 90^\circ</math>. It then follows that triangles <math>BFS</math> and <math>BFA</math> are congruent by <math>ASA</math>, and so <math>AF = FS</math>. Thus, <math>J</math> lies on the perpendicular bisector of <math>AS</math>. Similarly, <math>J</math> lies on the perpendicular bisector of <math>AT</math>, and so <math>J</math> is the circumcenter of <math>ATS</math>. In particular, <math>J</math> lies on the perpendicular bisector of <math>ST</math>, and so, because <math>JM</math> is perpendicular to <math>ST</math>, <math>M</math> must be the midpoint of <math>ST</math>, as desired. | ||
+ | |||
+ | --[[User:Suli|Suli]] 17:53, 8 February 2015 (EST) |
Revision as of 18:53, 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)