Difference between revisions of "2012 IMO Problems/Problem 1"
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− | Problem | + | == Problem == |
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Given triangle <math>ABC</math> the point <math>J</math> is the centre of the excircle opposite the vertex <math>A.</math> This excircle is tangent to the side <math>BC</math> at <math>M</math>, and to the lines <math>AB</math> and <math>AC</math> at <math>K</math> and <math>L</math>, respectively. The lines <math>LM</math> and <math>BJ</math> meet at <math>F</math>, and the lines <math>KM</math> and <math>CJ</math> meet at <math>G.</math> Let <math>S</math> be the point of intersection of the lines <math>AF</math> and <math>BC</math>, and let <math>T</math> be the point of intersection of the lines <math>AG</math> and <math>BC.</math> Prove that <math>M</math> is the midpoint of <math>ST</math>. | Given triangle <math>ABC</math> the point <math>J</math> is the centre of the excircle opposite the vertex <math>A.</math> This excircle is tangent to the side <math>BC</math> at <math>M</math>, and to the lines <math>AB</math> and <math>AC</math> at <math>K</math> and <math>L</math>, respectively. The lines <math>LM</math> and <math>BJ</math> meet at <math>F</math>, and the lines <math>KM</math> and <math>CJ</math> meet at <math>G.</math> Let <math>S</math> be the point of intersection of the lines <math>AF</math> and <math>BC</math>, and let <math>T</math> be the point of intersection of the lines <math>AG</math> and <math>BC.</math> Prove that <math>M</math> is the midpoint of <math>ST</math>. | ||
+ | |||
+ | == Solution == | ||
+ | |||
+ | First, <math>BK = BM</math> because <math>BK</math> and <math>BM</math> are both tangents from <math>B</math> to the excircle <math>J</math>. Then <math>BJ \bot KM</math>. Call the <math>X</math> the intersection between <math>BJ</math> and <math>KM</math>. Similarly, let the intersection between the perpendicular line segments <math>CJ</math> and <math>LM</math> be <math>Y</math>. We have <math>\angle XBM = \angle XBK = \angle FBA</math> and <math>\angle XMB = \angle XKB</math>. We then have, <math>\angle XBM + \angle XBK + \angle XMB + \angle XKB = \angle MBK + \angle XMB + \angle XKB</math> <math>= \angle MBK + \angle KMB + \angle MKB = 180^{\circ} </math>. So <math>\angle XBM = 90^{\circ} - \angle XMB</math>. We also have <math>180^{\circ} = \angle FBA + \angle ABC + \angle XBM = 2\angle XBM + \angle ABC = 180^{\circ} - 2\angle XMB</math> <math>+ \angle ABC</math>. Then <math>\angle ABC = 2\angle XMB</math>. Notice that <math>\angle XFM = 90^{\circ} - \angle XMB - \angle BMF = 90^{\circ} - \angle XMB - \angle YMC</math>. Then, <math>\angle ACB = 2\angle YMC</math>. <math>\angle BAC = 180^{\circ} - \angle ABC - \angle ACB = 180^{\circ} - 2(\angle XMB + \angle YMC)</math> <math>= 2(90^{\circ} - (\angle XMB + \angle YMC) = 2\angle XFM</math>. Similarly, <math>\angle BAC = 2\angle YGM</math>. Draw the line segments <math>FK</math> and <math>GL</math>. <math>\triangle FXK</math> and <math>\triangle FXM</math> are congruent and <math>\triangle GYL</math> and <math>\triangle GYM</math> are congruent. Quadrilateral <math>AFJL</math> is cyclic because <math>\angle JAL = \frac{\angle BAC}{2} = \angle XFM = \angle JFL</math>. Quadrilateral <math>AFKJ</math> is also cyclic because <math>\angle JAK = \frac{\angle BAC}{2} = \angle XFM = \angle XFK = \angle JFK</math>. The circumcircle of <math>\triangle AFJ</math> also contains the points <math>K</math> and <math>J</math> because there is a circle around the quadrilaterals <math>AFJL</math> and <math>AFKJ</math>. Therefore, pentagon <math>AFKJL</math> is also cyclic. Finally, quadrilateral <math>AGLJ</math> is cyclic because <math>\angle JAL = \frac{\angle BAC}{2} = \angle YGM = \angle YGL = \angle JGL</math>. Again, <math>\triangle AJL</math> is common in both the cyclic pentagon <math>AFKJL</math> and cyclic quadrilateral <math>AGLJ</math>, so the circumcircle of <math>\triangle AJL</math> also contains the points <math>F</math>, <math>K</math>, and <math>G</math>. Therefore, hexagon <math>AFKJLG</math> is cyclic. Since <math>\angle AKJ</math> and <math>\angle ALJ</math> are both right angles, <math>AJ</math> is the diameter of the circle around cyclic hexagon <math>AFKJLG</math>. Therefore, <math>\angle AFJ</math> and <math>\angle AGJ</math> are both right angles. <math>\triangle BFS</math> and <math>\triangle BFA</math> are congruent by ASA congruency, and so are <math>\triangle CGT</math> and <math>\triangle CGA</math>. We have <math>SB = AB</math>, <math>TC = AC</math>, <math>BM = BK</math>, and <math>CM = CL</math>. Since <math>AK</math> and <math>AL</math> are tangents from <math>A</math> to the circle <math>J</math>, <math>AK = AL</math>. Then, we have <math>AK = AL</math>, which becomes <math>AB + BK = AC + CL</math>, which is <math>SB + BM = TC + CM</math>, or <math>SM = TM</math>. This means that <math>M</math> is the midpoint of <math>ST</math>. | ||
+ | |||
+ | QED | ||
+ | |||
+ | --[[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) | ||
+ | |||
+ | ==Solution 3== | ||
+ | Same as Solution 2, except noticing that (letting <math>s = \dfrac{a + b + c}{2}</math> be the semi-perimeter): <cmath>FB = BM \cdot \frac{\sin \frac{C}{2}}{\sin \frac{A}{2}} = (s - c) \cdot \frac{\sqrt{\frac{(s-a)(s-b)}{ab}}}{\sqrt{\frac{(s-b)(s-c)}{bc}}} = c \sqrt{\frac{(s-a)(s-c)}{ac}} = c \sin \frac{B}{2}.</cmath> | ||
+ | |||
+ | --[[User:Suli|Suli]] 18:21, 8 February 2015 (EST) | ||
+ | |||
+ | ==Solution 4== | ||
+ | As before in Solution 2, we find that <math>\angle{JFL} = \dfrac{\angle{A}}{2}.</math> But it is clear that <math>AJ</math> bisects <math>\angle{KAL}</math>, so <math>\angle{JAL} = \dfrac{\angle{A}}{2} = \angle{JFL}</math> and hence <math>AFJL</math> is cyclic. In particular, <math>\angle{AFJ} = \angle{ALJ} = 90^\circ</math>, and continue as in Solution 2. | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{IMO box|year=2012|before=First Problem|num-a=2}} |
Latest revision as of 01:22, 19 November 2023
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)
Solution 3
Same as Solution 2, except noticing that (letting be the semi-perimeter):
--Suli 18:21, 8 February 2015 (EST)
Solution 4
As before in Solution 2, we find that But it is clear that
bisects
, so
and hence
is cyclic. In particular,
, and continue as in Solution 2.
See Also
2012 IMO (Problems) • Resources | ||
Preceded by First Problem |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |