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Difference between revisions of "2012 IMO Problems/Problem 1"

(Created page with "Problem 1: 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>...")
 
 
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Problem 1:
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== 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 $ST.
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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>.
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== Solution ==
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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>. 
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QED
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--[[User:Aopsqwerty|Aopsqwerty]] 21:19, 19 July 2012 (EDT)
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==Solution 2==
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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.
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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.
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Notice that <math>\angle{BJM} = \dfrac{\angle{B}}{2}</math> and <math>\angle{CJM} = \dfrac{\angle{C}}{2}</math>, and so
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<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>
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after converting tangents to sine and cosine. Thus,
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<cmath>R = a \cos \frac{B}{2} \cos \frac{C}{2} \sec \frac{A}{2}.</cmath>
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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
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<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>
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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.
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--[[User:Suli|Suli]] 17:53, 8 February 2015 (EST)
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==Solution 3==
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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>
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--[[User:Suli|Suli]] 18:21, 8 February 2015 (EST)
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==Solution 4==
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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.
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==See Also==
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{{IMO box|year=2012|before=First Problem|num-a=2}}

Latest revision as of 00:22, 19 November 2023

Problem

Given triangle $ABC$ the point $J$ is the centre of the excircle opposite the vertex $A.$ This excircle is tangent to the side $BC$ at $M$, and to the lines $AB$ and $AC$ at $K$ and $L$, respectively. The lines $LM$ and $BJ$ meet at $F$, and the lines $KM$ and $CJ$ meet at $G.$ Let $S$ be the point of intersection of the lines $AF$ and $BC$, and let $T$ be the point of intersection of the lines $AG$ and $BC.$ Prove that $M$ is the midpoint of $ST$.

Solution

First, $BK = BM$ because $BK$ and $BM$ are both tangents from $B$ to the excircle $J$. Then $BJ \bot KM$. Call the $X$ the intersection between $BJ$ and $KM$. Similarly, let the intersection between the perpendicular line segments $CJ$ and $LM$ be $Y$. We have $\angle XBM = \angle XBK = \angle FBA$ and $\angle XMB = \angle XKB$. We then have, $\angle XBM + \angle XBK + \angle XMB + \angle XKB = \angle MBK + \angle XMB + \angle XKB$ $= \angle MBK + \angle KMB + \angle MKB = 180^{\circ}$. So $\angle XBM = 90^{\circ} - \angle XMB$. We also have $180^{\circ} = \angle FBA + \angle ABC + \angle XBM = 2\angle XBM + \angle ABC = 180^{\circ} - 2\angle XMB$ $+ \angle ABC$. Then $\angle ABC = 2\angle XMB$. Notice that $\angle XFM = 90^{\circ} - \angle XMB - \angle BMF = 90^{\circ} - \angle XMB - \angle YMC$. Then, $\angle ACB = 2\angle YMC$. $\angle BAC = 180^{\circ} - \angle ABC - \angle ACB = 180^{\circ} - 2(\angle XMB + \angle YMC)$ $= 2(90^{\circ} - (\angle XMB + \angle YMC) = 2\angle XFM$. Similarly, $\angle BAC = 2\angle YGM$. Draw the line segments $FK$ and $GL$. $\triangle FXK$ and $\triangle FXM$ are congruent and $\triangle GYL$ and $\triangle GYM$ are congruent. Quadrilateral $AFJL$ is cyclic because $\angle JAL = \frac{\angle BAC}{2} = \angle XFM = \angle JFL$. Quadrilateral $AFKJ$ is also cyclic because $\angle JAK = \frac{\angle BAC}{2} = \angle XFM = \angle XFK = \angle JFK$. The circumcircle of $\triangle AFJ$ also contains the points $K$ and $J$ because there is a circle around the quadrilaterals $AFJL$ and $AFKJ$. Therefore, pentagon $AFKJL$ is also cyclic. Finally, quadrilateral $AGLJ$ is cyclic because $\angle JAL = \frac{\angle BAC}{2} = \angle YGM = \angle YGL = \angle JGL$. Again, $\triangle AJL$ is common in both the cyclic pentagon $AFKJL$ and cyclic quadrilateral $AGLJ$, so the circumcircle of $\triangle AJL$ also contains the points $F$, $K$, and $G$. Therefore, hexagon $AFKJLG$ is cyclic. Since $\angle AKJ$ and $\angle ALJ$ are both right angles, $AJ$ is the diameter of the circle around cyclic hexagon $AFKJLG$. Therefore, $\angle AFJ$ and $\angle AGJ$ are both right angles. $\triangle BFS$ and $\triangle BFA$ are congruent by ASA congruency, and so are $\triangle CGT$ and $\triangle CGA$. We have $SB = AB$, $TC = AC$, $BM = BK$, and $CM = CL$. Since $AK$ and $AL$ are tangents from $A$ to the circle $J$, $AK = AL$. Then, we have $AK = AL$, which becomes $AB + BK = AC + CL$, which is $SB + BM = TC + CM$, or $SM = TM$. This means that $M$ is the midpoint of $ST$.

QED

--Aopsqwerty 21:19, 19 July 2012 (EDT)

Solution 2

For simplicity, let $A, B, C$ written alone denote the angles of triangle $ABC$, and $a$, $b$, $c$ denote its sides.

Let $R$ be the radius of the A-excircle. Because $CM = CL$, we have $CML$ isosceles and so $\angle{CML} = \dfrac{\angle{C}}{2}$ by the Exterior Angle Theorem. Then because $\angle{FBS} = 90^\circ - \dfrac{B}{2}$, we have $\angle{BFM} = \dfrac{\angle{A}}{2}$, again by the Exterior Angle Theorem.

Notice that $\angle{BJM} = \dfrac{\angle{B}}{2}$ and $\angle{CJM} = \dfrac{\angle{C}}{2}$, and so \[a = R \tan \frac{B}{2} + R \tan \frac{C}{2} = R \frac{\sin \frac{B+C}{2}}{\cos \frac{B}{2} \cos \frac{C}{2}}\] after converting tangents to sine and cosine. Thus, \[R = a \cos \frac{B}{2} \cos \frac{C}{2} \sec \frac{A}{2}.\] It follows that $BM = a \sin \dfrac{B}{2} \cos \dfrac{C}{2} \sec \frac{A}{2}$. By the Law of Sines on triangle $BFM$ and $ABC$ and the double-angle formula for sine, we have \[BF = BM \cdot \frac{\sin \frac{C}{2}}{\sin \frac{A}{2}} = a \sin \frac{B}{2} \cdot \frac{\sin C}{\sin A} = c \sin \frac{B}{2}.\] Therefore, triangle $BFA$ is congruent to a right triangle with hypotenuse length $c$ and one angle of measure $90^\circ - \dfrac{B}{2}$ by SAS Congruence, and so $\angle{BFA} = 90^\circ$. It then follows that triangles $BFS$ and $BFA$ are congruent by $ASA$, and so $AF = FS$. Thus, $J$ lies on the perpendicular bisector of $AS$. Similarly, $J$ lies on the perpendicular bisector of $AT$, and so $J$ is the circumcenter of $ATS$. In particular, $J$ lies on the perpendicular bisector of $ST$, and so, because $JM$ is perpendicular to $ST$, $M$ must be the midpoint of $ST$, as desired.

--Suli 17:53, 8 February 2015 (EST)

Solution 3

Same as Solution 2, except noticing that (letting $s = \dfrac{a + b + c}{2}$ be the semi-perimeter): \[FB = BM \cdot \frac{\sin \frac{C}{2}}{\sin \frac{A}{2}} = (s - c) \cdot \frac{\sqrt{\frac{(s-a)(s-b)}{ab}}}{\sqrt{\frac{(s-b)(s-c)}{bc}}} = c \sqrt{\frac{(s-a)(s-c)}{ac}} = c \sin \frac{B}{2}.\]

--Suli 18:21, 8 February 2015 (EST)

Solution 4

As before in Solution 2, we find that $\angle{JFL} = \dfrac{\angle{A}}{2}.$ But it is clear that $AJ$ bisects $\angle{KAL}$, so $\angle{JAL} = \dfrac{\angle{A}}{2} = \angle{JFL}$ and hence $AFJL$ is cyclic. In particular, $\angle{AFJ} = \angle{ALJ} = 90^\circ$, 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
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