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2012 IMO Problems/Problem 1

Revision as of 20:19, 19 July 2012 by Aopsqwerty (talk | contribs)

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)

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