2012 IMO Problems/Problem 1


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 $ST$.


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 $ST$.


--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