IMO 2016 · Problem 1

Triangle \(BCF\) has a right angle at \(B\). Let \(A\) be the point on line \(CF\) such that \(FA = FB\) and \(F\) lies between \(A\) and \(C\). Point \(D\) is chosen such that \(DA = DC\) and \(AC\) is the bisector of \(\angle DAB\). Point \(E\) is chosen such that \(EA = ED\) and \(AD\) is the bisector of \(\angle EAC\). Let \(M\) be the midpoint of \(CF\). Let \(X\) be the point such that \(AMXE\) is a parallelogram (where \(AM \parallel EX\) and \(AE \parallel MX\)). Prove that lines \(BD\), \(FX\), and \(ME\) are concurrent.

Official text of the problem. Official problem paper (PDF)

Solution

Apply a similarity and, if necessary, a reflection so that \[ F=(0,0),\qquad A=(-1,0),\qquad C=(t,0), \] and \(B\) lies above the \(x\)-axis. Since \(FC\) is the hypotenuse of the right triangle \(BCF\), we have \(t>1\). Put \[ h=\sqrt{t^2-1}>0. \]

Writing \(B=(x,y)\), the conditions \(BF=1\) and \(BF\perp BC\) give \[ x^2+y^2=1,\qquad x(x-t)+y^2=0. \] Consequently, \[ B=\left(\frac1t,\frac ht\right). \]

Because \(AC\) bisects \(\angle DAB\), the ray \(AD\) is the reflection of the ray \(AB\) across the \(x\)-axis. Thus its direction is \((t+1,-h)\). Also, \(DA=DC\) places \(D\) on the perpendicular bisector of \(AC\), whose equation is \(x=(t-1)/2\). Hence \[ D=\left(\frac{t-1}{2},-\frac h2\right). \]

We next determine \(E\). The sum of two unit vectors points along their internal angle bisector. Since \[ \|(1,-h)\|=t,\qquad (1,0)+\frac1t(1,-h)=\frac1t(t+1,-h), \] the ray of direction \((t+1,-h)\) bisects the angle between the positive \(x\)-axis and the ray of direction \((1,-h)\). Therefore the given bisector condition implies \[ E=A+s(1,-h) \] for some \(s>0\). Now \(EA=ED\) yields \[ \begin{aligned} 0 &=\|E-A\|^2-\|E-D\|^2\\ &=2(E-A)\cdot(D-A)-\|D-A\|^2\\ &=s(t+1+h^2)-\frac{(t+1)^2+h^2}{4}\\ &=t(t+1)\left(s-\frac12\right). \end{aligned} \] Thus \[ E=\left(-\frac12,-\frac h2\right). \]

The midpoint and parallelogram conditions now give \[ M=\left(\frac t2,0\right),\qquad X=M+E-A=\left(\frac{t+1}{2},-\frac h2\right). \]

Define \[ \lambda=\frac{t}{2(t+1)},\qquad P=\left(\frac t4,-\frac{th}{4(t+1)}\right). \] Direct substitution into the coordinates above gives the three identities \[ P=(1-\lambda)F+\lambda X, \qquad P=(1-\lambda)M+\lambda E, \qquad P=\lambda B+(1-\lambda)D. \] The first places \(P\) on \(FX\), the second on \(ME\), and the third on \(BD\). Therefore the three lines are concurrent.

All denominators are nonzero because \(t>1\), and \(h>0\) ensures that each line is well-defined. Thus the argument covers every admissible configuration. \(\square\)

The solution was checked against the official IMO shortlist solutions.