IMO 1982 · Problem 2

A non-isosceles triangle \(A_1A_2A_3\) is given with sides \(a_1,a_2,a_3\) (\(a_i\) is the side opposite \(A_i\)). For all \(i=1,2,3\), \(M_i\) is the midpoint of side \(a_i\), and \(T_i\). is the point where the incircle touches side \(a_i\). Denote by \(S_i\) the reflection of \(T_i\) in the interior bisector of angle \(A_i\). Prove that the lines \(M_1,S_1\), \(M_2S_2\), and \(M_3S_3\) are concurrent.

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

Solution

Write \(A=A_1\), \(B=A_2\), \(C=A_3\), and let \[ a=BC,\qquad b=CA,\qquad c=AB,\qquad s=\frac{a+b+c}{2}. \] Put \[ x=s-a,\qquad y=s-b,\qquad z=s-c. \] The triangle inequalities give \(x,y,z>0\), and \[ a=y+z,\qquad b=z+x,\qquad c=x+y. \] We identify points with their position vectors relative to an arbitrary origin.

Equal tangent lengths from each vertex give \(BT_1=y\) and \(CT_1=z\). Consequently, \[ T_1-A=\frac{z}{a}(B-A)+\frac{y}{a}(C-A). \] Reflection in the internal bisector at \(A\) interchanges the unit vectors \[ \frac{B-A}{c}\quad\text{and}\quad\frac{C-A}{b}. \] It therefore gives \[ S_1-A=\frac{zc}{ab}(C-A)+\frac{yb}{ac}(B-A), \] or \[ abc\,S_1=(abc-yb^2-zc^2)A+yb^2B+zc^2C. \] Using the expressions for \(a,b,c\) above, \[ abc-yb^2-zc^2 =x\bigl((y+z)^2-4yz\bigr) =x(b-c)^2. \] Thus, together with the two cyclic versions, \[ \begin{aligned} abc\,S_1&=x(b-c)^2A+yb^2B+zc^2C,\\ abc\,S_2&=xa^2A+y(c-a)^2B+zc^2C,\\ abc\,S_3&=xa^2A+yb^2B+z(a-b)^2C. \end{aligned} \]

Define \[ u=x(b-c)^2,\qquad v=y(c-a)^2,\qquad w=z(a-b)^2,\qquad k=4xyz. \] Because the triangle is non-isosceles, its side lengths are pairwise distinct, so \(u,v,w>0\). Moreover, \[ \begin{aligned} xa^2-u&=x\bigl((y+z)^2-(y-z)^2\bigr)=k,\\ yb^2-v&=y\bigl((z+x)^2-(z-x)^2\bigr)=k,\\ zc^2-w&=z\bigl((x+y)^2-(x-y)^2\bigr)=k. \end{aligned} \] The formulas for the reflected points become \[ \begin{aligned} abc\,S_1&=uA+(v+k)B+(w+k)C,\\ abc\,S_2&=(u+k)A+vB+(w+k)C,\\ abc\,S_3&=(u+k)A+(v+k)B+wC. \end{aligned} \] The identity \(u=abc-yb^2-zc^2\) also shows that \[ abc=u+v+w+2k. \]

Let \(D=u+v+w\), and define the point \[ P=\frac{uA+vB+wC}{D}. \] Since \[ M_1=\frac{B+C}{2},\qquad M_2=\frac{C+A}{2},\qquad M_3=\frac{A+B}{2}, \] the preceding formulas yield, for every \(i=1,2,3\), \[ S_i=\frac{D}{D+2k}\,P+\frac{2k}{D+2k}\,M_i. \] Both coefficients are strictly positive and sum to \(1\). Also, \(P\) lies strictly inside \(ABC\), because \(u,v,w>0\), whereas each \(M_i\) lies on a side. Hence \(P\ne M_i\), and \(S_i\) lies strictly between \(P\) and \(M_i\).

Therefore all three lines \(M_1S_1\), \(M_2S_2\), and \(M_3S_3\) are well defined and pass through \(P\). \(\square\)

The solution was checked against John Scholes' IMO solution archive (Kalva).