IMO 2014 · Problem 3

Convex quadrilateral \(ABCD\) has \(\angle ABC=\angle CDA=90^\circ\). Point \(H\) is the foot of the perpendicular from \(A\) to \(BD\). Points \(S\) and \(T\) lie on sides \(AB\) and \(AD\), respectively, such that \(H\) lies inside triangle \(SCT\) and \[ \angle CHS-\angle CSB=90^\circ,\qquad \angle THC-\angle DTC=90^\circ. \] Prove that line \(BD\) is tangent to the circumcircle of triangle \(TSH\).

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

Solution

We use coordinates, taking care of the signs of the angles.

Since triangle \(SCT\) is contained in the convex quadrilateral, its interior point \(H\) is also interior to the quadrilateral. Thus \(H\) lies strictly between \(B\) and \(D\). After scaling and choosing axes, write \[ H=(0,0),\qquad A=(0,1),\qquad B=(-b,0),\qquad D=(d,0), \] where \(b,d>0\).

The perpendicularities \(BC\perp AB\) and \(DC\perp AD\) give \[ C=(d-b,-bd). \] Indeed, the equations of \(BC\) and \(DC\) are respectively \[ y=-b(x+b),\qquad y=d(x-d). \]

The given angles require \(S\ne B\) and \(T\ne D\). Consequently, we may write \[ S=(-bu,1-u),\qquad T=(dv,1-v), \qquad 0\le u,v<1. \] This also allows either point to equal \(A\), so those possible endpoint cases have not been excluded.

For vectors \(X,Y\), put \[ [X,Y]=x_Xy_Y-y_Xx_Y. \] We have \[ [S,T]=-bu(1-v)-dv(1-u)<0, \] because \(S,T\) cannot both equal \(A\). Moreover, since \(H=0\) lies strictly inside triangle \(SCT\), there are positive numbers \(\lambda,\mu,\nu\) such that \[ 0=\lambda S+\mu C+\nu T. \] Taking determinants with \(S\) and with \(T\) yields \[ [S,C]>0,\qquad [C,T]>0. \]

In the right triangle \(SBC\), \[ \tan\angle CSB=\frac{BC}{BS}=\frac{d}{1-u}. \] Using the first angle condition and the positive determinant just established, we obtain \[ \frac{b^2u-bd}{b^2du-(d-b)(1-u)} =\frac{C\cdot S}{[S,C]} =\cot\angle CHS =-\tan\angle CSB =-\frac{d}{1-u}. \] Multiplying out gives \[ (b^2u-d^2)(1-u)+b^2d^2u=0, \] or \[ b^2u^2-Lu+d^2=0, \qquad L=b^2+d^2+b^2d^2. \tag{1} \]

Likewise, the right triangle \(TDC\) gives \[ \tan\angle DTC=\frac{b}{1-v}. \] The second angle condition therefore implies \[ \frac{d^2v-bd}{bd^2v+(d-b)(1-v)} =\frac{T\cdot C}{[C,T]} =-\frac{b}{1-v}, \] which simplifies to \[ d^2v^2-Lv+b^2=0. \tag{2} \]

Multiply (1) by \(v\), multiply (2) by \(u\), and subtract. This gives \[ (uv-1)(b^2u-d^2v)=0. \] Since \(uv<1\), we conclude that \[ b^2u=d^2v. \tag{3} \] Combining this with (1), we find \[ \begin{aligned} d^2(1-u)(1-v) &=d^2-(b^2+d^2)u+b^2u^2\\ &=b^2d^2u. \end{aligned} \] Thus \[ (1-u)(1-v)=b^2u=d^2v. \tag{4} \]

Now (4) gives \[ \begin{aligned} |S|^2 &=b^2u^2+(1-u)^2\\ &=u(1-u)(1-v)+(1-u)^2\\ &=(1-u)(1-uv). \end{aligned} \] Similarly, \[ |T|^2=(1-v)(1-uv). \] Hence \(H,S,T\) all lie on the circle \[ x^2+y^2=(1-uv)y. \] Because \(1-uv>0\), this is a nondegenerate circle with centre \[ \left(0,\frac{1-uv}{2}\right). \] Also, \(H,S,T\) are noncollinear, since \(H\) lies strictly inside triangle \(SCT\). The displayed circle is therefore their circumcircle.

Its radius at \(H=(0,0)\) is vertical, so its tangent at \(H\) is the horizontal line \(y=0\), which is precisely \(BD\). \(\square\)

The solution was checked against the official IMO shortlist solutions.