Problem
GEO-B1-M02-P025 A Hidden Orthocenter in a Parallelogram
Inside parallelogram \(PQRS\), point \(X\) is chosen so that \(PX=SX\) and \(\angle PQX=90^\circ\). Point \(T\) is the midpoint of side \(QR\). Prove that \(XT\perp ST\).
C. Hint 1. Take the midpoint of side \(PS\).
D. Hint 2. Then look for two altitudes of one triangle passing through \(X\).
E. Full solution.
Let \(U\) be the midpoint of side \(PS\). Since \(PX=SX\), in isosceles triangle \(PXS\) the median \(XU\) to the base is also an altitude. Hence \(XU\perp PS\). Since \(PS\parallel QR\), we get \(XU\perp QT\).
Because \(U\) and \(T\) are midpoints of opposite sides of the parallelogram, quadrilaterals \(PQTU\) and \(QTSU\) are parallelograms. Thus \(UT\parallel PQ\) and \(QU\parallel ST\).
By the condition \(QX\perp PQ\), and since \(UT\parallel PQ\), we have \(QX\perp UT\). Therefore in triangle \(QUT\), the lines \(XU\) and \(XQ\) are two altitudes. Hence \(X\) is the orthocenter of triangle \(QUT\).
The third altitude of this triangle also passes through \(X\), so \(XT\perp QU\). But \(QU\parallel ST\), hence \(XT\perp ST\), as required.
A. Source analysis. Main objects: a parallelogram, midpoints of opposite sides, and an isosceles triangle. The obvious approach is to prove \(XT\perp ST\) directly, but the line \(ST\) is hidden as a parallel to \(QU\). The hidden observation is that \(X\) becomes the orthocenter of auxiliary triangle \(QUT\). Number of key ideas: 3.
F. Difficulty justification. This is Level 7: a regional-style problem with two auxiliary parallelograms and recognition of an orthocenter.
G. Check. This is not a one-step exercise: one must add a midpoint, obtain two parallelisms, and only then see the orthocenter.