#4 Simson Line and Pedal Geometry

E. Full Solution.

Since \(A_1\in BC\) and \(PA_1\perp BC\), we have \(\angle PA_1C=90^\circ\). Similarly, \(B_1\in CA\) and \(PB_1\perp CA\), so \(\angle PB_1C=90^\circ\).

Thus \(A_1\) and \(B_1\) lie on the circle with diameter \(PC\). Therefore \(P,A_1,C,B_1\) are cyclic.

E. Full Solution.

By the previous fact, \(A_1\) and \(B_1\) lie on the circle with diameter \(PC\), so \(A_1B_1\) is a chord of this circle. Similarly, \(B_1\) and \(C_1\) lie on the circle with diameter \(PA\), and \(C_1\) and \(A_1\) lie on the circle with diameter \(PB\).

Thus all three sides of the pedal triangle are naturally connected with the circles on \(PA,PB,PC\) as diameters.

E. Full Solution.

Since \(H\) lies on the altitude from \(A\), the line \(AH\) is perpendicular to \(BC\). Therefore the foot of the perpendicular from \(H\) to \(BC\) coincides with the foot of the altitude from \(A\). Similarly, the projections of \(H\) onto \(CA\) and \(AB\) coincide with the feet of the altitudes from \(B\) and \(C\).

E. Full Solution.

The side \(BC\) is a chord of the circumcircle. The perpendicular from the center \(O\) to the chord \(BC\) bisects it. Hence the projection of \(O\) onto \(BC\) is the midpoint of \(BC\). The same applies to \(CA\) and \(AB\).

E. Full Solution.

The right angles give cyclic quadrilaterals \(P,A_1,C,B_1\) and \(P,B_1,A,C_1\). Hence \(\angle A_1B_1C=\angle A_1PC\), and \(\angle C_1B_1A=\angle C_1PA\) in the directed sense.

Since \(A,B,C,P\) lie on one circle, angles subtending the corresponding arcs give equality of these two directed angles. Thus the rays \(B_1A_1\) and \(B_1C_1\) lie on one line. Therefore \(A_1,B_1,C_1\) are collinear.

E. Full Solution.

As in the direct theorem, the right angles give cyclic quadrilaterals \(P,A_1,C,B_1\) and \(P,B_1,A,C_1\). The collinearity of \(A_1,B_1,C_1\) gives equality of the corresponding directed angles.

Transferring these angles to the circles with diameters, we get \(\angle PCA=\angle PBA\). This is the cyclicity criterion for \(P,C,A,B\). Hence \(P\) lies on the circumcircle of triangle \(ABC\).

E. Full Solution.

The pedal triangle has zero area if and only if its vertices \(A_1,B_1,C_1\) are collinear.

If \(P\) lies on the circumcircle, then by the Simson theorem \(A_1,B_1,C_1\) are collinear. Conversely, if they are collinear, then by the converse Simson theorem \(P\) lies on the circumcircle. The claim follows.

E. Full Solution.

Let \(A_1,B_1,C_1\) be the ordinary perpendicular feet from \(P\). By the Simson theorem, they are collinear. The new points \(A_2,B_2,C_2\) on \(BC,CA,AB\) are obtained from \(A_1,B_1,C_1\) by the same change of angle at \(P\): from the perpendicular direction to the direction making angle \(\alpha\) with the side.

For each side, the correspondence \(A_1\mapsto A_2\), \(B_1\mapsto B_2\), \(C_1\mapsto C_2\) is given by one spiral homothety centered at \(P\) with angle \(90^\circ-\alpha\). Such a transformation sends a line to a line. Therefore \(A_2,B_2,C_2\) are collinear.

E. Full Solution.

Let \(A_1,B_1,C_1\) be the projections of \(P\) onto \(BC,CA,AB\). The direction of the Simson line can be expressed, for instance, through the angle between \(B_1C_1\) and \(AB\). From the cyclicity of \(P,B_1,A,C_1\), this angle equals the corresponding angle between \(PA\) and \(PC\).

Since \(PQ\perp BC\), the inscribed angles defining the direction of \(AQ\) coincide with the angles defining the direction of \(B_1C_1\). Therefore \(B_1C_1\parallel AQ\). Hence the whole Simson line of \(P\) is parallel to \(AQ\).

E. Full Solution.

Let \(M\) be the midpoint of \(PH\), and let \(B_1,C_1\) be the projections of \(P\) onto \(CA\) and \(AB\). Since \(BH\perp AC\), and \(PB_1\perp AC\), we have \(PB_1\parallel BH\). Similarly, \(PC_1\parallel CH\).

Under the homothety with center \(P\) and ratio \(\frac12\), the point \(H\) maps to \(M\), while the directions \(BH\) and \(CH\) correspond to the directions through \(B_1\) and \(C_1\). From the parallelism of corresponding segments, we get \(M\in B_1C_1\). But \(B_1C_1\) is the Simson line, so it passes through \(M\).

E. Full Solution.

The direction of the Simson line changes by half of the angular displacement of the point on the circumcircle. Points \(P\) and \(Q\) are antipodal, so the radii \(OP\) and \(OQ\) differ by a rotation of \(180^\circ\).

Therefore the directions of their Simson lines differ by \(90^\circ\). Hence the two lines are perpendicular.

E. Full Solution.

Let \(H\) be the orthocenter, and let \(M,N\) be the midpoints of \(PH\) and \(QH\). By the Simson line property, the Simson line of \(P\) passes through \(M\), and the Simson line of \(Q\) passes through \(N\). The points \(M,N\) lie on the nine-point circle, because this circle is the image of the circumcircle under the homothety with center \(H\) and ratio \(\frac12\).

By the previous problem, the Simson lines are perpendicular. In the standard nine-point circle configuration, the angle \(MXN\) subtends a diameter \(MN\) of this circle, so \(X\) lies on it. Equivalently, check by angles: \(\angle MXN=90^\circ\), and \(MN\) is the corresponding diameter.

E. Full Solution.

When a point moves along the circumcircle by an arc of angular measure \(2\varphi\), the direction of the Simson line changes by \(\varphi\). Thus if the Simson lines of \(P\) and \(Q\) are parallel, their directions coincide modulo \(180^\circ\). Hence the arcs between \(P\) and \(Q\) have angular measure \(0^\circ\) or \(360^\circ\), so \(P=Q\).

The converse is immediate: if \(P=Q\), the Simson lines coincide, hence are parallel.

E. Full Solution.

By earlier facts, the projections of the circumcenter \(O\) onto the sides of \(ABC\) are the side midpoints, while the projections of the orthocenter \(H\) are the altitude feet.

The nine-point circle of triangle \(ABC\) passes through the side midpoints and the altitude feet. Therefore both pedal triangles have all their vertices on the same nine-point circle.

E. Full Solution.

Let \(A_1\) and \(B_1\) be the projections of \(P\) onto \(CB\) and \(CA\). The points \(A_1\) and \(B_1\) lie on the circle with diameter \(PC\). The angle \(\angle A_1CB_1\) equals \(\angle ACB\), which is constant by assumption.

Therefore the chord \(A_1B_1\) of the circle with diameter \(PC\) has constant length. The Simson line is the line \(A_1B_1\). All chords of a fixed circle with the same length are tangent to a concentric circle of fixed radius. Hence all these Simson lines are tangent to one fixed circle.

E. Full Solution.

By the Simson theorem, \(A_1,B_1,C_1\) are collinear. Consider the angle between the Simson line and \(AB\). Through the cyclic quadrilateral \(P,B_1,A,C_1\), it equals the angle between \(PB_1\) and \(PA\), that is, the angle between a perpendicular to \(AC\) and a line parallel to \(BC\).

This angle equals the angle between the altitude from \(A\) and \(AB\). Therefore the Simson line has the same direction as the altitude from \(A\), so it is parallel to it.

E. Full Solution.

For triangle \(BCD\), the Simson line of \(A\) passes through the midpoint of \(AH_A\), where \(H_A\) is the orthocenter of \(BCD\). The other three midpoints are defined similarly. In a cyclic quadrilateral, these four orthocenters form a system in which the corresponding midpoints lie on one nine-point circle of the complete quadrilateral.

Moreover, the directions of the four Simson lines are pairwise coordinated through the same circle: passing from one vertex to another changes the triangle and the Simson point, but preserves the common Miquel point of the orthopedal configuration. Hence all four lines pass through this point.

The last step can also be checked by angles: the intersection of \(l_A\) and \(l_B\) lies on \(l_C\) and \(l_D\), because the corresponding angles with \(AB,BC,CD,DA\) are expressed through the same arcs of the original circle.

E. Full Solution.

The homothety with center \(H\) and ratio \(\frac12\) sends \(P\) to the midpoint \(M\) of \(PH\). Since \(P\) moves on the circumcircle, the image of this circle under the homothety is the nine-point circle. Hence \(M\) moves on the nine-point circle.

By the Simson line midpoint property, for each point \(P\), this same midpoint \(M\) lies on the Simson line of \(P\). Thus both statements hold simultaneously.

E. Full Solution.

The pedal triangle has area \(0\) if and only if its vertices, the three projections of \(P\), are collinear.

By the Wallace-Simson theorem, this happens if and only if \(P\) lies on the circumcircle of triangle \(ABC\). Therefore the desired locus is the circumcircle of \(ABC\).

E. Full Solution.

The direction of the Simson line of \(P\) can be expressed through the angle between this line and a fixed side, say \(AB\). In the proof of the Simson theorem, this angle equals an inscribed angle depending on an arc with endpoint \(P\).

When \(P\) moves from \(P_1\) to \(P_2\), the corresponding central angle changes by \(2\varphi\), while the inscribed angle changes by \(\varphi\). Hence the direction of the Simson line changes by \(\varphi\). Therefore the angle between the two Simson lines is \(\varphi\).

E. Full Solution.

The direction of the Simson line of \(A\) with respect to \(BCD\) is expressed through arcs of the circle \(ABCD\). The direction of the Euler line of triangle \(BCD\) also depends only on the positions of \(B,C,D\) on this circle and their orthocenter.

When the perpendicularity condition is written in directed angles, it becomes an equality of arcs symmetric under exchanging \(A\) and \(B\). Therefore the same equality gives perpendicularity of the Simson line of \(B\) with respect to \(ACD\) and the Euler line of triangle \(ACD\).

In other words, the condition is a property of the four points on one circle, not of the chosen vertex. The exchange \(A\leftrightarrow B\) preserves it.

E. Full Solution.

For each triangle obtained by deleting one vertex, the Simson line of \(P\) passes through the three projections of \(P\) onto the sides of that triangle. The projection of \(P\) onto such a Simson line can be described through the circle with diameter between \(P\) and the corresponding vertex of the deleted configuration.

Consider three triangles, for example \(BCD,CDA,DAB\). The angle expressions for the projections of \(P\) onto their Simson lines depend on the same arcs of the circle \(ABCDP\). Comparing directed angles shows that three such projections are collinear. The fourth projection satisfies the same angle condition by the cyclic symmetry.

Therefore all four projections lie on one line. This line is called the Simson line of the cyclic quadrilateral with respect to \(P\).

E. Full Solution.

Let \(M\) be the midpoint of \(PH\). Then \(M\) lies on the nine-point circle and also on the Simson line of \(P\). Under a small motion of \(P\), the point \(M\) moves on the nine-point circle, while the direction of the Simson line changes in a way coordinated with the tangent direction of the path of \(M\).

The contact point of the envelope is obtained as the limiting position of the intersection of two nearby Simson lines. It can be constructed from \(M\): draw the Simson line through \(M\) and use that the angular speed of this line is half the angular speed of \(P\). The resulting envelope is the classical Steiner deltoid.

For this module, the important part is not the name of the curve, but the mechanism: the Simson line always passes through a point of the nine-point circle, while its direction changes so that a fixed envelope appears.

E. Full Solution.

In the complex plane, for the line through two unit-circle points \(u,v\), the foot of the perpendicular from \(p\) can be written as a linear expression in \(p,u,v\) and their conjugates. Since \(\overline{u}=1/u\), \(\overline{v}=1/v\), the formula simplifies to a rational expression.

Subtracting the coordinates of two perpendicular feet, for example on \(CA\) and \(AB\), gives a direction number for the Simson line. After cancellation, it is proportional to \((p-a)(p-b)(p-c)/p\).

If \(p\) is replaced by the antipodal point \(q=-p\), the direction number is multiplied by a purely imaginary factor. This means a rotation of direction by \(90^\circ\). Hence the Simson lines of antipodal points are perpendicular.