# Amena Warrayat

Problem four states:

Find all conditions in which the three vertices of the Pedal Triangle are collinear (that is, the Pedal Triangle is a degenerate triangle). This line is called the Simson Line.

To define a pedal triangle, begin with triangle ABC. Then, extend the segments AB, AC, and BC into lines in the plane. Now, for any point P in the plane, take the perpendiculars from P to lines AB, AC, and BC. Each perpendicular will meet the triangle line at one point, which we'll call X, Y, and Z. The pedal triangle is triangle XYZ.
In the figure below, triangle ABC is orange. Point P is the base point for the pedal triangle. The three perpendiculars are drawn in blue, the foots of the perpendiculars at X, Y, and Z. The pedal triangle is gray.

The orthic triangle is the pedal triangle through the orthocenter O. For an acute triangle, the orthic triangle is the inscribed triangle of minimum perimeter. Also, as the purple angles denote, there are special angle equality properties for an orthic triangle, which do not hold for general pedal triangles.

No matter where P lies in the plane, there is always a pedal triangle. Whenever P lies inside triangle ABC, at least part of the pedal triangle will also lie inside ABC. But when P lies outside, the entire pedal triangle might also be outside, as in this example.

If we play around with P, A, B, and C, we find various locations for triangles. Sometimes the pedal triangle is completely inside ABC; sometimes part of it is inside; sometimes it's completely outside. There's one case of special interest, though. Sometimes it looks like X, Y, and Z are co-linear, and the triangle becomes a degenerate line. Here's one example.

Is this really true, that the triangle becomes degenerate? Yes, sometimes it does. When this occurs, the line is called the Simson line. Robert Simson was a Scottish professor who published a translation of Euclid's Elements.
Let's demonstrate that the pedal triangle becomes degenerate if and only if P lies on the circumcircle drawn around triangle ABC. So, we have to show points Y, X, and Z are on the same line. Looking at quadrilateral PACB, we have four points on the circumcircle. It's a cyclic quadrilateral. (If P does not lie on the circumcircle, PABC is not a cyclic quadrilateral and this does not work.)
For a cyclic quadrilateral, the sum of opposite angles is 180 degrees. Here, angle ACB + angle APB = 180. Angle APB equals angle APY plus angle YPB, so we have angle ACB + angle APY + angle YPB = 180.

We also have another cyclic quadrilateral PZCY. In this quadrilateral, one pair of opposite angles are YCZ and YPZ. Angle YCZ + angle YPZ = 180. Since Y lies on AC, and Z lies on CB, angle YCZ equals angle ACB.
Using the equations, this means angle APB = angle YPZ. Since angle YPB is part of both APB and YPZ, we have shown the equivalence of two small angles, angle APY equaling angle BPZ. Those angles are purple.

Next, we use cyclic quadrilateral PXYA to show that angle APY equals angle AXY, because points P and X are both on the circle and they both subtend the same arc. We also need cyclic quadrilateral PXBZ (outlined in yellow) to show that angle ZPB equals angle ZXB. We now have four purple angles, all equal to each other.

Most importantly, angles ZXB and YXA are equal. Since AB is a straight line segment, this can only occur if ZXB and YXA are vertical opposite angles, which means that Y, X, and Z are all on the same line. This completes the proof, since collinearity is achieved if P is on the circumcircle with A, B, and C, while the proof does not work if P is not on the circumcircle.