Assignment 9: Pedal Triangles

**July 18, 2003**

Click the **GSP File **to run this problem.

Let triangle ABC be any triangle. Then if P is any point in
the plane, then the triangle formed by constructing perpendiculars to the
sides of ABC (extended if necessary) locate three points R, S, and T that are
the intersections. Triangle RST is the Pedal
Point Script tool is created for the general construction of a
pedal triangle to triangle ABC where P is Original triangle is blue and the Pedal triangle is green. |

A single construction has been used for the following investigations. It can get confusing to see H's, I's, G's and construction lines flying all over, but my theory is that it will help to see the relationships. Pictures are not taken of the various investigations because it is so easy to open the GSP file to tab 2 and just do it. |

Legend: The following shorthand notation is used:

PedT - Pedal triangle

MainT. – Main triangle ABC

2. What if pedal point P is the centroid of triangle ABC?

The dynamics depend largely on the shape of the triangle itself:

* For equilateral when all key points converge at the 'center', the Pedal triangle (Abbreviated PedT) is the same as connecting the midpoints of the sides.

* For right triangle, the PedT has a side parallel to the hypotenuse. If P is anywhere along the Euler line it has a side that is parallel to the hypotenuse because the perpendiculars bisect the legs in equal ratios given that the Euler line bisects the right angle.

* For an extreme obtuse triangle, the PedT will extend beyond the bounds of the base triangle, but it always partially overlapping (never completely outside the bounds.

3. What if . . . P is the incenter . . . ?

* As above, for equilateral when all key points converge at the 'center', the PedT is the same as connecting the midpoints of the sides.

* For general acute triangle, the PedT doesn't appear to be anything special, (note below discovery)

* For right triangle, the side of the Ped T near the 90 deg. vertex forms an isosceles right triangle that when reflected over the hypotenuse, forms a square. Asking why highlights that the incenter is equidistant from the sides and therefore all sides of the PedT form isosceles triangles with the corresponding vertices of the base triangle.

* For an obtuse triangle, the PedT also forms isosceles triangles with the base of it's feet.

4. What if . . . P is the Orthocenter . . . ? Even if
outside ABC?

* As above, for equilateral when all key points converge at the 'center', the PedT is the same as connecting the midpoints of the sides.

* For general acute triangle, the PedT doesn't at first blush appear to have special properties, but we know the construction of the orthocenter (altitudes of triangle) and the Pedal point is very similar. The construction lines for the PedT overlay those for H and the PedT is formed by the feet of the perpendiculars of the altitudes.

* For right triangle, the PedT reduces to a line that forms the altitude of the triangle.

* The obtuse triangle behaves like an acute, the construction lines for the PedT overlay those for H and the PedT is formed by the feet of the perpendiculars of the altitudes.

5. What if . . . P is the Circumcenter . . . ? Even if outside ABC?

* As above, for equilateral when all key points converge at the 'center', the PedT is the same as connecting the midpoints of the sides.

* For general acute triangle, the PedT is concurrent with the midpoints of the sides, because the construction is the same.

* For right triangle, the circumcenter is at the midpoint of the hypotenuse, so the PedT simply connects the midpoints of all sides.

* For an obtuse triangle, the PedT has the same behavior as for an acute.

6. What if . . . P is the Center of the nine point circle for triangle ABC?

* As above, for isosceles when all key points converge at the 'center', the PedT is the same as connecting the midpoints of the sides.

* For general acute triangle, the PedT

* For right triangle, the PedT

* For an extreme obtuse triangle, the PedT

7. What if P is on a side of the triangle?

Point P is concurrent with one of the vertices of the PedT of course. The angle formed at the PedT vertex concurrent with P is also complementary with the opposite angle of the main triangle (they add to 180). The legs of the PedT intersect with the legs of the main triangle at right angles because P is concurrent with the side.

8. What if P is one of the vertices of triangle ABC?

as P approaches a vertex of the main triangle, the PedT reduces to a line.

9. Find all conditions in which the three vertices of the Pedal triangle are collinear

(that is, it is a
degenerate triangle). This line segment is called the *Simson**
Line.*

* if P is one of the vertices of triangle ABC.

* if P lies on the circumcircle

10. Locate the midpoints of the sides of the Pedal Triangle.

Construct a circle with center at the circumcenter
of triangle ABC such that the radius is __larger__ than the radius of the circumcircle. Trace the locus of the midpoints of the sides
of the Pedal Triangle as the Pedal Point P is animated around the circle you
have constructed. What are the three paths?

-> three midpoints appear to follow elliptical paths with their focal points approximately at the midpoints and vertices of Main triangle ABC.

11. Repeat where the path is the circumcircle?

11a. Construct lines (not segments) on the sides of the Pedal triangle.

Trace the lines as the Pedal point is moved along different paths.

11b. In particular, find the envelope of the Simson line as the Pedal point is moved along the circumcircle. Note, you will need to trace the image of the line, not the segment.

12. Repeat where the path is a circle with center at the circumcenter but radius __less__ than the radius of the circumcircle.

-> forms three 'bundles of wheat' by each line of triangle because it moves side to side less and less as the radius gets smaller.

13. Is there a point on the circumcircle for P that has side AC as its Simson line? AB? BC?

Yes – simson line is concurrent with the each of the sides of MainT - dunno wht at the point.

14. Construct the Simson line of a point P (on the circumcircle) and construct the segment connecting P to the Orthocenter. How do the two segments intersect?

->Always at the intersection of PH.

15. Select two pedal points on the circumcircle and construct their Simson lines. Compare the angle of intersection of the two Simson lines with the angular measure of the arc between the two pedal points.

16. Animate the Pedal point P about the incircle of ABC. Trace the loci of the midpoints of the sides. What curves result? Repeat if ABC is a right triangle.

17. Construct an excircle of triangle ABC. Animate the Pedal point P about the excircle and trace the loci of the midpoints of the sides of the Pedal triangle. What curves result? Look at the angle bisectors through the center of the excircle. How are the loci positioned with respect to the angle bisectors?

18. Other investigations. Have you found any observations about pedal points and pedal triangles other than the ones from the previous suggestions? If so, discuss them. If not, try something else. What about specials cases for right triangles, isosceles triangles, or equilateral triangles?