Jackie Gammaro

Assignment 9 – Pedal Triangles

Let triangle ABC be any triangle. If P is any point in the plane, then the triangle formed by constructing perpendiculars to the sides of ABC (Extended if necessary) locates three points R, S, an T that are the intersections. Triangle RST is the pedal triangle for Pedal Point P.

What if P is the centroid of triangle ABC?

Triangle RST is the Pedal triangle to triangle ABC at point P, the centroid.

In geometer sketchpad, one could drag vertex A, B, or C to investigate the effects of the pedal triangle, when ABC is a

a) equilateral triangle?

b) isosceles triangle?

c) right triangle?

Investigate the pedal triangle when the Pedal Point, P, is the centroid for equilateral triangle ABC.

Here, one can see, once the pedal triangle is formed, that neither ratio of areas for right triangles or isosceles triangles holds here. One probably notices immediately that there appears to be four congruent triangles, thus the ratio of areas for an equilateral triangle to its pedal triangle, whose pedal point is the centroid, is 4 to 1. Now I wonder is this true for any triangle, or just equilateral triangles, lets see for right triangles and isosceles triangles, I'm thinking this will hold true for any triangle.

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Investigate the pedal triangle when the Pedal Point, P, is the centroid for right triangle ABC.

So, my hypothesis was wrong and I notice no matter the position of the vertex A or vertex B, the ratio of areas of the original triangle to the pedal triangle is 4.5 or expressed as a ratio (9/2). Thus, the pedal triangle area's is always (2/9) that of the area of the original triangle.

Investigate the pedal triangle when the Pedal Point, P, is the centroid for Isosceles triangle ABC, (the scary case)

Let ABC be an isosceles triangle with base AC, altitude BT, and legs, AB, BC. Let AB>AC . One can investigate to find the pedal triangle formed is also an isosceles triangle with bases parallel to each other. Here the ratio of area was in between the ratio of areas for an equilateral triangles and right triangles, respectively. I wonder if this will hold true for all isosceles triangles.

Here pedal triangle DEF is an isosceles triangle to isosceles triangle ABC. Here AB=BC < AC. AC is the base of the triangle with altitude BD. Notice that a vertex of the pedal trianlge lies along the height of triangle ABC. We notice the ratio of areas is still greater than 4 but less than 4.5. I wondered, what if I make angle B, greater than 90degrees.

When vertex angle B > 90 degrees, the ratio of areas is greater than 4.5. Notice the pedal triangle is an isosceles triangle whose height lies along the height of original triangle ABC.

Here, I discovered that when vertex angle B increases, the ratio of areas to original triangle ABC to Pedal triangle DEF increases, and gets larger than 5? I wonder what the limit of ratio of areas is when triangle ABC is an isosceles triangle? Lets discover when Angle B limits toward 180 degrees.

When Angle B, gets very close to 180 degrees, we see the area of the pedal triangle DEF, decreases dramatically. WE can barely see it. The ratio of areas increases dramatically as one can see, even when the area of original triangle ABC is less than one.

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