Given a $\triangle ABC$ construct another triangle with sides measuring the inverses of the altitudes of $\triangle ABC$

67 Views Asked by At

Given a $\triangle ABC$, we want to construct the $\triangle XYZ$ whose sides are the inverses of the altitudes of $\triangle ABC$ . If we denote the altitudes by $h_a,h_b,h_c$ then the sides of $\triangle XYZ$ are $\frac{1}{h_a},\frac{1}{h_b},\frac{1}{h_c}$

We can find the lenght of $\frac{1}{h_a}$ using circle invertion with a circle of radii 1 and centered in the feet of the altitudes , but how to find the angle needed in order to solve it? Is there any other more pratical answer than this ?

1

There are 1 best solutions below

0
On

What is your definition of $1$ here? There is no inherent unit in the geometric plane.

If you just want line segments that are inversely proportional to the given sides, one way is to choose an arbitrary point and three line segments from that point with lengths of your given sides. Then construct the circle through the other endpoints and extend the segments into chords. The extended segments will be inversely proportional to the given lengths.

In the diagram, the red segments are the lengths of the sides of the triangle, and the blue segments are inversely proportional to the red ones.

enter image description here

If you do have a unit, the following diagram shows how to take three reciprocals. Use an arbitrary angle with vertex $O$. The distances from $O$ to the red points $A'$, $B'$ and $C'$ are the lengths for which to find the reciprocals, and the distance from $O$ to $U'$ is the unit. These points are all on the same ray. A circle determines $U''$ on the other ray. Line $\overline{U'A''}$ is parallel to segment $\overline{U''A'}$, and so on. The distance $OA''$ is then the reciprocal of distance $OA'$, and so on.

enter image description here

The first method here is more elegant and does not rely on a unit but does not give reciprocals.