implies the others. Further, under this equivalence, the orthic triangle of the parent triangle is necessarily isosceles.
Observe that , and so a triangle that satisfies the entire chain of equivalence in (1) cannot be equilateral. Also, a right triangle with hypotenuse satisfies both and , but not the rest. Hence, the non-right triangle requirement is essential.
As the post’s title gives away, the first obvious thing to get out of the way.
This is evident from .
Alternatively, one can compute the cosine of or explicitly, using and the cosine formula:
Similarly, . Now implies either or . In turn, either or , showing that one of the two angles is obtuse.
Now, let’s attempt to make our simple equivalence even simpler by splitting the proof into six parts.
From , isolate and then use the fact that :
By the extended law of sines, we have . So:
You’ll notice the semblance with our previous post. Repeated for emphasis.
From , we have:
Since we’re in a non-right triangle setting, we choose . Then compute :
From , isolate . Then:
Note that gives . So:
The triangle obtained by joining the feet of the three altitudes of a parent triangle is called its orthic triangle.
The lengths of the sides of an orthic triangle are given by , , (one of these may require an absolute value if the parent triangle is obtuse). One of our equivalent statements is . Thus, the orthic triangle has two sides equal. And we can even make it three.
First suppose that . Using the cosine formula:
Thus, . Next, suppose that the parent triangle is isosceles with . We’ll show that its orthic triangle is equilateral. Indeed, we already had . It now suffices to show that . By the cosine formula:
Thus, the orthic triangle is equilateral. Finally, suppose that the orthic triangle is equilateral. Consider :
Discard since we’re dealing with a non-right triangle. The second possibility then yields . Now use in the equation :
All this trouble just for a single triangle: the one in which , , . That’s the triangle that satisfies the equivalent conditions just established.
The feet of the altitudes from are , , and , as shown above. By the distance formula:
Notice that the foot of the altitude from coincides with our favourite point .
For any non-right triangle satisfying the chain of equivalence in (1), the three statements below are equivalent:
- the orthic triangle is equilateral.
Thus, there’s only “one” strictly isosceles triangle whose orthic triangle is equilateral.
- Let be the interior angles of a triangle , and let be the side-lengths. Relative to these measurements, PROVE that:
- the interior angles of the orthic triangle are , , , if the parent triangle is acute
- the interior angles of the orthic triangle are , , , if the parent triangle contains an obtuse angle
- the side-lengths of the orthic triangle are , , (one of these will be negative if the parent triangle contains an obtuse angle, in which case an absolute value will be needed).
- Let be a real number. Consider with interior angles , , and .
- What restrictions, if any, must be imposed on ?
- PROVE that the orthic triangle associated with is isosceles.
- PROVE that the two statements below are equivalent for a triangle :
- , , .
- Consider with vertices at , , . Find:
- the coordinates of the feet of the three altitudes
- the lengths of the sides of the orthic triangle
- a condition for which the orthic triangle will be equilateral.
- PROVE that the following statements are equivalent for any triangle:
- the orthic triangle is equilateral
- the parent triangle is equilateral or the parent triangle is isosceles with base angles of and an apex angle of .