(1)
implies the others. Further, under this equivalence, the orthic triangle of the parent triangle is necessarily isosceles.
Obvious things
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.
Obtuse triangle
As the post’s title gives away, the first obvious thing to get out of the way.
This is evident from .
Other techniques
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.
Oversimple theorem
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:


From , we have:
Since we’re in a non-right triangle setting, we choose . Then compute
:


From , isolate
. Then:


Note that gives
. So:
Six. Perfect.
Orthic triangle
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
.
Takeaway
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.
Tasks
- 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).
- the interior angles of the orthic triangle are
- 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.
- What restrictions, if any, must be imposed on
- PROVE that the two statements below are equivalent for a triangle
:
and
,
,
.
- 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
.