(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.
![Rendered by QuickLaTeX.com h_C=R\cos C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-48f6421c44bfc48ace49843232b3043e_l3.png)
![Rendered by QuickLaTeX.com \cos C=\frac{2ab}{a^2+b^2}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-164a93ee5f46c68d5c57a018067d33c8_l3.png)
From , isolate
and then use the fact that
:
![Rendered by QuickLaTeX.com \cos C=\frac{2ab}{a^2+b^2}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-164a93ee5f46c68d5c57a018067d33c8_l3.png)
![Rendered by QuickLaTeX.com a^2+b^2=4R^2](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-4da13955b46a55a23cd01625dafbace8_l3.png)
By the extended law of sines, we have . So:
![Rendered by QuickLaTeX.com h_A^2+h_B^2=AB^2](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-d232a2a214d71d679aaf9aa979bc0296_l3.png)
![Rendered by QuickLaTeX.com (a^2-b^2)^2=(ac)^2+(cb)^2](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-6a663d57e999fe0e03e37ad5e103f29b_l3.png)
From , we have:
Since we’re in a non-right triangle setting, we choose . Then compute
:
![Rendered by QuickLaTeX.com (a^2-b^2)^2=(ac)^2+(cb)^2](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-6a663d57e999fe0e03e37ad5e103f29b_l3.png)
![Rendered by QuickLaTeX.com a\cos A+b\cos B=0](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-718bd5a2fab6ebf987f1da462aabc067_l3.png)
From , isolate
. Then:
![Rendered by QuickLaTeX.com a\cos A+b\cos B=0](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-718bd5a2fab6ebf987f1da462aabc067_l3.png)
![Rendered by QuickLaTeX.com h_C=R\cos C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-48f6421c44bfc48ace49843232b3043e_l3.png)
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.
![Rendered by QuickLaTeX.com a^2=3b^2\iff](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-4b483399fcbf5ca39b29df79358ac3f1_l3.png)
![Rendered by QuickLaTeX.com b=c\iff](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b9c5711e402d3fa2d5ade077e4b9e906_l3.png)
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.
![Rendered by QuickLaTeX.com ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b5002f05fd90c80f81a9e0e9c845e02d_l3.png)
![Rendered by QuickLaTeX.com A(-6,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-79c16eb208dc820695dfd9a30fa1c847_l3.png)
![Rendered by QuickLaTeX.com B(0,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-7045877de833cb7e53ba1ef4ddfaeb24_l3.png)
![Rendered by QuickLaTeX.com C(2,4)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-e2fb9eee47c72cdf6ab23e648c29df8c_l3.png)
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
.