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# An upside-down Pythagorean identity

Towards the end of our post on June 28, we stated that if the side-lengths of a triangle satisfy the modified Pythagorean identity

(1)

then the following statements are equivalent:

• is the golden ratio

Today’s post will now establish the above equivalence. (Note that the fourth condition, , which motivated today’s title, was coined by the author of this article.)

If is the golden ratio, PROVE that equation (1) becomes .

Suppose that is the golden ratio, then , and so . This then gives . In turn:

In addition to equation (1), suppose that . PROVE that: .

Let be the circumradius. Then, by the extended law of sines, we have:

The assumption implies . (Henceforth we’ll frequently use the fact that is equivalent to equation (1).) So:

In addition to equation (1), suppose that . PROVE that .

Using the extended law of sines as in the previous example, the condition gives . Since , we obtain

So-called upside-down Pythagorean identity. Ours works under an added condition, namely equation (1).

In addition to equation (1), suppose that . PROVE that , where is the length of the altitude from vertex and as per the usual notation.

Re-arrange the given condition as

and obtain . The length of the altitude from vertex is given by

Suppose that , in addition to equation (1). PROVE that is the golden ratio.

If , then . Now use this in equation (1) to get

Thus, is the golden ratio.

## Takeaway

Let be a triangle whose side-lengths satisfy equation (1). Then the following statements are equivalent:

1. is the golden ratio
2. .

The last condition fixes the interior angles of the given triangle : , , .

• (Early sixties) In a non-right triangle , let be the side-lengths, the altitudes, the feet of the altitudes from the respective vertices, the circumradius, the circumcenter, the nine-point center, the orthocenter, the midpoint of side , the reflection of over side , the reflection of over side , and the reflection of over side . PROVE that the following sixty-one statements are equivalent:
1. or
2. is congruent to
3. is congruent to
4. is isosceles with
5. is isosceles with
6. is right angled at
7. is the circumcenter of
8. is right-angled at
9. is right-angled at
11. the points are concyclic with as diameter
12. the reflection of over lies internally on
13. the reflection of over lies externally on
14. radius is parallel to side
15. is the reflection of over side
16. the nine-point center lies on
17. the orthic triangle is isosceles with
18. the geometric mean theorem holds
19. the bisector of has length , where
20. the orthocenter is a reflection of vertex over side
21. segment is tangent to the circumcircle at point
22. median has the same length as the segment
23. the bisector of is tangent to the nine-point circle at
24. is a convex kite with diagonals and
25. altitude is tangent to the nine-point circle at
26. segment is tangent to the nine-point circle at .
( short of the target.)
• (Extra feature) If satisfies equation (??), PROVE that its nine-point center divides in the ratio .