Lean Napoleon's Triangles: What is this about?
A Mathematical Droodle
Explanation
Copyright © 1996-2008 Alexander Bogomolny
The applet attempts to suggest the following statement [Garfunkel, Stahl]:
| |
On the middle thirds of ΔABC construct (similarly oriented) equilateral triangle, KLC', MNB', and PQA'. Then ΔA'B'C' is equilateral.
|
Proof
The proof is computational but not tedious. ΔAKC' is isosceles, with AKC' = 120°. So its base angles (KAC' and AC'K) are both 30°. It follows that AC'L = 90° which makes ΔAC'L right. If, as usual, AB = c, then in ΔAC'L, AL = 2c/3 and LC' = c/3. By the Pythagorean theorem, (AC')² = c²/3.
Similarly, ΔAMB' is right and (AB')² = b²/3.
We are now in a position to apply the Law of Cosines in ΔAB'C' to determine the length of B'C'. Note that B'AC' = A + 60°.
| |
| (B'C')² | = b²/3 + c²/3 - 2bc/3·cos(A + 60°) |
| | = b²/3 + c²/3 - 2bc/3·cos(A)cos(60°) + 2bc/3·sin(A)sin(60°) |
| | = b²/3 + c²/3 - 2bc·cos(A)/6 + 2bc/3·sin(A)√3/2 |
| | = b²/3 + c²/3 - (b² + c² - a²)/6 + 2√3/3·S |
| | = (a² + b² + c²)/6 + 2√3/3·S, |
|
where S is the area of ΔABC. (We again used Law of Cosines and also a formula for the area of a triangle.) The expression is symmetric in a, b, c so that (A'B')² and (A'C')² are bound to be equal to the same quantity, proving that indeed ΔA'B'C' is equilateral.
Observe that the above derivation only works when triangles KLC', MNB', and PQA' are drawn outwardly on the sides of ΔABC. If they are drawn inwardly, then B'AC' may equal either A - 60° or 60° - A. In both cases we would obtain
| |
| (B'C')² | = (a² + b² + c²)/6 - 2√3/3·S, |
|
with the same conclusion. Thus we are led to two equilateral triangles related to a base ΔABC. From the above relations it follows that the difference in areas of the two triangles is exactly that of ΔABC.
References
- J. Garfunkel, S. Stahl, The Triangle Reinvestigated, Am Math Monthly, Vol. 72, No. 1. (Jan., 1965), pp. 12-20
Napoleon's Theorem
Copyright © 1996-2008 Alexander Bogomolny
|