Sorry, no pictures. Explanation here.
Proposition: Let $ABC$ be an equilateral triangle and $P$ be an arbitrary point in its plane. Then $PC \le PA + PB$.
This was brought up in
a message
on the geometry.puzzles newsgroup on November 11, 2001.
Go to that message and scroll to the bottom of the page to see the discussion
thread.
On November 22, 2001 Dan Hoey offered a particularly nice solution. He also commented that he had learned that problem may be related to the Van Schooten Theorem, which indeed it is. See Van Schooten's and Pompeiu's Theorems: What are these? for much detail and historical background.
Here is Dan Hoey's proof of the proposition as stated above.
On the line segment $AP$ construct the equilateral triangle $APD$, as shown in the diagram below, then add the line segment $DC$.
Let us show that the triangles $APB$ and $ADC$ are congruent. For this, Let us observe that the sides $AP$ and $AB$ in the triangle $APB$ equal the sides $AD$ and $AC$ in the triangle $ADC$, by the construction. Moreover, the angles $BAP$ and $CAD$ are equal because each equals the difference of a 60 degree angle and the angle $DAB$. Therefore, the triangles $APB$ and $ADC$ are congruent by the side-angle-side equality. We conclude, in particular, that $PB = DC$.
In the triangle $PDC$ we have $PC \le PD + DC$. In this inequality replace $PD$ and $DC$ by their equivalents $PA$ and $PB$ to arrive at $PC \le PA + PB$. QED
This applet was created by
Rouben Rostamian
using
David Joyce's
Geometry
Applet
on November 23, 2001.
Cosmetic revisions on June 23, 2010.
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