Monthly Archives: July 2018

A brain teaser with an Euclidean origin

It’s time for a brain teaser:

There is a triangle \triangle ABC, and D is an arbitrary interior point of this triangle (see Fig. 1). Prove that AD + DB < AC + CB.Screen Shot 2018-07-27 at 10.22.28 AM.png

Fig. 1

Here is my solution:

Extend line AD to point E on CB (see Fig. 2),

Screen Shot 2018-07-27 at 11.13.54 AM.png

Fig. 2

we have

AD + DB < AD + (DE + EB)\quad\quad\because \triangle DEB: DB < DE + EB

= (AD + DE) + EB

=AE + EB\quad\quad\because AD + DE = AE

< (AC +CE) +EB\quad\quad\because \triangle ACE: AE < AC +CE

= AC + (CE + EB)

= AC +CB\quad\quad\because CE + EB = CB

\implies AD +DB < AC +CB

My solution relies on a well known theorem:

Given a triangle ABC, the sum of the lengths of any two sides is greater than the length of the third side.

In the words of Euclid:

Screen Shot 2018-07-27 at 12.09.35 PM.png

“In any triangle two sides taken together in any manner are greater than the remaining one” (The Elements: Book I: Proposition 20)

I have conjured up the following algebraic proof of Euclid’s proposition:

Any \triangle ABC can be put in a rectangular coordinate system where x_1 > 0, x_2 \ge 0, y_2 > 0 (see Fig. 3)

Screen Shot 2018-07-27 at 2.34.18 PM.png

Fig. 3

It follows that

AB + AC = \sqrt{(x_2-(-x_1))^2+y_2^2} + \sqrt{(x_2-x_1)^2+y_2^2}

= \sqrt{(x_2+x_1)^2+y_2^2} + \sqrt{(x_2-x_1)^2+y_2^2}

> \sqrt{(x_2+x_1)^2} + \sqrt{(x_2-x_1)^2}\quad\quad\quad\because y_2 >  0

\geq \sqrt{(x_1)^2} + \sqrt{(-x_1)^2}\quad\quad\quad\because x_2 \geq 0

= 2 |x_1|

= BC

\implies AB+AC > BC

 

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