A brain teaser with an Euclidean origin

July 26, 2018July 28, 2018 / Michael Xue / Leave a comment

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

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

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

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

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

$= 2 |x_1|$

$= BC$

$\implies AB+AC > BC$