Prove that the diagonals are the longest lines in a square.

Question

I just pondered this question and have tried out several methods to solve it(mainly using trigonometry). However, I am not satisfied with my trigonometrical proof and is looking for better proofs.

Give an elegant proof that the diagonals are the longest lines in a square. It would rather be nice if the proof is by "reductio ad absurdum" method. Thank you :).

Bonus : Also prove that the diagonals of a rectangle are the longest lines.

On the Euclidean plane, the shortest distance between two points is a straight line. Traversing with any other line segments is always a greater distance. — CopyPasteIt, Jun 10 '17 at 12:07
@wid I think we both jumped to quickly. Be my guest to fill in the details and provide answer. — CopyPasteIt, Jun 10 '17 at 12:15
Yes, but are there any straight lines in the square which are longer that the diagonals. The straight lines can be in the square... — , Jun 10 '17 at 12:20
One can see this post: https://math.stackexchange.com/questions/2316550/distance-between-interior-points-of-triangle — szw1710, Jun 10 '17 at 12:43
Gee, I was hoping to get the rectangle bonus points, but now my answer may be deleted! Sure hope someone appreciates my diligence and clicks the undelete! — CopyPasteIt, Jun 11 '17 at 14:33

score 2 · Answer 1 · answered Jun 10 '17 at 12:12

2

Draw a circle circumscribing the square and with end points at opposite corners . Like $A,C $. This circle has diameter as $AC$ and the diameter is the longest chord of circle but this diameter is also the diagonal of the square. Thus the diagonals are the longest lines in a square.

answered Jun 10 '17 at 12:12

Archis Welankar

15,910

Ok, I get this. But I believe there are more simpler ways to get the answer. – Jun 10 '17 at 12:19
What is simple than geometry?. You look at what you have. – Archis Welankar Jun 10 '17 at 12:25
Don't feel bad @arch - my answer (under DELETE review) puts the rectangle in the circle! – CopyPasteIt Jun 11 '17 at 14:38

score 1 · Answer 2 · answered Jun 10 '17 at 12:31

Consider the axis-oriented rectangle $R$ consisting of all points whose Cartesian coordinates $(x, y)$ satisfy $$ x_{0} \leq x \leq x_{1},\qquad y_{0} \leq y \leq y_{1}. $$ If $(x, y)$ and $(x', y')$ are points of $R$, then $|x' - x| \leq |x_{1} - x_{0}|$ and $|y' - y| \leq |y_{1} - y_{0}|$, so their distance satisfies $$ \sqrt{(x' - x)^{2} + (y' - y)^{2}} \leq \sqrt{(x_{1} - x_{0})^{2} + (y_{1} - y_{0})^{2}}. $$ The right-hand upper bound is the distance between the corners $(x_{0}, y_{0})$ and $(x_{1}, y_{1})$, and between the corners $(x_{0}, y_{1})$ and $(x_{1}, y_{0})$.

score 0 · Accepted Answer · answered Jun 10 '17 at 12:56

It is sufficient to consider the two cases (since others are repetetive): without loss of generality, take two arbitrary points on two sides such that $0\le x\le y\le a$:

Case 1: From the Pythagoras: $$\sqrt{x^2+y^2}\le \sqrt{a^2+a^2}=d.$$ Case 2: From the Pythagoras: $$\sqrt{a^2+(y-x)^2}\le \sqrt{a^2+a^2}=d,$$because $y-x$ is maximum, when $y=a$ and $x=0$.

Prove that the diagonals are the longest lines in a square.

3 Answers3