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Excuse me if this is a silly question, but my plane geometry is very rusty. When I re-read Jack D'Aurizio's answer to the question "How can we prove that $\pi > 3$ using this definition", I wondered why, when viewed from the perspective of high school plane geometry, the area of the regular hexagon of unit spoke length is smaller than the area of the unit disc. Obviously, this follows if the regular hexagon entirely lies inside the unit circle, but why is this true? My first thought was that this is because the distance from the centre of the hexagon to any point on its perimeter is at most $1$. Put it another way, this is a consequence of the following (stronger) statement:

  • Given $\bigtriangleup ABC$, if $X$ lies on the line segment $\overline{BC}$, then the length of $\overline{AX}$ is $\le$ the larger length of $\overline{AB}$ or $\overline{AC}$.

enter image description here

The above statement in turn follows from another statement:

  • Let $\overline{AP}$ be a fixed line segment and $\bigtriangleup APX$ be a right-angled triangle, where $\overline{AP}\perp\overline{PX}$. The longer the base $\overline{PX}$, the longer the hypotenuse $\overline{AX}$.

This follows directly from Pythagoras theorem. However, all proofs of Pythagoras theorem that I knew, such as the ancient Chinese proof or Einstein's proof, make use of the notion of area. Why must the proof of a statement about length involve the concept of area?

So, here is my question:

  • Can the statement in the second bullet point be proved, within the framework of Eucliedan plane geometry, without using Pythagoras theorem or the notion of area?
user1551
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    Sorry if this is a poor question, but can you clarify on what is "acceptable" to use here? When you say "within the framework of Euclidean plane geometry," things like ruler and compass come to mind (since we're excluding Pythagorean's Theorem, area, etc.). Is this the intent, or do you have other things in mind? – Clayton Apr 11 '18 at 17:19
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    FYI: Proving the Pythagorean Theorem doesn't require calculating area. Suppose, for the purpose of this comment, that your $\triangle ABC$ has a right angle at $A$. Then $\overline{AP}$ creates similar triangles, and we have $$\frac{|\overline{PB}|}{|\overline{AB}|} = \frac{|\overline{AB}|}{|\overline{BC}|} \qquad \frac{|\overline{PC}|}{|\overline{AC}|} = \frac{|\overline{AC}|}{|\overline{BC}|}$$ so that $$|\overline{AB}|^2+|\overline{AC}|^2=|\overline{BC}||\overline{PB}|+|\overline{BC}||\overline{PC}|=|\overline{BC}|\left(;|\overline{PB}|+|\overline{PB}|;\right)=|\overline{BC}|^2$$ – Blue Apr 11 '18 at 17:25
  • @Clayton I haven't a crystal clear criterion, but I am expecting something that doesn't use analytic geometry (i.e. coordinates, trigo functions or the like) and also anything that hinges on the notion of area. If you have a plane geometry proof of Pythagoras theorem that doesn't use areas, that would be acceptable too. – user1551 Apr 11 '18 at 17:26
  • Exercise 22 on age 198, edition 4 of Euclidean and Non-Euclidean Geometries by Marvin Jay Greenberg. He says use Proposition 4.5 on page 168 – Will Jagy Apr 11 '18 at 17:28
  • @Blue Thanks. I don't know this proof. You can make it an answer. – user1551 Apr 11 '18 at 17:31

5 Answers5

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One of Euclid's useful propositions is that the lengths of sides are in the same order as the opposite angles. I.e., in $\triangle PQR$ we have $PQ>PR$ if and only if $\angle R>\angle Q$. (This follows from the fact that an exterior angle of a triangle is greater than either remote interior angle.) Most students these days understand this in terms of the law of sines, but it's far more elementary.

So, in your picture, take $X'$ between $B$ and $X$, so that $PX'>PX$. Consider $\triangle AXX'$. Then you can conclude $AX'>AX$ because $\angle X>\angle X'$.

Ted Shifrin
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  • Hi, Ted. This is Proposition 4.5 in Greenberg, Euclidean and Non-Euclidean Geometries, in the chapter on Neutral geometry. The application to the question above is Exercise 22 on page 198. Oh, 4th edition, although I think this chapter stayed fairly stable through the editions – Will Jagy Apr 11 '18 at 17:34
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    Thanks, @WillJagy. When I retired I gave away almost my entire collection of math texts. :( But I did keep Euclid's Elements, which was a gift from a student long ago. This is propositiions 18 and 19 in Book I of Euclid :) – Ted Shifrin Apr 11 '18 at 17:37
  • "This follows from the fact that an exterior angle of a triangle is greater than either remote interior angle." Ah, yes, that's prop. 18. Thanks! – user1551 Apr 11 '18 at 17:38
  • @user1551: I just looked for you. It's Proposition 16. :) – Ted Shifrin Apr 11 '18 at 17:40
  • @user1551, Marvin says this version combines 18 and 19 – Will Jagy Apr 11 '18 at 17:40
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Migrating a comment to an answer, as requested.

Proving the Pythagorean Theorem doesn't require calculating area. Suppose, for the purpose of this comment, that your $\triangle ABC$ has a right angle at $A$. Then $\overline{AP}$ creates similar triangles, and we have

$$\frac{|\overline{PB}|}{|\overline{AB}|}=\frac{|\overline{AB}|}{|\overline{BC}|} \qquad \qquad \frac{|\overline{PC}|}{|\overline{AC}|}=\frac{|\overline{AC}|}{|\overline{BC}|}$$

so that $$|\overline{AB}|^2 + |\overline{AC}|^2 = |\overline{BC}||\overline{PB}|+|\overline{BC}||\overline{PC}|= |\overline{BC}|\;\left(\;|\overline{PB}|+|\overline{PC}|\;\right) = |\overline{BC}|^2$$

Blue
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Let extend $PX$ and take point $X_1$ such as $PX_1>PX$. Then it's easy to see that $\angle{AXX_1}$ is obtuse and $AX_1$ is the biggest side in $\triangle{AXX_1}$

Vasili
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  • So, you're using the fact that the side opposite the larger angle is longer. I took me awhile to understand what you meant. – user1551 Apr 11 '18 at 17:52
  • @user1551: Yes, exactly. There can be only one obtuse angle in a triangle and side opposite the obtuse angle is the biggest.. – Vasili Apr 11 '18 at 18:08
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for Ted's answer, these are pages 168 and 198 from Marvin Jay Greenberg, Euclidean and Non-Euclidean Geometry, 4th edition. In the third edition it was pages 121 and 142.

enter image description here enter image description here

Will Jagy
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    Wow, that's exactly the same statement. – user1551 Apr 11 '18 at 18:02
  • @user1551 I think you would like the book. He is doing Hilbert's axioms. At this point he has not defined area, and he finally introduces length measurement (using real numbers for lengths) and angle measurement on page 169, maybe you can read that. To make a good image I need to go to the copy place to get a good image on a piece of paper, then place that page on my home scanner. – Will Jagy Apr 11 '18 at 18:09
  • Thanks for going through all these troubles. I was actually reading your answer to What is the modern axiomatization of (Euclidean) plane geometry? before I drafted this question. Is this book rigorous enough? (The price tag is a bit scary, btw.) – user1551 Apr 11 '18 at 18:23
  • @user, yes. I helped with it, just a little. Marvin accused me of thinking geometrically, meaning not really in the axiomatic manner. This is true. I always thought in terms of compass and straightedge constructions. Anyway, the other modern one on the Hilbert axioms is Hartshorne, Geometry: Euclid and Beyond. In preparing my article, however, I used George E. Martin, The Foundations of Geometry and the Non-Euclidean Plane. See if you can borrow the book before purchasing. Also, it seems there are sites where you can download a pdf. You need to log in somehow. – Will Jagy Apr 11 '18 at 18:29
  • @user, i should add that the third edition (1993) is exactly the same in this chapter, and is available used at reasonable cost. This website has versions in Europe as well... https://www.abebooks.com/servlet/BookDetailsPL?bi=4713392749&searchurl=tn%3Deuclidean%2Bnon-euclidean%2Bgeometries%2Bdevelopment%26sortby%3D17%26an%3Dmarvin%2Bgreenberg&cm_sp=snippet--srp2--title6 – Will Jagy Apr 11 '18 at 19:05
  • +1 for Greenberg's book. – Blue Apr 11 '18 at 23:42
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Note: $$\tan \measuredangle PXA=\frac{AP}{PX}; \\ \sin \measuredangle PXA = \frac{AP}{AX}; \\ 0<\measuredangle PXA<90.$$ When $\measuredangle PXA$ decreases, both $\sin \measuredangle PXA$ and $\tan \measuredangle PXA$ will decrease, hence both $PX$ and $AX$ will increase (since $AP$ is fixed).

farruhota
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