Ivan Shishkin, Rye (1878)

Problems/GeometryReviewed

A simpler Pythagorean theorem

by Ancient Tree·
19
Difficulty scaleÉchelle de difficulté

This score reflects both the level of the required concepts and the difficulty of the solution.Ce score tient compte à la fois du niveau des notions nécessaires et de la difficulté de la résolution.

  1. 110First steps / middle schoolPremiers pas / collège
  2. 1125Beginner / high schoolDébutant / lycée
  3. 2650Intermediate / undergraduateIntermédiaire / licence
  4. 5170Advanced / graduateAvancé / master
  5. 7190Expert / specializedExpert / spécialisé
  6. 91100Research levelNiveau recherche
These levels are approximate guides.Ces niveaux sont des repères approximatifs.
·
English
EnglishFrançais

When I was in middle school, while studying the Pythagorean theorem, I came across the following right triangle:
image
Trying to guess a relationship between the hypotenuse and the two other sides - the short one and the middle one - I noticed that 5=4+13×35=4+\frac{1}{3}\times 3.

I made the following conjecture: in a right triangle, the hypotenuse equals the middle side plus one third of the short side.

  1. Is this true for every right triangle?
  2. Can you find all right triangles whose side lengths are integers and for which this relation holds?
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Solutions

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Solution by SarahK(roche:)FR

Discussions0 useful votes
  1. Contrex: (5,12,13)(5,12,13)

  2. On a le système suivant avec abca\le b\le c :

{a2+b2=c2a/3+b=c\left\{a^2+b^2=c^2 \atop a/3+b=c \right.

on obtient aisément que 3b=4a3b=4a et donc que les seules solutions entières possibles sont multiples du triplet
(3,4,5)(3,4,5), réciproquement ces triplets sont solutions, ce qui clôt.

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