5
$\begingroup$

Four points on the plane are vertices of three different quadrilaterals. How can this happen?

The problem is taken from "Kiselev's Geometry - Book I : Planimetry"

At first, I thought it could be like this :

enter image description here

But, the way diagonals are defined in the book:

enter image description here

Makes me think that the 3 figures I drew, are the same quadrilateral.

How do you think four points on the plane can be vertices of three different quadrilaterals

$\endgroup$
  • 1
    $\begingroup$ Let A, B, C, D be the vertices. Then ABCD, ACDB, ADBC are three different quadrilaterals. The roles of sides and diagonals are not the same in them: there are 6 segments, and you can choose any two of them as diagonals, as long as they share no vertex (three possibilities to do so). $\endgroup$ – Jean-Claude Arbaut Dec 30 '17 at 20:11
18
$\begingroup$

Is this what you mean by three different quadrilaterals? enter image description here

$\endgroup$
10
$\begingroup$

One point (say D) is inside the triangle formed by the other three (ABC). Then the possible quadrilaterals are ABCD, ABDC or ADBC.

$\endgroup$
  • $\begingroup$ Your first sentence is not necessarily true. It fails (at least, but not only) for any convex quadrilateral. $\endgroup$ – Jean-Claude Arbaut Dec 30 '17 at 20:12
  • 1
    $\begingroup$ It is if the "three quadrilaterals" aren't allowed to be self-intersecting, which I imagine is the intended interpretation. $\endgroup$ – Especially Lime Dec 30 '17 at 20:20
  • $\begingroup$ Ah, yes, this is a possible interpretation. I would see no problem with a self-intersecting quadrilateral, but I don't know the intent of the author. So +1 after all :) $\endgroup$ – Jean-Claude Arbaut Dec 30 '17 at 20:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.