Timeline for Proofs without words
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| May 22, 2021 at 0:27 | comment | added | lukeuser | @StevenStadnicki You don't need the arc (since the tics are used to show equidistance), but surely the arc is by far the easiest way to construct the vertex at the correct position? You do after all need it to be possible to find such a point—I realise this could be taken as given, but that would make the proof slightly harder to follow in my opinion. | |
| Nov 2, 2020 at 18:03 | comment | added | Steven Stadnicki | It might be worth noting that one doesn't need to introduce the circle here; drawing the line between the top vertex of the triangle and the interior point on the bottom edge shows that the kite-shaped quadrilateral is bilaterally symmetric (the two triangles it's split into are congruent by side-angle-side equivalence) | |
| Jan 22, 2020 at 16:32 | comment | added | Hans-Peter Stricker | So happy to hear that! But your enjoyment could not have been greater than mine was about your answers. | |
| Jan 22, 2020 at 16:29 | comment | added | Joel David Hamkins | Oh yes, I've enjoyed many of your questions on MO over the years. | |
| Jan 22, 2020 at 16:21 | history | edited | Hans-Peter Stricker | CC BY-SA 4.0 |
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| Jan 22, 2020 at 16:21 | comment | added | Hans-Peter Stricker | @JoelDavidHamkins: Happy to hear from you again after all those years;-) I made the correction. Thanks for the hint. | |
| Jan 22, 2020 at 16:19 | comment | added | Joel David Hamkins | I think you mean to refer to the impossibility of an integer isosceles right triangle, since clearly one can have isosceles triangles with integer sides, such as an equilateral unit triangle. | |
| Sep 5, 2018 at 21:24 | history | edited | Hans-Peter Stricker | CC BY-SA 4.0 |
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| S Sep 5, 2018 at 20:57 | history | answered | Hans-Peter Stricker | CC BY-SA 4.0 | |
| S Sep 5, 2018 at 20:57 | history | made wiki | Post Made Community Wiki by Hans-Peter Stricker |