Timeline for What fraction of the larger semicircle is filled?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jan 5, 2023 at 21:23 | vote | accept | Simd | ||
| Aug 14, 2021 at 1:19 | comment | added | CELLSecret | Why is $\angle JHA=90^\circ$? It feels like it on the diagram, but bisect is not enough to the argument. For example, $HI$ also bisects $AB$ but $\angle AHI\not=90^\circ$ | |
| S Feb 2, 2021 at 22:44 | history | suggested | CiaPan | CC BY-SA 4.0 |
minor formatting fix – making parens around a fraction big enough and a middle fraction line a bit longer
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| Feb 2, 2021 at 22:19 | review | Suggested edits | |||
| S Feb 2, 2021 at 22:44 | |||||
| Dec 24, 2020 at 16:24 | comment | added | Paul Sinclair | +1 for coming up with the most complicated expression for $\sqrt 3$ I've seen. | |
| Dec 23, 2020 at 19:56 | history | edited | user37842 | CC BY-SA 4.0 |
added 571 characters in body
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| Dec 23, 2020 at 16:44 | history | edited | user37842 | CC BY-SA 4.0 |
added 29 characters in body
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| Dec 23, 2020 at 10:38 | comment | added | Chris | I think bubbler's explanation of why JHA is a right triangle should be added to the explanation. It isn't immediately obvious at a glance. | |
| Dec 23, 2020 at 0:14 | comment | added | Bubbler | Wow, I failed to spot that JHA is a right triangle (which works because H bisects BA and A and B are on the outer circle, whose center is J). But then actually you don't need cosine rules; using Pythagoras twice on JIH and JHA would have been enough (which gives $(1+\sqrt{3}-r)^2+1^2+1^2=r^2$, which reduces to a nice linear equation). | |
| Dec 22, 2020 at 22:14 | history | answered | user37842 | CC BY-SA 4.0 |