Timeline for Areas of a triangle divided by 2 cevians into 4 regions, values of $A+B+C+D$ if $C=\frac{(2A+B+D)BD}{A^2-BD}$
Current License: CC BY-SA 4.0
16 events
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| 1 hour ago | history | edited | Bill Dubuque | CC BY-SA 4.0 |
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| 11 hours ago | history | edited | Dan | CC BY-SA 4.0 |
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| 22 hours ago | comment | added | SmileyCraft | @Andre More generally, you can always multiply $A$, $B$, $C$, and $D$ by the same value to show that any multiple of a realizable number is also realizable. | |
| 22 hours ago | audit | First questions | |||
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| yesterday | answer | added | Blue | timeline score: 6 | |
| yesterday | history | became hot network question | |||
| yesterday | history | edited | Dan | CC BY-SA 4.0 |
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| yesterday | comment | added | Dan | @Blue In my exhaustive search for $A+B+C+D=2026$, I checked all combinations of $1\le A\le 2026$ and $1\le D\le 2026$ (assuming $C=\frac{(2A+B+D)BD}{A^2-BD}$). There were no integer solutions for $B$. | |
| yesterday | comment | added | Andre | I'll leave it as a comment: Let me start with the easiest case where $A=2k$ and $B=D=k$, with $k\in\mathbb Z^+$. We can verify that in this case $$C=\frac{(2A+B+D)BD}{A^2-BD}=\frac{6\times k^3}{3\times k^2}=2k.$$ Hence, $$A+B+C+D=6k$$ and any year divisible by $6$ can be realized (as 2022). | |
| yesterday | answer | added | SmileyCraft | timeline score: 8 | |
| yesterday | comment | added | Blue | @Dan: Thinking out loud ... Choosing positive-integer areas $A$, $B$, $D$, the total area is $$T := A+B+C+D=\frac{A(A+B)(A+D)}{A^2-BD}$$ So, valid solutions in general are those such that $A^2>BD$ and $T$ is an integer. For $T=2026$, we need each of its prime factors ($2$ and $1013$) to divide at least one of $A$, $A+B$, $A+D$, with $A^2-BD$ dividing-out all other factors (and any "extra" $2$s and $1013$s). | |
| yesterday | comment | added | Blue | @Dan: How exhaustive is your exhaustive search? That is, what range of $A$, $B$, $C$, $D$ values have you tested? | |
| yesterday | comment | added | Dan | @Andre Suppose my target sum is $2025$. I make a column of all $1$s; these are $A$ values. The second column is $1,2,3,\dots$; these are $D$ values. In the third column I calculate $B$ using a rearrangement of $C=\frac{(2A+B+D)BD}{A^2-BD}=2025-A-B-D$. I check for integer values of $B$ using the floor function. I create many copies of these three columns, where each first column is the same number repeated. | |
| yesterday | comment | added | Andre | @Dan: can you briefly explain how your Excel finds the numbers? | |
| yesterday | comment | added | Bowei Tang | $2026$ is of the form $2p$, where $p=2013$ is a prime. A naive guess is that the possible areas are all positive integers except those of the form $2p$, where $p$ is a prime. (maybe with some additional conditions) :) [Of course, this is probably wrong] | |
| yesterday | history | asked | Dan | CC BY-SA 4.0 |