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 Triangulation of Numbers (Posted on 2008-03-18)
```       W------------------------X
|                      * |
|    A              *    |
|                O       |
|             *  * *  B  |
|          *     *  *    |
|       *        *   *   |
|    *           *    *  |
| *      D       *  C  * |
Z----------------Q-------Y
```
What is the minimum area of rectangle WXYZ if all lengths are whole numbers, as are the areas of the similar triangles, denoted by A, B, C & D?

 Submitted by brianjn Rating: 3.6667 (3 votes) Solution: (Hide) I have labeled the line segments: ``` W------------a-----------X | * | | A * c | b z O | | * * * B b | * * d | | h * * | | * e * | | * D * C * | Z-------g--------Q---f---Y ``` and then assigned values to them. For convenience the multiplication signs are omitted. In the top line of the diagram, for instance, please read 554 as 5*5*4. ``` W-----------554----------X | * | | A *533 | 553 555 O * | | * * * B 554 | * * 543 | | *454 * * | | * 433 * | | * D * C * | Z------444-------Q--343--Y ``` Each triangle has its own 345 digit coloration for the respective side within the set of 3 digits assigned to a side. ```Consider this table: TRIANGLE A B C D C & D Hypotenuse z, 555 b, 535 d, 435 h, 445 g+f, 545 (125) (75) (60) (80) (100) Short Side b 553 c, 533 f, 334 e, 443 d, 543 (75) (45) (36) (48) (60) Long Side a 554 d, 534 e, 443 g, 444 h, 544 (100) (60) (48) (64) (80) ``` Consider Δ C and Δ D which are both 3,4,5 triangles. If g has the value 4 and f has the value of 3, then the diagram is invalid as e cannot be either 3 or 4. If we multiply the values of 3, 4 and 5 in Δ D by 4 we create the right triangle 12,16,20. Similarly if we multiply the same values for Δ C we get 9,12,15. In both of those triangles e is common and now has the value of 12. Following this approach will lead in further disparities which may be resolved in the same manner. In will become obvious that the dimensions created (above) will have to be further adjusted. These then are the areas: A. 3750, B. 1350, C. 864 and D. 1536 with a total of 7500. This was basically resolved by trial and error. I must commend readers to the succinctness of: Charlie, Dej Mar, Bractals andBrian Smith I have chosen to leave the above, rather than defer to any of my cited respondents as I think there may be those who might gain from even my approach. Note:Though Charlie has a computer listing it is not there for the sake of finding a solution, rather it is to identify/eliminate some other configuration that might have arisen within the criteria of the problem.

 Subject Author Date No Subject Brian Smith 2008-03-20 00:56:39 re(2): solution Charlie 2008-03-19 09:22:57 re: solution brianjn 2008-03-19 08:16:23 Solution Dej Mar 2008-03-19 00:08:05 solution Charlie 2008-03-18 10:50:27 Solution Bractals 2008-03-18 10:45:53

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